diff --git "a/ncert_chapters.csv" "b/ncert_chapters.csv" new file mode 100644--- /dev/null +++ "b/ncert_chapters.csv" @@ -0,0 +1,161982 @@ +standard,chapter_id,chapter_name,file_path,data +class_6,1,Patterns in Mathematics,ncert_books/class_6/Ganita_Prakash/fegp101.pdf,"PATTERNS IN + MATHEMATICS +1 +1.1 What is Mathematics? +Mathematics is, in large part, the search for patterns, and for +the explanations as to why those patterns exist. +Such patterns indeed exist all around us — in nature, in +our homes and schools, and in the motion of the sun, moon, +and stars. They occur in everything that we do and see, from +shopping and cooking, to throwing a ball and playing games, to +understanding weather patterns and using technology. +The search for patterns and their explanations can be a fun +and creative endeavour. It is for this reason that mathematicians +think of mathematics both as an art and as a science. This year, we +hope that you will get a chance to see the creativity and artistry +involved in discovering and understanding mathematical +patterns. +It is important to keep in mind that mathematics +aims to not just find out what patterns exist, but also the +explanations for why they exist. Such explanations can +often then be used in applications well beyond the context in +which they were discovered, which can then help to propel +humanity forward. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +2 +For example, the understanding of patterns in the motion of stars, +planets, and their satellites led humankind to develop the theory of +gravitation, allowing us to launch our own satellites and send rockets +to the Moon and to Mars; similarly, understanding patterns in genomes +has helped in diagnosing and curing diseases — among thousands of +other such examples. +  Figure it Out +1. Can you think of other examples where mathematics helps +us in our everyday lives? +2. +How has mathematics helped propel humanity forward? (You +might think of examples involving: carrying out scientific +experiments; running our economy and democracy; building +bridges, houses or other complex structures; making TVs, +mobile phones, computers, bicycles, trains, cars, planes, +calendars, clocks, etc.) +1.2 Patterns in Numbers +Among the most basic patterns that occur in mathematics are +patterns of numbers, particularly patterns of whole numbers: +0, 1, 2, 3, 4, ... +The branch of Mathematics that studies patterns in whole +numbers is called number theory. +Number sequences are the most basic and among the most +fascinating types of patterns that mathematicians study. +Table 1 shows some key number sequences that are studied in +Mathematics. +Math +Talk +Reprint 2025-26 + +Patterns in Mathematics +3 +Table 1: Examples of number sequences + +1, 1, 1, 1, 1, 1, 1, ... +(All 1’s) + +1, 2, 3, 4, 5, 6, 7, ... +(Counting numbers) + +1, 3, 5, 7, 9, 11, 13, ... +(Odd numbers) + +2, 4, 6, 8, 10, 12, 14, ... +(Even numbers) + +1, 3, 6, 10, 15, 21, 28, ... +(Triangular numbers) + +1, 4, 9, 16, 25, 36, 49, ... +(Squares) + +1, 8, 27, 64, 125, 216, ... +(Cubes) + +1, 2, 3, 5, 8, 13, 21, ... +(Virahānka numbers) + +1, 2, 4, 8, 16, 32, 64, ... +(Powers of 2) + +1, 3, 9, 27, 81, 243, 729, ... +(Powers of 3) +  Figure it Out +1. Can you recognise the pattern in each of the sequences in +Table 1? +2. Rewrite each sequence of Table 1 in your notebook, along +with the next three numbers in each sequence! After +each sequence, write in your own words what is the rule +for forming the numbers in the sequence. +1.3 Visualising Number Sequences +Many number sequences can be visualised using pictures. Visualising +mathematical objects through pictures or diagrams can be a very +fruitful way to understand mathematical patterns and concepts. +Let us represent the first seven sequences in Table 1 using the +following pictures. +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +4 +Table 2: Pictorial representation of some number sequences +Squares +1 +4 +9 +16 +25 +Triangular +numbers +6 +3 +1 +10 +15 +Cubes +27 +8 +1 +64 +125 + + + + + +All 1’s + 1 +1 +1 +1 +1  + + + + + +Counting + 1 +2 +3 +4 +5 +numbers + + + + + +Odd + 1 +3 +5 +7 +9 +numbers + + + + + +Even + 2 +4 +6 +8 +10 +numbers +Reprint 2025-26 + +Patterns in Mathematics +5 +  Figure it Out +1. Copy the pictorial representations of the number sequences +in Table 2 in your notebook, and draw the next picture for +each sequence! +2. Why are 1, 3, 6, 10, 15, … called triangular numbers? Why +are 1, 4, 9, 16, 25, … called square numbers or squares? +Why are 1, 8, 27, 64, 125, … called cubes? +3. You will have noticed that 36 is both a triangular number and a +square number! That is, 36 dots can be arranged perfectly both +in a triangle and in a square. Make pictures in your notebook +illustrating this! + +This shows that the same number can be represented differently, +and play different roles, depending on the context. Try +representing some other numbers pictorially in different ways! +4. +What would you call the following sequence of numbers? + +1 +7 +19 +37 + +That’s right, they are called hexagonal numbers! Draw these in +your notebook. What is the next number in the sequence? +5. Can you think of pictorial ways to visualise the sequence of +Powers of 2? Powers of 3? +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +6 + +Here is one possible way of thinking about Powers of 2: +1 +16 +8 +4 +2 +1.4 Relations among Number Sequences +Sometimes, number sequences can be related to each other in +surprising ways. +Example: What happens when we start adding up odd numbers? + + + + +1 = 1 + + + + +1 + 3 = 4 + + + + +1 + 3 + 5 = 9 + + + + +1 + 3 + 5 + 7 = 16 + + + + +1 + 3 + 5 + 7 + 9 = 25 + + + + +1 + 3 + 5 + 7 + 9 + 11 = 36 + + + + + + +... +This is a really beautiful pattern! +  Why does this happen? Do you think it will happen forever? +The answer is that the pattern does happen forever. But why? +As mentioned earlier, the reason why the pattern happens is just as +important and exciting as the pattern itself. +A picture can explain it +Visualising with a picture can help explain the phenomenon. Recall +that square numbers are made by counting the number of dots in a +square grid. +  How can we partition the dots in a square grid into odd +numbers of dots: 1, 3, 5, 7,... ? +Think about it for a moment before reading further! +Math +Talk +Reprint 2025-26 + +Patterns in Mathematics +7 +Here is how it can be done: +This picture now makes it evident that +1 + 3 + 5 + 7 + 9 + 11 = 36. +Because such a picture can be made for a square of any size, this +explains why adding up odd numbers gives square numbers. +  By drawing a similar picture, can you say what is the sum of the +first 10 odd numbers? +  Now by imagining a similar picture, or by drawing it partially, as +needed, can you say what is the sum of the first 100 odd numbers? +Another example of such a relation between sequences: +Adding up and down +Let us look at the following pattern: + + + +1 = 1 + + + +1 + 2 + 1 = 4 + + + +1 + 2 + 3 + 2 + 1 = 9 + + + +1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 + + + +1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 25 + + + +1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 36 +This seems to be giving yet another way of getting the square numbers— +by adding the counting numbers up and then down! +... +Reprint 2025-26 + +Ganita Prakash | Grade 6 +8 +  Can you find a similar pictorial explanation? +  Figure it Out +1. Can you find a similar pictorial explanation for why adding +counting numbers up and down, i.e., 1, 1 + 2 + 1, 1 + 2 + 3 + +2 + 1, …, gives square numbers? +2. By imagining a large version of your picture, or drawing +it partially, as needed, can you see what will be the value of +1 + 2 + 3 + ... + 99 + 100 + 99 + ... + 3 + 2 + 1? +3. Which sequence do you get when you start to add the All 1’s +sequence up? What sequence do you get when you add the All 1’s +sequence up and down? +4. +Which sequence do you get when you start to add the counting +numbers up? Can you give a smaller pictorial explanation? +5. +What happens when you add up pairs of consecutive triangular +numbers? That is, take 1 + 3, 3 + 6, 6 + 10, 10 + 15, … Which sequence +do you get? Why? Can you explain it with a picture? +6. +What happens when you start to add up powers of 2 starting with +1, i.e., take 1, 1 + 2, 1 + 2 + 4, 1 + 2 + 4 + 8, … ? Now add 1 to each +of these numbers — what numbers do you get? Why does this +happen? +Try +This +Reprint 2025-26 + +Patterns in Mathematics +9 +7. +What happens when you multiply the triangular numbers by 6 +and add 1? Which sequence do you get? Can you explain it with a +picture? +8. +What happens when you start to add up hexagonal numbers, i.e., +take 1, 1 + 7, 1 + 7 + 19, 1 + 7 + 19 + 37, … ? Which sequence do you +get? Can you explain it using a picture of a cube? + + +9. +Find your own patterns or relations in and among the sequences +in Table 1. Can you explain why they happen with a picture or +otherwise? +1.5 Patterns in Shapes +Other important and basic patterns that occur in Mathematics +are patterns of shapes. These shapes may be in one, two, or three +dimensions (1D, 2D, or 3D) — or in even more dimensions.  The +branch of Mathematics that studies patterns in shapes is called +geometry. +Shape sequences are one important type of shape pattern that +mathematicians study. Table 3 shows a few key shape sequences +that are studied in Mathematics. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +10 +Table 3: Examples of shape sequences +Stacked +Triangles +Complete +Graphs +K2 +K3 +K4 +K6 +K5 +Stacked +Squares +Koch +Snowflake +Triangle +Heptagon +Hexagon +Decagon +Pentagon +Nonagon +Octagon +Quadrilateral +Regular +Polygons +Reprint 2025-26 + +Patterns in Mathematics +11 + +  Figure it Out  +1. +Can you recognise the pattern in each of the sequences in +Table 3? +2. Try and redraw each sequence in Table 3 in your notebook. +Can you draw the next shape in each sequence? Why or why +not? After each sequence, describe in your own words what +is the rule or pattern for forming the shapes in the sequence. +1.6 Relation to Number Sequences +Often, shape sequences are related to number sequences in +surprising ways. Such relationships can be helpful in studying and +understanding both the shape sequence and the related number +sequence. +Example: The number of sides in the shape sequence of Regular +Polygons is given by the counting numbers starting at 3, i.e., 3, 4, 5, 6, +7, 8, 9, 10, .... That is why these shapes are called, respectively, regular +triangle, quadrilateral (i.e., square), pentagon, hexagon, heptagon, +octagon, nonagon, decagon, etc., respectively. +The word ‘regular’ refers to the fact that these shapes have +equal-length sides and also equal ‘angles’ (i.e., the sides look the same +and the corners also look the same). We will discuss angles in more +depth in the next chapter. +The other shape sequences in Table 3 also have beautiful +relationships with number sequences. +  Figure it Out +1. Count the number of sides in each shape in the sequence +of Regular Polygons. Which number sequence do you get? +What about the number of corners in each shape in the +sequence of Regular Polygons? Do you get the same number +sequence? Can you explain why this happens? +2. Count the number of lines in each shape in the sequence +of Complete Graphs. Which number sequence do you get? +Can you explain why? +Math +Talk +Try +This +Reprint 2025-26 + +Ganita Prakash | Grade 6 +12 +3. How many little squares are there in each shape of the +sequence of Stacked Squares? Which number sequence +does this give? Can you explain why? +4. +How many little triangles are there in each shape of the +sequence of Stacked Triangles? Which number sequence +does this give? Can you explain why? (Hint: In each shape in +the sequence, how many triangles are there in each row?) +5. To get from one shape to the next shape in the Koch +Snowflake sequence, one replaces each line segment­ ‘—’ +by a ‘speed bump’ +. As one does this more and more +times, the changes become tinier and tinier with very very +small line segments. How many total line segments are +there in each shape of the Koch Snowflake? What is the +corresponding number sequence? (The answer is 3, 12, 48, ..., +i.e., 3 times Powers of 4; this sequence is not shown in Table 1.) +S u m m a r y + + +Mathematics may be viewed as the search for patterns and for the +explanations as to why those patterns exist. + + +Among the most basic patterns that occur in mathematics are number +sequences. + + +Some important examples of number sequences include the counting +numbers, odd numbers, even numbers, square numbers, triangular +numbers, cube numbers, Virahānka numbers, and powers of 2. + + +Sometimes number sequences can be related to each other in beautiful +and remarkable ways. For example, adding up the sequence of odd +numbers starting with 1 gives square numbers. + + +Visualising number sequences using pictures can help to understand +sequences and the relationships between them. + + +Shape sequences are another basic type of pattern in mathematics. +Some important examples of shape sequences include regular polygons, +complete graphs, stacked triangles and squares, and Koch snowflake +iterations. Shape sequences also exhibit many interesting relationships +with number sequences. +Try +This +Reprint 2025-26 + +[1] + +CHAPTER 1 — SOLUTIONS +Patterns in Mathematics +Page No. 2. +Section 1.1 + Figure it Out +Q1. Can you think of other examples where mathematics helps us in our +everyday lives? +Ans. Some examples are paying for fruits, vegetables, groceries etc. calculation of +speed of vehicles, designs or patterns in different buildings, finding area of any +plot or out own home. There could be many more such contexts in our everyday +lives. Discuss for other examples also. + +Q2. How has mathematics helped propel humanity forward? (You might think of +examples involving: carrying out scientific experiments; running our +economy and democracy; building bridges, houses or other complex +structures; making TVs, mobile phones, computers, bicycles, trains, cars, +planes, calendars, clocks, etc.) +Ans. Teacher student discussion is required. + +Page No. 3. +Section 1.2 + Figure it Out +Q.1. Can you recognize the pattern in each of the sequences in Table 1? +Ans. Yes, There is a pattern in each case. + +Powers of 2- 1,2,4 = 2 x 2, 8 = 2 x 2 x2 ,16 = 2 x 2 x2 x 2 x 2… + +Powers of 3 - 1, 3, 9 = 3 x 3 = 27 =3 x 3 x3,… + +Virahanka numbers- 1,2, 3, 5 = 2 + 3, 8 = 3 + 5, 13 = 5 + 8,…. + Rest of the patterns have been shown on page 4 ,Table 2 + +Page no. 5 +Section 1.3 + Figure it out +Q.2. Why are 1, 3, 6, 10, 15, … called triangular numbers? Why are 1, 4, 9, 16, 25, … +called square numbers or squares? Why are 1, 8, 27, 64, 125, … called cubes? +Ans. +Refer Table 2, page 4, and check for yourself. + +Q.3. You will have noticed that 36 is both a triangular number and a square number! +That is, 36 dots can be arranged perfectly both in a triangle and in a square. Make +pictures in your notebook illustrating this! +Ans. Refer Table 2, page 4, and draw. + +[2] + +Q.4. What would you call the following sequence of numbers? +Ans. 61. +Q.5. Can you think of pictorial ways to visualise the sequence of powers of 2? powers of +3? +Ans. Sequence of powers of 2—given on page 6 + Sequence of powers of 3- One of the ways could be – + + + +Page No. 7 +Section 1.4 + By drawing a similar picture, can you say what is the sum of the first 10 odd +numbers? +Ans. 100 + +3 +9 +2 +1 + +[3] + + Now by imagining a similar picture, or by drawing it partially, as needed, can +you say what is the sum of the first 100 odd numbers? +Ans.10000 + + Figure it out +Q.1. Can you find a similar pictorial explanation for why adding counting numbers up +and down, i.e., 1, 1 + 2 + 1, 1 + 2 + 3 + 2 + 1, …, gives square numbers? +Ans. One of the ways is- + +1, 1 + 2 + 1, 1 + 2 + 3 + 2 + 1, …, +Q.2. By imagining a large version of your picture, or drawing it partially, as needed, can +you see what will be the value of 1 + 2 + 3 + ... + 99 + 100 + 99 + ... + 3 + 2 + 1? +Ans. 10,000 +Q.3. Which sequence do you get when you start to add the All 1’s sequence up? What +sequence do you get when you add the All 1’s sequence up and down? +Ans. + +• 1 + (1 +1) + (1 +1 +1) + (1 + 1+1+1) … +• 1, 1+(1+1) + 1, 1+(1+1) +(1+1+1) + (1+1) +1, 1+(1+1) +(1+1+1) + (1+1+1+1) + +(1+1+1) + (1+1) +1, ………. +Q.4. Which sequence do you get when you start to add the Counting numbers up? Can +you give a smaller pictorial explanation? +Ans. +1, 1+2, 1+2+3, 1+2+3+4, ……… +Which is triangular number sequence. For pictorial representation refer Table 2 on +page 4 +(Try it for isosceles right triangle also.) + + +[4] + +Q.5. What happens when you add up pairs of consecutive triangular numbers? That is, +take 1 + 3, 3 + 6, 6 + 10, 10 + 15, … ? Which sequence do you get? Why? Can you +explain it with a picture? +Ans. We get: 4, 9, 16, 25, ……… (Square numbers). + For the pictorial representation refer Table 2 on page 4 + +Q.6. What happens when you start to add up powers of 2 starting with 1, i.e., take 1, 1 + +2, 1 + 2 + 4, 1 + 2 + 4 + 8, …? Now add 1 to each of these numbers — what +numbers do you get? Why does this happen? +Ans. +• We get, 1, 3, 7, 15, 31, ………… +• After adding 1 to each number, we get:2, 4, 8, 16, 32, …… +Refer the picture on page 6. + +Q.7. What happens when you multiply the triangular numbers by 6 and add 1? Which +sequence do you get? Can you explain it with a picture? +Ans. +(1×6) +1, (3×6) +1, (6×6) +1, (10×6) +1, (15×6) +1, … + += 7, 19, 37, 61, 91, …. + +For picture refer picture in Q.4 page 5. +Q.8. What happens when you start to add up hexagonal numbers, i.e., take 1, 1 + 7, +1 + 7 + 19, 1 + 7 + 19 + 37, … ? Which sequence do you get? Can you explain it +using a picture of a cube? +Ans. 1, 8, 27, 64, ………We get cube numbers. + +For picture, refer Table 2, page 4 + +Page no. 10 +Section 1.5 + Figure it out +Q.1. Can you recognise the pattern in each of the sequences in Table 3? +Ans. Yes. +➢ 3, 4, 5, 6, 7, 8, 9, 10 +One of the ways to interpret this is that we get a sequence of number of sides +of the pictures of the shapes +➢ 1, 3, 6, 10, 15 +➢ 1, 4, 9, 16, 25 +➢ 1, 4, 9, 16, 25 +➢ 3, 3×4, 3×4×4, 3×4×4×4, 3×4×4×4×4 + +[5] + +Page no. 11 +Section 1.6 + Figure it out +Q.1. Count the number of sides in each shape in the sequence of Regular Polygons. Which +number sequence do you get? What about the number of corners in each shape in +the sequence of Regular Polygons? Do you get the same number sequence? Can you +explain why this happens? +Ans. + +• Number of sides = 3,4,5,6,7,8,9,10… We get counting number sequence, starting +with 3. +• Number of Corners = 3,4,5,6,7,8,9,10… Yes, we get the same number sequence +• in any closed figure, number of sides = number of corners (vertices). +Q.2. Count the number of lines in each shape in the sequence of Complete Graphs. +Which number sequence do you get? Can you explain why? +Ans. + +• 1, 3, 6, 10, 15. This is a Triangle number sequence + +Q.3. How many little squares are there in each shape of the sequence of Stacked +Squares? Which number sequence does this give? Can you explain why? +Ans. + +• +1, 4, 9, 16, 25. This is a Square number sequence. +• +Squares can be drawn using these number of dots. + + +Q.4. How many little triangles are there in each shape of the sequence of Stacked +Triangles? Which number sequence does this give? Can you explain why? (Hint: In +each shape in the sequence, how many triangles are there in each row?) + +Ans. 1, 4, 9, 16, 25 += 1, 1+2+1, 1+2+3+2+1, ………. +Square number sequence (As adding up and down) gives us square number sequence. + + +Q.5. To get from one shape to the next shape in the Koch Snowflake sequence, one +replaces each line segment‘—’ by a ‘speed bump’ . As one does this more and +more times, the changes become tinier and tinier with very very small line +segments. How many total line segments are there in each shape of the Koch +Snowflake? What is the corresponding number sequence? (The answer is 3, 12, 48, +..., i.e. 3 times Powers of 4; this sequence is not shown in Table 1) +Ans. +• +Total line segments in each shape: 3, 12, 48, 192, 768 +• +Corresponding sequence: 3, 3×4, 3×4×4, 3×4×4×4×, 3×4×4×4×4,…" +class_6,2,Lines and Angles,ncert_books/class_6/Ganita_Prakash/fegp102.pdf,"Lines and Angles +2 +In this chapter, we will explore some of the most basic ideas of +geometry including points, lines, rays, line segments and angles. +These ideas form the building blocks of ‘plane geometry’, and will +help us in understanding more advanced topics in geometry such as +the construction and analysis of different shapes. + 2.1 Point +Mark a dot on the paper with a sharp tip of a pencil. The sharper the +tip, the thinner will be the dot. This tiny dot will give you an idea of +a point. A point determines a precise location, but it has no length, +breadth or height. Some models for a point are given below. +The tip of a +compass +The sharpened +end of a pencil +The pointed +end of a needle +If you mark three points on a piece of paper, +you may be required to distinguish these three +points. For this purpose, each of the three points +may be denoted by a single capital letter such as +P +Z +T +Reprint 2025-26 + +Ganita Prakash | Grade 6 +14 +Z, P and T. These points are read as ‘Point Z’, ‘Point P’ and ‘Point T’. Of +course, the dots represent precise locations and must be imagined to be +invisibly thin. + 2.2 Line Segment +Fold a piece of paper and unfold it. Do you +see a crease? This gives the idea of a line +segment. It has two end points, A and B. +Mark any two points A and B on a sheet of +paper. Try to connect A to B by various +routes (Fig. 2.1). +What is the shortest route from A to B? +This shortest path from point A to Point B +(including A and B) as shown here is called +the line segment from A to B. It is denoted by +either AB or BA. The points A and B are called +the end points of the line segment AB. + 2.3 Line +Imagine that the line segment from A to B (i.e., +AB) is extended beyond A in one direction and +beyond B in the other direction without any +end (see Fig. 2.2). This is a model for a line. Do +you think you can draw a complete picture of +a line? No. Why? +A line through two points A and B is written as AB. It extends +forever in both directions. Sometimes a line is denoted by a letter like +l or m. +Observe that any two points determine a unique line that passes +through both of them. +A +B +B +A +Fig. 2.1 +A +B +m +Fig. 2.2 +Reprint 2025-26 + +Lines and Angles +15 +2.4 Ray +A ray is a portion of a line that starts at one point (called the starting +point or initial point of the ray) and goes on endlessly in a direction. +The following are some models for a ray: +Beam of light from a +lighthouse +Ray of light from a torch +Sun rays +Look at the diagram (Fig. 2.3) of a ray. Two points are +marked on it. One is the starting point A and the other +is a point P on the path of the ray. We then denote the +ray by AP. + Figure it Out +1. + +Can you help Rihan and Sheetal find their answers? +A +P +Fig. 2.3 +Rihan marked a point +on a piece of paper. +How many lines can he +draw that pass through +the point? +Sheetal marked two points +on a piece of paper. How +many different lines can +she draw that pass through +both of the points? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +16 +2. Name the line segments in Fig. 2.4. Which of the five marked +points are on exactly one of the line segments? Which are on two +of the line segments? +L +M +P +Q +R +Fig. 2.4 +3. Name the rays shown in Fig. 2.5. Is T the starting point of each of +these rays? +A +T +Fig. 2.5 +N +B +4. Draw a rough figure and write labels appropriately to illustrate +each of the following: +a. OP and OQ meet at O. +b. XY and PQ intersect at point M. +c. Line l contains points E and F but not point D. +d. Point P lies on AB.  +5. In Fig. 2.6, name: +a. Five points +b. A line +c. Four rays +d. Five line segments +D +E +O +C +B +Fig. 2.6 +Reprint 2025-26 + +Lines and Angles +17 +6. Here is a ray OA (Fig. 2.7). It starts at O and +passes through the point A. It also passes +through the point B. +a. Can you also name it as OB ? Why? +b. Can we write OAas AO? Why or why not? +2.5 Angle +An angle is formed by two rays having a +common starting point. Here is an angle +formed by rays BD and BE where B is +the common starting point (Fig. 2.8). +The point B is called the vertex of the +angle, and the rays BD and BE are called +the arms of the angle. How can we name +this angle? We can simply use the vertex and say that it is the Angle B. +To be clearer, we use a point on each of the arms together with the +vertex to name the angle. In this case, we name the angle as Angle DBE +or Angle EBD. The word angle can be replaced by the symbol ‘∠’, i.e., +∠DBE or ∠EBD. Note that in specifying the angle, the vertex is always +written as the middle letter. +To indicate an angle, we use a small curve at the vertex (refer +to Fig. 2.9). +Vidya has just opened her book. Let us observe her opening the +cover of the book in different scenarios. +Case 1 +Case 2 +Case 3 +Case 4 +Case 5 +Case 6 +O +B +A +Fig. 2.7 +Fig. 2.8 +B +D +E +vertex +arm +arm +Reprint 2025-26 + +Ganita Prakash | Grade 6 +18 +  Do you see angles being made in each of these cases? Can you mark +their arms and vertex? +Which angle is greater—the angle in Case 1 or the angle in Case 2? +Just as we talk about the size of a line based on its length, we also +talk about the size of an angle based on its amount of rotation. +So, the angle in Case 2 is greater as in this case she needs to rotate +the cover more. Similarly, the angle in Case 3 is even larger than that +of Case 2, because there is even more rotation, and Cases 4, 5, and 6 +are successively larger angles with more and more rotation. +The size of an angle is the amount of rotation or turn that is needed +about the vertex to move the first ray to the second ray. +Fig. 2.9 +Final position of ray +Amount of turn is the size of +the angle +Vertex +Initial position of ray +Let’s look at some other examples where angles arise in real life +by rotation or turn: +• In a compass or divider, we turn the arms to form +an angle. The vertex is the point where the two +arms are joined. Identify the arms and vertex of +the angle. +• A pair of scissors has two blades. When we open +them (or ‘turn them’) to cut something, the blades +form an angle. Identify the arms and the vertex of +the angle. +Reprint 2025-26 + +Lines and Angles +19 +• Look at the pictures of spectacles, wallet and other common +objects. Identify the angles in them by marking out their arms +and vertices. +Do you see how these angles are formed by turning one arm with +respect to the other? +Teacher needs to organise various activities with the students to +recognise the size of an angle as a measure of rotation. +Teacher’s Note + Figure it Out +1. Can you find the angles in the given pictures? Draw the rays +forming any one of the angles and name the vertex of the angle. +A +B +C +D +Reprint 2025-26 + +Ganita Prakash | Grade 6 +20 +2. Draw and label an angle with arms ST and SR. +3. Explain why ∠APC cannot be labelled as ∠P. +A +P +B +C +4. Name the angles marked in the given figure. +P +T +R +Q +5. +Mark any three points on your paper that are not on one line. Label +them A, B, C. Draw all possible lines going through pairs of these +points. How many lines do you get? Name them. How many angles +can you name using A, B, C? Write them down, and mark each of +them with a curve as in Fig. 2.9. +Math +Talk +Reprint 2025-26 + +Lines and Angles +21 +6. +Now mark any four points on your paper so that no three of them +are on one line. Label them A, B, C, D. Draw all possible lines going +through pairs of these points. How many lines do you get? Name +them. How many angles can you name using A, B, C, D? Write them +all down, and mark each of them with a curve as in Fig. 2.9. +2.6 Comparing Angles +Look at these animals opening their mouths. Do you see any angles +here? If yes, mark the arms and vertex of each one. Some mouths are +open wider than others; the more the turning of the jaws, the larger +the angle! Can you arrange the angles in this picture from smallest to +largest? +  Is it always easy to compare two angles? +Here are some angles. Label each of the angles. How will you +compare them? +Draw a few more angles; label them and compare. +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +22 +Comparing angles by superimposition +Any two angles can be compared by placing them one over the other, +i.e., by superimposition. While superimposing, the vertices of the +angles must overlap. +After superimposition, it becomes clear which angle is smaller +and which is larger. +B +A +C +Q +P +R +Q (B) +P +A +C +R +The picture shows the two angles superimposed. It is now clear +that ∠PQR is larger than ∠ABC. +Equal angles. Now consider ∠AOB and ∠XOY in the figure. Which +is greater? +Y +O +X +O +B +Y +X +A +O +B +A +The corners of both of these angles match and the arms overlap with +each other, i.e., OA ↔ OX and OB ↔ OY. So, the angles are equal in size. +The reason these angles are considered to be equal in size is +because when we visualise each of these angles as being formed out +of rotation, we can see that there is an equal amount of rotation +needed to move OB to OA and OY to OX. +From the point of view of superimposition, when two angles +are superimposed, and the common vertex and the two rays of +both angles lie on top of each other, then the sizes of the angles +are equal. +Reprint 2025-26 + +Lines and Angles +23 + Where else do we use superimposition to compare? + Figure it Out +1. Fold a rectangular sheet of paper, then draw a line along the fold +created. Name and compare the angles +formed between the fold and the sides +of the paper. Make different angles by +folding a rectangular sheet of paper and +compare the angles. Which is the largest +and smallest angle you made? +2. In each case, determine which angle is +greater and why. +a. ∠AOB or ∠XOY +b. ∠AOB or ∠XOB +c. ∠XOB or ∠XOC + +Discuss with your friends on how +you decided which one is greater. +3. Which angle is greater: ∠XOY or ∠AOB? Give reasons. +X +O +B +A +Y +Comparing angles without superimposition +Two +cranes +are +arguing +about who can open their +mouth wider, i.e., who is +making a bigger angle. +Let us first draw their +angles. How do we know +which one is bigger? As seen +Math +Talk +A +O +B +C +X +Y +Math +Talk +Fig. 2.10 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +24 +before, one could trace these angles, superimpose them and then +check. But can we do it without superimposition? +Suppose we have a transparent circle which can be moved and +placed on figures. Can we use this for comparison? +Let us place the circular paper on the angle made by the first +crane. The circle is placed in such a way that its centre is on the +vertex of the angle. Let us mark the points A and B on the edge circle +at the points where the arms of the angle pass through the circle. +O +O +B +B +A +A +Can we use this to find out if this angle is greater than, or equal to +or smaller than the angle made by the second crane? +Let us place it on the angle made by the second crane so that the +vertex coincides with the centre of the circle and one of the arms +passes through OA. +B +Y +A +O +X +Can you now tell +which angle is bigger? +Reprint 2025-26 + +Lines and Angles +25 +Which crane was making the bigger angle? +If you can make a circular piece of transparent paper, try this method +to compare the angles in Fig. 2.10 with each other. + Teacher’s Note +A teacher needs to check the understanding of the students around +the notion of an angle. Sometimes students might think that +increasing the length of the arms of the angle increases the angle. +For this, various situations should be posed to the students to check +their understanding on the same. +2.7 Making Rotating Arms +Let us make ‘rotating arms’ using two paper straws and a paper clip +by following these steps: +1. Take two paper straws and a paper clip.  +2. Insert the straws into the arms of the paper +clip. +3. Your rotating arm is ready! +Make several ‘rotating arms’ with different angles between the +arms. Arrange the angles you have made from smallest to largest by +comparing and using superimposition. +Passing through a slit: Collect a number of rotating arms with different +angles; do not rotate any of the rotating arms during this activity. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +26 +Take a cardboard and make an angle-shaped slit as shown below +by tracing and cutting out the shape of one of the rotating arms. +Now, shuffle and mix up all the rotating arms. Can you identify +which of the rotating arms will pass through the slit? +The correct one can be found by placing each of the rotating arms +over the slit. Let us do this for some of the rotating arms: +Slit angle is greater than +the arms’ angle. The arms +will not go through the +slit. +Slit angle is less than the +arms’ angle. The arms +will not go through the +slit. +Slit angle is equal to the +arms’ angle. The arms will +go through the slit. +Only the pair of rotating arms where the angle is equal to that of the +slit passes through the slit. Note that the possibility of passing through +the slit depends only on the angle between the rotating arms and not +on their lengths (as long as they are shorter than the length of the slit). +Reprint 2025-26 + +Lines and Angles +27 +Challenge: Reduce +this angle. +Angle +reduced. +The angle is still +the same! +2.8 Special Types of Angles +Let us go back to Vidya’s +notebook and observe her +opening the cover of the book +in different scenarios. +She makes a full turn of the +cover when she has to write +while holding the book in +her hand. +She makes a half turn of the +cover when she has to open +it on her table. In this case, +observe the arms of the angle formed. They lie in a straight line. +Such an angle is called a straight angle. +A +O +B +Let us consider a straight angle ∠AOB. Observe that any ray OC +divides it into two angles, ∠AOC and ∠COB. +Fig. 2.11 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +28 +  Is it possible to draw OC such that the two angles are +equal to each other in size? +Let’s Explore +We can try to solve this problem using a piece of paper. Recall that when +a fold is made, it creates a crease which is straight. +Take a rectangular piece of paper and on one of its sides, mark +the straight angle AOB. By folding, try to get a line (crease) passing +through O that divides ∠AOB into two equal angles. +How can it be done? +Fold the paper such that OB overlaps with OA. Observe the crease +and the two angles formed. +Math +Talk +Reprint 2025-26 + +Lines and Angles +29 +Justify why the two angles are equal. Is there a way to +superimpose and check? Can this superimposition be done by +folding? +Each of these equal angles formed are called right angles. So, a +straight angle contains two right angles. +Why shouldn't you +argue with a 90 ̊ +angle? +Because they're +always right. + If a straight angle is formed by half of a full turn, how much of a +full turn will form a right angle?  +Observe that a right angle resembles the shape of an ‘L’. An angle +is a right angle only if it is exactly half of a straight angle. Two lines +that meet at right angles are called perpendicular lines. + Figure it Out +1. How many right angles do the windows of your classroom +contain? Do you see other right angles in your classroom? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +30 +2. Join A to other grid points in the figure by a straight line to get a +straight angle. What are all the different ways of doing it? + +A +B +A +B +3. Now join A to other grid points in the figure by a straight line to +get a right angle. What are all the different ways of doing it? +A +B +A +B + + +Hint: Extend the line further as shown in the figure below. To get +a right angle at A, we need to draw a line through it that +divides the straight angle CAB into two equal parts. + +A +C +B +Reprint 2025-26 + +Lines and Angles +31 +4. Get a slanting crease on the paper. Now, try to get another crease +that is perpendicular to the slanting crease. +a. How many right angles do you have now? Justify why the +angles are exact right angles. +b. Describe how you folded the paper so that any other person +who doesn’t know the process can simply follow your +description to get the right angle. +Classifying Angles +Angles are classified in three groups as shown below. Right angles +are shown in the second group. What could be the common feature +of the other two groups? +In the first group, all angles are less than a right angle or in other +words, less than a quarter turn. Such angles are called acute angles. +In the third group, all angles are greater than a right angle but +less than a straight angle. The turning is more than a quarter turn +and less than a half turn. Such angles are called obtuse angles. + Figure it Out +1. +Identify acute, right, obtuse and straight angles in the previous +figures. +2. +Make a few acute angles and a few obtuse angles. Draw them in +different orientations. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +32 +3. +Do you know what the words acute and obtuse mean? Acute means +sharp and obtuse means blunt. Why do you think these words have +been chosen? +4. Find out the number of acute angles in each of the figures below. +What will be the next figure and how many acute angles will it have? +Do you notice any pattern in the numbers? +2.9 Measuring Angles +We have seen how to compare two angles. But can we actually +quantify how big an angle is using a number without having to +compare it to another angle? +We saw how various angles can be compared using a circle. +Perhaps a circle could be used to assign measures for angles? +Fig. 2.12 +To assign precise measures to angles, mathematicians came up +with an idea. They divided the angle in the centre of the circle into +Reprint 2025-26 + +Lines and Angles +33 +360 equal angles or parts. The angle measure of each of these unit +parts is 1 degree, which is written as 1°. +This unit part is used to assign measure to any angle: the measure +of an angle is the number of 1° unit parts it contains inside it. +For example, see this figure: +30 units +It contains 30 units of 1° angle and so we say that its angle measure is 30°. +Measures of different angles: What is the measure of a full turn in +degrees? As we have taken it to contain 360 degrees, its measure is 360°. +  What is the measure of a straight angle in degrees? A straight +angle is half of a full turn. As a full-turn is 360°, a half turn is 180°. +What is the measure of a right angle in degrees? Two right angles +together form a straight angle. As a straight angle measures 180°, a +right angle measures 90°. +A +O +180 units +90 units +B +A +O +B +A +O +B +A +O +B +A pinch of history +A full turn has been divided into 360°. Why 360? The reason why +we use 360° today is not fully known. The division of a circle into +Reprint 2025-26 + +Ganita Prakash | Grade 6 +34 +360 parts goes back to ancient times. The Rigveda, one of the very +oldest texts of humanity going back thousands of years, speaks of +a wheel with 360 spokes (Verse 1.164.48). Many ancient calendars, +also going back over 3000 years—such as calendars of India, Persia, +Babylonia and Egypt—were based on having 360 days in a year. In +addition, Babylonian mathematicians frequently used divisions of 60 +and 360 due to their use of sexagesimal numbers and counting by 60s. +Perhaps the most important and practical answer for why +mathematicians over the years have liked and continued to use 360 +degrees is that 360 is the smallest number that can be evenly divided +by all numbers up to 10, aside from 7. Thus, one can break up the +circle into 1, 2, 3, 4, 5, 6, 8, 9 or 10 equal parts, and still have a whole +number of degrees in each part! Note that 360 is also evenly divisible +by 12, the number of months in a year, and by 24, the number of +hours in a day. These facts all make the number 360 very useful. + The circle has been divided into 1, 2, 3, 4, 5, 6, 8, 9 10 and 12 parts +below. What are the degree measures of the resulting angles? Write +the degree measures down near the indicated angles. +Degree measures of different angles +How can we measure other angles in degrees? It is for this purpose +that we have a tool called a protractor that is either a circle divided +into 360 equal parts as shown in Fig. 2.12 (on page 32), or a half +circle divided into 180 equal parts. +Reprint 2025-26 + +Lines and Angles +35 +Unlabelled protractor +Here is a protractor. Do you see the straight angle at the center +divided into 180 units of 1 +degree? Only part of the +lines dividing the straight +angle are visible, though! +Starting +from +the +marking on the rightmost +point of the base, there +is a long mark for every +10°. From every such long +mark, there is a medium +sized mark after 5°. + Figure it out +1. +Write the measures of the +following angles: +a. ∠ KAL +Notice that the vertex of this +angle coincides with the centre of +the protractor. So the number of +units of 1 degree angle between KA +and AL gives the measure of ∠KAL. +By counting, we get — + + +∠KAL = 30° +Making use of the medium sized +and large sized marks, is it possible +to count the number of units in 5s +or 10s? +b. ∠WAL +c. ∠TAK +A +T +W +L +K +Reprint 2025-26 + +Ganita Prakash | Grade 6 +36 +Labelled protractor +This is a protractor that you find in your geometry box. It would +appear similar to the protractor above except that there are numbers +written on it. Will these make it easier to read the angles? +0 +180 +10 +170 +20 +160 +30 +150 +40 +140 +50 +130 +60 +120 +70 +110 +80 +100 +100 +80 +110 +70 +120 +60 +130 +50 +140 +40 +150 +30 +160 +20 +170 +10 +180 +0 +90 +90 +There are two sets of numbers on the protractor: one increasing +from right to left and the other increasing from left to right. Why +does it include two sets of numbers? + Name the different angles in the figure and write their measures. +0 +180 +10 +170 +20 +160 +30 +150 +40 +140 +50 +130 +60 +120 +70 +110 +80 +100 +100 +80 +110 +70 +120 +60 +130 +50 +140 +40 +150 +30 +160 +20 +170 +10 +180 +0 +90 +90 +P +O +Q +R +S +T +U +Reprint 2025-26 + +Lines and Angles +37 +Did you include angles such as ∠TOQ? +Which set of markings did you use — inner or outer? +What is the measure of ∠TOS? +Can you use the numbers marked to find the angle without +counting the number of markings? +Here, OT and OS pass through the numbers 20 and 55 on the outer +scale. How many units of 1 degree are contained between these +two arms? +Can subtraction be used here? +How can we measure angles directly without having to subtract? +Place the protractor so the center is on the vertex of the angle. +Align the protractor so that one the arms passes through the 0º +mark as in the picture below. +0 +180 +10 +170 +20 +160 +30 +150 +40 +140 +50 +130 +60 +120 +70 +110 +80 +100 +100 +80 +110 +70 +120 +60 +130 +50 +140 +40 +150 +30 +160 +20 +170 +10 +180 +0 +90 +90 +A +O +B +What is the degree measure of ∠AOB? +Make your own Protractor! +You may have wondered how the different equally spaced markings are +made on a protractor. We will now see how we can make some of them! +1. +Draw a circle of a convenient radius on a sheet of paper. Cut out +the circle (Fig. 2.13). A circle or one full turn is 360°. +2. +Fold the circle to get two equal halves and cut it through the +crease to get a semicircle. Write ‘0°’ in the bottom right corner of +the semicircle. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +38 +Fig. 2.13 +Fig. 2.14 +A +O +180 units +B +The measure of half +a circle is 1 +2 of a full +turn. (Fig. 2.14) +So, the measure of +half a turn = 1 +2 of ____ += 180°. +Thus, write 180° in +the left bottom corner +of the semicircle. +3. +Fold the semi-circular sheet in half as shown in Fig. 2.15 to form +a quarter circle. +Fig. 2.15 +90 units +A +O +B +The measure of a +quarter circle is 1 +4 of a +full turn. +The measure of a +1 +4 turn = 1 +4 of 360° = +________. +Or, the measure of a 1 +4 +turn = 1 +2 of a half turn += 1 +2 of 180° = ______. +Thus, mark 90° at the +top of the semicircle. +Reprint 2025-26 + +Lines and Angles +39 +4. +Fold the sheet again as shown in Figs. 2.16 and 2.17: +Fig. 2.16 +Fig. 2.17 +180O +0O +45O +90O +135O +When folded, this is 1 +8 of the circle, or 1 +8 of a turn, or 1 +8 of 360°, +or 1 +4 of 180° or 1 +2 of 90° = ________________________. +The new creases formed give us measures of 45° and +180°− 45° = 135° as shown. Write 45° and 135° at the correct +places on the new creases along the edge of the semicircle. +5. +Continuing with another half fold as shown in Fig. 2.18, we get +an angle of measure ________________________. +Fig. 2.18 +6. +Unfold and mark the creases as OB, OC, ..., etc., as shown in +Fig. 2.19 and Fig. 2.20. +Fig. 2.20 +180O +0O +45O +67.5O +112.5O +135O +22.5O +90O +157.5O +Fig. 2.19 +O +A +B +C +D +E +F +G +H +I +Reprint 2025-26 + +Ganita Prakash | Grade 6 +40 + Think! + +In Fig. 2.19, we have ∠AOB = ∠BOC = ∠COD = ∠DOE = ∠EOF = ∠FOG = +∠GOH = ∠HOI=_____. Why? +Angle Bisector +At each step, we folded in halves. This process of getting half of a +given angle is called bisecting the angle. The line that bisects a given +angle is called the angle bisector of the angle. +Identify the angle bisectors in your handmade protractor. Try to make +different angles using the concept of angle bisector through paper folding. + Figure it Out +1. +Find the degree measures of the following angles using your +protractor. +H +I +J +H +I +J +G +K +I +H +J +2. Find the degree measures of different angles in your classroom +using your protractor. +Teacher’s Note +It is important that students make their own protractor and use it to +measure different angles before using the standard protractor so that +they know the concept behind the marking of the standard protractor. +Reprint 2025-26 + +Lines and Angles +41 +3. Find the degree measures for the angles given below. Check if +your paper protractor can be used here! + +c +I +H +J +H +I +J +4. How can you find the degree measure of the angle given below +using a protractor? +5. Measure and write the degree measures for each of the following +angles: +a. +b. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +42 +d. +c. +e. +f. +6. Find the degree measures of ∠BXE, ∠CXE, ∠AXB and ∠BXC. +0 +180 +10 +170 +20 +160 +30 +150 +40 +140 +50 +130 +60 +120 +70 +110 +80 +100 +100 +80 +110 +70 +120 +60 +130 +50 +140 +40 +150 +30 +160 +20 +170 +10 +180 +0 +90 +90 +A +X +E +C +B +7. Find the degree measures of ∠PQR, ∠PQS and ∠PQT. +P +Q +R +S +T +Reprint 2025-26 + +Lines and Angles +43 +8. Make the paper craft as per the given instructions. Then, unfold +and open the paper fully. Draw lines on the creases made and +measure the angles formed. +1 +2 +3 +7 +6 +5 +4 +8 +9. +Measure all three angles of the triangle shown in Fig. 2.21 (a), and +write the measures down near the respective angles. Now add up +the three measures. What do you get? Do the same for the triangles +in Fig. 2.21 (b) and (c). Try it for other triangles as well, and then +make a conjecture for what happens in general! We will come back +to why this happens in a later year. +A +B +C +(a) +A +B +C +(b) +A +B +C +(c) +Fig. 2.21 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +44 +Mind the Mistake, Mend the Mistake! +A student used a protractor to measure the angles as shown below. +In each figure, identify the incorrect usage(s) of the protractor and +discuss how the reading could have been made and think how it can +be corrected. +0 +180 +10 +170 +20 +160 +30 +150 +40 +140 +50 +130 +60 +120 +70 +110 +80 +100 +100 +80 +110 +70 +120 +60 +130 +50 +140 +40 +150 +30 +160 +20 +170 +10 +180 +0 +90 +90 +∠U = 35⁰ +∠V = 80⁰ +0 +180 +10 +170 +20 +160 +30 +150 +40 +140 +50 +130 +60 +120 +70 +110 +80 +100 +100 +80 +110 +70 +120 +60 +130 +50 +140 +40 +150 +30 +160 +20 +170 +10 +180 +0 +90 +90 +0 +180 +10 +170 +20 +160 +30 +150 +40 +140 +50 +130 +60 +120 +70 +110 +80 +100 +100 +80 +110 +70 +120 +60 +130 +50 +140 +40 +150 +30 +160 +20 +170 +10 +180 +0 +90 +90 +∠W = 70⁰ +U +V +0 +180 +10 +170 +20 +160 +30 +150 +40 +140 +50 +130 +60 +120 +70 +110 +80 +100 +100 +80 +110 +70 +120 +60 +130 +50 +140 +40 +150 +30 +160 +20 +170 +10 +180 +0 +90 +90 +∠X = 150⁰ +∠Z = 85⁰ +0 +180 +10 +170 +20 +160 +30 +150 +40 +140 +50 +130 +60 +120 +70 +110 +80 +100 +100 +80 +110 +70 +120 +60 +130 +50 +140 +40 +150 +30 +160 +20 +170 +10 +180 +0 +90 +90 +∠Y = 120⁰ +0 +180 +10 +170 +20 +160 +30 +150 +40 +140 +50 +130 +60 +120 +70 +110 +80 +100 +100 +80 +110 +70 +120 +60 +130 +50 +140 +40 +150 +30 +160 +20 +170 +10 +180 +0 +90 +90 +Y +Reprint 2025-26 + +Lines and Angles +45 + Figure it Out + +Where are the angles? +1. Angles in a clock: +a. The hands of a clock make different +angles at different times. At 1 o’clock, +the angle between the hands is 30°. +Why? +b. What will be the angle at 2 o’clock? +And at 4 o’clock? 6 o’clock? +c. Explore other angles made by +the hands of a clock. +2. The angle of a door: + +Is it possible to express the amount by which a door is opened +using an angle? What will be the vertex of the angle and what will +be the arms of the angle? +3. Vidya is enjoying her time +on the swing. She notices +that the greater the angle +with which she starts the +swinging, +the +greater +is +the speed she achieves on +her swing. But where is the +angle? Are you able to see +any angle? +12 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +46 +4. Here is a toy with slanting slabs attached to +its sides; the greater the angles or slopes of +the slabs, the faster the balls roll. Can angles +be used to describe the slopes of the slabs? +What are the arms of each angle? Which +arm is visible and which is not? +5. +Observe the images below where there is +an insect and its rotated version. Can angles +be used to describe the amount of rotation? +How? What will be the arms of the angle and +the vertex? + +Hint: Observe the horizontal line touching the insects. +It is important that students see the application of each mathematical +concept in their daily lives. Teacher can organise some activities where +students can appreciate the practical applications of angles in real-life +situations, e.g., clocks, doors, swings, concepts of uphill and downhill, +location of the sun, the giving of directions, etc. +Teacher’s Note +2.10 Drawing Angles +Vidya wants to draw a 30° angle and name it ∠TIN using a protractor. +In will be the vertex, IT and IN will be the arms of the angle. +Keeping one arm, say IN, as the reference (base), the other arm IT should +take a turn of 30°. +Reprint 2025-26 + +Lines and Angles +47 +Step 1: We begin with the base and draw 𝐼𝑁: + +I +N +Step 2: We will place the centre point of the protractor on I and align +IN to the 0 line. +0 +180 +10 +170 +20 +160 +30 +150 +40 +140 +50 +130 +60 +120 +70 +110 +80 +100 +100 +80 +110 +70 +120 +60 +130 +50 +140 +40 +150 +30 +160 +20 +170 +10 +180 +0 +90 +90 +I +N +Step 3: Now, starting from 0, count your degrees (0, 10, 30) up to 30 +on the protractor. Mark point T at the label 30°. +0 +180 +10 +170 +20 +160 +30 +150 +40 +140 +50 +130 +60 +120 +70 +110 +80 +100 +100 +80 +110 +70 +120 +60 +130 +50 +140 +40 +150 +30 +160 +20 +170 +10 +180 +0 +90 +90 +I +N +T +Step 4: Using a ruler join the point I and T. +∠TIN = 30° is the required angle. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +48 +I +N +30º +T + Let’s Play a Game #1 +This is an angle guessing game! Play this game with your classmates +by making two teams, Team 1 and Team 2. Here are the instructions +and rules for the game: +• Team 1 secretly choose an angle measure, for example, 49° and +makes an angle with that measure using a protractor without +Team 2 being able to see it. +• Team 2 now gets to look at the angle. They have to quickly +discuss and guess the number of degrees in the angle (without +using a protractor!). +• Team 1 now demonstrates the true measure of the angle with +a protractor. +• Team 2 scores the number of points that is the absolute +difference in degrees between their guess and the correct +measure. For example, if Team 2 guesses 39°, then they score +10 points (49°–39°). +• Each team gets five turns. The winner is the team with the +lowest score! + Let’s Play a Game #2 +We now change the rules of the game a bit. Play this game with your +classmates by again making two teams, Team 1 and Team 2. Here +are the instructions and rules: +Fig. 2.22 +Reprint 2025-26 + +Lines and Angles +49 +• Team 1 announces to all, an angle measure, e.g., 34°. +• A player from Team 2 must draw that angle on the board without +using a protractor. Other members of Team 2 can help the player +by speaking words like ‘Make it bigger!’ or ‘Make it smaller!’. +• A player from Team 1 measures the angle with a protractor +for all to see. +• Team 2 scores the number of points that is the absolute +difference in degrees between Team 2’s angle size and the +intended angle size. For example, if player’s angle from Team  2 +is measured to be 25°, then Team 2 scores 9 points (34°–25°). +• Each team gets five turns. The winner is again the team with the +lowest score. +Teacher’s Note +These games are important to play to build intuition about angles +and their measures. Return to this game at least once or twice on +different days to build practice in estimating angles. Note that +these games can also be played between pairs of students. + Figure it Out +1. In Fig. 2.23, list all the angles possible. Did you find them all? Now, +guess the measures of all the angles. Then, measure the angles +with a protractor. Record all your numbers in a table. See how +close your guesses are to the actual measures. +A +C +D + L +S +P +R +B +Fig. 2.23 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +50 +2. Use a protractor to draw angles having the following degree +measures: +a. 110° + b. 40° +c. 75° +d. 112° +e. 134° +3. Draw an angle whose degree measure is the same as the angle +given below: +H +I +J + +Also, write down the steps you followed to draw the angle. +2.11 Types of Angles and their Measures +We have read about different types of angles in this chapter. We +have seen that a straight angle is 180° and a right angle is 90°. How +can other types of angles — acute and obtuse — be described in terms +of their degree measures? +Acute Angle: Angles that are smaller than the right angle, i.e., less +than 90° and are greater than 0°, are called acute angles. +40⁰ +P +R +Q +50⁰ +S +T +R +75⁰ +E +F +Q +Examples of acute angles +Reprint 2025-26 + +Lines and Angles +51 +Obtuse Angle: Angles that are greater than the right angle and less +than the straight angle, i.e., greater than 90° and less than 180°, are +called obtuse angles. +T +W +S +130º +S +I +X +110º +Examples of obtuse angles +Have we covered all the possible measures that an angle can take? +Here is another type of angle. +Reflex angle: Angles that are greater than the straight angle and less +than the whole angle, i.e., greater than 180° and less than 360°, are +called reflex angles. +S +C +P +A +B +T +M +Examples of reflex angles + Figure it Out +1. In each of the below grids, join A to other grid points in the figure +by a straight line to get: +a. An acute angle +Reprint 2025-26 + +Ganita Prakash | Grade 6 +52 +b. An obtuse angle +c. A reflex angle + +Mark the intended angles with curves to specify the angles. One +has been done for you. +2. Use a protractor to find the measure of each angle. Then classify +each angle as acute, obtuse, right, or reflex. + +a. ∠PTR +b. ∠PTQ +c. ∠PTW +d. ∠WTP +T +Q +R +P +W +Reprint 2025-26 + +Lines and Angles +53 + Let’s Explore +In this figure, ∠TER = 80°. What is +the measure of ∠BET? What is the +measure of ∠SET? +80 +o90 +o +B +E +S +T +R +Hint: Observe that ∠REB is a straight angle. Hence, the degree measure of +∠REB = 180° of which 80° is covered by ∠TER. A similar argument +can be applied to find the measure of ∠SET. + Figure it Out +1. Draw angles with the following degree measures: +a. 140°   b. 82°   c. 195°   d. 70°   e. 35° +2. Estimate the size of each angle and then measure it with a +protractor: +a. +b. +c. +d. +e. +f. + +Classify these angles as acute, right, obtuse or reflex angles. +3. Make any figure with three acute angles, one right angle and two +obtuse angles. +4. Draw the letter ‘M’ such that the angles on the sides are 40° each +and the angle in the middle is 60°. +5. Draw the letter ‘Y’ such that the three angles formed are 150°, 60° +and 150°. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +54 +6. The Ashoka Chakra has 24 spokes. What is the +degree measure of the angle between two spokes +next to each other? What is the largest acute angle +formed between two spokes? +7. Puzzle: I am an acute angle. If you double my +measure, you get an acute angle. If you triple my measure, you +will get an acute angle again. If you quadruple (four times) my +measure, you will get an acute angle yet again! But if you multiply +my measure by 5, you will get an obtuse angle measure. What are +the possibilities for my measure? +S u m m a r y + + +A point determines a location. It is denoted by a capital letter. + + +A line segment corresponds to the shortest distance between two +points. The line segment joining points S and T is denoted by ST. + + +A line is obtained when a line segment like ST is extended on both +sides indefinitely; it is denoted by ST or sometimes by a single small +letter like m. + + +A ray is a portion of a line starting at a point D and going in one direction +indefinitely. It is denoted by DP where P is another point on the ray. + + +An angle can be visualised as two rays starting from a common starting +point. Two rays OP and OM form the angle ∠POM (also called ∠MOP); +here, O is called the vertex of the angle, and the rays OP and OM are +called the arms of the angle. + + +The size of an angle is the amount of rotation or turn needed about the +vertex to rotate one ray of the angle onto the other ray of the angle. + + +The sizes of angles can be measured in degrees. One full rotation or +turn is considered as 360 degrees and denoted as 360°. + + +Degree measures of angles can be measured using a protractor. + + +Angles can be straight (180°), right (90°), acute (more than 0° and less +than 90°), obtuse (more than 90° and less than 180°), and reflex (more +than 180° and less than 360°). +Reprint 2025-26 + +[1] + +CHAPTER 2 — SOLUTIONS +Lines and Angles + +Section 2.4 +Page No. 15 +Figure it Out +Q.1. + + +Can you help Rihan and Sheetal find their answers? +Ans. +Rihan can draw many/uncountable number of lines through the given point. +Sheetal can draw only one line through the two given points. +Q.2. Name the line segments in Fig. 2.4. Which of the five marked points are on exactly +one of the line segments? Which are on two of the line segments? +Ans. +LM, MP, PQ, QR +Points L and R are exactly on one line segment. Points M, P and Q are on two line +segments. +Q.3. Name the rays shown in Fig. 2.5. Is T the starting point of each of these rays? +Ans. + TA, TB, TN and NB + +No, T is the starting point of TB, TN and TA but not of NB. +Q.4. Draw a rough figure and write labels appropriately to illustrate each of the +following: +a. OP and OQ meet at O. +b. XY and PQ intersect at point M. +c. Line l contains points E and F but not point D. +d. Point P lies on AB. + +Ans. + +Rihan marked a point on a piece of +paper. How many lines can he draw +that pass through the point? +Sheetal marked two points on a piece +of paper. How many different lines +can she draw that pass through +both of the points? + +[2] + + +Q.5. In Fig. 2.6, name: +a. Five points +b. A line +c. Four rays +d. Five line segments +Ans. + +a) D, E, O, B and C + +b) DE or DO or DB or EO or EB or OB + +c) OC, OB, OE, OD (Try for other rays) + +d) DE, DO, DB, EO, EB (OB; OC are also possible) + +Q.6. Here is a ray OA (Fig. 2.7). It starts at O and passes through the point A. It also +passes +through the point B. +a. Can you also name it as OB? Why? +b. Can we write OA as AO? Why or why not? +Ans. +a) Yes, O is the starting point and point B lies on the rays that goes endlessly in the +direction of A. OA is the extension of OB. +b) No, OA is a ray with starting point O whereas AO is a ray with starting point A. + +[3] + +Section 2.5 +Page – 19 +Figure it Out +Q.1. Can you find the angles in the given pictures? Draw the rays forming any one of the +angles and name the vertex of the angle. +Ans. Yes, one of the angles is ∠BDC. It’s vertex is D. One ray is DC and the other rays DB. +Try for other pictures. +Q.2. Draw and label an angle with arms ST and SR. + +Ans. + +Q.4. Name the angles marked in the given figure. +Ans. +∠RTQ, ∠RTP + +Q.5. Mark any three points on your paper that are not on one line. Label them A, B, C. +Draw all possible lines going through pairs of these points. How many lines do you +get? Name them. How many angles can you name using A, B, C? Write them down, +and mark each of them with a curve as in Fig. 2.9. +Ans. +We get three lines AB, BC, CA. + +Using A, B & C we can name three angles: ∠ABC or ∠CBA, ∠BCA or ∠ACB & ∠CAB +or ∠BAC. + +[4] + +Q.6. Now mark any four points on your paper so that no three of them are on one line. Label them +A, B, C, D. Draw all possible lines going through pairs of these points. How many lines do +you get? Name them. How many angles can you name using A, B, C, D? Write them all +down, and mark each of them with a curve as in Fig. 2.9. + +Ans. +We get six lines AB, BC, CD, DA, AC & BD + +Using A, B, C & D we can name the following angles ∠BAC, ∠CAD, ∠BAD, ∠ADB, +∠BDC, ∠ADC, ∠DCA, ∠ACB, ∠DCB, ∠CBD, ∠DBA & ∠CBA + +Section 2.6 +Page 20 + Is it always easy to compare two angles? +No. it is not always easy to compare two angles. For eg. 89º & 91º angles cannot be +compared without measuring or overlapping. But for the given figures, comparison is easy. +Page 23 + Where else do we use superimposition to compare? +A few examples are- line segments, squares and circles. Think of more. + Figure it out + +Q.1. Fold a rectangular sheet of paper, then draw a line along the fold created. Name and +compare the angles formed between the fold and the sides of the paper. Make +different angles by folding a rectangular sheet of paper and compare the angles. +Which is the largest and smallest angle you made? + +[5] + +Ans. + +Angles Formed: ∠AEF, ∠BEF, ∠DFE, ∠CFE +Here ∠AEF & ∠CFE are larger than ∠BEF & ∠DFE +Try more cases by folding rectangular sheets in different ways. +Q.2. In each case, determine which angle is greater and why. +a. ∠AOB or ∠XOY +b. ∠AOB or ∠XOB +c. ∠XOB or ∠XOC +Discuss with your friends on how you decided which one is greater. +Ans. (a)- ∠AOB; ∠XOY is an acute angle and ∠AOB = ∠AOX + ∠XOY+∠YOB + (b)- ∠AOB + (c) – None. ∠XOB = ∠XOC + +Q.3. Which angle is greater: ∠XOY or ∠AOB? Give reasons. +Ans. By looking at the figure we cannot say. Superimposition or measurement is necessary here. + +Section 2.8 +Page 28 +Q. Is it possible to draw OC such that the two angles are equal to each other in size? +Ans. Yes, when OA and OB overlap each other on folding the Vidya’s notebook, the crease +OC will divide ∠AOB in two equal sized angles. + +Page 29 +Q. If a straight angle is formed by half of a full turn, how much of a full turn will form a right +angle? +Ans. +1 +4 of a full turn. + +[6] + + +Section 2.8 +Page No. 29 +Figure it Out +Q.4. Get a slanting crease on the paper. Now, try to get another crease that is +perpendicular to the slanting crease. +a. How many right angles do you have now? Justify why the angles are exact right +angles. +b. Describe how you folded the paper so that any other person who doesn’t know +the process can simply follow your description to get the right angle. +Ans. a. Four right angle. Each angle is ¼ the of the complete angle. +b. Explore different ways of doing it. + +Page 31 +Figure it Out +Q.2. Make a few acute angles and a few obtuse angles. Draw them in different +orientations. +Ans. +Acute angle + + +Obtuse angle + +Q.3. Do you know what the words acute and obtuse mean? Acute means sharp and obtuse +means blunt. Why do you think these words have been chosen? +Ans. In acute angles the opening of the edges is lesser than the obtuse angle which have larger +openings. +Q.4. Find out the number of acute angles in each of the figures below. What will be the +next figure and how many acute angles will it have? Do you notice any pattern in +the numbers? +Ans. +(i) three +(ii) Twelve +(iii) Twenty-one +The next figure will have thirty acute angles. +Yes, the pattern is 3 × 0 + 1, 3 × 1 + 1, 3 × 2 + 1, 3 × 3 + 1,… +The number 0,1,2,3,4,….. are number of inner triangles. + +[7] + +Section 2.9 +Page No. 35 +Figure it out +Q.1. Write the measures of the following angles: +a. ∠ KAL +b. ∠WAL +c. ∠TAK +Ans. +a. ∠KAL = 30° +Yes, it is possible to count the number of units in 5s or 10s. + +b. ∠WAL = 50° + +c. ∠TAK = 120° + +Page No. 36 +Q. Name the different angles in the figure and write their measures. +Ans. +∠POQ = 35° + +∠POR = 95° + +∠POS = 125° + +∠POT = 160° + +∠QOR = 60° + +∠QOS = 90° + +∠QOT = 125° + +∠QOU = 145° + +∠ROS = 30° + +∠ROT = 65° + +∠ROU = 85° + +∠SOT = 35° + +∠SOU = 55° + +∠TOU = 20° + + + + +[8] + +Page No. 40 +Think! +Q. In Fig. 2.20, we have ∠AOB = ∠BOC = ∠COD = ∠DOE = ∠EOF = ∠FOG = +∠GOH = ∠HOI=_____. Why? +Ans. +Each of the angles = 22.5º +As the straight angle of 180º is divided into eight equal parts so each of the right angles +will be of measure = +180º +8 = 22.5º + +Figure it Out +Q.1. Find the degree measures of the following angles using your protractor. +Ans. ∠IHJ = ∠JHI = 47º +∠GHK = ∠IHJ = 23º +∠IHJ = ∠JHI = 108º +Q.3. Find the degree measures for the angles given below. Check if your paper protractor +can be used here! +Ans. ∠IHJ = 42º, +∠IHJ = 116º + +No, paper protractor cannot work here. +Q.4. How can you find the degree measure of the angle given below using a protractor? +Ans. +Measure of marked angle = 360º – Measure of unmarked angle += 360º - 100º = 260º +Try other ways to find the marked angle +Q.5. Measure and write the degree measures for each of the following angles: +Ans. +a. 80º +b. 120º +c. 60º +d. 130º +e. 130º +f. 60º +Q.6. Find the degree measures of ∠BXE, ∠CXE, ∠AXB and ∠BXC. +Ans. ∠BXE = 115º, +∠CXE = 85º, +∠AXB = 65º +∠BXC = 30º +Q.7. Find the degree measures of ∠PQR, ∠PQS and ∠PQT. +Ans. ∠PQR =45º +∠PQS=100º + + ∠PQT=150º. + +[9] + +Page 45 +Figure it Out +Q.1. Angles in a clock: +a. The hands of a clock make different angles at different times. At 1 o’clock, the +angle between the hands is 30°. Why? +b. What will be the angle at 2 o’clock? And at 4 o’clock? 6 o’clock? +c. Explore other angles made by the hands of a clock. +Ans. (a) The angles at the centre of the clock is 360º which is divided into 12 equal parts. So. +The angle between two successive numbers = +360 +12 = 30º +(b) At 2’O clock = 60º = 2×30 +At 4’O clock = 120º = 4×30 +At 6’O clock = 180º = 6×30 +(c) At 3’O clock = 90º +At 9’O clock = 270º +Try for other angles made by the hands of a clock. +Q.2. The angle of a door: +Is it possible to express the amount by which a door is opened using an angle? What +will be the vertex of the angle and what will be the arms of the angle? +Ans. Yes, the vertex of the angle will be the point where door meets the wall. The arms will +be the edges of the door and the wall. + +Q.3. Vidya is enjoying her time on the swing. She notices that the greater the angle with +which she starts the swinging, the greater is the speed she achieves on her swing. +But where is the angle? Are you able to see any angle? +Ans. Student may not see the angle, but when the starting arm is fixed as the position where +she starts swinging. The angle can be thought to be between positions where she starts +the swinging (the initial position) and the position where she attains the greatest position +of the swing at any one side. +Q.4. Here is a toy with slanting slabs attached to its sides; the greater the angles or slopes +of the slabs, the faster the balls roll. Can angles be used to describe the slopes of the +slabs? What are the arms of each angle? Which arm is visible and which is not? +Ans. Yes, angles can be used directly to describe slopes of the slab, larger the angle, greater +the slope of the slab. For each angle, one arm is a side and one arm is the slope. +The vertical arm is not visible, whereas the other arm is visible. +Here in this toy, edges of the slabs are the arms of the angles. Top horizontal ray is not +visible, other arms in the form of edges of the slab are visible. +Teacher should motivate students to get other possible answers. + + +[10] + +Page 49 +Section 2.10 +Figure it Out +Q.1. In Fig. 2.23, list all the angles possible. Did you find them all? Now, guess the +measures of all the angles. Then, measure the angles with a protractor. Record all +your numbers in a table. See how close your guesses are to the actual measures. +Ans. ∠CAP, ∠ACD, ∠APL, ∠DLP, ∠RPL, ∠SLP, ∠PRS, ∠LSR, ∠BRS, ∠CLP Try more! + +Page 52 +Section 2.11 +Figure it Out +Q.2. Use a protractor to find the measure of each angle. Then classify each angle as +acute, obtuse, right or reflex. +Ans. +a. ∠PTR = 30º (Acute angle) +b. ∠PTQ = 60º (Acute angle) +c. ∠PTW = 102º (Obtuse angle) +d. ∠WTP = 258º (Reflex angle) + +Let’s Explore: +Q. In this figure, ∠TER = 80°. What is the measure of ∠BET? What is the measure of +∠SET? +Ans. ∠BET = 100º, ∠SET = 10º + +Page – 53 +Figure it out +Q.3. Make any figure with three acute angles, one right angle and two obtuse angles. +Ans. +∠A, ∠B & ∠C are three acute angles +∠D, ∠F are obtuse angle +∠E is right angle + + +[11] + + +Q.4. Draw the letter ‘M’ such that the angles on the sides are 40° each and the angle in +the middle is 60°. +Ans. + + + Q.5. Draw the letter ‘Y’ such that the three angles formed are 150°, 60° and 150°. +Ans. + +Q.6. The Ashoka Chakra has 24 spokes. What is the degree measure of the angle between +two spokes +next to each other? What is the largest acute angle formed between two spokes? +Ans. The angle between two spokes next to each other is 15º. Largest acute angle between +spokes is 75º. + +Q.7. Puzzle: I am an acute angle. If you double my measure, you get an acute angle. If +you triple my measure, you will get an acute angle again. If you quadruple (four +times) my measure, you will get an acute angle yet again! But if you multiply my +measure by 5, you will get an obtuse angle measure. What are the possibilities for +my measure? +Ans. The acute angle can be 19º, 20º, 21º & 22º." +class_6,3,Number Play,ncert_books/class_6/Ganita_Prakash/fegp103.pdf,"NUMBER PLAY +3 +Numbers are used in different contexts and in many different ways +to organise our lives. We have used numbers to count, and have +applied the basic operations of addition, subtraction, multiplication +and division on them, to solve problems related to our daily lives. +In this chapter, we will continue this journey, by playing with +numbers, seeing numbers around us, noticing patterns, and learning +to use numbers and operations in new ways. +  Think about various situations where we use numbers. +List five different situations in which numbers are used. See what +your classmates have listed, share, and discuss. +3.1 Numbers can Tell us Things +What are these numbers telling us? +Some children in a park are standing in a line. Each one says a number. +  What do you think these numbers mean? + +The children now rearrange themselves, and again each one +says a number based on the arrangement. +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +56 +Did you figure out what these numbers represent? +Hint: Could their heights be playing a role? +A child says ‘1’ if there is only one taller child standing next to them. +A child says ‘2’ if both the children standing next to them are taller. +A child says ‘0’, if neither of the children standing next to them are taller. +That is each person says the number of taller neighbours they have. +  Try answering the questions below and share your reasoning. +1. Can the children rearrange themselves so that the children +standing at the ends say ‘2’? +2. Can we arrange the children in a line so that all would say +only 0s? +3. Can two children standing next to each other say the same +number? +4. There are 5 children in a group, all of different heights. Can +they stand such that four of them say ‘1’ and the last one says +‘0’? Why or why not? +5. For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible? +6. Is the sequence 0, 1, 2, 1, 0 possible? Why or why not? +7. How would you rearrange the five children so that the +maximum number of children say ‘2’? +Math +Talk +Reprint 2025-26 + +Number Play +57 +3.2 Supercells +Observe the numbers written in the table below. Why are some +numbers coloured? Discuss. +200 +577 +626 +345 +790 +694 +109 +43 +79 +75 +63 +10 +29 +28 +34 +198 +A cell is coloured if the number in it is larger than its adjacent +cells. The number 626 is coloured as it is larger than 577 and 345, +whereas 200 is not coloured as it is smaller than 577. The number +198 is coloured as it has only one adjacent cell with 109 in it, and 198 +is larger than 109. +  Figure it Out +1. Colour or mark the supercells in the table below. +6828 +670 +9435 +3780 +3708 +7308 +8000 +5583 +52 +2. Fill the table below with only 4-digit numbers such that the +supercells are exactly the coloured cells. +5346 +1258 +9635 +3. Fill the table below such that we get as many supercells as possible. +Use numbers between 100 and 1000 without repetitions. +4. Out of the 9 numbers, how many supercells are there in the table +above? ___________ +5. Find out how many supercells are possible for different +numbers of cells. +Do you notice any pattern? What is the method to fill a given +table to get the maximum number of supercells? Explore and +share your strategy. +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +58 +6. Can you fill a supercell table without repeating numbers such +that there are no supercells? Why or why not? +7. Will the cell having the largest number in a table always be a +supercell? Can the cell having the smallest number in a table +be a supercell? Why or why not? +8. Fill a table such that the cell having the second largest number +is not a supercell. +9. Fill a table such that the cell having the second largest +number is not a supercell but the second smallest number is +a supercell. Is it possible? +10. Make other variations of this puzzle and challenge your +classmates. +Let’s do the supercells activity with more rows. +Here the neighbouring cells are those that are immediately to the +left, right, top and bottom. +The rule remains the same­: a +cell becomes a supercell if the +number in it is greater than all +the numbers in its neighbouring +cells. In Table 1, 8632 is greater +than all its neighbours 4580, +8280, 4795 and 1944. +  Complete Table 2 with 5-digit +numbers whose digits are ‘1’, +‘0’, ‘6’, ‘3’, and ‘9’ in some order. +Only a coloured cell should +have a number greater than all +its neighbours. +The biggest number in the table +is ____________ . +Try +This +2430 +7500 +7350 +9870 +3115 +4795 +9124 +9230 +4580 +8632 +8280 +3446 + 5785 +1944 +5805 +6034 +Table 1 +Table 2 +96,301 +36,109 +13,609 +60,319 +19,306 +60,193 +10,963 +Reprint 2025-26 + +Number Play +59 +The smallest even number in the table is ____________. +The smallest number greater than 50,000 in the table is ____________. +Once you have filled the table above, put commas appropriately +after the thousands digit. +3.3 Patterns of Numbers on the Number Line +  We are quite familiar with number lines now. Let’s see if we can +place some numbers in their appropriate positions on the number +line. Here are the numbers: 2180, 2754, 1500, 3600, 9950, 9590, 1050, +3050, 5030, 5300 and 8400. +1000 +2000 +2180 +2754 +3000 +4000 +5000 +6000 +7000 +8000 +9000 +10,000 +  Figure it Out + +Identify the numbers marked on the number lines below, and label +the remaining positions. +b. +9996 +9997 +a. +2010 +2020 +15,077 15,078 +15,083 +c. +86,705 87,705 +d. + +Put a circle around the smallest number and a box around the +largest number in each of the sequences above. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +60 +. +3.4 Playing with Digits +We start writing numbers from 1, 2, 3 … and so on. There are nine +1-digit numbers. +  Find out how many numbers have two digits, three digits, four +digits, and five digits. +1-digit +numbers +From 1–9 +2-digit +numbers +3-digit +numbers +4-digit +numbers +5-digit +numbers +9 +Digit sums of numbers +Komal observes that when she adds up +digits of certain numbers the sum is the +same. +For example, adding the digits of the +number 68 will be same as adding the +digits of 176 or 545. + Figure it Out +1. Digit sum 14 +a. Write other numbers whose digits add up to 14. +b. What is the smallest number whose digit sum is 14? +c. What is the largest 5-digit whose digit sum is 14? +d. How big a number can you form having the digit sum +of 14? Can you make an even bigger number? +2. Find out the digit sums of all the numbers from 40 to 70. +Share your observations with the class. +3. Calculate the digit sums of 3-digit numbers whose digits are +consecutive (for example, 345). Do you see a pattern? Will this +pattern continue? +Math +Talk +Reprint 2025-26 + +Number Play +61 +Digit Detectives +After writing numbers from 1 to 100, +Dinesh wondered how many times he +would have written the digit ‘7’! +  Among the numbers 1–100, how +many times will the digit ‘7’ occur? +Among the numbers 1–1000, how +many times will the digit ‘7’ occur? +3.5 Pretty Palindromic Patterns +What pattern do you see in these numbers: 66, 848, 575, 797, 1111? +These numbers read the same from left to right and from right to left. +Try and see. Such numbers are called palindromes or palindromic +numbers. + +All palindromes using 1, 2, 3 + +The numbers 121, 313, 222 are some examples of palindromes +using the digits ‘1’, ‘2’, 3’. +  Write all possible 3-digit palindromes using these digits. +Reverse-and-add palindromes +Now, look at these additions. Try to figure out what is happening. +Steps to follow: Start with a +2-digit number. Add this number +to its reverse. Stop if you get a +palindrome or else repeat the +steps of reversing the digits and +adding. +Try +the +same +procedure +for some other numbers, and +perform the same steps. Stop if +Reprint 2025-26 + +Ganita Prakash | Grade 6 +62 +you get a palindrome. There are numbers for which you have to +repeat this a large number of times. +Are there numbers for which you do not reach a palindrome +at all? + Explore +Will reversing and adding numbers repeatedly, starting with +a 2-digit number, always give a palindrome? Explore and find +out.* + Puzzle time +tth +Write the number in words: +h +th +u +t + +I am a 5-digit palindrome. + +I am an odd number. + +My ‘t’ digit is double of my ‘u’ digit. + +My ‘h’ digit is double of my ‘t’ digit. + +Who am I? _________________ +3.6 The Magic Number of Kaprekar +D.R. Kaprekar was a mathematics teacher in a +government school in Devlali, Maharashtra. He liked +playing with numbers very much and found many +beautiful patterns in numbers that were previously +unknown. +In 1949, he discovered a fascinating and magical +phenomenon when playing with 4-digit numbers. +Math +Talk +*The answer is yes! For 3-digit numbers the answer is unknown. It is suspected that +starting with 196 never yields a palindrome! +Reprint 2025-26 + +Number Play +63 +Follow these steps and experience the magic for yourselves! Pick +any 4-digit number having at least two different digits, say 6382. +Make the largest number from these +digits. Call it A. +Make the smallest number from these +digits. Call it B. +Subtract B from A. Call it C. +C = A – B +Take a 4-digit number. +Use digits +of C +What happens if we continue +doing this? +A = 8632 + + +A = 6642 + + +A = 7641 + + +A = +B = 2368 + + +B = 2466 + + +B = 1467 + + +B = +C = 8632–2368 +C = 6642–2466 +C = 7641–1467 +C = + = 6264 + + + = 4176 + + + = 6174 + Explore +Take different 4-digit numbers and try carrying out these steps. Find +out what happens. Check with your friends what they got. +You will always reach the magic number ‘6174’! The number +‘6174’ is now called the ‘Kaprekar constant’. +Carry out these same steps with a few 3-digit numbers. What +number will start repeating? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +64 +3.7 Clock and Calendar Numbers +On the usual 12-hour clock, there are timings with different patterns. +For example, 4:44, 10:10, 12:21. +  Try and find out all possible times on a 12-hour clock of each of +these types. +Manish +has +his +birthday +on +20/12/2012 where the digits ‘2’, ‘0’, ‘1’, +and ‘2’ repeat in that order. +  Find some other dates of this form +from the past. +His +sister, +Meghana, +has +her +birthday on 11/02/2011 where the +digits read the same from left to right and from right to left. +  Find all possible dates of this form from the past. +Jeevan was looking at this year’s calendar. He started wondering, +“Why should we change the calendar every year? Can we not reuse a +calendar?”. What do you think? +You might have noticed that last year’s calendar was different +from this year’s. Also, next year’s calendar will also be different from +the previous years. +  But, will any year’s calendar repeat again after some +years? Will all dates and days in a year match exactly with +that of another year? + Figure it Out +1. Pratibha uses the digits ‘4’, ‘7’, ‘3’ and ‘2’, and makes the smallest and +largest 4-digit numbers with them: 2347 and 7432. The difference +between these two numbers is 7432 – 2347 = 5085. The sum of these +two numbers is 9779. Choose 4 - digits to make: +a. the difference between the largest and smallest numbers +greater than 5085. +Try +This +Reprint 2025-26 + +Number Play +65 +b. the difference between the largest and smallest numbers less +than 5085. +c. the sum of the largest and smallest numbers greater than 9779. +d. the sum of the largest and smallest numbers less than 9779. +2. What is the sum of the smallest and largest 5-digit palindrome? +What is their difference? +3. The time now is 10:01. How many minutes until the clock shows +the next palindromic time? What about the one after that? +4. How many rounds does the number 5683 take to reach the Kaprekar +constant? +3.8 Mental Math +Observe the figure below. What can you say about the numbers and +the lines drawn? +38,800 +3,400 +28,000 +63,000 +61,600 +19,500 +31,000 +20,900 +25,000 +400 +13,000 +1,500 +60,000 +Numbers in the middle column are added in different ways to get +the numbers on the sides (1500 + 1500 + 400 = 3400). The numbers in +the middle can be used as many times as needed to get the desired +sum. Draw arrows from the middle to the numbers on the sides to +obtain the desired sums. +Two examples are given. It is simpler to do it mentally! +38,800 = 25,000 + 400 × 2 + 13,000 + 3400 = 1500 + 1500 + 400 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +66 + Can we make 1,000 using the numbers in the middle? Why not? +What about 14,000, 15,000 and 16,000? Yes, it is possible. Explore how. +What thousands cannot be made? +Adding and Subtracting +Here, using the numbers in the boxes, we are allowed to use both addition +and subtraction to get the required number. An example is shown. +39,800 = 40,000 – 800 + 300 + 300 +45,000 = + 5,900 = +17,500 = +21,400 = +Digits and Operations +An example of adding two 5-digit numbers to get another 5-digit +number is 12,350 + 24,545 = 36,895. +An example of subtracting two 5-digit numbers to get another +5-digit number is 48,952 – 24,547 = 24,405. + Figure it Out +1. Write an example for each of the below scenarios whenever +possible. +5-digit + +5-digit to +give a 5-digit +sum more +than 90,250 +5-digit – +5-digit to give +a difference +less than +56,503 +4-digit ++ 4-digit +to give a +6-digit sum +5-digit +− 4-digit +to give +a 4-digit +difference +5-digit ++ 3-digit +to give a +6-digit sum +5-digit +– 3-digit +to give +a 4-digit +difference +5-digit ++ 5-digit +to give a +6-digit sum +5-digit +− 5-digit +to give +a 3-digit +difference +5-digit + +5-digit to +give 18,500 +5-digit − +5-digit to +give 91,500 +Math +Talk +40,000 +7,000 +300 +1,500 +12,000 +800 +Reprint 2025-26 + +Number Play +67 +Could you find examples for all the cases? If not, think and +discuss what could be the reason. Make other such questions +and challenge your classmates. +2. Always, Sometimes, Never? +Below are some statements. Think, explore and find out if +each of the statement is ‘Always true’, ‘Only sometimes true’ +or ‘Never true’. Why do you think so? Write your reasoning +and discuss this with the class. +a. 5-digit number + 5-digit number gives a 5-digit number +b. 4-digit number + 2-digit number gives a 4-digit number +c. 4-digit number + 2-digit number gives a 6-digit number +d. 5-digit number – 5-digit number gives a 5-digit number +e. 5-digit number – 2-digit number gives a 3-digit number +3.9 Playing with Number Patterns +Here are some numbers arranged in some patterns. Find out the +sum of the numbers in each of the below figures. Should we add +them one by one or can we use a quicker way? + Share and discuss in class the different methods each one of you +used to solve these questions. +40 +40 +40 +50 +50 +50 +50 +50 +50 +50 +50 +50 +50 +40 +40 +40 +40 +40 +40 +40 +40 +40 +a. +b. +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +68 +32 32 32 32 32 32 32 32 +32 32 32 32 32 32 32 32 +32 32 32 32 32 32 32 32 +32 32 32 32 32 32 32 32 +64 64 64 +64 +64 64 64 +64 +64 64 64 +64 +64 64 64 +64 +c. +d. +125 +250 +500 +500 +500 +500 +1000 +250 +250 +250 +250 +250 +250 +250 +125 +125 +125 +125 +125 +125 +125 +125 +125 +125 +125 +125 +125 +125 +125 +125 +125 +15 +15 +25 +25 +25 +25 +35 +25 +25 +25 +25 +25 +25 +25 +25 +25 +25 +25 +25 +25 +25 35 +35 +35 +35 +35 +35 +35 +35 +35 +35 +35 +35 +35 +15 +15 +15 +15 +15 +35 +35 +35 +25 +25 +25 +25 +35 +35 +35 +35 +35 +15 +15 +15 +15 +15 +15 +15 +15 +15 +15 +15 +15 +15 +15 +15 +e. +f. +3.10 An Unsolved Mystery — the Collatz Conjecture! +Look at the sequences below—the same rule is applied in all the +sequences: +a. 12, 6, 3, 10, 5, 16, 8, 4, 2, 1 +b. 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 +c. 21, 64, 32, 16, 8, 4, 2, 1 +d. 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 +Do you see how these sequences were formed? +Reprint 2025-26 + +Number Play +69 +The rule is: one starts with any number; if the number is even, +take half of it; if the number is odd, multiply it by 3 and add 1; +repeat. +Notice that all four sequences above eventually reached the +number 1. In 1937, the German mathematician, Lothar Collatz +conjectured that the sequence will always reach 1, regardless of +the whole number you start with. Even today — despite many +mathematicians working on it — it remains an unsolved problem as +to whether Collatz’s conjecture is true! Collatz’s conjecture is one of +the most famous unsolved problems in mathematics. + Make some more Collatz sequences like those above, starting +with your favourite whole numbers. Do you always reach 1? +Do you believe the conjecture of Collatz that all such sequences +will eventually reach 1? Why or why not? +3.11 Simple Estimation +At times, we may not know or need an exact count of things and +an estimate is sufficient for the purpose at hand. For example, +your school headmaster might know the exact number of students +enrolled in your school, but you may only know an estimated +count. How many students are in your school? About 150? 400? A +thousand? +Paromita’s class section has 32 children. The other 2 sections of +her class have 29 and 35 children. So, she estimated the number of +children in her class to be about 100. Along with Class 6, her school +also has Classes 7–10 and each class has 3 sections each. She assumed +a similar number in each class and estimated the number of students +in her school to be around 500. + Figure it Out + +We shall do some simple estimates. It is a fun exercise, and you may +find it amusing to know the various numbers around us. Remember, +Reprint 2025-26 + +Ganita Prakash | Grade 6 +70 +we are not interested in the exact numbers for the following questions. +Share your methods of estimation with the class. +1. Steps you would take to walk: +a. From the place you are sitting to the classroom door +b. Across the school ground from start to end +c. From your classroom door to the school gate +d. From your school to your home +2. Number of times you blink your eyes or number of breaths you +take: +a. In a minute +b. In an hour +c. In a day +3. Name some objects around you that are: +a. a few thousand in number +b. more than ten thousand in number + Estimate the answer + +Try to guess within 30 seconds. Check your guess with your friends. +1. Number of words in your maths textbook: +a. More than 5000 +b. Less than 5000 +2. Number of students in your school who travel to school by bus: +a. More than 200 +b. Less than 200 +3.  Roshan wants to buy milk and 3 types of fruit to make fruit +custard  for 5 people. He estimates the cost to be ₹100. Do you +agree with him? Why or why not? +4. Estimate the distance between Gandhinagar (in Gujarat) to +Kohima (in Nagaland). + +Hint: Look at the map of India to locate these cities. +Reprint 2025-26 + +Number Play +71 +5.  Sheetal is in Grade 6 and says she has spent around 13,000 hours +in school till date. Do you agree with her? Why or why not? +6. Earlier, people used to walk long distances as they had no other +means of transport. Suppose you walk at your normal pace. +Approximately, how long would it take you to go from: +a. Your current location to one of your favourite places nearby. +b. Your current location to any neighbouring state’s capital city. +c. The southernmost point in India to the northernmost point in +India. +7. Make some estimation questions and challenge your classmates! +3.12 Games and Winning Strategies +Numbers can also be used to play games and develop winning +strategies. +Here is a famous game called 21. Play it with a classmate. Then +try it at home with your family! + Rules for Game #1: The first player says 1, 2 or 3. Then the two +players take turns adding 1, 2, or 3 to the previous number said. The +first player to reach 21 wins! +Play this game several times with your classmate. Are you starting +to see the winning strategy? +Which player can always win if they play correctly? What is the +pattern of numbers that the winning player should say? +There are many variations of this game. Here is another common +variation: + Rules for Game #2: The first player says a number between 1 and +10. Then the two players take turns adding a number between 1 and +10 to the previous number said. The first player to reach 99 wins! +Play this game several times with your classmate. See if you can +figure out the corresponding winning strategy in this case! Which +Reprint 2025-26 + +Ganita Prakash | Grade 6 +72 +player can always win? What is the pattern of numbers that the +winning player should say this time? +Make your own variations of this game — decide how much one +can add at each turn, and what number is the winning number. Then +play your game several times, and figure out the winning strategy +and which player can always win! + Figure it Out +1. There is only one supercell +(number greater than all its +neighbours) in this grid. If you +exchange two digits of one of +the numbers, there will be 4 +supercells. Figure out which +digits to swap. +2. How many rounds does your year of birth take to reach the +Kaprekar constant? +3. We are the group of 5-digit numbers between 35,000 and 75,000 +such that all of our digits are odd. Who is the largest number in our +group? Who is the smallest number in our group? Who among us +is the closest to 50,000? +4. Estimate the number of holidays you get in a year including +weekends, festivals and vacation. Then, try to get an exact number +and see how close your estimate is. +5. Estimate the number of liters a mug, a bucket and an overhead +tank can hold. +6. Write one 5-digit number and two 3-digit numbers such that their +sum is 18,670. +7. Choose a number between 210 and 390. Create a number pattern +similar to those shown in Section 3.9 that will sum up to this number. +16,200 +39,344 +29,765 +45,306 +23,609 +19,381 +50,319 +38,408 +62,871 +Try +This +Reprint 2025-26 + +Number Play +73 + 8. Recall the sequence of Powers of 2 from Chapter 1, Table 1. Why is +the Collatz conjecture correct for all the starting numbers in this +sequence? + 9. Check if the Collatz Conjecture holds for the starting number 100. +10. Starting with 0, players alternate adding numbers between 1 and 3. +The first person to reach 22 wins. What is the winning strategy now? +Su m m a r y + + +Numbers can be used for many different purposes including, to convey +information, make and discover patterns, estimate magnitudes, pose +and solve puzzles, and play and win games. + + +Thinking about and formulating set procedures to use numbers for +these purposes is a useful skill and capacity (called ‘computational +thinking’). + + +Many problems about numbers can be very easy to pose, but very +difficult to solve. Indeed, numerous such problems are still unsolved +(for example, Collatz’s Conjecture). +Reprint 2025-26 + +[1] + +CHAPTER 3 — SOLUTIONS +Number Play + +Q. Think about various situations where we use numbers. List five different situations in +which numbers are used. See what your classmates have listed, share, and discuss. +Ans. +Five different possible situations in which numbers are used - +1. +Time +2. +Calendar +3. +Counting objects/Marks +4. +Measurement of height & weight +5. +Money +There could many more. + +Section 3.1 +Page No. 55 +Q. What do you think these numbers mean? +Ans. +Refer page 56. + Page No. 56 +Q1. Can the children rearrange themselves so that the children standing at the ends say +‘2’? +Ans. No; There will be no one standing on the other side of the child standing at the end. +Q2. Can we arrange the children in a line so that all would say only 0s? +Ans. Yes; All the children in the line should be of same height. +Q3. Can two children standing next to each other say the same number? +Ans. Yes; Refer picture on page 55. +Q4. There are 5 children in a group, all of different heights. Can they stand such that four +of them say ‘1’ and the last one says ‘0’? Why or why not? +Ans. Yes, they can, if they are standing in ascending order of height. + + + + +[2] + +Q5. For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible? +Ans. +No; the tallest child at the end cannot say1. + +Q6. Is the sequence 0, 1, 2, 1, 0 possible? Why or why not? +Ans. +Yes, it is possible. + +Q7. How would you rearrange the five children so that the maximum number of children +say ‘2’? +Ans. +At the most only 2 children can say 2 as given is the following arrangement. + +Section 3.2 +Page No. 57 +Figure it out +Q1. Colour or mark the supercells in the table below. +6828 +670 +9435 +3780 +3708 +7308 +8000 +5583 +52 + +Ans. + +6828 +670 +9435 +3780 +3708 +7308 +8000 +5583 +52 + + +[3] + +Q2. Fill the table below with only 4-digit numbers such that the supercells are exactly the +coloured cells. +5346 + + +1258 + + + +9635 + + +Ans. One of the ways could be-5346; 9636.Try more +5346 +5347 +1000 +1258 +1100 +1200 +1300 +9635 +9636 + +Q3. Fill the table below such that we get as many supercells as possible. Use numbers +between 100 and 1000 without repetitions. + + + + + + + + + +Ans. + +Q4. Out of the 9 numbers, how many supercells are there in the table above? ___________ +Ans. 5 +Q5 Find out how many supercells are possible for different numbers of cells. +Do you notice any pattern? What is the method to fill a given table to get the +maximum number of supercells? Explore and share your strategy. +Ans. +For even number of cells say,2,4,6,… the number of supercells would be respectively, + 2/2 =1,4/2 =2,6/2=3,… +For odd number of cells , say 1,3,5,7,… the number of supercells would be respectively +(1+1)/2= 1, (3+1)/2 = 2, (5+1)/2= 3,(7+1)/2 = 4,… +To get the maximum number ofsupercells, we have to start by filling the first cell as +super cell & then fill alternately. +Q6. Can you fill a supercell table without repeating numbers such that there are no +supercells? Why or why not? +Ans. No; the cell which is filled by the greatest number among the given numbers chosen, will +become super cell irrespective of its position in the table. +Q7. Will the cell having the largest number in a table always be a supercell? Can the cell +having the smallest number in a table be a supercell? Why or why not? +Ans. +Yes, the largest number in a table will always be a supercell. +No, the smallest number in a table can never be a supercell as the number in all the +adjacent cells will be greater than it. +Q8. Fill a table such that the cell having the second largest number is not a supercell. +Ans. +One of the ways could be- +1 2 3 4 5 6 7 9 8 + +110 +100 +150 +130 +280 +200 +230 +210 +270 + +[4] + +Q9. Fill a table such that the cell having the second largest number is not a supercell but +the second smallest number is a supercell. Is it possible? +Ans. +One of the ways is- +Second smallest number +a super cell + +2 1 3 4 5 6 7 9 8 +Second largest number +8 is not a supercell. + + +Q10. Make other variations of this puzzle and challenge your classmates. +Ans. Some of these could be- +Can you fill the table with 9 cells such that there are more than 5 super cells? +Can you fill the table with 9 cells such that there are exactly 4 super cells? + +Page No. 58 +Q. Complete Table 2 with 5-digit numbers whose digits are ‘1’, ‘0’, ‘6’, ‘3’, and ‘9’ in +some order. Only a coloured cell should have a number greater than all its +neighbours. +Ans. Table 2 (One of the ways) – +96,310 96,301 36,109 39,160 +96,103 13,609 60,319 19,306 +13,906 10,396 60,193 60,931 +10,369 10,963 10,936 69,031 + +Q. The biggest number in the table is ____________. +Ans. The biggest number in the table is 96,310 +Q. The smallest even number in the table is ____________. +Ans. The smallest even number in the table is 10,396 +Q. The smallest number greater than 50,000 in the table is ____________. +Ans. The smallest number greater than 50,000 in the table is 60,193. + +Section 3.3 +Page no.59 + Q. We are quite familiar with number lines now. Let’s see if we can place some numbers +in their appropriate positions on the number line. Here are the numbers: 2180, 2754, +1500, 3600, 9950, 9590, 1050, 3050, 5030, 5300 and 8400. +Ans. + + +1000 +2000 +3000 +4000 +5000 +6000 +7000 +8000 +9000 +10000 +1050 +1500 +2180 2754 +3600 +5030 +5300 +8400 +9590 +9950 + +[5] + + Q. Identify the numbers marked on the number lines below, and label the remaining +positions. + + (a). + + (b). + + + +(c). + + +(d). + + +Put a circle around the smallest number and a box around the largest number in each of the +sequences above. + +Section 3.4 +Page no. 60 + +Q. Find out how many numbers have two digits, three digits, four digits, and five digits: +1 +Digit +numbers +2 +Digit +numbers +3 +Digit +numbers +4 +Digit +numbers +5 +Digit +numbers +9 + + + + + +Ans. +1 Digit numbers 2 Digit numbers 3 Digit numbers 4 Digit numbers 5 Digit numbers +9 +90 +900 +9,000 +90,000 + + + + + 1990 + 1995 +2000 + 2005 + 2010 + 2015 +2020 +2025 +2030 + 2035 + 9993 + 9994 +9995 + 9996 + 9997 + 9998 +9999 +10000 +10001 + 10002 + 15077 15078 +15079 + 15080 +15081 + 15082 +15083 +15084 +15085 + 15086 + 83705 84705 +85705 + 86705 + 87705 88705 +89705 +90705 +91705 +92705 + +[6] + + Figure it out +Q.1. Digit sum 14 +a. Write other numbers whose digits add up to 14. +b. What is the smallest number whose digit sum is 14? +c. What is the largest 5-digit number whose digit sum is 14? +d. How big a number can you form having the digit sum 14? Can you make an even +bigger number? +Ans. +a. Some such numbers are:248, 653, 356, 815, 833, 12335, 23351. +b. 59 +c. 95000 +d. 95, 9005, 900005, 90000005, 9000000005, 90000000000005 … + +Q.3. Calculate the digit sums of 3-digit numbers whose digits are consecutive (for +example, 345). Do you see a pattern? Will this pattern continue? +Ans. +123 → 1+2+3 = 6 +234 → 2+3+4 = 9 +345 → 3+4+5 = 12 +456 → 4+5+6 = 15 +567 → 5+6+7 = 18 +678 → 6+7+8 = 21 +789 → 7+8+9 = 24 +• Yes, there is a pattern, all the sums are multiples of 3. +• No. + +Page no. 61 + Q. Among the numbers 1–100, how many times will the digit ‘7’ occur? Among the +numbers 1–1000, how many times will the digit ‘7’ occur? +Ans. + +• 20 times. +• 300 times. + + + + + + + +[7] + +Section 3.5 +Page no. 61 + Q. Write all possible 3-digit palindromes using these digits. +Ans. Palindromes: +111, 121, 131 + + + +222, 212, 232 + + + +313, 323, 333 + +Explore +Page no. 62 + Q. Will reversing and adding numbers repeatedly, starting with a 2-digit number, +always give a palindrome? Explore and find out.* + Some of these are- + 12 ++21 + 33 + + 47 ++74 +121 + +Try more +Yes, it will always give a palindrome. + + Puzzle time +Q. I am a 5-digit palindrome. +I am an odd number. +My ‘t’ digit is double of my ‘u’ digit. +My ‘h’ digit is double of my ‘t’ digit. +Who am I? _________________ +Ans. +tth + th + h + t + u + + +Twelve thousand four hundred twenty one. + +Section 3.6 +Page no. 63 + + +Q. Carry out these same steps with a few 3-digit numbers. What number will start +repeating? +Ans. +Take a 3- Digit number say, 321. +321 +-123 +198 + 981 + -189 + 792 + 972 + -279 + 693 + 963 + -369 + 594 +954 + -459 +495 +954 + -459 +495 +The number 495 starts repeating. + +Try for other 3-digit numbers. +1 +2 +4 +2 +1 + +[8] + +Section 3.7 +Page no. 64 +Q. Try and find out all possible times on a 12-hour clock of each of these types. +Ans. + 4:44 +2:22 + +3:33 + + + +10:10 +11:11 +12:12 +09:09 + +12:21 +05:50 +10:01 + Think of some more! + +Q. Find some other dates of this form from the past. +Ans. +20/04/2004, +20/06/2006, + +Try for yourself! + +Q. Find all possible dates of this form from the past. +Ans. +01/02/2001, +02/02/2002, +Think of some more! + +Q. Will any year’s calendar repeat again after some years? Will all dates and days in a +year match exactly with that of another year? +Ans. +Yes, +The calendar repeats itself after 6 years if only one leap year is included in these 6 years. +If 2 leap years are included, then it will repeat after 5 years. + + + +Page no. 64 + +Figure it out +Q.1. Pratibha uses the digits ‘4’, ‘7’, ‘3’ and ‘2’, and makes the smallest and largest 4- +digit numbers with them: 2347 and 7432. The difference between these two +numbers is 7432 – 2347 = 5085. The sum of these two numbers is 9779. Choose 4– +digits to make: +a. the difference between the largest and smallest numbers greater than 5085. +b. the difference between the largest and smallest numbers less than 5085. +c. the sum of the largest and smallest numbers greater than 9779. +d. the sum of the largest and smallest numbers less than 9779. +Ans. Some of the possibilities are– +a. 7431 – 1347 = 6084 +b. 7433 – 3347 = 4086 +c. 7433 + 3347 = 10780 +d. 7431 + 1347 = 8778 + +Q.2. What is the sum of the smallest and largest 5-digit palindrome? What is their +difference? +Ans. Smallest 5 digit palindrome = 10001 +largest 5 digit palindrome = 99999 + +Sum = 10001 + 99999 = 110,000 + +Difference = 99999 – 10001 = 89,998 + + + +[9] + +Q.3. The time now is 10:01. How many minutes until the clock shows the next palindromic +time? What about the one after that? +Ans. Time Now → 10:01 +Next palindrome time → 11:11 +After 1 hr. 10 min = 70 min the clock will show next palindrome time. +Next palindrome time = 12:21 which will occur after 2 hr. 20 min = 140 min from +10:01. + +Q.4. How many rounds does the number 5683 take to reach the Kaprekar constant? +Ans. 5683 + 8653 +-3568 + 5085 + 8550 +-5058 + 3492 + 9432 +-2349 + 7083 + 8730 +-3078 + 5652 + 6552 +-2556 + 3996 + 9963 +-3699 + 6264 + 6642 +-2466 + 4176 + 7641 +-1467 + 6174 +1 +2 +3 +4 +5 +6 +7 +8 +It will take 8 rounds to reach the Kaprekar constant. + +Page no. 66 +Section 3.8 +Q. Can we make 1,000 using the numbers in the middle? Why not? What about 14,000, +15,000 and 16,000? Yes, it is possible. Explore how. What thousands cannot be made? +Ans. No; the only number which is smaller than 1000 is 400 and 1000 is not a multiple of 400. +14000 = 1500 × 8 + 400 × 5 + += 12000 + 2000 + += 14000 +15000 = 13000 + 400 × 5 + += 13000 + 2000 + += 15000 +16000 = 1500 × 8 + 400 × 10 + += 12000 + 4000 + += 16000 +Only one thousand (1000) cannot be made. + + +Figure it out + +Q.1. Write an example for each of the below scenarios whenever possible. +Could you find examples for all the cases? If not, think and discuss what could be +the reason. Make other such questions and challenge your classmates. +Ans. +• 5 digit + 5 digit > 90,250 +45,000 + 45,400 = 90,400 > 90,250 +• 5 digit + 3 digit = 6 digit sum +99,999 + 999 = 100,998 +• 4 digit + 4 digit = 6 digit sum +Not possible as even the sum of the greatest 4 digit numbers will not give a six +digit sum. (9999 + 9999 = 19,998) + +[10] + +• 5 digit + 5 digit = 6 digit sum +60,000 + 40,000 = 1,00,000 +• 5 digit + 5 digit = 18,500 +Not possible as smallest 5-digit number is 10,000. +If both the numbers are 10,000 then the sum is 20,000, which is more than +18,500. +• 5 digit – 5 digit < 56,503 +80,000 – 50,000 < 56,503 + + + < 56,503 +• 5 digit – 3 digit = 4 digit difference +10,000 – 999 = 9001 +• 5 digit – 4 digit = 4 digit difference +12,000 – 2,500 = 9,500 +• 5 digit – 5 digit = 3 digit difference +50,999 – 50,000 = 999 +• 5 digit – 5 digit = 91,500 +Not possible as the difference of the greatest and the smallest 5 digit numbers, +the maximum difference, can be 99,999 – 10,000 = 89,999 +Some examples of other such questions are- +1. 5 digit + 5 digit = 7 digit sum +2. 4 digit + 4 digit = 2900 +More such examples can be made. + +Q.2. Always, Sometimes, Never? +Below are some statements. Think, explore and find out if each of the statement is +‘Always true’, ‘Only sometimes true’ or ‘Never true’. Why do you think so? Write +your reasoning; discuss this with the class. +a. 5-digit number + 5-digit number gives a 5-digit number +b. 4-digit number + 2-digit number gives a 4-digit number +c. 4-digit number + 2-digit number gives a 6-digit number +d. 5-digit number – 5-digit number gives a 5-digit number +e. 5-digit number – 2-digit number gives a 3-digit number +Ans. a. Only sometimes true. + 20,000 + 80,000 = 1,00,000 not a 5 digit number +b. Only sometimes true + 9,999 + 99 = 10,098 not a 4 digit number +c. Never true + 9,999 + 99 = 10,098 +On adding the greatest 4 digit and the greatest 2 digit numbers, we can reach only 5 +digit number 10,098. So there is no possibility of getting a 6 digit number. +d. Only sometimes true +Ex. 12,000 – 10,000 = 2,000 +5 digit – 5 digit = 4 digit +e. Never true + Ex. 10,000 – 99 = 9901 + +[11] + +Even if the greatest 2 digit numbers is subtracted from the smallest 5 digit number, 4 +digit number will be obtained. + +Page no. 69 +Section 3.10 + Q. Make some more Collatz sequences like those above, starting with your favourite +whole numbers. Do you always reach 1? +Do you believe the conjecture of Collatz that all such sequences will eventually +reach 1? Why or why not? +Ans. a) 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 + b) 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 +Yes we always reach 1. +The even numbers are halved and when we have an odd number we convert it into +an even number by multiplying by 3 and adding 1 so that eventually it can be halved +again. The smallest even number is 2 so we will reach 1 for sure. + +Q.3. Name some objects around you that are: +a. a few thousand in number +b. more than ten thousand in number +Ans. +A few thousands: car numbers, 4 digit pin +More than ten thousands: Salary, Mobile numbers. + +Q.3. Roshan wants to buy milk and 3 types of fruit to make fruit custard for 5 people. He +estimates the cost to be ₹ 100. Do you agree with him? Why or why not? +Ans. +Yes, it is possible with less quantity of serving and with purchase of 1-1-1 fruit of each +type. +However, it is not possible with costly fruits and more quantity of serving. + +Q.4. Estimate the distance between Gandhinagar (in Gujarat) to Kohima (in Nagaland). +Ans. +2500 kilometer + +Q. 5. Sheetal is in Grade 6 and says she has spent around 13,000 hours in school till date. +Do you agree with her? Why or why not? +Ans. +No, I do not agree with her. +There are 6 school hours in a day and around 200 working days in a year. +𝟏𝟑,𝟎𝟎𝟎 +𝟔×𝟐𝟎𝟎 = 10.8 years +(Nursery, KG, 1,2,3,4,5,6) 8 years +She is in school for 8 years, 13,000 hours is way too high. + +Q.7. Make some estimation questions and challenge your classmates! +Ans. Some such are +• How many students are there in your school? +• How many hours does a person sleep in his lifetime on an average? +(More such questions can be made) + + + +[12] + +Section 3.12 +Page No. 72 +Figure it Out +Q.1. There is only one supercell (number greater than all its neighbours) in this grid. If +you +exchange two digits of one of the numbers, there will be 4 supercells. Figure out +which digits to swap. +16,200 +39,344 +29,765 +23,609 +62,871 +45,306 +19,381 +50,319 +38,408 + +Ans. +If I exchange the digits 1 and 6 in the number 62,871 then there will be 4 Super cells. + +16,200 +39,344 +29,765 +23,609 +12,876 +45,306 +19,381 +50,319 +38,408 + +Q.2. How many rounds does your year of birth take to reach the Kaprekar constant? +Ans. +Suppose the birth year is 1980, then, - + 9810 +-1089 + 8721 + 8721 +-1278 + 7443 + 7443 +-3447 + 3996 + 9963 +-3699 + 6264 + 6642 +-2466 + 4176 + 7641 +-1467 + 6174 + + +It takes 6 rounds. (Try now for your year of birth.) + +Q.3. We are the group of 5-digit numbers between 35,000 and 75,000 such that all of our +digits are odd. Who is the largest number in our group? Who is the smallest number +in our group? Who among us is the closest to 50,000? +Ans. + + +With repeating digit +With non repeating digit +Largest number → +73,999 + +73,951 +Smallest number → +35,111 +35,179 + + +With repeating digit +With non repeating digit +Closest to 50000 +51,111 +51,379 + +Q.6. Write one 5-digit number and two 3-digit numbers such that their sum is 18,670. +Ans. +18000 + 300 + 370 = 18670. Try for more. + + + + + + +[13] + +Q.7. Choose a number between 210 and 390. Create a number pattern similar to those +shown in Section 3.9 that will sum up to this number. +Ans. +Number Chosen: 250 + + + +25 + +25 + +50 + +50 + +50 + + +25 + +25 + + + Or + + +10 + +10 + +10 + +10 + +10 +10 + +10 + +10 + +10 + +10 +10 + +10 + +10 + +10 + +10 +10 + +10 + +10 + +10 + +10 +10 + +10 + +10 + +10 + +10 + +Q.8. Recall the sequence of Powers of 2 from Chapter 1, Table 1. Why is the Collatz +conjecture correct for all the starting numbers in this sequence? +Ans. +when we divide 28 = 2x2x2x2x2x2x2x2 by 2 it become 27 and every time you divide +by 2 the same will continue happening, until you are left with 2 which when divided by +2 will leave 1. + +Q.9. Check if the Collatz Conjecture holds for the starting number 100. +Ans. 100, 50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, +2, 1. + +Q.10 Starting with 0, players alternate adding numbers between 1 and 3. The first person +to reach 22 wins. What is the winning strategy now? +Ans. +Winning strategy is to be the first player." +class_6,4,Data Handling and Presentation,ncert_books/class_6/Ganita_Prakash/fegp104.pdf,"If you ask your classmates about their favourite colours, you will +get a list of colours. This list is an example of data. Similarly, if you +measure the weight of each student in your class, you would get a +collection of measures of weight—again data. + +Any collection of facts, numbers, measures, observations or other +descriptions of things that convey information about those things is +called data. + +We live in an age of information. We constantly see large amounts +of data presented to us in new and interesting ways. In this chapter, +we will explore some of the ways that data is presented, and how we +can use some of those ways to correctly display, interpret and make +inferences from such data! +4.1 Collecting and Organising Data +Navya and Naresh are discussing their favourite games. +Cricket is my +favourite game! +I play cricket sometimes +but hockey is the game I +like the most. +I think cricket is the +most popular game in +our class. +I am not sure. How can we find +the most popular game in our +class? +Data Handling and +Presentation +4 +Reprint 2025-26 + +Data Handling and Presentation +75 +To figure out the most popular game in +their class, what should Navya and Naresh +do? Can you help them? + Naresh and Navya decided to go to each student in the class and +ask what their favourite game is. Then they prepared a list. + +Navya is showing the list: + +She says (happily), “I have collected the data. I can figure out the +most popular game now!”. + +A few other children are looking at the list and wondering, “We +can’t yet see the most popular game. How can we get it from this list?”. + Figure it Out +1. +What would you do to find the most popular game among Naresh’s +and Navya’s classmates? +2. +What is the most popular game in their class? +3. + Try to find out the most popular game among your classmates. +4. + Pari wants to respond to the questions given below. Put a tick () +for the questions where she needs to carry out data collection and +Mehnoor – Kabaddi +Pushkal – Satoliya (Pittu) +Anaya – Kabaddi +Jubimon – Hockey +Densy – Badminton +Jivisha – Satoliya (Pittu) +Simran – Kabaddi +Jivika – Satoliya (Pittu) +Rajesh – Football +Nand – Satoliya (Pittu) +Leela – Hockey +Thara – Football +Ankita – Kabaddi +Afshan – Hockey +Soumya – Cricket +Imon – Hockey +Keerat – Cricket +Navjot – Hockey +Yuvraj – Cricket +Gurpreet – Hockey +Hemal – Satoliya (Pittu) +Rehana – Hockey +Arsh – Kabaddi +Debabrata – Football +Aarna – Badminton +Bhavya – Cricket +Ananya – Hockey +Kompal – Football +Sarah – Kabaddi +Hardik – Cricket +Tahira – Cricket +75 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +76 +put a cross () for the questions where she doesn’t need to collect +data. Discuss your answers in the classroom. +a. +What is the most popular TV show among her classmates? +b. +When did India get independence?  +c. + How much water is getting wasted in her locality?  +d. +What is the capital of India?  + +Shri Nilesh is a teacher. He decided to bring sweets to the class +to celebrate the new year. The sweets shop nearby has jalebi, gulab +jamun, gujiya, barfi, and rasgulla. He wanted to know the choices of +the children. He wrote the names of the sweets on the board and +asked each child to tell him their preference. He put a tally mark +‘|’ for each student and when the count reached 5, he put a line +through the previous four and marked it as ||||. +Sweets +Tally Marks +No. of Students + Jalebi +| +|||| +6 +Gulab jamun +|||| +|||| +9 +Gujiya +||| +|||| |||| +____________ +Barfi +||| +____________ +Rasgulla +|| +|||| +____________ + Figure it Out + + 1. +Complete the table to help Shri Nilesh to purchase the correct +numbers of sweets: +a. +How many students chose jalebi? +b. Barfi was chosen by + students? +c. +How many students chose gujiya? + +d. Rasgulla was chosen by + students? +e. +How many students chose gulab jamun? +Reprint 2025-26 + +Data Handling and Presentation +77 + +Shri Nilesh requested one of the staff members to bring the sweets as +given in the table. The above table helped him to purchase the correct +numbers of sweets. + + 2. +Is the above table sufficient to distribute each type of sweet to +the correct student? Explain. If it is not sufficient, what is the +alternative? +To organise the data, we can write the name of each sweet in one +column and using tally signs, note the number of students who prefer +that sweet. The numbers 6, 9, … are the frequencies of the sweet +preferences for jalebi, gulab jamun … respectively. + +Sushri Sandhya asked her students about the sizes of the shoes +they wear. She noted the data on the board. +4 +5 +3 +4 +3 +4 +5 +5 +4 +5 +5 +4 +5 +6 +4 +3 +5 +6 +4 +6 +4 +5 +7 +5 +6 +4 +5 + +She then arranged the shoe sizes of the students in ascending order — + +3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7 + Figure it Out + + 1. +Help her to figure out the following: +a. +The largest shoe size in the class is _________. +b. The smallest shoe size in the class is _________. +c. +There are _________ students who wear shoe size 5. +d. There are _________ students who wear shoe sizes larger +than 4. + + 2. +How did arranging the data in ascending order help to +answer these questions? + + 3. +Are there other ways to arrange the data? +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +78 + +4. +Write the names of a few trees you see around you. When you +observe a tree on the way from your home to school (or while +walking from one place to another place), record the data and fill +in the following table: +Tree +No. of Trees +Peepal +Neem +… +…. +a. +Which tree was found in the greatest number? +b. +Which tree was found in the smallest number? +c. +Were there any two trees found in the same numbers? + +5. +Take a blank piece of paper and paste any small news item from +a newspaper. Each student may use a different article. Now, +prepare a table on the piece of paper as given below. Count the +number of each of the letters ‘c’, ‘e’, ‘i’, ‘r’, and ‘x’ in the words of +the news article, and fill in the table. +Letter +c +e +i +r +x +Any other letter +of your choice +Number of times +found in the news item +a. +The letter found the most number of times is ________. +b. +The letter found the least number of times is ________. +c. +List the five letters ‘c’, ‘e’, ‘i’, ‘r’, ‘x’ in ascending order of +frequency. Now, compare the order of your list with that of +your classmates. Is your order the same or nearly the same +as theirs? (Almost everyone is likely to get the order ‘x, c, r, +i, e’.) Why do you think this is the case? +Reprint 2025-26 + +Data Handling and Presentation +79 +d. +Write the process you followed to complete this task. +e. +Discuss with your friends the processes they followed. +f. +If you do this task with another news item, what process +would you follow? +Provide more opportunities to collect and organise data. Ask students +to guess what is the most popular colour, game, toy, school subject, etc., +amongst the students in their classroom and then collect the data for it. +It can be a fun activity in which they also learn about their classmates. +Discuss how they can organise the data in different ways, each way +having its own advantages and limitations. For all these tasks and the +tasks under ‘Figure it Out’, discuss the tasks with the children and let +them understand the tasks, and then let them plan and present their +research processes and conclusions in the class. +Teacher’s Note +4.2 Pictographs +Pictographs are one visual and suggestive way to represent data +without writing any numbers. Look at this picture — you may be +familiar with it from previous classes. + Modes of Travelling + Number of Students + = 1 Student +Private car + + + + +Public bus + + + + + +School bus + + + + + + + + + + + +Cycle + + + +Walking + + + + + + + + +This picture helps you understand at a glance the different +modes of travel used by students. Based on this picture, answer the +following questions: +Reprint 2025-26 + +Ganita Prakash | Grade 6 +80 +•• +Which mode of travel is used by the most number of students? +•• +Which mode of travel is used by the least number of students? + +A pictograph represents data through pictures of objects. It helps +answer questions about data with just a quick glance. + +In the above pictograph, one unit or symbol ( ) is used to represent +one student. There are also other pictographs where one unit or +symbol stands for many people or objects. + Example: Nand Kishor collected responses from the children of +his middle school in Berasia regarding how often they slept at least +9 hours during the night. He prepared a pictograph from the data: +Response +Number of Children ( + = 10 Children) +Always + + + + +Sometimes + + +Never + + + + +Answer the following questions using the pictograph: +1. +What is the number of children who always slept at least 9 +hours at night? +2. +How many children sometimes slept at least 9 hours at night? +3. +How many children always slept less than 9 hours each night? +Explain how you got your answer. +Solutions +1. + In the table, there are 5 pictures + for ‘Always’. Each picture + represents 10 children. Therefore, 5 pictures indicate +5 × 10 = 50 children. +2. +There are 2 complete pictures + (2 × 10 = 20) and a half +picture + (half of 10 = 5). Therefore, the number of children +who sleep at least 9 hours only sometimes is 20 + 5 = 25. +Reprint 2025-26 + +Data Handling and Presentation +81 +3. +There are 4 complete pictures for ‘Never’. Hence, 4 × 10 = +40 children never sleep at least 9 hours in a night, i.e., they +always sleep less than 9 hours. +Drawing a Pictograph +One day, Lakhanpal collected data on how many students were +absent in each class: +Class +I +II +III +IV +V +VI +VII +VIII +No. of +students +3 +5 +4 +2 +0 +1 +5 +7 + +He created a pictograph to present this data and decided to show +1 student as + in the pictograph — +VIII +VII +VI +V +IV +III +No. of students absent +Classes += 1 student +II +I + +Meanwhile, his friends Jarina and Sangita collected data on how +many students were present in each class: +Class +I +II +III +IV +V +VI +VII +VIII +No. of +students +30 +35 +20 +25 +30 +25 +30 +20 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +82 + If they want to show their data through a pictograph, where they +also use one symbol + for each student, as Lakhanpal did, what are +the challenges they might face? + +Jarina made a plan to address this — since there were many +students, she decided to use + to represent 5 students. She figured +that would save time and space too. +VIII +VII +VI +V +IV +III +No. of students present +Classes += 5 students +II +I + +Sangita decided to use one + to represent 10 students instead. + +Since she used one + to show 10 students, she had a problem in +showing 25 students and 35 students in the pictograph. Then, she +realised she could use +to show 5 students. +VIII +VII +VI +V +IV +III +No. of students present +Classes += 10 students +II +I +Reprint 2025-26 + +Data Handling and Presentation +83 +  What could be the problems faced in preparing such a +pictograph, if the total number of students present in a +class is 33 or 27? +•• +Pictographs are a nice visual and suggestive way to represent +data. They represent data through pictures of objects. +•• +Pictographs can help answer questions and make inferences +about data with just a quick glance (in the examples above— +about favourite games, favourite colours, most common modes +of conveyance, number of students absent, etc.). +•• +By reading a pictograph, we can quickly understand the +frequencies of the different categories (for example, cricket, +hockey, etc.) and the comparisons of these frequencies. +•• +In a pictograph, the categories can be arranged horizontlly or +vertically. For each category, simple pictures and symbols are +then drawn in the designated columns or rows according to the +frequency of that category. +•• +A scale or key (for example, + : 1 student or + : 5 students) +is added to show what each symbol or picture represents. Each +symbol or picture can represent one unit or multiple units. +•• +It can be more challenging to prepare a pictograph when the +amount of data is large or when the frequencies are not exact +multiples of the scale or key. + Figure it Out +1. The following pictograph shows the number of books borrowed +by students, in a week, from the library of Middle School, +Ginnori: +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +84 +Day +Number of Books Borrowed +( +=1 Book ) +Monday + + + + + +Tuesday + + + + +Wednesday + +Thursday +Friday + + + + + +Saturday + + + + + + + + +a. +On which day were the minimum number of books borrowed? +b. +What was the total number of books borrowed during the +week? +c. +On which day were the maximum number of books borrowed? +What may be the possible reason? +2. Magan Bhai sells kites at Jamnagar. Six shopkeepers from nearby +villages come to purchase kites from him. The number of kites he +sold to these six shopkeepers are given below — +Shopkeeper +Number of Kites Sold +Chaman +250 +Rani +300 +Rukhsana +100 +Jasmeet +450 +Jetha Lal +250 +Poonam Ben +700 +Reprint 2025-26 + +Data Handling and Presentation +85 + +Prepare a pictograph using the symbol + to represent 100 kites. + +Answer the following questions: +a. +How many symbols represent the kites that Rani purchased? +b. +Who purchased the maximum number of kites? +c. +Who purchased more kites, Jasmeet or Chaman? +d. +Rukhsana says Poonam Ben purchased more than double the +number of kites that Rani purchased. Is she correct? Why? +4.3 Bar Graphs +Have +you +seen +graphs +like this on TV or in a +newspaper? + +Like pictographs, such +bar graphs can help us to +quickly understand and +interpret +information, +such as the highest value, +the comparison of values +of different categories, etc. +However, when the amount +of data is large, presenting +it by a pictograph is not +only time consuming but +at times difficult too. Let us +see how data can be presented instead using a bar graph. + +Let’s take the data collected by Lakhanpal earlier, regarding the +number of students absent on one day in each class: +Class +I +II +III +IV +V +VI +VII +VIII +No. of +students +3 +5 +4 +2 +0 +1 +5 +7 +Source: https://www.statista.com/chart/17122/ +number-of-threatened-species-red-list/ +Reprint 2025-26 + +Ganita Prakash | Grade 6 +86 + +He presented the same data using a bar graph: + +1 unit length = 1 student +Number of students +Class +No. of students absent in each class +1 +2 +3 +4 +5 +6 +7 +8 +0 +1 +2 +3 +4 +5 +6 +7 +8 +If the students have not noticed, please point out the equally spaced +horizontal lines. Explain that this means that each pair of consecutive +numbers on the left has the same gap. +Teacher’s Note + Answer the following questions using the bar graph: +1. +In Class 2, ___________ students were absent that day. +2. +In which class were the maximum number of students absent? +___________ +3. +Which class had full attendance that day? ___________ + +When making bar graphs, bars of uniform width can be drawn +horizontally or vertically with equal spacing between them; then the +Reprint 2025-26 + +Data Handling and Presentation +87 +length or height of each bar represents the given number. As we saw in +pictographs, we can use a scale or key when the frequencies are larger. + +Let us look at an example of vehicular traffic at a busy road crossing +in Delhi, which was studied by the traffic police on a particular day. +The number of vehicles passing through the crossing each hour from +6 a.m. to 12:00 noon is shown in the bar graph. One unit of length +stands for 100 vehicles. +100 +600 +200 +700 +300 +800 +400 +900 +1100 +500 +1000 +1200 +Time intervals +Number of vehicles +11–12 +10–11 +9–10 +8–9 +7–8 +6–7 + +We can see that the maximum traffic at the crossing is shown +by the longest bar, i.e., for the time interval 7–8 a.m. The bar graph +shows that 1200 vehicles passed through the crossing at that time. +The second longest bar is for 8–9 a.m. During that time, 1000 vehicles +passed through the crossing. Similarly, the minimum traffic is shown +by the smallest bar, i.e., the bar for the time interval 6–7 a.m. During +that time, only about 150 vehicles passed through the crossing. The +second smallest bar is that for the time interval 11 a.m.–12 noon, +when about 600 vehicles passed through the crossing. + +The total number of cars passing through the crossing during the +two-hour interval 8.00–10.00 a.m. as shown by the bar graph is about +1000 + 800 = 1800 vehicles. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +88 + Figure it Out +1. How many total cars passed through the crossing between 6 a.m. +and noon? +2. Why do you think so little traffic occurred during the hour of +6–7 a.m., as compared to the other hours from 7 a.m.–noon? +3. Why do you think the traffic was the heaviest between 7–8 a.m.? +4. Why do you think the traffic was lesser and lesser each hour after +8 a.m. all the way until noon? +Example: +Years +Population of India in crores +110 +36 +44 +54 +68 +84 +102 +1951 +1961 +1971 +1981 +1991 +2001 +100 +90 +80 +70 +60 +50 +40 +30 +20 +10 +Population of India in crores +This bar graph shows the population of India in each decade over a +period of 50 years. The numbers are expressed in crores. If you were +to take 1 unit length to represent one person, drawing the bars will +Reprint 2025-26 + +Data Handling and Presentation +89 +be difficult! Therefore, we choose the scale so that 1 unit represents +10 crores. The bar graph for this choice is shown in the figure. So a +bar of length 5 units represents 50 crores and of 8 units represents +80 crores. +•• +On the basis of this bar graph, what may be a few questions you +may ask your friends? +•• +How much did the population of India increase over 50 years? +How much did the population increase in each decade? +4.4 Drawing a Bar Graph +In a previous example, Shri Nilesh prepared a frequency table +representing the sweet preferences of the students in his class. Let’s +try to prepare a bar graph to present his data — +1. +First, we draw a horizontal +line and a vertical line. On +the horizontal line, we will +write the name of each of +the sweets, equally spaced, +from which the bars will +rise in accordance with +their frequencies; and on +the vertical line we will write the frequencies representing the +number of students. +2. +We must choose a scale. That means we must decide how many +students will be represented by a unit length of a bar so that +it fits nicely on our paper. Here, we will take 1 unit length to +represent 1 student. +3. +For jalebi, we therefore need to draw a bar having a height of 6 +units (which is the frequency of the sweet jalebi), and similarly +for the other sweets we have to draw bars as high as their +frequencies. +Sweet +No. of Students +Jalebi +6 +Gulab jamun +9 +Gujiya +13 +Barfi +3 +Rasgulla +7 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +90 +4. +We, therefore, get a bar graph as shown below — +Sweets +Sweet preferences of students +14 +13 +12 +11 +10 +9 +8 +7 +6 +5 +4 +3 +2 +1 +0 +Number of students +Jalebi +Gulab jamun +Gujia +Barfi +Rasgulla + +When the frequencies are larger and we cannot use the scale of +1 unit length = 1 number (frequency), we need to choose a different +scale like we did in the case of pictographs. +Example: The number of runs scored by Smriti in each of the +8 matches are given in the table below: +Match +Match +1 +Match +2 +Match +3 +Match +4 +Match +5 +Match +6 +Match +7 +Match +8 +Runs +80 +50 +10 +100 +90 +0 +90 +50 + +In this example, the minimum score is 0 and the maximum score +is 100. Using a scale of 1 unit length = l run would mean that we +have to go all the way from 0 to 100 runs in steps of l. This would +be unnecessarily tedious. Instead, let us use a scale where 1 unit +length = 10 runs. We mark this scale on the vertical line and draw +the bars according to the scores in each match. We get the following +bar graph representing the above data. +Reprint 2025-26 + +Data Handling and Presentation +91 +Match 1 +Match 2 +Match 3 +Match 4 +Match 5 +Match 6 +Match 7 +Match 8 +Matches +100 +0 +10 +20 +30 +40 +50 +60 +70 +80 +90 +Runs +Runs scored by Smriti +Example: The following table shows the monthly expenditure of +Imran’s family on various items: +Items +Expenditure (in `) +House rent +Food +Education +Electricity +Transport +Miscellaneous +3000 +3400 + 800 + 400 + 600 + 1200 + +To represent this data in the form of a bar graph, here are the steps — +•• +Draw two perpendicular lines, one horizontal and one vertical. +•• +Along the horizontal line, mark the ‘items’ with equal spacing +between them and mark the corresponding expenditures along +the vertical line. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +92 +•• +Take bars of the same width, keeping a uniform gap between them. +•• +Choose a suitable scale along the vertical line. Let, 1 unit length = +` 200, and then mark and write the corresponding values (` 200, +` 400, etc.) representing each unit length. + +Finally, calculate the heights of the bars for various items as +shown below — +House rent +Food +Education +Electricity +Transport +Miscellaneous +3000 ÷ 200 +3400 ÷ 200 + 800 ÷ 200 + 400 ÷ 200 + 600 ÷ 200 + 1200 ÷ 200 +15 units +17 units +4 units +2 units +3 units +6 units + +Here is the bar graph that we obtain based on the above steps: +Expenditure ( in `) +House rent +Education +Electricity +Transport +Miscellaneous +Food +3600 +3400 +3200 +3000 +2800 +2600 +2400 +2200 +2000 +1800 +1600 +1400 +1200 +1000 +800 +600 +400 +200 +Item +Reprint 2025-26 + +Data Handling and Presentation +93 + Use the bar graph to answer the following questions: +1. +On which item does Imran’s family spend the most and the +second most? +2. +Is the cost of electricity about one-half the cost of education? +3. +Is the cost of education less than one-fourth the cost of food? + Figure it Out +1. +Samantha visited a tea garden, and collected data of the insects +and critters she saw there. Here is the data she collected: +Mites +Caterpillars +Beetles +Butterflies +Grasshoppers +6 +10 +5 +3 +2 + +Help her prepare a bar graph representing this data. +2. +Pooja collected data on the number of tickets sold at the Bhopal +railway station for a few different cities of Madhya Pradesh over a +two-hour period. +City +Vidisha +Jabalpur +Seoni +Indore +Sagar +Number of +tickets +24 +20 +16 +28 +16 + +She used this data and prepared a bar graph on the board to discuss +the data with her students, but someone erased a portion of the graph. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +94 +Vidisha +Jabalpur +Seoni +Indore +Sagar +City +No. of Tickets +a. +Write the number of tickets sold for Vidisha above the bar. +b. +Write the number of tickets sold for Jabalpur above the bar. +c. +The bar for Vidisha is 6 unit lengths and the bar for Jabalpur +is 5 unit lengths. What is the scale for this graph? +d. +Draw the correct bar for Sagar. +e. +Add the scale of the bar graph by placing the correct numbers +on the vertical axis. +f. +Are the bars for Seoni and Indore correct in this graph? If +not, draw the correct bar(s). +3. Chinu listed the various means of transport that passed across +the road in front of his house from 9 a.m. to 10 a.m.: +bike +car +bike +bus +bike +bike +bike +auto rickshaw +bicycle +bullock +cart +bicycle +auto +rickshaw +car +scooter +car +auto +rickshaw +bicycle +bike +car +auto rickshaw +bike +scooter +bike +car +bicycle +scooter +bicycle +scooter +bike +bus +auto rickshaw +auto rickshaw +bike +bicycle +bus +bike +bicycle +scooter +bus +scooter +auto +rickshaw +bike +scooter +bicycle +bike +bullock +cart +auto +rickshaw +scooter +car +scooter +Reprint 2025-26 + +Data Handling and Presentation +95 +a. +Prepare a frequency distribution table for the data. +b. +Which means of transport was used the most? +c. +If you were there to collect this data, how could you do it? +Write the steps or process. +4. Roll a die 30 times and record the number you obtain each time. +Prepare a frequency distribution table using tally marks. Find the +number that appeared: +a. +The minimum number of times. +b. +The maximum number of times. +c. +Find numbers that appeared an equal number of times. +5. Faiz prepared a frequency distribution table of data on the number +of wickets taken by Jaspreet Bumrah in his last 30 matches: +Wickets Taken +Number of Matches +0 +2 +1 +4 +2 +6 +3 +8 +4 +3 +5 +5 +6 +1 +7 +1 +a. +What information is this table giving? +b. +What may be the title of this table? +c. +What caught your attention in this table? +d. +In how many matches has Bumrah taken 4 wickets? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +96 +e. +Mayank says, “If we want to know the total number of +wickets he has taken in his last 30 matches, we have to add +the numbers 0, 1, 2, 3 …, up to 7.” Can Mayank get the total +number of wickets taken in this way? Why? +f. +How would you correctly figure out the total number of +wickets taken by Bumrah in his last 30 matches, using this +table? +6. The following pictograph shows the number of tractors in five +different villages. +Villages + + +Number of Tractors +( + = 1 Tractor ) +Village A + + + + + +Village B + + + + + +Village C + + + + + + + +Village D + + +Village E + + + + + + + +Observe the pictograph and answer the following questions— +a. +Which village has the smallest number of tractors? +b. +Which village has the most tractors? +c. +How many more tractors does Village C have than Village B? +d. +Komal says, “Village D has half the number of tractors as +Village E.” Is she right? +Reprint 2025-26 + +Data Handling and Presentation +97 +7. The number of girl students in each class of a school is depicted +by the pictograph: +Classes + Number of Girl Students ( + = 4 Girls ) +1 + + + + + + +2 + + + + +3 + + + + + +4 + + + +5 + + +6 + + + + +7 + + + +8 + + + +Observe this pictograph and answer the following questions: +a. +Which class has the least number of girl students? +b. +What is the difference between the number of girls in Classs +5 and 6? +c. +If two more girls were admitted in Class 2, how would the +graph change? +d. +How many girls are there in Class 7? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +98 +8. Mudhol Hounds (a type of breed of Indian dogs) are largely found +in North Karnataka’s Bagalkote and Vijaypura districts. The +government took an initiative to protect this breed by providing +support to those who adopted these dogs. Due to this initiative, the +number of these dogs increased. The number of Mudhol dogs in six +villages of Karnataka are as follows — + + +Village A : 18, Village B : 36, Village C : 12, Village D : 48, Village E : 18, +Village F : 24 + + +Prepare a pictograph and answer the following questions: +a. +What will be a useful scale or key to draw this pictograph? +b. +How many symbols will you use to represent the dogs in +Village B? +c. +Kamini said that the number of these dogs in Village B and +Village  D together will be more than the number of these +dogs in the other 4  villages. Is she right? Give reasons for +your response. +9. A survey of 120 school students was conducted to find out which +activity they preferred to do in their free time: +Preferred Activity +Number of Students +Playing +Reading story books +Watching TV +Listening to music +Painting +45 +30 +20 +10 +15 + + +Draw a bar graph to illustrate the above data taking the scale of +1 unit length = 5 students. Which activity is preferred by most +students other than playing? +Reprint 2025-26 + +Data Handling and Presentation +99 +10. Students and teachers of a primary school decided to plant tree +saplings in the school campus and in the surrounding village +during the first week of July. Details of the saplings they planted +are as follows — +70 +60 +50 +40 +30 +20 +10 +0 +Monday +Tuesday +Wednesday +Day +Thursday +Friday +Saturday +Sunday +Number of saplings planted +a. +The total number of saplings planted on Wednesday and +Thursday is ___________. +b. +The total number of saplings planted during the whole week +is ___________. +c. +The greatest number of saplings were planted on ___________ +and the least number of saplings were planted on ___________. +Why do you think that is the case? Why were more saplings +planted on certain days of the week and less on others? Can +you think of possible explanations or reasons? How could you +try and figure out whether your explanations are correct? +11. The number of tigers in India went down drastically between 1900 +and 1970. Project Tiger was launched in 1973 to track and protect +the tigers in India. Starting in 2006, the exact number of tigers +in India was tracked. Shagufta and Divya looked up information +about the number of tigers in India between 2006 and 2022 in four- +year intervals. They prepared a frequency table for this data and a +bar graph to present this data, but there are a few mistakes in the +graph. Can you find those mistakes and fix them? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +100 +Year +Number of Tigers +(approx.) +2006 +1400 +2010 +1700 +2014 +2200 +2018 +3000 +2022 +3700 +Number of Tigers in India +Year +Number of Tigers +2022 +2018 +2014 +2010 +2006 +0 +1000 +2000 +3000 +4000 +• +Like pictographs, bar graphs give a nice visual way to represent +data. They represent data through equally-spaced bars, each of +equal width, where the lengths or heights give frequencies of the +different categories. +• +Each category is represented by a bar where the length or height +depicts the corresponding frequency (for example, cost) or +quantity (for example, runs). +• +The bars have uniform spaces between them to indicate that they +are free standing and represent equal categories. +• +The bars help in interpreting data much faster than a frequency +table. By reading a bar graph, we can compare frequencies of +different categories at a glance. +• +We must decide the scale (for example, 1 unit length = 1 student +or 1 unit length = ` 200) for a bar graph on the basis of the data +including the minimum and maximum frequencies, so that the +resulting bar graph fits nicely and looks visually appealing on the +paper or poster we are preparing. The markings of the unit lengths +as per the scale must start from zero. +Teacher’s Note +The main focus of this chapter is to learn how to handle data to find +answers to specific questions or inquiries, to test hypotheses or to +take specific decisions. This should be kept in mind when providing +practice opportunities to collect, organise and analyse data. +Reprint 2025-26 + +Data Handling and Presentation +101 +4.5 Artistic and Aesthetic Considerations +In addition to the steps described in previous sections, there are +also some other more artistic and aesthetic aspects one can consider +when preparing visual presentations of data to make them more +interesting and effective. First, when making a visual presentation +of data such as a pictograph or bar graph, it is important to make +it fit in the intended space; this can be controlled, for example, by +choosing the scale appropriately, as we have seen earlier. It is also +desirable to make the data presentation visually appealing and +easy-to-understand, so that the intended audience appreciates the +information being conveyed. + +Let us consider an example. Here is a table naming the tallest +mountain on each continent, along with the height of each mountain +in meters. +Continent +Asia +South +America +North +America +Africa +Europe Antarctica +Australia +Tallest +Mountain Everest Aconcagua +Denali +Kiliman- +jaro +Elbrus +Vinson +Massif +Koscuiszko +Height +8848m +6962m +6194m +5895m +5642m +4892m +2228m + +How much taller is Mount Everest than Mount Koscuiszko? Are +Mount Denali and Mount Kilimanjaro very different in height? This +is not so easy to quickly discern from a large table of numbers. + +As we have seen earlier, we can convert the table of numbers +into a bar graph, as shown on the right. Here, each value is drawn +as a horizontal box. These are longer or shorter depending on the +number they represent. This makes it easier to compare the heights +of all these mountains at a glance. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +102 +Asia — Everest +South America — Aconcagua   +North America — Denali +Africa — Kilimanjaro +Europe — Elbrus +Antarctica — Vinson Massif +Australia — Koscuiszko +0 +1000 +2000 +3000 4000 +5000 6000 7000 8000 9000 + +However, since the boxes represent heights, it is better and more +visually appealing to rotate the picture, so that the boxes grow +upward, vertically from the ground like mountains. A bar graph +with vertical bars is also called a column graph. Columns are the +pillars you find in a building that hold up the roof. + +Below is a column graph for our table of tallest mountains. From +this column graph, it becomes easier to compare and visualise the +heights of the mountains. +0 +10000 +9000 +8000 +7000 +6000 +5000 +4000 +3000 +2000 +1000 +Asia +Everest +South America +Aconcagua   +North America +Denali +Africa +Kilimanjaro +Europe +Elbrus +Antarctica +Vinson Massif +Australia +Koscuiszko +Reprint 2025-26 + +Data Handling and Presentation +103 + +In general, it is more intuitive, suggestive and visually appealing +to represent heights, that are measured upwards from the ground, +using bar graphs that have vertical bars or columns. Similarly, +lengths that are parallel to the ground (for example, distances +between location on Earth) are usually best represented using bar +graphs with horizontal arcs. + Figure it Out +1. +If you wanted to visually represent the data of the heights of the +tallest persons in each class in your school, would you use a graph +with vertical bars or horizontal bars? Why? +2. +If you were making a table of the longest rivers on each continent +and their lengths, would you prefer to use a bar graph with +vertical bars or with horizontal bars? Why? Try finding out this +information, and then make the corresponding table and bar +graph! Which continents have the longest rivers? +Infographics +When data visualisations such as bar graphs are further beautified +with more extensive artistic and visual imagery, they are called +information graphics or infographics for short. The aim of +infographics is to make use of attention-attracting and engaging +visuals to communicate information even more clearly and quickly, +in a visually pleasing way. + +As an example of how infographics can be used to communicate +data even more suggestively, let us go back to the table above listing +the tallest mountain on each continent. We drew a bar graph with +vertical bars (columns) rather than horizontal bars, to be more +indicative of mountains. But instead of rectangles, we could use +triangles, which look a bit more like mountains. And, we can add a +splash of colour as well. Here is the result. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +104 +8000m +7000m +6000m +5000m +4000m +3000m +2000m +1000m +Everest +8848m + Asia +Aconcagua +6962m +S America +Denali +6194m +North +America +Kilimanjaro +5895m +Africa +Elbrus +5642m +Europe +Vinson Massif +4892m +Antarctica +Koscuiszko +2228m +Australia + +While this infographic might look more appealing and suggestive at +first glance, it does have some issues. The goal of our bar graph earlier +was to represent the heights of various mountains — using bars of the +appropriate heights but the same widths. The purpose of using the +same widths was to make it clear that we are only comparing heights. +However, in this infographic, the taller triangles are also wider! Are +taller mountains always wider? The infographic is implying additional +information that may be misleading and may or may not be correct. +Sometimes going for more appealing pictures can also accidentally +mislead. + +Taking this idea further, and to make the picture even more visually +stimulating and suggestive, we can further change the shapes of the +mountains to make them look even more like mountains, and add +other details, while attempting to preserve the heights. For example, +we can create an imaginary mountain range that contains all these +mountains. + +Is the infographic below better than the column graph with +rectangular columns of equal width? The mountains look more +realistic, but is the picture accurate? + +For example, Everest appears to be twice as tall as Elbrus. +Reprint 2025-26 + +Data Handling and Presentation +105 +The seven highest peaks on +the seven continents. + +What is 5642 × 2? + +While preparing visually-appealing presentations of data, we +also need to be careful that the pictures we draw do not mislead us +about the facts. In general, it is important to be careful when making +or reading infographics, so that we do not mislead our intended +audiences and we, ourselves, are not misled. +Su m m a r y + + +Facts, numbers, measures, observations and other descriptions of +things that convey information about those things is called data. + + +Data can be organised in a tabular form using tally marks for easy +analysis and interpretation. + + +Frequencies are the counts of the occurrences of values, measures or +observations. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +106 + + +Pictographs represent data in the form of pictures, or objects or parts +of objects. Each picture represents a frequency which can be 1 or more +than 1 — this is called the scale and it must be specified. + + +Bar graphs have bars of uniform width; the length or height that +indicates the total frequency of occurrence. The scale that is used to +convert length or height to frequency again, must be specified. + + +Choosing the appropriate scale for a pictograph or bar graph is +important to accurately and effectively convey the desired information +or data and to also make it visually appealing. + + +Other aspects of a graph also contribute to its effectiveness and visual +appeal such as how colours are used, what accompanying pictures +are drawn, and whether the bars are horizontal or vertical. These +aspects correspond to the artistic and aesthetic side of data handling +and presentation. + + +However, making visual representations of data ‘fancy’ can also +sometimes be misleading. + + +By reading pictographs and bar graphs accurately, we can quickly +understand and make inferences about the data presented. +Reprint 2025-26 + +CHAPTER 4 — SOLUTIONS +Data Handling and Presentation +Section 4.1 +Page No. 75 +Figure it Out +Q.1. What would you do to find the most popular game among Naresh’s and Navya’s +classmates? +Ans. +One of the ways could be to arrange and organize the given data in a table. Think of +some other ways. +Q.2. What is the most popular game in their class? +Ans. +Hockey (frequency – 8). +Q.4. Pari wants to respond to the questions given below. Put a tick () for the questions +where she needs to carry out data collection and put a cross (⨉) for the questions +where she doesn’t need to collect data. Discuss your answers in the classroom. +a. What is the most popular TV show among her classmates? + +  + +b. When did India get independence? + + + + +  + +c. How much water is getting wasted in her locality? + + +  + +d. What is the capital of India? + + + + + +  + + +Ans. +a. + +b. + +c. + +d. + + +Page No. 76 +Figure it Out +Q.1. Complete the table to help Shri Nilesh to purchase the correct numbers of sweets: +• How many students chose jalebi? + +• Barfi was chosen by + students? +• How many students chose gujiya? + +• Rasgulla was chosen by + students? + +⨉ + +⨉ + +• How many students chose gulab jamun? + +Shri Nilesh requested one of the staff members to bring the sweets as given in the table. +The above table helped him to purchase the correct numbers of sweets. +Ans. + +• 6 +• 3 +• 13 +• 7 +• 9 + +Q.2. Is the above table sufficient to distribute each type of sweet to the correct student? +Explain. If it is not sufficient, what is the alternative? +Ans. +No. +It shows only the number of students who chose a particular sweet, and not about the +sweet chosen by a particular student. +One of the alternatives could be, students should be categorized according to their +choice of sweet. + +Section 4.1 +Page No. 77 +Figure it Out +Q.1. Help her to figure out the following – +• The largest shoe size in the class is _________ +• The smallest shoe size in the class is _________ +• There are _________ students who wear shoe size 5. +• There are _________ students who wear shoe sizes larger than 4. +Ans. +• 7 +• 3 +• 10 +• 15 + +Q.2. How did arranging the data in ascending order help to answer these questions? +Ans. +Ordered data is helpful because in this form, frequency of any data is easy to count and +the given information can be used easily. + + + +Q.3. Are there other ways to arrange the data? +Ans. +Yes, the data can be arranged in a frequency table. + +Section 4.2 +Page No. 83 +Q. What could be the problems faced in preparing such a pictograph, if the total number +of students present in a class is 33 or 27? +Ans. + represents 10 students and + represent 5 students. But to represent 3 or 7 students +accurately by dividing these symbols is not possible. + +Figure it Out +Q.1. The following pictograph shows the number of books borrowed by students, in a +week, from the library of Middle School, Ginnori — +a. On which day were the minimum number of books borrowed? +b. What was the total number of books borrowed during the week? +c. On which day were the maximum number of books borrowed? What may be the +possible reason? +Ans. +a. Thursday +b. 24 +c. Saturday. One of the reasons could be that Sunday being a holiday students can read +the borrowed books on Sunday. +Q.2. Magan Bhai sells kites at Jamnagar. Six shopkeepers from nearby villages come to +purchase kites from him. The number of kites he sold to these six shopkeepers are +given below — +Prepare a pictograph using the symbol +to represent 100 kites. + +Shopkeeper +No. of Kites sold ( +=100 +Kits) +Chaman +Rani + + +Rukhsana +Jasmeet +Jetha Lal +Poonam Ben + +Answer the following questions: +a. How many symbols represent the kites that Rani purchased? +b. Who purchased the maximum number of kites? +c. Who purchased more kites, Jasmeet or Chaman? +d. Rukhsana says Poonam Ben purchased more than double the number of kites +that Rani purchased. Is she correct? Why? +Ans. +a. 3 symbols +b. Poonam Ben +c. Jasmeet +d. Yes. Number. of kites purchased by Poonam Ben = 700 = 2 x 300 +100 + +Section 4.3 +Page no. 86 +Q1. In Class 2, ___________ students were absent that day. +Ans. +5 +Q2. In which class were the maximum number of students absent? ___________ +Ans. +Class 8 +Q3. Which class had full attendance that day? ___________ +Ans. +Class 5 + + + + +Page no. 88 +Figure it Out +Q.1. How many total cars passed through the crossing between 6 am and noon? +Ans. +4450 cars + +Section 4.4 +Page No. 93 +Q. On which item does Imran’s family spend the most and the second most? +Ans. +Imran’s family spends the most on food and the second most on house rent. +Q.2. Is the cost of electricity about one-half the cost of education? +Ans. +Yes, +Q.3. Is the cost of education less than one-fourth the cost of food? +Ans. +Yes. Cost of education = Rs. 800 and cost of food = Rs 3400 =4 x Rs 850 +So, cost of education = less than one-fourth of cost of food. + +Figure it Out +Q.1. Samantha visited a tea garden and collected data of the insects and critters she saw +there. Here is the data she collected — +Help her prepare a bar graph representing this data. +Ans. + + +Q.2. Pooja collected data on the number of tickets sold at the Bhopal railway station for +a few different cities of Madhya Pradesh over a 2-hour period. +She used this data and prepared a bar graph on the board to discuss the data with +her students, but someone erased a portion of the graph. +a. Write the number of tickets sold for Vidisha above the bar. +b. Write the number of tickets sold for Jabalpur above the bar. +c. The bar for Vidisha is 6 unit lengths and the bar for Jabalpur is 5 unit lengths. +What is the scale for this graph? +d. Draw the correct bar for Sagar. +e. Add the scale of the bar graph placing the correct numbers on the vertical axis. +f. Are the bars for Seoni and Indore correct in this graph? If not, draw the correct +bar(s). +Ans. +a. 24 +b. 20 +c. 1unit length = 4 tickets +d. + + + + + + + + + + + + + +e. + + + + + + + + + + + + + +f. The bar for Seoni is correct but for Indore is incorrect. + + +Q.3. Chinu listed the various means of transport that passed across the road in front of +his house from 9 am to 10 am: +a. Prepare a frequency distribution table for the data. +b. Which means of transport was used the most? +c. If you were there to collect this data, how could you do it? Write the steps or +process. +Ans. +a. +Means of Transport + +Number +Bike +13 +Car +6 +Bicycle +8 +Auto Rickshaw +8 +Scooter +9 +Bus +4 +Bullock Cart +2 + +b. Bike +c. To collect the data of various means of transport, one of the ways could be – +(i) +Prepare a table with 2 columns. First for means of transport and second for +its frequency. +(ii) +List the various means of transport as they pass across the road. +(iii) +Frequency should be presented using tally marks. + + +Q.5. Faiz prepared a frequency distribution table of data on the number of wickets taken +by Jaspreet Bumrah in his last 30 matches: +a. What information is this table giving? +b. What may be the title of this table? +c. What caught your attention in this table? +d. In how many matches has Bumrah taken 4 wickets? +e. Mayank says “If we want to know the total number of wickets he has taken in his +last 30 matches, we have to add the numbers 0, 1, 2, 3 …, up to 7.” Can Mayank +get the total number of wickets taken in this way? Why? +f. How would you correctly figure out the total number of wickets taken by Bumrah +in his last 30 matches, using this table? +Ans. +a. This table is giving the information about different number of wickets taken by +Jaspreet Bumrah in different number of matches. +b. The title of this table may be: Wickets Taken by Jaspreet Bumrah in Last 30 +Matches.(think of more!) +c. In this table, attention seeking point is that in 1 match 7 wickets were taken by +Jaspreet Bumrah only.(think of more!) +d. In 3 matches Bumrah has taken 4 wickets. +e. No, Mayank can’t get the total number of wickets taken in this way. For example, 3 +wickets per match were taken in 8 matches, which means total 24 wickets were taken. +f. To know the total number of wickets, prepare a next column of total wickets .It will +have products of corresponding numbers of the two given columns. Then the sum of +the numbers in the 3rd column, will be the total number of wickets. Total number of +wickets =90 +Q.6. The following pictograph shows the number of tractors in five different villages. +Observe the pictograph and answer the following questions— +a. Which village has the smallest number of tractors? +b. Which village has the most tractors? +c. How many more tractors does Village C have than Village B? +d. Komal says, “Village D has half the number of tractors as Village E.” Is she right? +Ans. +a. Village D. +b. Village C. +c. 3. +d. Yes +Q.7. The number of girl students in each class of a school is depicted by a pictograph: +Observe this pictograph and answer the following questions: +a. Which class has the least number of girl students? +b. What is the difference between the number of girls in Classs 5 and 6? +c. If 2 more girls were admitted in Class 2, how would the graph change? +d. How many girls are there in Class 7? +Ans. +a. Class 8. +b. 6. + +c. If 2 more girls were admitted in class 2, the last half symbol will be converted into +full symbol. +d. 12. +Q.8. Mudhol Hounds (a type of breed of Indian dogs) are largely found in North +Karnataka’s Bagalkote and Vijaypura districts. The government took an initiative +to protect this breed by providing support to those who adopted these dogs. Due to +this initiative, the number of these dogs increased. The number of Mudhol dogs in +six villages of Karnataka are as follows — +Village A : 18, Village B : 36, Village C : 12, Village D : 48, Village E : 18, Village F : 24 +Prepare a pictograph and answer the following questions: +a. What will be a useful scale or key to draw this pictograph? +b. How many symbols will you use to represent the dogs in Village B? +c. Kamini said that the number of dogs in Village B and Village D together will be +more than the number of dogs in the other 4 villages. Is she right? Give reasons +for your response. +Ans. + +a. Scale is += 6 dogs, as numbers of dogs are multiples of 6. +b. 6 +symbols. +c. Yes, Kamini is right as the number of dogs in village B and village D together is +84 which is more than the number of dogs in the other 4 villages (72). + + + + + + +Village +Number of Mudhol Hounds + ) += 6 dogs) +Village A + +Village B + +Village C + +Village D + +Village E + +Village F + + +Q.9 A survey of 120 school students was conducted to find out which activity they +preferred to do in their free time. +Draw a bar graph to illustrate the above data taking the scale of 1 unit length = 5 +students. Which activity is preferred by most students other than playing? +Ans. + + +Other than playing, reading story books is preferred by most students. +Q.10 Students and teachers of a primary school decided to plant tree saplings in the school +campus and in the surrounding village during the first week of July. Details of the +saplings they planted are as follows — +a. The total number of saplings planted on Wednesday and Thursday is +___________. +b. The total number of saplings planted during the whole week is ___________. +c. The greatest number of saplings were planted on ___________, and the least +number of saplings were planted on ___________. Why do you think that is the +case? Why were more saplings planted on certain days of the week and less on +others? Can you think of possible explanations or reasons? How could you try +and figure out whether your explanations are correct? +Ans. +a. 70 +b. 310 +c. Saturday, Wednesday +The variation in the number of saplings planted on different days may be because of +rainy season or different number of students present on different days. (Discuss for +other reasons.) + + + +0 +5 +10 +15 +20 +25 +30 +35 +40 +45 +50 +Playing +Reading +Story Books +Watching +TV +Listening to +music +Painting +Number of Students +Preferred Activity + +Q.11. The number of tigers in India went down drastically between 1900 and 1970. +Project Tiger was launched in 1973 to track and protect tigers in India. Starting +in 2006, the exact number of tigers in India was tracked. Shagufta and Divya +looked up information about the number of tigers in India between 2006 and 2022 +in 4-year intervals. They prepared a frequency table for this data and a bar graph +to present this data, but there are a few mistakes in the graph. Can you find those +mistakes and fix them? +Ans. +There are mistakes in the bars of 2006, 2010, 2014, 2018. + +Section 4.5 +Page no. 103 +Figure it out +Q.1. If you wanted to visually represent the data of the heights of the tallest persons in +each class in your school, would you use a graph with vertical bars or horizontal +bars? Why? +Ans. +To present the data of the heights of the tallest persons in each class in a school, vertical +bar graph is useful as heights of persons can be compared by heights of bars easily. +(Although both bar graphs can be used.) (Think of more reasons!) +Q.2. If you were making a table of the longest rivers on each continent and their +lengths, would you prefer to use a bar graph with vertical bars or with horizontal +bars? Why? Try finding out this information, and then make the corresponding +table and bar graph! Which continents have the longest rivers? +Ans. +In this case, bar graph with horizontal bars is useful as length of a river is a horizontal +feature, so lengths of rivers can be compared with lengths of horizontal bars easily. +(Although both bar graphs can be used.) (can you think of any other reasons?)" +class_6,5,Prime Time,ncert_books/class_6/Ganita_Prakash/fegp105.pdf,"Children sit in a circle and play a game of numbers. +One of the children starts by saying ‘1’. The second +player says ‘2’, and so on. But when it is the turn of 3, 6, +9, … (multiples of 3), the player should say ‘idli’ instead +of the number. When it is the turn of 5, 10, … (multiples +of 5), the player should say ‘vada’ instead of the number. +When a number is both a multiple of 3 and a multiple +of 5, the player should say ‘idli-vada’! If a player makes +any mistake, they are out. +The game continues in rounds till only one person +remains. +For which numbers should the players say ‘idli’ +instead of saying the number? These would be 3, 6, 9, +12, 18, … and so on. +For which numbers should the players say ‘vada’? +These would be 5, 10, 15, 20, … and so on. +Which is the first number for which the players +should say, ‘idli-vada’? It is 15, which is a multiple of 3, +and also a multiple of 5. Find out other such numbers +that are multiples of both 3 and 5. These numbers are +called _____________________________. +Idli-Vada Game +PRIME TIME +5 +5.1 Common Multiples and Common Factors +Reprint 2025-26 + +Ganita Prakash | Grade 6 +108 + Figure it Out +1. At what number is ‘idli-vada’ said for the 10th time? +2. If the game is played for the numbers 1 to 90, find out: +a. +How many times would the children say ‘idli’ (including the +times they say ‘idli-vada’)? +b. +How many times would the children say ‘vada’ (including the +times they say ‘idli-vada’)? +c. +How many times would the children say ‘idli-vada’? +3. +What if the game was played till +900? How would your answers +change? +4. Is this figure somehow related to +the ‘idli-vada’ game? +Hint: Imagine playing the game +till 30. Draw the figure if +the game is played till 60. +  Let us now play the ‘idli-vada’ game +with different pairs of numbers: +a. 2 and 5, +b. 3 and 7, +c. 4 and 6. +We will say ‘idli’ for multiples of the smaller number, ‘vada’ for +multiples of the larger number and ‘idli-vada’ for common multiples. +Draw a figure similar to Fig. 5.1 if the game is played up to 60. +Yesterday, we played this game +with two numbers. We ended +up saying just ‘idli’ or ‘idli-vada’ +and nobody said just ‘vada’! +Oh, what could +those numbers be!? +One of the numbers +was 4. +Fig. 5.1 +Multiples +of 3 +Multiples +of 5 +21 +24 +30 +15 +10 +20 +5 +25 +12 +18 +27 +9 +3 +Common multiples +of 3 and 5 +Reprint 2025-26 + +Prime Time +109 + Which of the following could be the other number: + +2, 3, 5, 8, 10? +Jump Jackpot +Jumpy and Grumpy play a game. +• Grumpy places a treasure on some number. For example, he +may place it on 24. +• Jumpy chooses a jump size. If he chooses 4, then he has to +jump only on multiples of 4, starting at 0. +• Jumpy gets the treasure if he lands on the number where +Grumpy placed it. +Which jump sizes will get Jumpy to land on 24? +If he chooses 4: Jumpy lands on 4 → 8 → 12 → 16 → 20 → 24 → 28 → ... +Other successful jump sizes are 2, 3, 6, 8 and 12. +24 +23 +22 +21 +20 +19 +18 +17 +16 +15 +14 +13 +12 +11 +10 +9 +8 +7 +6 +5 +4 +3 +2 +1 +0 +What about the jump of sizes 1 and 24? Yes, they also will land +on 24. +The numbers 1, 2, 3, 4, 6, 8, 12, 24 all divide 24 exactly. Recall that +such numbers are called factors or divisors of 24. +Grumpy increases the level of the game. Two treasures are kept +on two different numbers. Jumpy has to choose a jump size and stick +to it. Jumpy gets the treasures only if he lands on both the numbers +with the chosen jump size. As before, Jumpy starts at 0. +Grumpy has kept the treasures on 14 and 36. And, Jumpy chooses +a jump size of 7. +Will Jumpy land on both the treasures? Starting from 0, he jumps +to 7 → 14 → 21 → 28 → 35 → 42 … We see that he landed on 14 but +Reprint 2025-26 + +Ganita Prakash | Grade 6 +110 +did not land on 36, so he does not get the treasure. What jump size +should he have chosen? +The factors of 14 are: 1, 2, 7, 14. So, these jump sizes will land on 14. +The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18 and 36. These jump sizes +will land on 36. +So, the jump sizes of 1 or 2 will land on both 14 and 36. Notice that +1 and 2 are the common factors of 14 and 36. +The jump sizes using which both the treasures can be reached are +the common factors of the two numbers where the treasures are +placed. + What jump size can reach both 15 and 30? There are multiple +jump sizes possible. Try to find them all. + Look at the table below. What do you notice? +31 +32 +33 +34 +35 +36 +37 +38 +39 +40 +41 +42 +43 +44 +45 +46 +47 +48 +49 +50 +51 +52 +53 +54 +55 +56 +57 +58 +59 +60 +61 +62 +63 +64 +65 +66 +67 +68 +69 +70 +In the table, +1. Is there anything common among the shaded +numbers? +2. Is there anything common among the circled +numbers? +3. Which numbers are both shaded and circled? What +are these numbers called? + Figure it Out + +1. +Find all multiples of 40 that lie between 310 and 410. +Math +Talk +Reprint 2025-26 + +Prime Time +111 + +2. +Who am I? +a.   I am a number less than 40. One of my factors is 7. + +The sum of my digits is 8. +b.   I am a number less than 100. Two of my factors are 3 and 5. +One of my digits is 1 more than the other. +3. A number for which the sum of all its factors is equal to twice the +number is called a perfect number. The number 28 is a perfect +number. Its factors are 1, 2, 4, 7, 14 and 28. Their sum is 56 which +is twice 28. Find a perfect number between 1 and 10. +4. Find the common factors of: +a.  20 and 28 + b.  35 and 50 +c.  4, 8 and 12 + d.  5, 15 and 25 +5. Find any three numbers that are multiples of 25 but not multiples +of 50. +6. Anshu and his friends play the ‘idli-vada’ game with two numbers, +which are both smaller than 10. The first time anybody says ‘idli- +vada’ is after the number 50. What could the two numbers be +which are assigned ‘idli’ and ‘vada’? +7. In the treasure hunting game, Grumpy has kept treasures on 28 +and 70. What jump sizes will land on both the numbers? +8. In the diagram below, Guna has erased all the numbers except +the common multiples. Find out what those numbers could be +and fill in the missing numbers in the empty regions. + +Multiples of ____ +Common multiples +72 +48 +24 +Multiples of ____ +9. +Find the smallest number that is a multiple of all the numbers +from 1 to 10, except for 7. +10. +Find the smallest number that is a multiple of all the numbers +from 1 to 10. +Math +Talk +Try +This +Reprint 2025-26 + +Ganita Prakash | Grade 6 +112 +5.2 Prime Numbers +Guna and Anshu want to pack figs (anjeer) that grow in their farm. +Guna wants to put 12 figs in each box and Anshu wants to put 7 figs +in each box. +How many arrangements are possible? +Think and find out the different ways how — + 1.  Guna can arrange 12 figs in a rectangular manner. + 2.  Anshu can arrange 7 figs in a rectangular manner. +Guna has listed out these possibilities. +Observe the number of rows and +columns in each of the arrangements. +How are they related to 12? +In the second arrangement, for +example, 12 figs are arranged in two +columns of 6 each or 12 = 2 × 6. +Anshu could make only one +arrangement: 7 × 1 or 1 × 7. There are +no other rectangular arrangements +possible. +In each of Guna’s arrangements, +multiplying the number of rows by +the number of columns gives the +number 12. So, the number of rows +or columns are factors of 12. +We saw that the number 12 can be arranged in a rectangle in +more than one way as 12 has more than two factors. The number 7 +can be arranged in only one way, as it has only two factors — 1 and 7. +Numbers that have only two factors are called prime numbers or +primes. Here are the first few primes — 2, 3, 5, 7, 11, 13, 17, 19. Notice +that the factors of a prime number are 1 and the number itself. +What about numbers that have more than two factors? They +are called composite numbers. The first few composite numbers +are — 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20. +Reprint 2025-26 + +Prime Time +113 +What about 1, which has only one factor? The number 1 is neither +a prime nor a composite number. + How many prime numbers are there from 21 to 30? How many +composite numbers are there from 21 to 30? +Can we list all the prime numbers from 1 to 100? +Here is an interesting way to find prime numbers. Just follow the +steps given below and see what happens. +Step 1: Cross out 1 because it is neither prime nor composite. +Step 2: Circle 2, and then cross +out all multiples of 2 after that, +i.e., 4, 6, 8, and so on. +Step 3: You will find that the +next uncrossed number is 3. +Circle 3 and then cross out all +the multiples of 3 after that, +i.e., 6, 9, 12, and so on. +Step 4: The next uncrossed +number is 5. Circle 5 and then +cross out all the multiples of 5 +after that, i.e., 10, 15, 20, and +so on. +Step 5: Continue this process till all the numbers in the list are either +circled or crossed out. +All the circled numbers are prime +numbers. All the crossed out numbers, +other than 1, are composite numbers. This +method is called the Sieve of Eratosthenes. +This procedure can be carried on +for numbers greater than 100 also. +Eratosthenes was a Greek mathematician +who lived around 2200 years ago and +developed this method of listing primes. +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +13 +14 +15 +16 +17 +18 +19 +20 +21 +22 +23 +24 +25 +26 +27 +28 +29 +30 +31 +32 +33 +34 +35 +36 +37 +38 +39 +40 +41 +42 +43 +44 +45 +46 +47 +48 +49 +50 +51 +52 +53 +54 +55 +56 +57 +58 +59 +60 +61 +62 +63 +64 +65 +66 +67 +68 +69 +70 +71 +72 +73 +74 +75 +76 +77 +78 +79 +80 +81 +82 +83 +84 +85 +86 +87 +88 +89 +90 +91 +92 +93 +94 +95 +96 +97 +98 +99 100 +It is definitely not some +magic; there should be a +reason why it works. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +114 +Guna and Anshu started wondering how this simple method is able +to find prime numbers! Think how this method works. Read the steps +given above again and see what happens after each step is carried out. + Figure it Out +1. We see that 2 is a prime and also an even number. Is there any +other even prime? +2. Look at the list of primes till 100. What is the smallest difference +between two successive primes? What is the largest difference? +3. Are there an equal number of primes occurring in every row in the +table on the previous page? Which decades have the least number +of primes? Which have the most number of primes? +Primes through the Ages +Prime numbers are the building blocks of all whole numbers. Starting +from the time of the Greek civilisation (more than 2000 years ago) to +this day, mathematicians are still struggling to uncover their secrets! +Food for thought: is there a largest prime number? Or does the +list of prime numbers go on without an end? A mathematician named +Euclid found the answer and so will you in a later class! +Fun fact: The largest prime number that anyone has ‘written down’ +is so large that it would take around 6500 pages to write it! So they +could only write it on a computer! +4. Which of the following numbers are prime: 23, 51, 37, 26? +5. Write three pairs of prime numbers less than 20 whose sum is a +multiple of 5. +6. The numbers 13 and 31 are prime numbers. Both these numbers have +same digits 1 and 3. Find such pairs of prime numbers up to 100. +7. +Find seven consecutive composite numbers between 1 and 100. +8. Twin primes are pairs of primes having a difference of 2. For +example, 3 and 5 are twin primes. So are 17 and 19. Find the other +twin primes between 1 and 100. +Reprint 2025-26 + +Prime Time +115 +9. Identify whether each statement is true or false. Explain. +a. +There is no prime number whose units digit is 4. +b. +A product of primes can also be prime. +c. +Prime numbers do not have any factors. +d. +All even numbers are composite numbers. +e. +2 is a prime and so is the next number, 3. For every other +prime, the next number is composite. +10. Which of the following numbers is the product of exactly three +distinct prime numbers: 45, 60, 91, 105, 330? +11. How many three-digit prime numbers can you make using each +of 2, 4 and 5 once? +12. Observe that 3 is a prime number, and 2 × 3 + 1 = 7 is also a prime. +Are there other primes for which doubling and adding 1 gives +another prime? Find at least five such examples. +5.3 Co-prime numbers for safekeeping treasures +Which pairs are safe? +Let us go back to the treasure finding game. This time, treasures are +kept on two numbers. Jumpy gets the treasures only if he is able to +reach both the numbers with the same jump size. There is also a +new rule — a jump size of 1 is not allowed. + Where should Grumpy place the treasures so that Jumpy cannot +reach both the treasures? +Will placing the treasure on 12 and 26 work? No! If the jump size +is chosen to be 2, then Jumpy will reach both 12 and 26. +What about 4 and 9? Jumpy cannot reach both using any jump size +other than 1. So, Grumpy knows that the pair 4 and 9 is safe. +Check if these pairs are safe: +a. 15 and 39 +b. 4 and 15 +c. 18 and 29 +d. 20 and 55 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +116 +What is special about safe pairs? They don’t have any common +factor other than 1. Two numbers are said to be co-prime to each +other if they have no common factor other than 1. +Example: As 15 and 39 have 3 as a common factor, they are not +co-prime. But 4 and 9 are co-prime. + Which of the following pairs of numbers are co-prime? +a. 18 and 35 +b. 15 and 37 +c. 30 and 415 +d. 17 and 69 +e. 81 and 18 +  While playing the ‘idli-vada’ game with different number +pairs, Anshu observed something interesting! +1. Sometimes the first common multiple was the same +as the product of the two numbers. +2. At other times the first common multiple was less +than the product of the two numbers. +Find examples for each of the above. How is it related to the +number pair being co-prime? +Co-prime art + Observe the following thread art. The first diagram has 12 +pegs and the thread is tied to every fourth peg (we say that +the thread-gap is 4). The second diagram has 13 pegs and the +thread-gap is 3. What about the other diagrams? Observe these +pictures, share and discuss your findings in class. +12 +11 +10 +9 +8 +7 +6 +5 +4 +3 +2 +1 +13 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +16 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +13 +14 +15 +24 1 2 3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +13 +14 +15 +16 +17 +18 +19 +20 +21 +22 23 +In some diagrams, the thread is tied to every peg. In some, it is +not. Is it related to the two numbers (the number of pegs and the +thread-gap) being co-prime? +Math +Talk +Math +Talk +Reprint 2025-26 + +Prime Time +117 +Make such pictures for the following: +a.  15 pegs, thread-gap of 10 + +b.  10 pegs, thread-gap of 7 +c.  14 pegs, thread-gap of 6 + +d.  8 pegs, thread-gap of 3 +5.4 Prime Factorisation +Checking if two numbers are co-prime +Teacher: +Are 56 and 63 co-prime? +Anshu and Guna: If they have a common factor other than 1, then +they are not co-prime. Let us check. +Anshu: +I can write 56 = 14 × 4 and 63 = 21 × 3. So, 14 and 4 +are factors, of 56. Further, 21 and 3 are factors of +63. So, there are no common factors. The numbers +are co-prime. +Guna: +Hold on. I can also write 56 = 7 × 8 and 63 = 9 × 7. +We see that 7 is a factor of both numbers, so, they +are not co-prime. +Clearly Guna is right, as 7 is a common factor. + But where did Anshu go wrong? +Writing 56 = 14 × 4 tells us that 14 and 4 are both factors of 56, but it +does not tell all the factors of 56. The same holds for the factors of 63. +Try another example: 80 and 63. There are many ways to factorise +both numbers. +80 = 40 × 2 = 20 × 4 = 10 × 8 = 16 × 5 = ??? +63 = 9 × 7 = 3 × 21 = ??? +We have written ‘???’ to say that there may be more ways to +factorise these numbers. But if we take any of the given factorisations, +for example, 80 = 16 × 5 and 63 = 9 × 7, then there are no common +factors. Can we conclude that 80 and 63 are co-prime? As Anshu’s +mistake above shows, we cannot conclude that as there may be other +ways to factorise the numbers. +What this means is that we need a more systematic approach to +check if two numbers are co-prime. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +118 +Prime factorisation +Take a number such as 56. It is composite, as we saw that it can be +written as 56 = 4 × 14 . So, both 4 and 14 are factors of 56. Now take one +of these, say 14. It is also composite and can be written as 14 = 2 × 7. +Therefore, 56 = 4 × 2 × 7. Now, 4 is composite and can be written as +4 = 2 × 2. Therefore, 56 = 2 × 2 × 2 × 7. All the factors appearing here, 2 +and 7, are prime numbers. So, we cannot divide them further. +In conclusion, we have written 56 as a product of prime numbers. +This is called prime factorisation of 56. The individual factors are +called prime factors. For example, the prime factors of 56 are 2 and 7. +Every number greater than 1 has a prime factorisation. The idea +is the same: keep breaking the composite numbers into factors till +only primes are left. +The number 1 does not have any prime factorisation. It is not +divisible by any prime number. +What is the prime factorisation of a prime number like 7? It is just +7 (we cannot break it down any further). +Let us see a few more examples. +By going through different ways of breaking down the number, +we wrote 63 as 3 × 3 × 7 and as 3 × 7 × 3. Are they different? Not +really! The same prime numbers 3 and 7 occur in both cases. Further, +3 appears two times in +both and 7 appears once. +Here, you see four +different ways to get +prime factorisation of 36. +Observe that in all four +cases, we get two 2s and +two 3s. +Multiply back to see +that you get 36 in all four cases. +For any number, it is a remarkable fact that there is only one prime +factorisation, except that the prime factors may come in different +36 +2×18 +3×12 +4×9 +6×6 +2×2×9 +3×3×4 +2×2×9 +2×3×6 +2×2×3×3 +3×3×2×2 +2×2×3×3 +2×3×2×3 +2×3×2×3 +2×2×3×3 +Reprint 2025-26 + +Prime Time +119 +orders. As we explain below, the order is not important. However, +as we saw in these examples, there are many ways to arrive at the +prime factorisation! +Does the order matter? +Using this diagram, +can you explain why 30 = 2 × 3 × 5, no matter which way you multiply +2, 3, and 5? +When multiplying numbers, we can do so in any order. The +end result is the same. That is why, when two 2s and two 3s are +multiplied in any order, we get 36. In a later class, we shall study +this under the names of commutativity and associativity of +multiplication. +Thus, the order does not matter. Usually we write the prime +numbers in increasing order. For example, 225 = 3 × 3 × 5 × 5 or 30 = +2 × 3 × 5. +Prime factorisation of a product of two numbers +When we find the prime factorisation of a number, we first write +it as a product of two factors. For example, 72 = 12 × 6. Then, we +find the prime factorisation of each of the factors. In the above example, +12 = 2 × 2 × 3 and 6 = 2 × 3. Now, can you say what the prime factorisation +of 72 is? +The prime factorisation of the original number is obtained by +putting these together. +72 = 2 × 2 × 3 × 2 × 3 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +120 +We can also write this as 2 × 2 × 2 × 3 × 3. Multiply and check that +you get 72 back! +Observe how many times each prime factor occurs in the +factorisation of 72. +Compare it with how many times it occurs in the factorisations of +12 and 6 put together. + Figure it Out +1. Find the prime factorisations of the following numbers: 64, 104, +105, 243, 320, 141, 1728, 729, 1024, 1331, 1000. +2. The prime factorisation of a number has one 2, two 3s, and one +11. What is the number? +3. Find three prime numbers, all less than 30, whose product is 1955. +4. Find the prime factorisation of these numbers without multiplying +first  + +a.  56 × 25  b.  108 × 75  c.  1000 × 81 +5. What is the smallest number whose prime factorisation has: + +a.  three different prime numbers?  + +b.  four different prime numbers? +Prime factorisation is of fundamental importance in the study of +numbers. Let us discuss two ways in which it can be useful. +Using prime factorisation to check if two numbers are co-prime +Let us again take the numbers 56 and 63. How can we check if they +are co-prime? We can use the prime factorisation of both numbers — +56 = 2 × 2 × 2 × 7 and 63 = 3 × 3 × 7 +Now, we see that 7 is a prime factor of 56 as well as 63. Therefore, +56 and 63 are not co-prime. +What about 80 and 63? Their prime factorisations are as follows: +80 = 2 × 2 × 2 × 2 × 5 and 63 = 3 × 3 × 7 +There are no common prime factors. Can we conclude that they +are co-prime? Suppose they have a common factor that is composite. +Would the prime factors of this composite common factor appear in +the prime factorisation of 80 and 63? +Reprint 2025-26 + +Prime Time +121 +Therefore, we can say that if there are no common prime factors, +then the two numbers are co-prime. +Let us see some examples. +Example: Consider 40 and 231. Their prime factorisations are as +follows: +40 = 2 × 2 × 2 × 5 and 231 = 3 × 7 × 11 +We see that there are no common primes that divide both 40 and +231. Indeed, the prime factors of 40 are 2 and 5 while, the prime +factors of 231 are 3, 7, and 11. Therefore, 40 and 231 are co-prime! +Example: Consider 242 and 195. Their prime factorisations are as +follows: +242 = 2 × 11 × 11 and 195 = 3 × 5 × 13 +The prime factors of 242 are 2 and 11. The prime factors of 195 +are 3, 5, and 13. There are no common prime factors. Therefore, 242 +and 195 are co-prime. +Using prime factorisation to check if one number is divisible +by another +We can say that if one number is divisible by another, the prime +factorisation of the second number is included in the prime +factorisation of the first number. +We say that 48 is divisible by 12 because when we divide 48 by 12, +the remainder is zero. How can we check if one number is divisible +by another without carrying out long division? +Example: Is 168 divisible by 12? Find the prime factorisations of both: +168 = 2 × 2 × 2 × 3 × 7 and 12 = 2 × 2 × 3 +Since we can multiply in any order, now it is clear that, + 168 = 2 × 2 × 3 × 2 × 7 = 12 × 14 +Therefore, 168 is divisible by 12. +Example: Is 75 divisible by 21? Find the prime factorisations of both: +75 = 3 × 5 × 5 and 21 = 3 × 7 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +122 +As we saw in the discussion above, if 75 was a multiple of 21, then +all prime factors of 21 would also be prime factors of 75. However, 7 +is a prime factor of 21 but not a prime factor of 75. Therefore, 75 is +not divisible by 21. +Example: Is 42 divisible by 12? Find the prime factorisations of both: +42 = 2 × 3 × 7 and 12 = 2 × 2 × 3 +All prime factors of 12 are also prime factors of 42. But the prime +factorisation of 12 is not included in the prime factorisation of 42. +This is because 2 occurs twice in the prime factorisation of 12 but +only once in the prime factorisation of 42. This means that 42 is not +divisible by 12. +We can say that if one number is divisible by another, then the +prime factorisation of the second number is included in the prime +factorisation of the first number. + Figure it Out +1. Are the following pairs of numbers co-prime? Guess first and +then use prime factorisation to verify your answer. +a. +30 and 45 +b. +57 and 85  +c. +121 and 1331 +d. +343 and 216 +2. Is the first number divisible by the second? Use prime factorisation. +a. +225 and 27 +b. +96 and 24 +c. +343 and 17 +d. +999 and 99 +3. The first number has prime factorisation 2 × 3 × 7 and the second +number has prime factorisation 3 × 7 × 11. Are they co-prime? +Does one of them divide the other? +4. Guna says, “Any two prime numbers are co-prime?”. Is he right? +5.5 Divisibility Tests +So far, we have been finding factors of numbers in different contexts, +including to determine if a number is prime or not, or if a given pair +of numbers is co-prime or not. +Reprint 2025-26 + +Prime Time +123 +It is easy to find factors of small numbers. How do we find factors +of a large number? +Let us take 8560. Does it have any factors from 2 to 10 (2, 3, 4, +5, ..., 9, 10)? +It is easy to check if some of these numbers are factors or not +without doing long division. Can you find them? +Divisibility by 10 +Let us take 10. Is 8560 divisible by 10? This is another way of +asking if 10 is a factor of 8560. +For this, we can look at the pattern in the multiples of 10. +The first few multiples of 10 are: 10, 20, 30, 40, … Continue this +sequence and observe the pattern. +Is 125 a multiple of 10? Will this number appear in the previous +sequence? Why or why not? +Can you now answer if 8560 is divisible by 10? + Consider this statement: +Numbers that are divisible by 10 are those that end with +‘0’. Do you agree? +Divisibility by 5 +The number 5 is another number whose divisibility can easily be +checked. How do we do it? +Explore by listing down the multiples: 5, 10, 15, 20, 25, ... What do you +observe about these numbers? Do you see a pattern in the last digit? +What is the largest number less than 399 that is divisible by 5? Is +8560 divisible by 5? + Consider this statement: +Numbers that are divisible by 5 are those that end with +either a ‘0’ or a ‘5’. Do you agree? +Divisibility by 2 +The first few multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ... +What do you observe? Do you see a pattern in the last digit? +Math +Talk +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +124 +Is 682 divisible by 2? Can we answer this without doing the long +division? +Is 8560 divisible by 2? Why or why not? + Consider this statement: +Numbers that are divisible by 2 are those that end with ‘0’, +‘2’, ‘4, ‘6’ or ‘8’. Do you agree? +What are all the multiples of 2 between 399 and 411? +Divisibility by 4 +Checking if a number is divisible by 4 can also be done easily! +Look at its multiples: 4, 8, 12, 16, 20, 24, 28, 32, … +Are you able to observe any patterns that can be used? The +multiples of 10, 5 and 2 have a pattern in their last digits which we +are able to use to check for divisibility. Similarly, can we check if a +number is divisible by 4 by looking at the last digit? +It does not work! Look at 12 and 22. They have the same last digit, +but 12 is a multiple of 4 while 22 is not. Similarly 14 and 24 have the +same last digit, but 14 is not a multiple of 4 while 24 is. Similarly, 16 +and 26 or 18 and 28. What this means is that by looking at the last +digit, we cannot tell whether a number is a multiple of 4. +Can we answer the question by looking at more digits? Make a list +of multiples of 4 between 1 and 200 and search for a pattern. +  Find numbers between 330 and 340 that are divisible by 4. Also, +find numbers between 1730 and 1740, and 2030 and 2040, that are +divisible by 4. What do you observe? +  Is 8536 divisible by 4? +���Consider these statements: +1. Only the last two digits matter when deciding if a given +number is divisible by 4. +2. If the number formed by the last two digits is divisible by 4, +then the original number is divisible by 4. +3. If the original number is divisible by 4, then the number +formed by the last two digits is divisible by 4. + +Do you agree? Why or why not? +Math +Talk +Reprint 2025-26 + +Prime Time +125 +Divisibility by 8 +Interestingly, even checking for divisibility by 8 can be simplified. +Can the last two digits be used for this? + Find numbers between 120 and 140 that are divisible by 8. Also +find numbers between 1120 and 1140, and 3120 and 3140, that are +divisible by 8. What do you observe? + Change the last two digits of 8560 so that the resulting number is +a multiple of 8. + Consider these statements: +1. Only the last three digits matter when deciding if a +given number is divisible by 8. +2. If the number formed by the last three digits is divisible +by 8, then the original number is divisible by 8. +3. If the original number is divisible by 8, then the number +formed by the last three digits is divisible by 8. +Do you agree? Why or why not? +We have seen that long division is not always needed to check if a +number is a factor or not. We have made use of certain observations +to come up with simple methods for 10, 5, 2, 4, 8. Do we have such +simple methods for other numbers as well? We will discuss simple +methods to test divisibility by 3, 6, 7, and 9 in later classes! + Figure it Out +1. +2024 is a leap year (as February has 29 days). Leap years occur in +the years that are multiples of 4, except for those years that are +evenly divisible by 100 but not 400. +a. From the year you were born till now, which years were leap +years? +b. From the year 2024 till 2099, how many leap years are there? +2. +Find the largest and smallest 4-digit numbers that are divisible by +4 and are also palindromes. +3. +Explore and find out if each statement is always true, sometimes +true or never true. You can give examples to support your reasoning. +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +126 +a. Sum of two even numbers gives a multiple of 4. +b. Sum of two odd numbers gives a multiple of 4. +4. Find the remainders obtained when each of the following +numbers are divided by (a) 10, (b) 5, (c) 2. +78, 99, 173, 572, 980, 1111, 2345 +5. The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and +10. Guna checked for divisibility of 14560 by only two of these +numbers and then declared that it was also divisible by all of +them. What could those two numbers be? +6. Which of the following numbers are divisible by all of 2, 4, 5, 8 +and 10: 572, 2352, 5600, 6000, 77622160. +7. Write two numbers whose product is 10000. The two numbers +should not have 0 as the units digit. +5.6 Fun with numbers +Special numbers +There are four numbers in this box. Which number looks special to +you? Why do you say so? +9 +16 +25 +43 +Look at the what Guna’s classmates have to share: +• +Karnawati says, “9 is special because it is a single-digit +number whereas all the other numbers are 2-digit numbers”. +• +Gurupreet says, “9 is special because it is the only number +that is a multiple of 3”. +• +Murugan says, “16 is special because it is the only even +number and also the only multiple of 4”. +• +Gopika says, “25 is special as it is the only multiple of 5”. +• +Yadnyikee says, “43 is special because it is the only prime +number”. +• +Radha says, “43 is special because it is the only number that +is not a square”. +Reprint 2025-26 + +Prime Time +127 +Math +Talk + Below are some boxes with four numbers in each box. +Within each box try to say how each number is special +compared to the rest. Share with your classmates and find +out who else gave the same reasons as you did. Did anyone +give different reasons that may not have occurred to you?! +5 +7 +12 +35 +3 +8 +11 +24 +27 +3 +123 +31 +17 +27 +44 +65 +A prime puzzle +The figure on the left shows the puzzle. The figure on the right +shows the solution of the puzzle. Think what the rules can be to +solve the puzzle. +75 +42 +102 +170 +30 +63 +5 +5 +3 +75 +2 +3 +7 +42 +17 +2 +3 +102 +170 +30 +63 +Rules +Fill the grid with prime numbers only so that the product of each row +is the number to the right of the row and the product of each column +is the number below the column. +105 +20 +30 +28 +125 +18 +8 +105 +70 +30 +70 +28 +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +128 +63 +27 +190 +45 +42 +171 +343 +66 +44 +28 +154 +231 + + +If a number is divisible by another, the second number is called a factor +of the first. For example, 4 is a factor of 12 because 12 is divisible by 4 +(12 ÷ 4 = 3). + + +Prime numbers are numbers like 2, 3, 5, 7, 11, … that have only two +factors, namely 1 and themselves. + + +Composite numbers are numbers like 4, 6, 8, 9, … that have more +than 2 factors, i.e., at least one factor other than 1 and themselves. For +example, 8 has the factor 4 and 9 has the factor 3, so 8 and 9 are both +composite. + + +Every number greater than 1 can be written as a product of prime +numbers. This is called the number’s prime factorisation. For example, +84 = 2 × 2 × 3 × 7. + + +There is only one way to factorise a number into primes, except for +the ordering of the factors. + + +Two numbers that do not have a common factor other than 1 are said +to be co-prime. + + +To check if two numbers are co-prime, we can first find their prime +factorisations and check if there is a common prime factor. If there is no +common prime factor, they are co-prime, and otherwise they are not. + + +A number is a factor of another number if the prime factorisation of +the first number is included in the prime factorisation of the second +number. +Reprint 2025-26 + +1 + +CHAPTER 5 — SOLUTIONS +Prime Time + +Section 5.1 +Page no. 108 + Figure it out +Q.1. At what number is ‘idli-vada’ said for the 10th time? +Ans. 150. +Q.2. If the game is played for the numbers from 1 till 90, find out: +a. How many times would the children say ‘idli’ (including the times they say ‘idli- +vada’)? +Ans. 30 times. +b. How many times would the children say ‘vada’ (including the times they say ‘idli- +vada’)? +Ans. 18 times. +c. How many times would the children say ‘idli-vada’? +Ans. 6 times. +Q.3. What if the game was played till 900? How would your answers change? +Ans. If the game was played till 900, idli-vada would be said 60 times by the children. ‘Vada’ + + will be said 180 time and ‘idli’ will be said 300 times. +Q.4. Is this figure somehow related to the ‘idli-vada’ game? Hint: Imagine playing the +game till 30. Draw the figure if the game is played till 60. +Ans. +Yes, the common numbers represent the numbers when to say ‘idli-vada’. + + + + +2 + +Page no. 109 +Which of the following could be the other number: 2, 3, 5, 8, 10? +Ans. The other number will be 8. +Page no. 110 + Q. What jump size can reach both 15 and 30? There are multiple jump sizes possible. +Try to find them all. +Ans. +Jump size of 3 or 5 will take us to 15 & 30. +Other possible jump sizes are 1, 15. + + Q.1. Is there anything common among the shaded numbers? +Ans. All shaded numbers are multiples of 3. + Q.2. Is there anything common among the circled numbers? +Ans. All circled numbers are multiples of 4. + Q.3. Which numbers are both shaded and circled? What are these numbers called? +Ans. 36, 48, 60. These numbers are called common multiples of 3 and 4. + +Section 5.1 +Page no. 110 + Figure it out +Q.1. Find all multiples of 40 that lie between 310 and 410. +Ans. +Multiples of 40 that lie between 310 and 410 are: +320, 360, 400. +Q.2. Who am I? +a. I am a number less than 40. One of my factors is 7. The sum of my digits is 8. +b. I am a number less than 100. Two of my factors are 3 and 5.One of my digits is 1 +more than the other. +Ans. +a. 35 +b. 45 + + +3 + +Q.3. A number for which the sum of all its factors is equal to twice the number is called a +perfect number. The number 28 is a perfect number. Its factors are 1, 2, 4, 7, 14 and +28. Their sum is 56 which is twice 28. Find a perfect number between 1 and 10. +Ans. +6. +Q.4. Find the common factors of: +a. 20 and 28 +b. 35 and 50 +c. 4, 8 and 12 +d. 5, 15 and 25 +Ans. +a. Common factors of 20 and 28 = 1, 2, 4. +b. Common factors of 35 and 50 = 1, 5. +c. Common factors of 4, 8 and 12 = 1, 2, 4. +d. Common factors of 5, 15 and 25 = 1, 5 +Q.5. Find any three numbers that are multiples of 25 but not multiples of 50. +Ans. + Some such numbers are: 25, 75, 125, 175. +Q.6. Anshu and his friends play the ‘idli-vada’ game with two numbers, which are both +smaller than 10. The first time anybody says ‘idlivada’ is after the number 50. What +could the two numbers be which are assigned ‘idli’ and ‘vada’? +Ans. +7, 8 + +8, 9 + +(Try for other possibility) +Q.7. In the treasure hunting game, Grumpy has kept treasures on 28 and 70. What jump +sizes will land on both the numbers? +Ans. +The jump sizes are 1, 2, 7 or 14 +Q.8. In the diagram below, Guna has erased all the numbers except the common +multiples. Find out what those numbers could be and fill in the missing numbers in +the empty regions. +Ans. +Multiples of 3 + + + +Multiples of 4 +Think of some more. +Q.9. Find the smallest number that is a multiple of all the numbers from 1 to 10 except +for 7. +Ans. +360 +Q.10. Find the smallest number that is a multiple of all the number from 1 to 10. +Ans. +2520 + + + +4 + +Section 5.2 +Page No. 113 + Q. How many prime numbers are there from 21 to 30? How many composite numbers +are there from 21 to 30? +Ans. +Prime numbers from 21 to 30: 2 (23,29) + +Composite numbers from 21 to 30: 8 +Page No. 114 + Figure it out +Q.1. We see that 2 is a prime and also an even number. Is there any other even prime? +Ans. +No. +Q.2. Look at the list of primes till 100. What is the smallest difference between two +successive primes? What is the largest difference? +Ans. +The smallest difference between two successive primes is one (3-2=1) the largest +difference is 8 (97-89). +Q.3. Are there an equal number of primes occurring in every row in the table on the +previous page? Which decades have the least number of primes? Which have the +most number of primes? +Ans. +No. The least number of primes occur in the decades 91 to 100 and most number of +primes and occur in the decades 1 to 10, 11 to 20. + +Q.4. Which of the following numbers are prime: 23, 51, 37, 26? +Ans. +23 and 37. +Q.5. Write three pairs of prime numbers less than 20 whose sum is a multiple of 5. +Ans. +Pairs of prime numbers less than 20 whose sum is a multiple of 5 are + (2,3), (3,7), (2,13) + (Try for other possibilities). +Q.6. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 +and 3. Find such pairs of prime numbers up to 100. +Ans. +17 and 71, 37 and 73, 79 and 97. + +Q.7. Find seven consecutive composite numbers between 1 and 100. +Ans. +90, 91, 92, 93, 94, 95, 96. + +5 + +Q.8. Twin primes are pairs of primes having a difference of 2. For example, 3 and 5 are +twin primes. So are 17 and 19. Find the other twin primes between 1 and 100. +Ans. +3 and 5, 5 and 7, 11 and 13, 17 and 19, 29 and 31, 41 and 43, 59 and 61, 71 and 73. +Q.9 Identify whether each statement is true or false. Explain. +a. There is no prime number whose units digit is 4. +b. A product of primes can also be prime. +c. Prime numbers do not have any factors. +d. All even numbers are composite numbers. +e. 2 is a prime and so is the next number, 3. For every other prime, the next number +is composite. +Ans. +a. True, unit digit 4 means even number. We know that 2 is the only even prime number. +b. False, product of primes becomes a composite number. +c. False, prime numbers have exactly two factors: 1 and the number itself. +d. False, all even numbers except 2 are composite numbers +e. True, as every pair, other than (2, 3), contains an even number which is a composite +number. +Q.10. Which of the following numbers is the product of exactly three distinct prime +numbers: 45, 60, 91, 105, 330? +Ans. +105 = 3 × 5 × 7 is the only number which is the product of three distinct prime numbers. +Q.11. How many three-digit prime numbers can you make using each of 2, 4 and 5 once? +Ans. +None +Q.12. Observe that 3 is a prime number, and 2 × 3 + 1 = 7 is also a prime. Are there other +primes for which doubling and adding 1 gives another prime? Find at least five such +examples. +Ans. +Yes, 11 = 2 × 5 + 1, 47 = 2 × 23 + 1, +83 = 2 × 41 + 1 +23 = 2 × 11 +1, + +59 = 2 × 29 +1 + +(Try other possibilities.) + + + + + + + +6 + +Section 5.3 +Page No. 115 + Where should Grumpy place the treasures so that Jumpy cannot reach both the +treasures? +Check if these pairs are safe: +a. 15 and 39 + +b. 4 and 15 +c. 18 and 29 + +d. 20 and 55 +Ans. +Safe pairs are: 4 and 15, 18 and 29. + +Page no. 116 + Q. Which of the following pairs of numbers are co-prime? +a. 18 and 35 +b. 15 and 37 +c. 30 and 415 +d. 17 and 69 +e. 81 and 18 +Ans. +The pairs of co-prime numbers are: +a. 18 and 35 +b. 15 and 37 +d. 17 and 69 + Q. While playing the ‘idli-vada’ game with different number pairs, Anshu observed +something interesting! +1. Sometimes the first common multiple was the same as the product of the two +numbers. +2. At other times the first common multiple was less than the product of the two +numbers. +Find examples for each of the above. How is it related to the number pair being co- +prime? +Ans. +1. Examples when the first common multiple is the product of the two numbers: + +3, 5; +3, 7 and 4, 9. + +2. Examples when the first common multiple is less than the product of the two +numbers: + +3, 6; +3, 12 and 6, 15 +Whenever the number pair is co-prime the first common multiple is the product of the +two numbers. + + +7 + +Section 5.4 +Page no. 120 + Figure it out +Q.1. Find the prime factorisations of the following numbers: 64, 104, 105, 243, 320, 141, +1728, 729, 1024, 1331, 1000. +Ans. +64 = 2 × 2 × 2 × 2 × 2 × 2 + +104 = 2 × 2 × 2 × 13 + +105 = 3 × 5 × 7 + +243 = 3 ×3 ×3 ×3 ×3 + +320 = 2 × 2 × 2 × 2 × 2 × 2 × 5 + +141 = 3 × 47 + +1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 + +729 = 3 ×3 ×3 × 3 ×3 ×3 + +1024 = 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 2 + +1331 = 11 × 11 × 11 + +1000 = 2 × 2 × 2 × 5 × 5 × 5 +Q.2. The prime factorisation of a number has one 2, two 3s, and one 11. What is the +number? +Ans. 198 = 2 × 3 × 3 × 11 +Q.3. Find three prime numbers, all less than 30, whose product is 1955. +Ans. 1955 = 5 × 17 × 23 +Q.4. Find the prime factorisation of these numbers without multiplying first + a. 56 × 25 +b. 108 × 75 + c. 1000 × 81 +Ans. +a. 56 × 25 = 2 × 2 × 2 × 7 × 5 × 5 +b. 108 × 75 = 2 × 2 × 3 × 3 × 3 ×3 × 5 × 5 +c. 1000 × 81 = 2 × 2 × 2 × 3 × 3 × 3 ×3 × 5 × 5 × 5 + +Q.5. What is the smallest number whose prime factorisation has: +a. three different prime numbers? +b. four different prime numbers? + +8 + +Ans. a. 2 × 3 × 5 = 30 + +b. 2 × 3 × 5 × 7 = 210 + +Page no. 122 + Figure it out +Q.1. Are the following pairs of numbers co-prime? Guess first and then use prime +factorisation to verify your answer. +a. 30 and 45 +b. 57 and 85 +c. 121 and 1331 +d. 343 and 216 +Ans. +a. No +30 = 2 × 3 × 5 +45 = 3 × 3 × 5 + +b. Yes +57 = 19 × 3 +85 = 17 × 5 + +c. No +121 = 11 × 11 +1331 = 11 × 11 × 11 +c. Yes +343 = 7 × 7 × 7 +216 = 2 × 2 × 2 × 3 × 3 × 3. + +Q.2. Is the first number divisible by the second? Use prime factorisation. +a. 225 and 27 +b. 96 and 24 +c. 343 and 17 +d. 999 and 99 +Ans. +a. No +( +3 × 3 × 5 × 5 +3 × 3 × 3 ) + +b. Yes +( +2 × 2 × 2 × 2 × 2 × 3 +2 × 2 ×2 ×3 +) + +c. No +( +7 × 7 × 7 +17 +) + +d. No +( +3 × 3 × 3 × 3 × 37 +3 × 3 × 11 +) + +9 + +Q.3. The first number has prime factorisation 2 × 3 × 7 and the second number has prime +factorisation 3 × 7 × 11. Are they co-prime? Does one of them divide the other? +Ans. +No, they are not co-prime. + +No, one of them is not dividing the other. +Q.4. Guna says, “Any two prime numbers are co-prime”. Is he right? +Ans. +Yes, for example 2, 3; 3,11 (Try other examples) + +Section 5.5 +Page no. 124 + Q. Is 8536 divisible by 4? +Ans. +Yes, as 36 is divisible by 4. +Q. Consider these statements: +1. Only the last two digits matter when deciding if a given number is divisible by 4. +2. If the number formed by the last two digits is divisible by 4, then the original +number is divisible by 4. +3. If the original number is divisible by 4, then the number formed by the last two +digits is divisible by 4. +Do you agree? Why or why not? +Ans. +1. Yes + +2. Yes + +3. Yes +Yes,we agree. +Consider numbers 124,364,4028 etc. These are divisible by 4.Check for more numbers. + +Page no. 125 + Q. Find numbers between 120 and 140 that are divisible by 8. Also find numbers +between 1120 and 1140, and 3120 and 3140, that are divisible by 8. What do you +observe? +Ans. +The numbers between 120 & 140 divisible by 8 are = 128, 136. + +The numbers between 1120 & 1140 divisible by 8 are = 1128, 1136. + +The numbers between 3120 & 3140 divisible by 8 are = 3128, 3136. + +10 + +if the number formed by one’s, ten’s and hundred’s digit is divisible by 8, then the +number is divisible by 8. + Q. Change the last two digits of 8560 so that the resulting number is a multiple of 8. +Ans. +8552 is a multiple of 8. + Q. Consider this statement: +1. Only the last three digits matter when deciding if a given number is divisible by +8. +2. If the number formed by the last three digits is divisible by 8, then the original +number is divisible by 8. +3. If the original number is divisible by 8, then the number formed by the last three +digits is divisible by 8. +Do you agree? Why or why not? +Ans. +a. Yes + +b. Yes + +c. Yes +Yes. Some examples are: 8576,7648,5024. Try some more. + + Figure it out +Q.1. 2024 is a leap year (as February has 29 days). Leap years occurs in the years that are +multiples of 4, except for those years that are evenly divisible by 100 but not 400. +a. From the year you were born till now, which years were leap years? +b. From the year 2024 till 2099, how many leap years are there? +Ans. +b. From the year 2024 till 2099, the number of leap years is 19 + +Q.2. Find the largest and smallest 4-digit numbers that are divisible by 4 and are also +palindromes. +Ans. +The largest 4-digit number divisible by 4 and also a palindrome: 9999 + +The smallest 4-digit number divisible by 4 and also a palindrome: 1001. + +Q.3. Explore and find out if each statement is always true, sometimes true or never true. +You can give examples to support your reasoning. +a. Sum of two even numbers gives a multiple of 4. +b. Sum of two odd numbers gives a multiple of 4. + +11 + +Ans. +a. Sometimes true +Example 2+6 = 8, a multiple of 4. +but 2+4 = 6, not a multiple of 4. +b. Sometimes true +Example 1+3 = 4, a multiple of 4. +but 1+5 = 6, not a multiple of 4. +Note: Think of more such examples. +Q.4. Find the remainders obtained when each of the following numbers are divided by i) +10, ii) 5, iii) 2. +78, 99, 173, 572, 980, 1111, 2345 +Ans. +Numbers +a) Remainder when +divided by 10 +b) Remainder +when +divided by 5 +c) Remainder when +divided by 2 +78 +99 +173 +572 +980 +1111 +2345 +8 +9 +3 +2 +0 +1 +5 +3 +4 +3 +2 +0 +1 +0 +0 +1 +1 +0 +0 +1 +1 + +Q.5. The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for +divisibility of 14560 by only two of these numbers and then declared that it was also +divisible by all of them. What could those two numbers be? +Ans. +The two numbers that are sufficient to declare the divisibility of 14560 by 2, 4, 5, 8 and +10 are 5 and 8. +Q.6. Which of the following numbers are divisible by all of 2, 4, 5, 8 and 10: 572, 2352, +5600, 6000, 77622160. +Ans. +The numbers divisible by 2, 4, 5, 8 and 10 are: + +5600, 6000 and 77622160. +Q.7. Write two numbers whose product is 10000. The two numbers should not have 0 as +their unit digit. +Ans. +10000 = 16 × 625." +class_6,6,Perimeter and Area,ncert_books/class_6/Ganita_Prakash/fegp106.pdf,"6.1 Perimeter +Do you remember what the perimeter of a closed plane figure is? +Let us refresh our understanding! + +The perimeter of any closed plane figure is the distance covered +along its boundary when you go around it once. For a polygon, i.e., +a closed plane figure made up of line segments, the perimeter is +simply the sum of the lengths of its all sides, i.e., the total distance +along its outer boundary. + +The perimeter of a polygon = the sum of the lengths of its all sides. + +Let us revise the formulas for the perimeter of rectangles, squares, +and triangles. +Perimeter of a rectangle +Consider a rectangle ABCD whose length and breadth are 12 cm and +8 cm, respectively. What is its perimeter? +Perimeter of the rectangle = Sum of the lengths of its four sides + + + + + + + += AB + BC + CD + DA +A +12 cm +B +8 cm +D +C +PERIMETER AND AREA +6 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +130 + + + + += AB + BC + AB + BC + + + + += 2 × AB + 2 × BC + + + + += 2 × (AB + BC) + + + + += 2 × (12 cm + 8 cm) + + + + += 2 × (20 cm) + + + + += 40 cm. +From this example, we see that — +Perimeter of a rectangle = length + breadth + length + breadth. +Perimeter of a rectangle = 2 × (length + breadth). +The perimeter of a rectangle is twice the sum of its length and breadth. +Perimeter of a square +Debojeet wants to put coloured tape all around +a square photo frame of side 1m as shown. +What will be the length of the coloured tape +he requires? Since Debojeet wants to put the +coloured tape all around the square photo +frame, he needs to find the perimeter of the +photo frame. + +Thus, the length of the tape required = +perimeter of the square + += sum of the lengths of all four sides of the square + += 1 m + 1 m + 1 m + 1 m = 4 m. + +Now, we know that all four sides of a square are equal in length. +Therefore, in place of adding the lengths of each side, we can simply +multiply the length of one side by 4. + +Thus, the length of the tape required = 4 × 1 m = 4 m. + +From this example, we see that +Perimeter of a square = 4 × length of a side. + +The perimeter of a square is quadruple the length of its side. +Opposite sides of a +rectangle are always +equal. So, AB = CD and +AD = BC +1 m +Reprint 2025-26 + +Perimeter and Area +131 +Perimeter of a triangle +Consider a triangle having three given sides of +lengths 4 cm, 5 cm and 7 cm. Find its perimeter. +Perimeter of the triangle = 4 cm + 5 cm + 7 cm + + + + + + +  = 16 cm. +Perimeter of a triangle = sum of the lengths of its three sides. +Example: Akshi wants to put lace all around a +rectangular tablecloth that is 3 m long and 2 m +wide. Find the length of the lace required. +Solution + + Length of the rectangular table cover = 3 m. + + Breadth of the rectangular table cover = 2 m. + + Akshi wants to put lace all around the +tablecloth. + + Therefore, the length of the lace required will be the perimeter of the +rectangular tablecloth. + + Now, the perimeter of the rectangular tablecloth = 2 × (length + breadth) + + + + + + + + + + + + += 2 × (3 m + 2 m) + + + + + + + + + + + + += 2 × 5 m = 10 m. + + Hence, the length of the lace required is 10 m. +Example: Find the distance travelled by Usha if she takes three rounds of +a square park of side 75 m. +Solution + + Perimeter of the square park = 4 × length +of a side = 4 × 75 m = 300 m. + + Distance covered by Usha in one +round = 300 m. + + Therefore, the total distance travelled by +Usha in three rounds = 3 × 300 m = 900 m. +5 cm +4 cm +7 cm +Reprint 2025-26 + +Ganita Prakash | Grade 6 +132 + Figure it Out +1. Find the missing terms: +a. +Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?. +b. +Perimeter of a square = 20 cm; side of a length = ?. +c. +Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?. +2. A rectangle having sidelengths 5 cm and 3 cm is made using a +piece of wire. If the wire is straightened and then bent to form a +square, what will be the length of a side of the square? +3. Find the length of the third side of a triangle having a perimeter +of 55 cm and having two sides of length 20 cm and 14 cm, +respectively. +4. What would be the cost of fencing a rectangular park whose length +is 150 m and breadth is 120 m, if the fence costs `40 per metre? +5. A piece of string is 36 cm long. What will be the length of each +side, if it is used to form: +a. +A square, +b. +A triangle with all sides of equal length, and +c. +A hexagon (a six sided closed figure) with sides of equal +length? +6. A farmer has a rectangular field having length 230 m and breadth +160 m. He wants to fence it with 3 rounds of rope as shown. What +is the total length of rope needed? +Reprint 2025-26 + +Perimeter and Area +133 +Starting Point +for Toshi +Akshi and Toshi start running along the rectangular +tracks as shown in the figure. Akshi runs along the +outer track and completes 5 rounds. Toshi runs along +the inner track and completes 7 rounds. Now, they +are wondering who ran more. Find out who ran the +longer distance. +Starting Point +for Akshi +60 m +70 m +30 m +40 m + +Each track is a rectangle. Akshi’s track has length 70 m and breadth +40 m. Running one complete round on this track would cover 220 m, +i.e., 2 × (70 + 40) m = 220 m. This is the distance covered by Akshi in +one round. + Figure it Out +1. Find out the total distance Akshi has covered in 5 rounds. +2. Find out the total distance Toshi has covered in 7 rounds. Who ran +a longer distance? +3. Think and mark the positions as directed— +a. + Mark ‘A’ at the point where Akshi will be after she ran 250 m. +b. +Mark ‘B’ at the point where Akshi will be after she ran 500 m. +c. +Now, Akshi ran 1000 m. How many full rounds has she finished +running around her track? Mark her position as ‘C’. +d. +Mark ‘X’ at the point where Toshi will be after she ran 250 m. +e. +Mark ‘Y’ at the point where Toshi will be after she ran 500 m. +Matha Pachchi! +Reprint 2025-26 + +Ganita Prakash | Grade 6 +134 +f. +Now, Toshi ran 1000 m. How many full rounds has she finished +running around her track? Mark her position as ‘Z’. + + Deep Dive: In races, usually there is a common finish line for all +the runners. Here are two square running tracks with the inner track +of 100 m each side and outer track of 150 m each side. The common +finishing line for both runners is shown by +the flags in the figure which are in the center +of one of the sides of the tracks. + +If the total race is of 350 m, then we have +to find out where the starting positions of the +two runners should be on these two tracks so +that they both have a common finishing line +after they run for 350 m. Mark the starting +points of the runner on the inner track as ‘A’ +and the runner on the outer track as ‘B’. + Estimate and Verify +Take a rough sheet of paper or a sheet of newspaper. +Make a few random shapes by cutting the paper +in different ways. Estimate the total length of the +boundaries of each shape then use a scale or measuring +tape to measure and verify the perimeter for each +shape. + Akshi says that the perimeter of this triangle shape is 9 units. +Toshi says it can’t be 9 units and the perimeter will be more than 9 +units. What do you think? +The perimeter is +9 units. +No, it will be more +than 9 units. +150 m +Common Finishing Line +100 m +Reprint 2025-26 + +Perimeter and Area +135 + +This figure has lines of two different unit lengths. Measure the +lengths of a red line and a blue line; are they same? We will call the +red lines — straight lines and the blue lines — diagonal lines. So, the +perimeter of this triangle is 6 straight units + 3 diagonal units. We +can write this in a short form as: 6s + 3d units. + Write the perimeters of the figures below in terms of straight and +diagonal units. +Perimeter of a regular polygon +Like squares, closed figures that have all sides and all angles equal +are called regular polygons. We studied the sequence of regular +polygons as ‘Shape Sequence’ #1 in Chapter 1. Examples of regular +polygons are the equilateral triangle (where all three sides and all +three angles are equal), regular pentagon (where all five sides and +all five angles are equal), etc. +Perimeter of an equilateral triangle +We know that for any triangle its perimeter +is sum of all three sides. + +Using this understanding, we can find the +perimeter of an equilateral triangle. +Perimeter of an equilateral triangle += AB + BC + AC = AB + AB + AB += 3 times length of one side. +Perimeter of an equilateral triangle = 3 × length of a side. +What is a similarity between a square and an equilateral triangle? +A +B +C +Reprint 2025-26 + +Ganita Prakash | Grade 6 +136 + Find various objects from your surroundings that have regular +shapes and find their perimeters. Also, generalise your understanding +for the perimeter of other regular polygons. +Teacher’s Note +Discuss more about regular polygons and encourage students to +come up with a general formula for the perimeter of a regular +polygon. +Split and rejoin +A rectangular paper chit of dimension 6 cm × 4 cm +is cut as shown into two equal pieces. These two +pieces are joined in different ways. + +a. +6 cm +6 cm +2 cm + + For example, the arrangement a. has a perimeter of 28 cm. +  Find out the length of the boundary (i.e., the perimeter) of each +of the other arrangements below. +b. +c. +d. +3 cm +2 cm +2 cm +2 cm + Arrange the two pieces to form a figure with a perimeter of 22 cm. +6 cm +4 cm +Reprint 2025-26 + +Perimeter and Area +137 +6.2 Area +We have studied the areas of closed figures (regular and irregular) +in previous grades. Let us recall some key points. +The amount of region enclosed by a closed figure is called its area. + +In previous grades, we arrived at the formula for the area of a +rectangle and a square using square grid paper. Do you remember? +Area of a square = ________ + Area of a rectangle = _______ +Teacher’s Note +Help students in recalling the method of finding the area of a +rectangle and a square using grid papers. Provide square grid +papers to students and let them come up with the formula. + +Let’s see some real-life problems related to these ideas. +Example: A floor is 5 m long and 4 m wide. A square carpet of sides 3 m +is laid on the floor. Find the area of the floor that is not carpeted. +Solution +Length of the floor = 5 m. +Width of the floor = 4 m. +Area of the floor = length × width = 5 m × 4 m = 20 sq m. +Length of the square carpet = 3 m. +Area of the carpet = length × length = 3 m × 3 m = 9 sq m. +Hence, the area of the floor laid with carpet is 9 sq m. + +Therefore, the area of the floor that is not carpeted is: area of the floor +minus the area of the floor laid with carpet = 20 sq m – 9 sq m = 11 sq m. +Example: Four square flower beds each of side 4 m are in four corners +on a piece of land 12 m long and 10 m wide. Find the area of the remaining +part of the land. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +138 +Solution +Length of the land (l) = 12 m. +Width of land (w) = 10 m. +Area of the whole land = l × w = 12 m × 10 m = 120 sq m. +The sidelength of each of the four square flower beds is (s) = 4 m. +Area of one flower bed = s × s = 4 m × 4 m = 16 sq m. +Hence, the area of the four flower beds = 4 × 16 sq m = 64 sq m. + +Therefore, the area of the remaining part of the land is: area of the complete +land minus the area of all four flower beds = 120 sq m – 64 sq m = 56 sq m. + Figure it Out +1. The area of a rectangular garden 25 m long is 300 sq m. What is +the width of the garden? +2. What is the cost of tiling a rectangular plot of land 500 m long and +200 m wide at the rate of ` 8 per hundred sq m? +3. A rectangular coconut grove is 100 m long and 50 m wide. If each +coconut tree requires 25 sq m, what is the maximum number of +trees that can be planted in this grove? +4. By splitting the following figures into rectangles, find their areas +(all measures are given in metres). + +4 +4 +3 +3 +5 +3 +3 +2 +2 +2 +1 +1 +1 +3 +a. +b. +Reprint 2025-26 + +Perimeter and Area +139 + Figure it Out + + Cut out the tangram pieces given at the end of your textbook. +A +B +D +C +G +F +E +1. Explore and figure out how many pieces have the same area. +2. How many times bigger is Shape D as compared to Shape C? What +is the relationship between Shapes C, D and E? +3. Which shape has more area: Shape D or F? Give reasons for your +answer. +4. Which shape has more area: Shape F or G? Give reasons for your +answer. +5. What is the area of Shape A as compared to Shape G? Is it twice as +big? Four times as big? + + +Hint: In the tangram pieces, by placing the shapes over each +other, we can find out that Shapes A and B have the same +area, Shapes C and E have the same area. You would have +also figured out that Shape D can be exactly covered using +Shapes C and E, which means Shape D has twice the area of +Shape C or shape E, etc. +6. Can you now figure out the area of the big square formed with all +seven pieces in terms of the area of Shape C? +7. Arrange these 7 pieces to form a rectangle. What will be the area +of this rectangle in terms of the area of Shape C now? Give reasons +for your answer. +8. Are the perimeters of the square and the rectangle formed from +these 7 pieces different or the same? Give an explanation for your +answer. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +140 +  Look at the figures below and guess which one of them has a +larger area. +a. +b. + +We can estimate the area of any simple closed shape by using a +sheet of squared paper or graph paper where every square measures +1 unit × 1 unit or 1 square unit. + +To estimate the area, we can trace the shape onto a piece of +transparent paper and place the same on a piece of squared or graph +paper and then follow the below conventions — +1. +The area of one full small square of the squared or graph paper +is taken as 1 sq unit. +2. +Ignore portions of the area that are less than half a square. +3. +If more than half of a square is in a region, just count it as 1 sq unit. +4. +If exactly half the square is counted, take its area as 1 +2 sq unit. +  Find the area of the following figures. +Reprint 2025-26 + +Perimeter and Area +141 +Let’s Explore! +Why is area generally measured +using squares? + +Draw a circle on a graph sheet with +diameter (breadth) of length 3. Count +the squares and use them to estimate +the area of the circular region. + +As you can see, circles can’t +be packed tightly without gaps in +between. So, it is difficult to get +an accurate measurement of area +using circles as units. Here, the +same rectangle is packed in two +different ways with circles — the +first one has 42 circles and the second one has 44 circles. + Try using different shapes (triangle and rectangle) to fill the +given space (without overlaps and gaps) and find out the merits +associated with using a square shape to find the area rather than +another shape. List out the points that make a square the best shape +to use to measure area. +1. Find the area (in square metres) of the floor outside of the +corridor. +2. Find the area (in square metres) occupied by your school +playground. +Let’s Explore! + On a squared grid paper (1 square = 1 square unit), make +as many rectangles as you can whose lengths and widths are +a whole number of units such that the area of the rectangle is +24 square units. +a. +Which rectangle has the greatest perimeter? +b. +Which rectangle has the least perimeter? +Why can’t we use +circles instead of +squares to find the +area? +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +142 +c. +If you take a rectangle of area 32 sq cm, what will your answers be? +Given any area, is it possible to predict the shape of the rectangle with +the greatest perimeter as well as the least perimeter? Give examples +and reasons for your answer. +6.3 Area of a Triangle +Draw a rectangle on a piece of paper and draw one of its diagonals. +Cut the rectangle along that diagonal and get two triangles. + Check! whether the two triangles overlap each other exactly. +Do they have the same area? + +Try this with more rectangles having different dimensions. You +can check this for a square as well. + Can you draw any inferences from this exercise? Please write it +here. + + + +Now, see the figures below. Is the area of the blue rectangle more +or less than the area of the yellow triangle? Or is it the same? Why? + Can you see some relationship between the blue rectangle and +the yellow triangle and their areas? Write the relationship here. + +Help students in articulating their inferences and in defining the +relationships they have observed in their own words, gradually +leading to a common statement for whole classroom. Recall the +definition of a diagonal in the classroom. +Teacher’s Note + +Draw suitable triangles on grid paper to verify your inferences +and relationships observed in the above exercises. +Reprint 2025-26 + +Perimeter and Area +143 + Use your understanding from previous +grades to calculate the area of any closed +figure using grid paper and — +1. +Find the area of blue triangle BAD. +__________ +2. +Find the area of red triangle ABE. +___________ + +Area of rectangle ABCD = ________________ + +So, the area of triangle BAD is half of the area of the rectangle ABCD. +Area of triangle ABE = Area of triangle AEF + Area of triangle BEF. +Here, the area of triangle AEF = half of the area of rectangle AFED. +Similarly, the area of triangle BEF = half of the area of rectangle BFEC. +Thus, the area of triangle ABE =  half of the area of rectangle AFED ++ half of the area of rectangle BFEC +Both the red and +blue triangles have +the same area but +they look very +different. +What about +triangle +ABE? +There are +two halves of +two different +rectangles. +E +D +C +F +A +B +Reprint 2025-26 + +Ganita Prakash | Grade 6 +144 + + + + + + + += half of the sum of the areas of the +rectangles AFED and BFEC + + + + + + + += half of the area of rectangle ABCD. +Conclusion ____________________________________________________ + + + +____________________________________________________ + Figure it Out +1. Find the areas of the figures below by dividing them into rectangles +and triangles. +c +b +a +e +d +Reprint 2025-26 + +Perimeter and Area +145 +Making it ‘More’ or ‘Less’ +Observe these two figures. Is there any similarity or difference +between the two? + +Using 9 unit squares (having an area of 9 sq units), we have +made figures with two different perimeters—the first figure has a +perimeter of 12 units and the second has a perimeter of 20 units. + +Arrange or draw different figures with 9 sq units to get other +perimeters. Each square should align with at least one other square +on at least one side completely and together all squares should form +a single connected figure with no holes. + Using 9 unit squares, solve the following. +1. What is the smallest perimeter possible? +2. What is the largest perimeter possible? +3. Make a figure with a perimeter of 18 units. +4. Can you make other shaped figures for each of the above three +perimeters, or is there only one shape with that perimeter? +What is your reasoning? +  Let’s do something tricky now! We have a figure below having +perimeter 24 units. + +Without +calculating +all +over +again,  observe, think and find +out what will be the change in +the perimeter if a new square is +attached as shown on the right. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +146 + +Experiment placing this new square at different places and think +what the change in perimeter will be. Can you place the square so +that the perimeter: a) increases; b) decreases; c) stays the same? + Below is the house plan of Charan. It is in a rectangular plot. +Look at the plan. What do you notice? +30 ft +Master Bedroom +(15 ft × 15 ft) +Area=225 sq ft +Toilet +(5ft × 10 ft) +Kitchen +(15 ft × 12 ft) +Area = 180 sq ft +Hall +Area = ____ +Small Bedroom +(15 ft × ____ ft) +Area = 180 sq ft +Garden +(____ ft × ____ ft) +Area = ____ +Parking +(____ ft × ____ ft) +Area = ____ +Utility +(____ ft × ____ ft) +Area = ____ + +Some of the measurements are given. +a. +Find the missing measurements. +b. Find out the area of his house. +Reprint 2025-26 + +Perimeter and Area +147 + +Now, find out the missing dimensions and area of Sharan’s home. +Below is the plan: +42 ft +Master Bedroom +(12 ft × 15 ft) +Area=180 sq ft +Toilet +(____ ft × +____ ft) +Area = +____ +Utility +(____ ft × ____ ft) +Area = 70 sq ft +Kitchen +(18 ft × 10 ft) +Area = 180 sq ft +Hall +(23 ft × ____ ft) +Area = ____ +Entrance +(____ ft × ____ ft) +Area = ____ +Small Bedroom +(12 ft × 10 ft) +Area = ____ + +Some of the measurements are given. +a. +Find the missing measurements. +b. Find out the area of his house. + +What are the dimensions of all the different rooms in Sharan’s +house? Compare the areas and perimeters of Sharan’s house and +Charan’s house. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +148 + Area Maze Puzzles +In each figure, find the missing value of either the length of a side or +the area of a region. +13 sq cm +15 sq cm +26 sq cm +? sq cm +a. +c. +b. +d. +2 cm +2 cm +3 cm +10 sq cm +10 sq +cm +? sq cm +3 cm +3 cm +60 sq cm +42 sq cm +? sq cm +15 cm +5 cm +6 cm +38 sq cm +18 sq cm +5 cm +4 cm +? cm +Reprint 2025-26 + +Perimeter and Area +149 + Figure it Out +1. Give the dimensions of a rectangle whose area is the sum of the +areas of these two rectangles having measurements: 5 m × 10 m +and 2 m × 7 m. +2. The area of a rectangular garden that is 50 m long is 1000 sq m. +Find the width of the garden. +3. The floor of a room is 5 m long and 4 m wide. A square carpet +whose sides are 3 m in length is laid on the floor. Find the area +that is not carpeted. +4. Four flower beds having sides 2 m long and 1 m wide are dug at +the four corners of a garden that is 15 m long and 12 m wide. How +much area is now available for laying down a lawn? +5. Shape A has an area of 18 square units and Shape B has an area +of 20 square units. Shape A has a longer perimeter than Shape B. +Draw two such shapes satisfying the given conditions. +6. On a page in your book, draw a rectangular border that is 1 cm +from the top and bottom and 1.5 cm from the left and right sides. +What is the perimeter of the border? +7. Draw a rectangle of size 12 units × 8 units. Draw another rectangle +inside it, without touching the outer rectangle that occupies +exactly half the area. +8. A square piece of paper is folded in half. The square is then cut +into two rectangles along the fold. Regardless of the size of the +square, one of the following statements is always true. Which +statement is true here? +a. +The area of each rectangle is larger than the area of the square. +b. The perimeter of the square is greater than the perimeters of +both the rectangles added together. +c. +The perimeters of both the rectangles added together is +always 11 +2 times the perimeter of the square. +d. The area of the square is always three times as large as the +areas of both rectangles added together. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +150 +S u m m a r y + + +The perimeter of a polygon is the sum of the lengths of all its sides. + + a. The perimeter of a rectangle is twice the sum of its length and +width. + + b. The perimeter of a square is four times the length of any one of its +sides. + + +The area of a closed figure is the measure of the region enclosed by +the figure. + + +Area is generally measured in square units. + + +The area of a rectangle is its length times its width. The area of a square +is the length of any one of its sides multiplied by itself. + + +Two closed figures can have the same area with different perimeters, +or the same perimeter with different areas. + + +Areas of regions can be estimated (or even determined exactly) by +breaking up such regions into unit squares, or into more general- +shaped rectangles and triangles whose areas can be calculated. +Reprint 2025-26 + +[1] + +CHAPTER 6 — SOLUTIONS +Perimeter and Area + +Section 6.1 +Page no. 132 + Figure it Out +Q.1. Find the missing terms: +a. Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ? +b. Perimeter of a square = 20 cm; side of a length = ? +c. Perimeter of a rectangle = 12 m; length = 3 m; breadth = ? +Ans. (a) length = 5 cm + +(b) length of a side = 5 cm + +(c) breadth = 3cm +Q.2. A rectangle having side lengths 5 cm and 3 cm is made using a piece of wire. If the +wire is straightened and then bent to form a square, what will be the length of a side +of the square? +Ans. Length of a side of square = 4 cm +Q.3. Find the length of the third side of a triangle having a perimeter of 55 cm and having +two sides of length 20 cm and 14 cm, respectively? +Ans. Length of the third side of triangle = 21 cm +Q.4. What would be the cost of fencing a rectangular park whose length is 150 m and +breadth is 120 m, if the fence costs Rs.40 per metre? +Ans. P = 2 × (150+120) = 540; So, 540 × 40 = Rs. 21,600 +Q.5. A piece of string is 36 cm long. What will be the length of each side, if it is used to +form: +a. A square, +b. A triangle with all sides of equal length, and +c. A hexagon (a six sided closed figure) with sides of equal length? +Ans. (a) Length of each side of square = 9 cm + +(b) Length of each side of regular triangle = 12 cm + +(c) Length of each side of regular hexagon = 6 cm + +[2] + +Q.6. A farmer has a rectangular field having length 230 m and breadth 160 m. He wants +to fence it with 3 rounds of rope as shown. What is the total length of rope needed? +Ans. Perimeter of the field = 780 m + +Total length of the rope needed = 2340 m + +Section 6.1 +Page No. 133 + Figure it out +Q.1. Find out the total distance Akshi has covered in 5 rounds. +Ans. +Perimeter of the outer track = 220 m + +Total distance Akshi has covered in 5 rounds = 1100 m +Q.2. Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance? +Ans. +Perimeter of the inner track = 180 m + +Total distance Toshi has covered in 7 rounds = 1260 m + +Toshi ran longer distance +Deep Dive + Page No. 134 +Ans. +From point A to flag: 100 + 100 + 100 + 50 + + + + += 350 m + +From point B to flag: 125 + 150 + 75 + + + + += 350 m + + + + +Section 6.1 +Page no. 134 + Q. Akshi says that the perimeter of this triangle shape is 9 units. Toshi says it can’t +be 9 units and the perimeter will be more than 9 units. What do you think? +Ans. The perimeter will be more than 9 units as the length of a diagonal of a square is always +greater than the sides of the square. + +[3] + +P.135 + Q. Write the perimeters of the figures below in terms of straight and diagonal units. +Ans. +The perimeters of the figures: 8s + 2d, 4s + 6d, 12s + 6d, 18s + 6d. + Q. Find various objects from your surroundings that have regular shapes and find +their perimeters. Also, generalise your understanding for the perimeter of other +regular polygons. +Ans. +In general perimeter of regular polygons = Number of sides × Length of a side. + +Section 6.2 +Page No. 138 + Figure it out +Q.1. The area of a rectangular garden 25 m long is 300 sq m. What is the width of the +garden? +Ans. +Width of the garden = 300  25 = 12 m. +Q.2. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the +rate of Rs.8 per hundred sq. m? +Ans. +Cost of tiling = Rs. 8000 +Q.3. A rectangular coconut grove is 100 m long and 50 m wide. If each coconut tree +requires 25 sq m, what is the maximum number of trees that can be planted in this +grove? +Ans. +200. +Q.4. By splitting the following figures into rectangles, find their areas (all measures are +given in meters): +Ans. +(a) +28 sq. m. + + +(b) +9 sq. m. + + + +Page No. 140 +Q. Find the area of the following figures. +Ans. +4 sq. units, 9 sq. units, 10 sq. units, 11 sq. units. + + + + +[4] + +Let’s Explore! +Page No. 141 +Section 6.3 +Figure it Out +Page No. 144 +Q.1. Find the areas of the figures below by dividing them into rectangles. +Ans. + a. 24 sq. units. +b. 30 sq. units. +c. 48 sq. units +d. 16 sq. units + + e. 12 sq. units. +Page No. 145 +Q. Using 9 unit squares, solve the following. +1. What is the smallest perimeter possible? +2. What is the largest perimeter possible? +3. Make a figure with a perimeter of 18 units. +4. Can you make other shaped figures for each of the above three perimeters, or is +there only one shape with that perimeter? +What is your reasoning? +Ans. 1. The smallest possible perimeter is 12 cm obtained from a square of side 3 cm +2. The largest possible perimeter is 20 cm. obtained from a rectangle of length 9 cm and +breadth 1 cm. + + +3. + +4. Yes, except for the smallest perimeter (12 sq. units). + + + + +[5] + +Page No. 146 +Q. Below is the house plan of Charan. It is in a rectangular plot. Look at the plan. What +do you notice? +Some of the measurements are given. +a. Find the missing measurements. +b. Find out the area of his house. +Ans. a. +i. +Small bedroom 15ft × 12ft +Area = 180 sq ft + +ii. +Utility 15ft × 3ft +Area = 45 sq ft + +iii. +Hall 20ft × 12ft +Area = 240 sq ft + +iv. +Parking 15ft × 3ft +Area = 45 sq ft + +v. +Garden 20ft × 3ft +Area = 60 sq ft + +b. +Area of his house = 35ft × 30ft +=1050 sq ft +Page No. 147 +Q. Now, find out the missing dimensions and area of Sharan’s home. Below is the plan: +Some of the measurements are given. +a. Find the missing measurements. +b. Find out the area of his house. +What are the dimensions of all the different rooms in Sharan’s house? Compare the +areas and perimeters of Sharan’s house and Charan’s house. +Ans. a. Dimensions of Sharan’s house. +i. +Utility 7ft × 10 ft +Area = 70 sq ft + +ii. +Hall 23ft × 15 ft +Area = 345 sq ft +iii. +Entrance 7 ft × 15ft +Area = 105 sq ft + +iv. +Small bedroom 12ft × 10ft +Area = 120 sq ft + +v. +Toilet 5ft × 10 ft + +[6] + +Area = 50 sq ft + +b. Area of his house = 42 ft × 25 ft + = 1050 sq ft + +Area of Charan’s house = 35ft × 30 ft = 1050 sq ft +Area of Sharan’s house = 42 ft × 25ft = 1050 sq ft +Area of both the houses are equal. +Now, Perimeter of Charan’s house = 130 ft +Perimeter of Sharan’s house = 134 ft +Perimeter of Sharan’s house is greater than the Perimeter of Charan’s house. + +Page No. 148 +Area Maze Puzzles +In each fig. find the missing value of either the length of the side or the area of the +region. +Ans. +a. 30 sq cm. +b. 9 sq cm. +c. 16 sq cm. +d. 5 cm. + +Section 6.3 +Page No. 149 +Figure it Out +Q.1. Give the dimensions of a rectangle whose area is the sum of the areas of these two +rectangles having measurements: 5 m × 10 m and 2 m × 7 m. +Ans. Possible dimensions are 16 m and 4 m Or 32 m and 2m +Or 8 m and 8 m +Q.2. The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of +the garden? +Ans. 20 m. +Q.3. The floor of a room is 5 m long and 4 m wide. A square carpet whose sides are 3 m +in length is laid on the floor. Find the area that is not carpeted. +Ans. Area of floor not carpeted = 11 sq. m. + +[7] + + +Q.4. Four flower beds having sides 2 m long and 1 m wide are dug at the four corners of +a garden that is 15 m long and 12 m wide. How much area is now available for laying +down a lawn? +Ans. Available area for lawn = 172 sq. m. + +Q.5. Shape A has an area of 18 square units and Shape B has an area of 20 square units. +Shape A has a longer perimeter than Shape B. Draw two such shapes satisfying the +given conditions. +Ans. Possible dimensions of shape A are 6m and 3m; 2m and 9m; 18m and 1m +Corresponding Perimeters =18 m, 22m, 38m respectively +Possible dimensions of shape B are 5m and 4m; 10m and 2m; 20m and 1m +Corresponding Perimeters =18m, 24m, 42m respectively +Given P(A) > P(B) +We may take P(A) = 22m or 38m and P(B) = 18 m and draw the figures accordingly. +Q.7. Draw a rectangle of size 12 units × 8 units. Draw another rectangle inside it, without +touching the outer rectangle that occupies exactly half the area. +Ans. Area of outer rectangle = 96 sq. units. +Area of inner rectangle = 48 sq. units. +Draw the rectangles accordingly. +Q.8. A square piece of paper is folded in half. The square is then cut into two rectangles +along the fold. Regardless of the size of the square, one of the following statements +is always true. Which statement is true here? +a. The area of each rectangle is larger than the area of the square. +b. The perimeter of the square is greater than the perimeters of both the rectangles +added together. +c. The perimeters of both the rectangles added together is always 𝟏 +𝟏 +𝟐 times the +perimeter of the square. +d. The area of the square is always three times as large as the areas of both +rectangles added together. +Ans. Only C is true." +class_6,7,Fractions,ncert_books/class_6/Ganita_Prakash/fegp107.pdf,"Recall that when some whole number of things are shared +equally among some number of people, fractions tell us how +much each share is. +Shabnam: Do you remember, if one roti is divided equally +between two children, how much +roti will each child get? +Mukta: +Each child will get half a roti. +Shabnam: The fraction ‘one half’ is written as +1 +2 . We also sometimes read this as +‘one upon two.’ +Mukta: +If one roti is equally shared among +4 children, how much roti will one child get? +Shabnam: Each child’s share is 1 +4 roti. +Mukta: +And which is more 1 +2 roti or 1 +4 roti? +Shabnam: When 2 children share 1 roti +equally, each child gets 1 +2 roti. When +4 children share 1 roti equally, each +child gets 1 +4 roti. Since, in the second +group more children share the +Fractions +7 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +152 +same one roti, each child gets a smaller share. So, 1 +2 roti is +more than 1 +4 roti. +7.1 Fractional Units and Equal Shares +Beni: +Which fraction is greater — 1 +5 or 1 +9? +Arvin: +9 is bigger than 5. So I would guess that 1 +9 is greater +than 1 +5. Am I right? +Beni: +No! That is a common mistake. Think of these fractions +as shares. +Arvin: +If one roti is shared among 5 children, each one gets a +share of 1 +5 roti. If one roti is shared among 9 children, +each one gets a share of 1 +9 roti? +Beni: +Exactly! Now think again - which share is higher? +Arvin: +If I share with more people, I will get less. So, 1 +9 < 1 +5. +Beni: +You got it! +Oh, so +1 +100 is bigger than 1 +200! +When one unit is divided into several equal parts, each part is +called a fractional unit. These are all fractional units: +1 +2 , 1 +3 , 1 +4 , 1 +5 , 1 +6 , …, 1 +10 , …, 1 +50 , …, 1 +100, etc. +We also sometimes refer to fractional units as ‘unit fractions.’ + Figure it Out + +Fill in the blanks with fractions. +1. Three guavas together weigh 1 kg. If they are roughly of the same +size, each guava will roughly weigh ____ kg. +2. A wholesale merchant packed 1 kg of rice in four packets +of equal weight. The weight of each packet is ___ kg. +3. Four friends ordered 3 glasses of sugarcane juice and +shared it equally among themselves. Each one drank ____ +glass of sugarcane juice. +1 +2 > 1 +4 +Math +Talk +Reprint 2025-26 + +Fractions +153 +4. The big fish weighs 1 +2 kg. The small one weighs 1 +4 kg. +Together they weigh ____ kg. +Knowledge from the past! +Fractions have been used and named in India since ancient times. +In the Rig Veda, the fraction 3 +4 is referred to as tri-pada. This has the +same meaning as the words for 3 +4 in many Indian languages today, +e.g., ʻteen paavʼ in colloquial Hindi and ‘mukkaal’ in Tamil. Indeed, +words for fractions used today in many Indian languages go back to +ancient times. +Find out and discuss the words for fractions that are used in the different +languages spoken in your home, city, or state. Ask your grandparents, +parents, teachers, and classmates what words they use for different +fractions, such as for one and a half, three quarters, one and a quarter, half, +quarter, and two and a half, and write them here: +___________ ___________ ___________ ___________ ___________ ___________ +5. Arrange these fraction words in order of size from the smallest to +the biggest in the empty box below: + +One and a half, three quarters, one and a quarter, half, quarter, +two and a half. +Write your answer here. +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +154 +7.2 Fractional Units as Parts of a Whole +The picture shows a whole chikki. + By dividing the whole chikki into 6 equal parts in different +ways, we get 1 +6 chikki pieces of different shapes. Are they of +the same size? +A whole chikki cut +into 6 equal pieces. +A whole chikki cut into 6 equal +pieces in a different way. +Math +Talk +A whole chikki +A picture of the chikki broken into 2 +pieces is shown below. How much of +the original chikki is each piece? +We can see that the bigger piece has 3 pieces of 1 +4 chikki in it. So, we +can measure the bigger piece using the fractional unit 1 +4. We see that +the bigger piece is 3 +4 chikki. +1 +4 +1 +6 +1 +6 +Reprint 2025-26 + +Fractions +155 +What is the fractional unit of chikki shown below? +We get this piece +by breaking the +chikki into 3 +equal pieces. So +this is 1 +3 chikki. + Figure it Out +The figures below show different fractional units of a whole chikki. +How much of a whole chikki is each piece? +a. +e. +f. +g. +h. +b. +c. +d. +A whole chikki +1 +3 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +156 +7.3 Measuring Using Fractional Units +Take a strip of paper. We consider this paper strip to be one unit long. +1 Strip Paper +Fold the strip into two equal parts and then open up the strip again. +Taking the strip to be one unit in length, what are the lengths of the +two new parts of the strip created by the crease? + 1 +2 +1 +2 +What will you get if you fold the previously-folded strip again +into two equal parts? You will now get four equal parts. +2 times 1 +4 = 2 +4 +3 times 1 +4 = 3 +4 +4 times 1 +4 = 4 +4 +1 +4 +Do it once more! Fill in the blank boxes. +Reprint 2025-26 + +Fractions +157 +2 times 1 +8 +4 times 1 +8 +6 times 1 +8 + += += += += += +8 times 1 +8 + 8 +8 += += +1 +Fractional quantities can be measured using fractional units. +Let us look at another example, + + + + +1 +2 += 1 times half +1 +2 + 1 +2 += 2 times half +1 +2 + 1 +2 + 1 +2 += 3 times half +1 +2 + 1 +2 + 1 +2 + 1 +2 += 4 times half +1 +2 + 1 +2 + 1 +2 + 1 +2 + 1 +2 += 5 times half +We can describe how much the quantity is by collecting together the +fractional units. +Represents a +full roti (whole) +Reprint 2025-26 + +Ganita Prakash | Grade 6 +158 + Figure it Out + +1. Continue this table of 1 +2 for 2 more steps. + +2. Can you create a similar table for 1 +4? + +3. Make 1 +3 using a paper strip. Can you use this to also make 1 +6 ? + +4. Draw a picture and write an addition statement as above to show: + + +a. 5 times 1 +4 of a roti +b. 9 times 1 +4 of a roti + +5. Match each fractional unit with the correct picture: +1 +3 +1 +5 +1 +8 +1 +6 +Reading Fractions +We usually read the fraction 3 +4 as ‘three quarters’ or ‘three upon four’, +but reading it as ‘3 times 1 +4’ helps us to understand the size of the +fraction because it clearly shows what the fractional unit is (1 +4) and +how many such fractional units (3) there are. +Recall what we call the top number and the bottom number of fractions. +In the fraction 5 +6 , 5 is the numerator and 6 is the denominator. +Teacher’s Note +Give several opportunities to the children to explore the idea of +fractional units with different shapes like circles, squares, rectangles, +triangles, etc. +Reprint 2025-26 + +Fractions +159 +7.4 Marking Fraction Lengths on the Number Line +We have marked lengths equal to 1, 2, 3, … units on the number line. +Now, let us try to mark lengths equal to fractions on the number line. +What is the length of the blue line? Write the fraction that gives +the length of the blue line in the box. +0 +1 +2 +The distance between 0 and 1 is one unit long. It is divided into +two equal parts. So, the length of each part is 1 +2 unit. So, this blue line +is 1 +2 unit long. + Now, can you find the lengths of the various blue lines shown +below? Fill in the boxes as well. +1. Here, the fractional unit is dividing a length of 1 unit into three +equal parts. Write the fraction that gives the length of the blue +line in the box or in your notebook. +0 +1 +2 +1 +3 +2. Here, a unit is divided into 5 equal parts. Write the fraction that +gives the length of the blue lines in the respective boxes or in +your notebook. +0 +1 +2 +1 +5 +3 +5 +3. Now, a unit is divided into 8 equal parts. Write the appropriate +fractions in your notebook. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +160 + Figure it Out +1. On a number line, draw lines of lengths 1 +10 , 3 +10 , and 4 +5 . +2. Write five more fractions of your choice and mark them on the +number line. +3. How many fractions lie between 0 and 1? Think, discuss with +your classmates, and write your answer. +4. What is the length of the blue line and black line shown below? The +distance between 0 and 1 is 1 unit long, and it is divided into two +equal parts. The length of each part is 1 +2. So the blue line is 1 +2 units +long. Write the fraction that gives the length of the black line in the +box. +1 +2 +0 +1 +2 +5. Write the fraction that gives the lengths of the black lines in the +respective boxes. +0 +1 +2 +1 +5 +2 +5 +3 +5 +4 +5 +Draw these lines on the board and ask the students to write the +answers in their notebooks. +Teacher’s Note +Math +Talk +Reprint 2025-26 + +Fractions +161 +7.5 Mixed Fractions +Fractions greater than one +You marked some fractions on the number line earlier. Did you +notice that the lengths of all the blue lines were less than one and +the lengths of all the black lines were more than 1? + Write down all the fractions you marked on the number line +earlier. + +Now, let us classify these in two groups: +Lengths less than 1 unit +Lengths more than 1 unit +  Did you notice something common between the fractions that +are greater than 1? +In all the fractions that are less than 1 unit, the numerator is +smaller than the denominator, while in the fractions that are more +than 1 unit, the numerator is larger than the denominator. +We know that 3 +2 , 5 +2 and 7 +2 are all greater than 1 unit. But can we +see how many whole units they contain? +3 +2 = 1 +2 + 1 +2 + 1 +2 = 1 + 1 +2 +5 +2 = 1 +2 + 1 +2 + 1 +2 + 1 +2 + 1 +2 = 2 + 1 +2 +I know that 1 +3 + 1 +3 + 1 +3 = 3 +3 = 1. If I add one more 1 +3, +I will get more than 1 unit! So, 4 +3 > 1. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +162 + Figure it Out +1. How many whole units are there in 7 +2 ? +2. How many whole units are there in 4 +3 and in 7 +3 ? +Writing fractions greater than one as mixed numbers + + We saw that: 3 +2 = 1 + 1 +2 . + + We can write other fractions in a similar way. For example, + + + +4 +3 = 1 +3 + 1 +3 + 1 +3 + 1 +3 = 1 + 1 +3 . + + + 3 × 1 +3 = 1 + Figure it Out +1. Figure out the number of whole units in each of the following +fractions: +a. 8 +3 + +b. 11 +5 + +c. 9 +4 +We saw that + +8 +3 = 2 + 2 +3 +Fraction Mixed number +2. Can all fractions greater than 1 be written as such mixed numbers? +A mixed number or mixed fraction contains a whole number +(called the whole part) and a fraction that is less than 1 (called the +fractional part). +3. Write the following fractions as mixed fractions (e.g., 9 +2 = 4 1 +2 ): +a. 9 +2 +b. 9 +5 +c. 21 +19 d. 47 +9 +e. 12 +11 +f. 19 +6 +Math +Talk +This number is thus also called ‘two and +two thirds’. We also write it as 2 2 +3 . +Reprint 2025-26 + +Fractions +163 +Can we write a mixed +number (mixed fraction) +as a regular fraction? +Yes! I figured out a way to +write a mixed number as a +regular fraction! +Jaya: When I have 3 + 3 +4 , this means 1 + 1 + 1 + 3 +4 . I know +1 = 1 +4 + 1 +4 + 1 +4 + 1 +4 . +So I get +( 1 +4 + 1 +4 + 1 +4 + 1 +4 ) + ( 1 +4 + 1 +4 + 1 +4 + 1 +4 ) + ( 1 +4 + 1 +4 + 1 +4 + 1 +4 ) + ( 1 +4 + 1 +4 + 1 +4 ) = 15 +4 . +Therefore, (4 × 1 +4 ) + (4 × 1 +4 ) + (4 × 1 +4 ) + (3 × 1 +4 ) = 15 +4 . + Figure it Out + +Write the following mixed numbers as fractions: + +a. 3 1 +4 +b. 7 2 +3 +c. +9 4 +9 + +d. 3 1 +6 +e. 2 3 +11 +f. 3 9 +10 +7.6 Equivalent Fractions +Using a fraction wall to find equal fractional lengths! +In the previous section, you used paper folding to represent various +fractions using fractional units. Let us do some more activities with +the same paper strips. +Math +Talk +Reprint 2025-26 + +Ganita Prakash | Grade 6 +164 +What do you +observe? +• Are the lengths 1 +2 and 2 +4 equal? +• Are the lengths 2 +4 and 4 +8 equal? + We can say that 1 +2 = 2 +4 = 4 +8 . +1 +2 +2 +4 +4 +8 +1 +These are ‘equivalent fractions’ that denote the same length, but +they are expressed in terms of different fractional units. +Now, check whether 1 +3 and 2 +6 are equivalent fractions or not, +using paper strips. +Make your own fraction wall using such strips as given in the +picture below! +  Answer the following questions after looking at the fraction wall: +1. +Are the lengths 1 +2 and 3 +6 equal? +2. +Are 2 +3 and 4 +6 equivalent +fractions? Why? +3. +How many pieces of +length 1 +6 will make a +length of 1 +2 ? +4. +How many pieces of length 1 +6 will make a length of 1 +3 ? +1 UNIT +1 +6 +2 +6 +3 +6 +4 +6 +5 +6 +6 +6 +1 +5 +2 +5 +3 +5 +4 +5 +5 +5 +1 +4 +2 +4 +3 +4 +4 +4 +1 +3 +2 +3 +3 +3 +1 +2 +2 +2 +Reprint 2025-26 + +Fractions +165 +We can extend this idea to make a fraction wall up to the fractional +unit 1 +10. (This fraction wall is given at the end of the book.) +1 UNIT +1 +2 +1 +3 +1 +4 +1 +5 +1 +6 +1 +7 +1 +8 +1 +9 +1 +10 +2 +10 +3 +10 +4 +10 +5 +10 +6 +10 +7 +10 +8 +10 +9 +10 +10 +10 +2 +9 +3 +9 +4 +9 +5 +9 +6 +9 +7 +9 +8 +9 +9 +9 +2 +8 +3 +8 +4 +8 +5 +8 +6 +8 +7 +8 +8 +8 +2 +7 +3 +7 +4 +7 +5 +7 +6 +7 +1 +7 +2 +6 +3 +6 +4 +6 +5 +6 +6 +6 +2 +5 +3 +5 +4 +5 +5 +5 +2 +4 +3 +4 +4 +4 +2 +3 +3 +3 +2 +2 + Figure it Out + +1. Are 3 +6 , 4 +8 , 5 +10 equivalent fractions? Why? + +2. Write two equivalent fractions for 2 +6 . + +3. 4 +6 = + = + = + = ............ (Write as many as you can) +Understanding equivalent fractions using equal shares +One roti was shared equally by four children. +What fraction of the whole did each child get? +The adjoining picture shows the division of a +roti among four children. +Fraction of roti each child got is 1 +4 . +The four shares must be +equal to each other! +Reprint 2025-26 + +Ganita Prakash | Grade 6 +166 +You can also express this event through division facts, addition +facts, and multiplication facts. +The division fact is 1 ÷ 4 = 1 +4 . +The addition fact is 1 = 1 +4 + 1 +4 + 1 +4 + 1 +4 . +The multiplication fact is 1 = 4 × 1 +4 . + Figure it Out +1. Three rotis are shared equally by four children. Show the division +in the picture and write a fraction for how much each child gets. +Also, write the corresponding division facts, addition facts, and, +multiplication facts. +Fraction of roti each child gets is ______. +Division fact: +Addition fact: +Multiplication fact: +Compare your picture and answers with your classmates! +2. Draw a picture to show how much each child gets when 2 rotis +are shared equally by 4 children. Also, write the corresponding +division facts, addition facts, and multiplication facts. +3. Anil was in a group where 2 cakes were divided equally among +5  children. How much cake would Anil get? +What if we put two such groups +together? one group where +2 cakes are divided equally +between 5 children, and another +group again with 4 cakes and +10 children. +Now, if there are 10 children +in my group, how many +cakes will I need so that they +get same amount of cake as +Anil? +Reprint 2025-26 + +Fractions +167 +Let us examine the shares of each child in the following situations. +• 1 roti is divided equally between 2 children. +• 2 rotis are divided equally among 4 children. +• 3 rotis are divided equally among 6 children. +Let us draw and share! + +Did you notice that in each situation the share of every child is +the same? So, we can say that 1 +2 = 2 +4 = 3 +6 . +So, the share of each child is the +same in both these situations! +So, 2 +5 = 4 +10! +Group 1 +Group 2 +Fractions where the shares are equal are called ‘equivalent fractions’. +1 roti is divided +equally between 2 +2 rotis are divided +equally among 4 +3 rotis are divided +equally among 6 +1 +2 +1 +2 +2 +4 +2 +4 +2 +4 +2 +4 +3 +6 +3 +6 +3 +6 +3 +6 +3 +6 +3 +6 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +168 +So, 1 +2 , 2 +4 , and 3 +6 are all equivalent fractions. +Find some more fractions equivalent to 1 +2 . Write them in the +boxes here: + + + + + +Equally divide the rotis in the situations shown below and write +down the share of each child. Are the shares in each of these cases +the same? Why? + Figure it Out + +Find the missing numbers: +a. 5 glasses of juice shared equally among 4 friends is +the same as ____ glasses of juice shared equally among +8 friends. + +So, 5 +4 = 8  . +b. 4 kg of potatoes divided equally in 3 bags is the same as +12 kgs of potatoes divided equally in ___ bags. + +So, 4 +3 = 12 +Do you notice anything about the relationship between the +numerator and denominator in each of these fractions? +2 rotis divided +equally among +3 children +4 rotis divided +equally among +6 children +6 rotis divided +equally among +9 children +2 +3 +2 +3 +2 +3 +2 +3 is also called the simplest form of 4 +6 . It is also the simplest form of 6 +9 as well. +Math +Talk +Reprint 2025-26 + +Fractions +169 +c. + 7 rotis divided among 5 children is the same as ____ rotis divided +among _____ children. + +So, 7 +5 = +. + In which group will each child get more chikki? +1 chikki divided between 2 children or 5 chikkis divided among +8 children. +Mukta: So, we must compare 1 +2 and 5 +8 . Which is more? +Shabnam: Well, we have seen that 1 +2 = 4 +8 ; and clearly 4 +8 < 5 +8 . So, the +children for whom 5 chikkis is divided equally among +8 will get more than those children for whom 1 chikki is +divided equally among 2. The children of the second +group will get more chikki each. + What about the following groups? In which group will each child +get more? +1 chikki divided between 2 children or 4 chikkis divided among +7 children. +Shabnam: The children of which group will get more chikki this +time? +Mukta: We must compare 1 +7 and 4 +7 . +Now +1 × 4 +2 × 4 = 4 +8 so, 1 +2 = 4 +8 . +Shabnam: But why did you multiply the numerator and denominator +by 4 again? +Mukta: You will see! +When 4 chikkis are divided equally among 7 children, each +one will get 4 +7 chikki. When 4 chikkis are divided equally +among 8 children, each one will get 4 +8 chikki. So 4 +7 > 4 +8. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +170 + Suppose the number of children is kept the same, but the number +of units that are being shared is increased? What can you say about +each child’s share now? Why? Discuss how your reasoning explains +1 +5 < 2 +5 , 3 +7 < 4 +7 , and 1 +2 < 5 +8 . + Now, decide in which of the two groups will each child get a +larger share: +1. Group 1 : 3 glasses of sugarcane juice divided equally +among 4 children. + Group 2: 7 glasses of sugarcane juice divided equally +among 10 children. +2. Group 1 : 4 glasses of sugarcane juice divided equally +among 7 children. + Group 2: 5 glasses of sugarcane juice divided equally +among 7 children. +Which groups were easier to compare? Why? +Shabnam: To compare the first two groups, +we have to find fractions +equivalent to the fractions +3 +4 and 7 +10. +Mukta: How about 6 +8 = 3 +4 and 21 +30 = 7 +10? +Math +Talk + If the number of units that are shared +is the same, but the number of +children among whom the units are +shared is more, then the share is less. +Therefore, 4 +7 > 4 +8 and 4 +8 = 1 +2 , so 4 +7 > 1 +2. +Now I understood why you multiplied +the numerator and denominator by 4. +When the number of +children is same, it is easier +to compare, isn’t it? +Reprint 2025-26 + +Fractions +171 +Shabnam: There is a condition. The fractional unit used for the two +fractions have to be the same! Like 2 +6 and 3 +6 both use the +same fractional unit 1 +6 (i.e., the denominators are the same). +But 6 +8 and 21 +30 do not use the same fractional units (they have +different denominators). +Mukta: Okay, so let us start making equivalent fractions then: + + + 3 +4 = 6 +8 = 9 +12 = 12 +16 = 15 +20 … But when do I stop? +Shabnam: Got it! How about we go on till 4 × 10 = 40. +Mukta: You mean the product of the two denominators? +Sounds good! + +We have 3 +4 and 7 +10. The product of the two denominators +(4 and 10) is 40. + + +3 +4 = 6 +8 = 9 +12 = 12 +16 = 15 +20 = 18 +24 = … = 27 +36 = 30 +40 . + + +7 +10 = 14 +20 = 21 +30 = 28 +40. +But notice that +15 +20 and +14 +20 also had +the same denominator! +Yes! We just needed to get the +same fractional units for each +fraction. +Shabnam: So, fractions equivalent to 3 +4 and 7 +10 with the same fractional +unit (same denominators) are 30 +40 and 28 +40, or 15 +20 and 14 +20. + + +  Since clearly 30 +40 > 28 +40, we conclude that 3 +4 > 7 +10. +Go till we reach the +denominator 40. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +172 + Find equivalent fractions for the given pairs of fractions such that the +fractional units are the same. +a. 7 +2 and 3 +5 +b. 8 +3 and 5 +6 +c. 3 +4 and 3 +5 +d. 6 +7 and 8 +5 +e. 9 +4 and 5 +2 +f. 1 +10 and 2 +9 +g. 8 +3 and 11 +4 +h. 13 +6 and 1 +9 +Expressing a fraction in lowest terms (or in its simplest +form) +In any fraction, if its numerator and denominator have no common +factor except 1, then the fraction is said to be in lowest terms or in +its simplest form. In other words, a fraction is said to be in lowest +terms if its numerator and denominator are as small as possible. + Any fraction can be expressed in lowest terms by finding an +equivalent fraction whose numerator and denominator are as small +as possible. +Let’s see how to express fractions in lowest terms. +Example: Is the fraction 16 +20 in lowest terms? No, 4 is a common factor +of 16 and 20. Let us reduce 16 +20 to lowest terms. +We know that both 16 (numerator) and 20 (denominator) are +divisible by 4. +So, 16 ÷ 4 +20 ÷ 4 = 4 +5. +Now, there is no common factor between 4 and 5. Hence, 16 +20 +expressed in lowest terms is 4 +5. So, 4 +5 is called the simplest form of 16 +20, +since 4 and 5 have no common factor other than 1. +Any fraction can be converted to +lowest terms by dividing both the +numerator and denominator by the +highest common factor between +them. +Reprint 2025-26 + +Fractions +173 +Expressing a fraction in lowest terms can also be done in steps. +Suppose we want to express 36 +60 in lowest terms. First, we notice +that both the numerator and denominator are even. So, we divide +both by 2, and see that 36 +60 = 18 +30. +Both the numerator and denominator are even again, so we can +divide them each by 2 again; we get 18 +30 = 9 +15. +We now notice that 9 and 15 are both multiples of 3, so we divide +both by 3 to get 9 +15 = 3 +5 . +Now, 3 and 5 have no common factor other than 1, so, 36 +60 in lowest +terms is 3 +5. +Alternatively, we could have noticed that in 36 +60 , both the numerator +and denominator are multiples of 12 : we see that 36 = 3 × 12 and +60 = 5 × 12. Therefore, we could have concluded that 36 +60 = 3 +5 straight away. +Either method works and will give the same answer! But +sometimes it can be easier to go in steps. + Figure it Out + +Express the following fractions in lowest terms: + +a. 17 +51 + +b. 64 +144 +e. 126 +147 +d. 525 +112 +7.7 Comparing Fractions +Which is greater, 4 +5 or 7 +9? It can be difficult to compare two such +fractions directly. However, we know how to find fractions equivalent +to two fractions with the same denominator. Let us see how we can +use it: + 4 +5 = 4×9 +5×9 = 36 +45 + 7 +9 = 7×5 +9×5 = 35 +45. +45 is a common multiple +of 5 and 9, so we can +use 45 as a common +denominator. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +174 +Clearly, 36 +45 > 35 +45 +So, 4 +5 > 7 +9 ! +Let us try this for another pair: 7 +9 and 17 +21. +63 is a common multiple of 9 and 21. We can then write: + 7 +9 = 7×7 +9×7 = 49 +63 , 17 +21 = 17×3 +21×3 = 51 +63. +Clearly, 49 +63 < 51 +63 . So, 7 +9 < 17 +21! +Let’s Summarise! +Steps to compare the sizes of two or more given fractions: +Step 1: Change the given fractions to equivalent fractions so that +they all are expressed with the same denominator or same +fractional unit. +Step 2: Now, compare the equivalent fractions by simply comparing +the numerators, i.e., the number of fractional units each has. + Figure it Out + 1. Compare the following fractions and justify your answers: +a. 8 +3 , 5 +2 +b. 4 +9 , 3 +7 +c. 7 +10 , 9 +14 +d. 12 +5 , 8 +5 +e. 9 +4 , 5 +2 + 2. Write the following fractions in ascending order. + + +a. 7 +10 , 11 +15 , 2 +5 + +b. 19 +24 , 5 +6 , 7 +12 + 3. Write the following fractions in descending order. + + a. 25 +16, 7 +8, 13 +4 , 17 +32 + +b. 3 +4 , 12 +5 , 7 +12, 5 +4 +Reprint 2025-26 + +Fractions +175 +7.8 Addition and Subtraction of Fractions +Meena’s father made some chikki. Meena ate 1 +2 +of it and her younger brother ate 1 +4 of it. How +much of the total chikki did Meena and her +brother eat together? +We can arrive at the answer by visualising it. Let us take a piece +of chikki and divide it into two halves first like this. +Meena ate 1 +2 of it as +shown in the picture. +Let us now divide the +remaining half into two further halves as shown. Each of these +pieces is 1 +4 of the whole chikki. +Meena’s brother ate 1 +4 +of the whole chikki, as is +shown in the picture. +The total chikki eaten +is 1 +2 (by Meena) and 1 +4 (by her +brother) +The total chikki eaten + = 1 +2 + 1 +4 + +  = 1 +4 + 1 +4 + 1 +4 + +  = 3 × 1 +4 = 3 +4. +How much of the total chikki is remaining? +Meena ate +Meena ate +Brother ate +Total chikki eaten +Reprint 2025-26 + +Ganita Prakash | Grade 6 +176 +Adding fractions with the same fractional unit or +denominator +Example: Find the sum of 2 +5 and 1 +5. +Let us represent both using the rectangular strips. In both fractions, +the fractional unit is the same 1 +5 , so, each strip will be divided into +5 equal parts. +So 2 +5 will be represented as — +And 1 +5 will be represented as — +Adding the two given fractions is the same as finding out the total +number of shaded parts, each of which represent the same fractional +unit 1 +5. +In this case, the total number of shaded parts is 3. Since, each +shaded part represents the fractional unit 1 +5 , we see that the 3 shaded +parts together represent the fraction 3 +5. +Therefore, 2 +5 + 1 +5 = 3 +5? +Example: Find the sum of 4 +7 and 6 +7. +Let us represent both again using the rectangular strip model. Here in +both fractions, the fractional unit is the same, i.e., 1 +7 , so each strip will +be divided into 7 equal parts. +Then 4 +7 will be represented as — +Reprint 2025-26 + +Fractions +177 +and 6 +7 will be represented as — +In this case, the total number of +shaded parts is 10, and each shaded part +represents the fractional unit 1 +7, so, the +10 shaded parts together represent the +fraction 10 +7 as seen here. + + + + + + +Therefore, 4 +7 + 6 +7 = 10 +7 + = 1 + 3 +7 + = 1 3 +7. + + + + Try adding 4 +7 + 6 +7 using a number line. Do you get the same answer? +Adding fractions with different fractional units or +denominators +Example: Find the sum of 1 +4 and 1 +3. +To add fractions with different fractional units, first convert the +fractions into equivalent fractions with the same denominator or + While adding fractions +with the same fractional +unit, just add the number of +fractional units from each +fraction. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +178 +fractional unit. In this case, the common denominator can be made +3 × 4 = 12, i.e., we can find equivalent fractions with fractional unit 1 +12. +Let us write the equivalent fraction for each given fraction. +1 +4 = 1 × 3 +4 × 3 = 3 +12 , 1 +3 = 1 × 4 +3 × 4 = 4 +12 . +Now, 3 +12 and 4 +12 have the same fractional unit, i.e., 1 +12 . +Therefore, 1 +4 + 1 +3 = 3 +12 + 4 +12 = 7 +12 . +This method of addition, which works for adding any number of +fractions, was first explicitly described in general by Brahmagupta +in the year 628 CE! We will describe the history of the development +of fractions in more detail later in the chapter. For now, we simply +summarise the steps in Brahmagupta’s method for addition of +fractions. +Brahmagupta’s method for adding fractions +1. Find equivalent fractions so that the fractional unit is common +for all fractions. This can be done by finding a common multiple +of the denominators (e.g., the product of the denominators, or the +smallest common multiple of the denominators). +2. Add these equivalent fractions with the same fractional units. +This can be done by adding the numerators and keeping the same +denominator. +3. Express the result in lowest terms if needed. +Let us carry out another example of Brahmagupta’s method. +Example: Find the sum of 2 +3 and 1 +5. +The denominators of the given fractions are 3 and 5. The lowest +common multiple of 3 and 5 is 15. Then we see that +2 +3 = 2 × 5 +3 × 5 = 10 +15 , 1 +5 = 1 × 3 +5 × 3 = 3 +15 . +Reprint 2025-26 + +Fractions +179 +Therefore, 2 +3 + 1 +5 = 10 +15 + 3 +15 = 13 +15 . +Example: Find the sum of 1 +6 and 1 +3. +The smallest common multiple of 6 and 3 is 6. +1 +6 will remain 1 +6 . +1 +3 = 1 × 2 +3 × 2 = 2 +6 +Therefore, 1 +6 + 1 +3 = 1 +6 + 2 +6 = 3 +6. +The fraction +3 +6 can now be re-expressed in lowest terms, if +desired. This can be done by dividing both the numerator and +denominator by 3 (the biggest common factor of 3 and 6): +3 +6 = 3 ÷ 3 +6 ÷ 3 = 1 +2. +Therefore, 1 +6 + 1 +3 = 1 +2. + Figure it Out +1. Add the following fractions using Brahmagupta’s method: +a.  2 +7 + 5 +7 + 6 +7 +b.  3 +4 + 1 +3 +c.  2 +3 + 5 +6 +d.  2 +3 + 2 +7 +e.  3 +4 + 1 +3 + 1 +5 +f.  2 +3 + 4 +5 + +g.  4 +5 + 2 +3 +h.  3 +5 + 5 +8 + i.  9 +2 + 5 +4 + j.  8 +3 + 2 +7 +k.  3 +4 + 1 +3 + 1 +5 +l.  2 +3 + 4 +5 + 3 +7 m.  9 +2 + 5 +4 + 7 +6 +2. Rahim mixes 2 +3 litres of yellow paint with 3 +4 litres of blue paint to +make green paint. What is the volume of green paint he has made? +3. Geeta bought 2 +5 meter of lace and Shamim bought 3 +4 meter of the +same lace to put a complete border on a table cloth whose perimeter +is 1 meter long. Find the total length of the lace they both have +bought. Will the lace be sufficient to cover the whole border? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +180 +Subtraction of fractions with the same fractional unit or +denominator +Brahmagupta’s method also applies when subtracting fractions! +Let us start with the problem of subtracting 4 +7 from 6 +7, i.e., what is +6 +7 – 4 +7? +To solve this problem, we can again use the rectangular strips. +In both fractions, the fractional unit is the same, i.e., 1 +7. Let us first +represent the bigger fraction using a rectangular strip model as +shown: +6 +7 +Each shaded part represents 1 +7. Now, we need to subtract 4 +7. To do +this let us remove 4 of the shaded parts: +So, we are left with 2 shaded parts, i.e., 6 +7 – 4 +7 = 2 +7. +Try doing this same exercise using the number line. +We can do this here directly +because both fractions have +the same fractional units. +Fractional parts to +be removed. +Reprint 2025-26 + +Fractions +181 + Figure it Out + +1. +5 +8 – 3 +8 + +2. +7 +9 – 5 +9 + +3. + 10 +27 – 1 +27 +Subtraction of fractions with different fractional units or +denominators +Example: What is 3 +4 – 2 +3? +As we already know the procedure for subtraction of fractions with +the same fractional units, let us convert each of the given fractions +into equivalent fractions with the same fractional units. +3 +4 = (3×3) +(4×3) = 9 +12 +Yes! By doing this we can easily +subtract the two fractions. +Think! Why did we choose to +multiply both the numerator and +denominator by 3? +Again! Why did we choose to multiply +both the numerator and denominator +here by 4? +and similarly, +2 +3 = (2×4) +(3×4) = 8 +12 . +Therefore, 3 +4 – 2 +3 = 9 +12 – 8 +12 = 1 +12 . +Brahmagupta’s method for subtracting two fractions — +1. +Convert the given fractions into equivalent fractions with the +same fractional unit, i.e., the same denominator. +2. +Carry out the subtraction of fractions having the same fractional +units. This can be done by subtracting the numerators and +keeping the same denominator. +3. +Simplify the result into lowest terms if needed. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +182 +  Figure it Out +1. Carry out the following subtractions using Brahmagupta’s method: +a. 8 +15 – 3 +15 + +b. 2 +5 – 4 +15 + + c. 5 +6 – 4 +9 +d. 2 +3 – 1 +2 +2. Subtract as indicated: +a. 13 +4 from 10 +3 +b. 18 +5 from 23 +3 +c. 29 +7 from 45 +7 +3. Solve the following problems: + + +a. +Jaya’s school is 7 +10 km from her home. She takes an auto for +1 +2  km from her home daily, and then walks the remaining +distance to reach her school. How much does she walk daily +to reach the school? + + +b. Jeevika takes 10 +3 minutes to take a complete round of the +park and her friend Namit takes 13 +4 minutes to do the same. +Who takes less time and by how much? +7.9 A Pinch of History +Do you know what a fraction was called in ancient India? It was +called bhinna in Sanskrit, which means ‘broken’. It was also called +bhaga or ansha meaning ‘part’ or ‘piece’. +The way we write fractions today, globally, originated in India. In +ancient Indian mathematical texts, such as the Bakshali manuscript +(from around the year 300 CE), when they wanted to write 1 +2, they +wrote it as 1 +2 which is indeed very similar to the way we write it +today! This method of writing and working with fractions continued +to be used in India for the next several centuries, including by +Aryabhata (499 CE), Brahmagupta (628 CE), Sridharacharya +(c. 750 CE), and Mahaviracharya (c. 850 CE), among others. The line +segment between the numerator and denominator in ‘1 +2’ and in other +Reprint 2025-26 + +Fractions +183 +fractions was later introduced by the Moroccan mathematician +Al-Hassar (in the 12th century). Over the next few centuries the +notation then spread to Europe and around the world. +Fractions had also been used in other cultures such as the ancient +Egyptian and Babylonian civilisations, but they primarily used only +fractional units, that is, fractions with a 1 in the numerator. More +general fractions were expressed as sums of fractional units, now +called ‘Egyptian fractions’. Writing numbers as the sum of fractional +units, e.g., 19 +24 = 1 +2 + 1 +6 + 1 +8, can be quite an art and leads to beautiful +puzzles. We will consider one such puzzle below. +General fractions (where the numerator is not necessarily 1) +were first introduced in India, along with their rules of arithmetic +operations like addition, subtraction, multiplication, and even +division of fractions. The ancient Indian treatises called the ‘Sulba- +sutras’ shows that even during Vedic times, Indians had discovered +the rules for operations with fractions. General rules and procedures +for working with and computing with fractions were first codified +formally and in a modern form by Brahmagupta. +Brahmagupta’s methods for working with and computing with +fractions are still what we use today. For example, Brahmagupta +described how to add and subtract fractions as follows: +“By the multiplication of the numerator and the denominator +of each of the fractions by the other denominators, the fractions +are reduced to a common denominator. Then, in case of addition, +the numerators (obtained after the above reduction) are added. +In case of subtraction, their difference is taken.’’ (Brahmagupta, +Brahmasphuṭasiddhānta, Verse 12.2, 628 CE) +The Indian concepts and methods involving fractions were +transmitted to Europe via the Arabs over the next few centuries and +they came into general use in Europe in around the 17th century +and then spread worldwide. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +184 +  Puzzle! + +It is easy to add up fractional units to obtain the sum 1, if one +uses the same fractional unit, for example, + +1 +2 + 1 +2 = 1, 1 +3 + 1 +3 + 1 +3 = 1, 1 +4 + 1 +4 + 1 +4 + 1 +4 = 1, etc. + + However, can you think of a way to add fractional units that +are all different to get 1? + + It is not possible to add two different fractional units to get 1. +The reason is that 1 +2 is the largest fractional unit, and 1 +2 + 1 +2 = 1. + + To get different fractional units, we would have to replace at +least one of the 1 +2’s with some smaller fractional unit - but then +the sum would be less than 1! Therefore, it is not possible for +two different fractional units to add up to 1. + + We can try to look instead for a way to write 1 as the sum of +three different fractional units. + +1.  Can you find three different fractional units that add +up to 1? + +  It turns out there is only one solution to this problem +(up to changing the order of the 3 fractions)! Can you +find it? Try to find it before reading further. +Here is a systematic way to find the solution. We know that  +1 +3 + 1 +3 + 1 +3 = 1. To get the fractional units to be different, we will have +to increase at least one of the 1 +3’s, and decrease at least one of the +other 1 +3’s to compensate for that increase. The only way to increase +1 +3 to another fractional unit is to replace it by 1 +2. So 1 +2 must be one of +the fractional units. +Now 1 +2 + 1 +4 + 1 +4 = 1. To get the fractional units to be different, we +will have to increase one of the 1 +4’s and decrease the other 1 +4 to +compensate for that increase. Now the only way to increase 1 +4 to +Try +This +Reprint 2025-26 + +Fractions +185 +another fractional unit, that is different from 1 +2, is to replace it by 1 +3. +So two of the fractions must be 1 +2 and 1 +3! What must be third fraction +then, so that the three fractions add up to 1? +This explains why there is only one solution to the above problem. +1 +2 + 1 +3 + 1 +6 = 1 +What if we look for four different fractional units that add up to 1? + +2.  Can you find four different fractional units that add +up to 1? + +  It turns out that this problem has six solutions! Can +you find at least one of them? Can you find them all? +You can try using similar reasoning as in the cases +of two and three fractional units — or find your own +method! +Once you find one solution, try to divide a circle into parts +like in the figure above to visualise it! +Try +This +Reprint 2025-26 + +Ganita Prakash | Grade 6 +186 + + +Fraction as equal share: When a whole number of units is divided +into equal parts and shared equally, a fraction results. + + +Fractional Units: When one whole basic unit is divided into equal +parts, then each part is called a fractional unit. + + +Reading Fractions: In a fraction such as 5 +6, 5 is called the numerator +and 6 is called the denominator. + + +Mixed fractions contain a whole number part and a fractional part. + + +Number line: Fractions can be shown on a number line. Every fraction +has a point associated with it on the number line. + + +Equivalent Fractions: When two or more fractions represent the +same share or number, they are called equivalent fractions. + + +Lowest terms: A fraction whose numerator and denominator have +no common factor other than 1 is said to be in lowest terms or in its +simplest form. + + +Brahmagupta’s method for adding fractions: When adding fractions, +convert them into equivalent fractions with the same fractional unit +(i.e., the same denominator), and then add the number of fractional +units in each fraction to obtain the sum. This is accomplished by +adding the numerators while keeping the same denominator. + + +Brahmagupta’s method for subtracting fractions: When subtracting +fractions, convert them into equivalent fractions with the same +fractional unit (i.e., the same denominator), and then subtract the +number of fractional units. This is accomplished by subtracting the +numerators while keeping the same denominator. +Reprint 2025-26 + +CHAPTER 7 — SOLUTIONS +Fractions + Section 7.1 +Page no. 152 +Figure it out +Q1. Three guavas together weigh 1 kg. If they are roughly of the same size, each guava +will roughly weigh ____kg. +Ans. +1 +3 +Q2. A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight +of each packet is_______ kg. +Ans. +1 +4 +Q3. Four friends ordered 3 glasses of sugarcane juice and shared it equally among +themselves. Each one drank ____ glass of sugarcane juice +Ans. +3 +4 +Q4. The big fish weighs +𝟏 +𝟐 kg. The small one weighs +𝟏 +𝟒 kg. Together they weigh ____ kg. +Ans. +3 +4 +Q5. Arrange these fraction words in order of size from the smallest to the biggest in the +empty box below: +One and a half, three quarters, one and a quarter, half, quarter, two and a half. +Ans. +1 +4, +1 +2, +3 +4, 1 +1 +4, 1 +1 +2, 2 +1 +2 + + Section 7.2 +Page No. 154 +Q. By dividing the whole chikki into 6 equal parts in different ways, we get 1/6 chikki +pieces of different shapes. Are they of the same size? +Ans. +Yes, they are of the same size. + + + + +Page No. 155 +Figure it out +Q. The figures below show different fractional units of a whole chikki. How much of a +whole chikki is each piece? +Ans. a. +1 +12 b. +1 +4 +c. +1 +8 d. +1 +6 +e. +1 +8 +f. +1 +6 +g. +1 +24 h. +1 +24 + Section 7.3 +Figure it out Page No. 158 +Q1. continue this table of +𝟏 +𝟐 for 2 more steps. +Ans. +1 +2 + +1 +2 + +1 +2 + +1 +2 + +1 +2 + +1 +2 = 6 times +1 +2 +1 +2 + +1 +2 + +1 +2 + +1 +2 + +1 +2 + +1 +2 + +1 +2 = 7 times +1 +2 +Q2. Can you create a similar table for +𝟏 +𝟒 ? +Ans. + + + + + + + + + +1 +4 = +1-time quarter + +1 +4 + +1 +4 = +2 times quarter + +1 +4 + +1 +4 + +1 +4 = +3 times quarter + +1 +4 + +1 +4 + +1 +4 + +1 +4= +4 times quarter + + +Try Further! + +Q4. Draw a picture and write an addition statement as above to show: + a. 5 times +𝟏 +𝟒 of a roti + b. 9 times +𝟏 +𝟒 of a roti +Ans. +a. + + +1 +4 + +1 +4 + +1 +4 + +1 +4 + +1 +4 = +5 +4 = 5 times quarter + +b. + + + + +1 +4 + +1 +4 + +1 +4 + +1 +4 + +1 +4 + +1 +4 + +1 +4 + +1 +4 + +1 +4 = +9 +4 = 9 times quarter = 2 + +1 +4 + +Q5. Match each fractional unit with the correct picture: +𝟏 +𝟑 +𝟏 +𝟓 +𝟏 +𝟖 +𝟏 +𝟔 +Ans. +1 +3 + + + +1 +5 + + + +1 +6 + + + +1 +8 + + Section 7.4 +Page no. 159 +Q1. Here, the fractional unit is dividing a length of 1 unit into three equal parts. Write +the fraction that gives the length of the blue line in the box or in your notebook. +Ans. +2 +3 +Q2. Here, a unit is divided into 5 equal parts. Write the fraction that gives the length of +the blue lines in the respective boxed or in your notebook. +Ans. +2 +5, +4 +5 + + +Q3. Now, a unit is divided into 8 equal parts. Write the appropriate fractions in your +notebook. +Ans. +1 +8, +2 +8, +3 +8 ,… + +Page no. 160 +Figure it out +Q1. On a number line, draw lines of lengths +𝟏 +𝟏𝟎, +𝟑 +𝟏𝟎, and +𝟒 +𝟓. +Ans. + + + +Q3. How many fractions lie between 0 and 1? Think, discuss with your classmates, and +write your answer. +Ans. +Uncountable number of fractions +Q4. What is the length of the blue line and black line shown below? The distance between +0 and 1 is 1 unit long, and it is divided into two equal parts. The length of each part +is +𝟏 +𝟐. So the blue line is +𝟏 +𝟐 units long. Write the fraction that gives the length of the +black line in the box. +Ans. +3 +2 +Q5. Write the fraction that gives the lengths of the black lines in the respective boxes. +Ans. +6 +5, +7 +5, +8 +5, +9 +5 + Section 7.5 +Page No. 162 +Figure it out +Q1. How many whole units are there in +𝟕 +𝟐? +Ans. +There are 3 whole units in +7 +2 +Q2. How many whole units are there in +𝟒 +𝟑 and in +𝟕 +𝟑? +Ans. +There is 1 whole unit in +4 +3 and 2 whole units in +7 +3. + + 0 + +𝟏 +𝟏𝟎 + +𝟑 +𝟏𝟎 +4 +5 +1 + +Figure it out +Q1. Figure out the number of whole units in each of the following fractions: +a. +𝟖 +𝟑 + +b. +𝟏𝟏 +𝟓 + c. +𝟗 +𝟒 +Ans. +a. 2 + +b. 2 + +c. 2 + +Q2. Can all fractions greater than 1 be written as such mixed numbers? +Ans. +No. For example: +8 +4 = 2 cannot be written as a mixed number. +Q3. Write the following fractions as mixed fractions (e.g., +𝟗 +𝟐 = 𝟒 +𝟏 +𝟐): +a. +𝟗 +𝟐 +b. +𝟗 +𝟓 +c. +𝟐𝟏 +𝟏𝟗 d. +𝟒𝟕 +𝟗 e. +𝟏𝟐 +𝟏𝟏 f. +𝟏𝟗 +𝟔 +Ans. a. +9 +2 = 4 +1 +2 + +b. +9 +5 = 1 +4 +5 + +c. +21 +19 = 1 +2 +19 + +d. +47 +9 = 5 +2 +9 + +e. +12 +11 = 1 +1 +11 + +f. +19 +6 = 3 +1 +6 + +Page No. 163 +Figure it out +Q1. Write the following mixed numbers as fractions: +a. 3 +𝟏 +𝟒 +b. 7 +𝟐 +𝟑 +c. 9 +𝟒 +𝟗 +d. 3 +𝟏 +𝟔 +e. 2 +𝟑 +𝟏𝟏 +f. 3 +𝟗 +𝟏𝟎 +Ans. a. 3 +1 +4 += 3 + +1 +4 += 1 + 1 + 1 + +1 +4 += ( +1 +4 + +1 +4 + +1 +4 + +1 +4) + ( +1 +4 + +1 +4 + +1 +4 + +1 +4) + ( +1 +4 + +1 +4 + +1 +4 + +1 +4) + +1 +4 + + += (4 × +1 +4) + (4 × +1 +4) + (4 × +1 +4) + +1 +4 + += +13 +4 +b. +23 +3 +c. +85 +9 +d. +19 +6 +e. +25 +11 +f. +39 +10 + + Section 7.6 +Page no. 164 +Q1. Are the lengths +𝟏 +𝟐 and +𝟑 +𝟔 equal? +Ans. +Yes. +Q2. Are +𝟐 +𝟑 and +𝟒 +𝟔 equivalent fractions? Why? +Ans. Yes, since they are of equal length which can be seen in fractional wall also. +Q3. How many pieces of length +𝟏 +𝟔 will make a length of +𝟏 +𝟐? +Ans. 3 pieces. +Q4. How many pieces of length +𝟏 +𝟔 will make a length of +𝟏 +𝟑? +Ans. 2 pieces. +Page no. 165 +Figure it out +Q1. Are +𝟑 +𝟔, +𝟒 +𝟖, and +𝟓 +𝟏𝟎 are equivalent fractions? Why? +Ans. Yes. In the fraction wall their lengths can be seen to be equal. +Q2. Write two equivalent fractions for +𝟐 +𝟔. +Ans. Two equivalent fractions for +2 +6 are +1 +3 and +3 +9. (Try for other equivalent fractions also) + +Q3 . +𝟒 +𝟔 = = = = = ……. (Write as many as you can) +Ans. +4 +6 = +2 +3 = +6 +9 = +8 +12 = +10 +15 = ……. + +Page no. 166 +Figure it out +Q1. Three rotis are shared equally by four children. Show the division in the picture and +write a fraction for how much each child gets. Also, write the corresponding division +facts, addition facts, and, multiplication facts. +Fraction of roti each child gets is ______. +Division fact: +Addition fact: +Multiplication fact: +Compare your picture and answers with your classmates! +Ans. Each child gets +3 +4 roti. +Division fact: 3  4 = +3 +4 + +Addition fact: 3 = +3 +4 + +3 +4 + +3 +4 + +3 +4 + +Multiplication fact: 3 = 4 × +3 +4 +Q2. Draw a picture to show how much each child gets when 2 rotis are shared equally by +4 children. Also, write the corresponding division facts, addition facts, and +multiplication facts. +gets +1 +2 +Ans. + 2 Rotis equally shared by 4 children. Each child +roti. +Division fact :2  4 = +2 +4 = +1 +2 +Addition fact: 2 = +1 +2 + +1 +2 + +1 +2 + +1 +2 +Multiplication fact: 2 = 4 × +1 +2 +Q3. Anil was in a group where 2 cakes were divided equally among 5 children. How much +cake would Anil get? +Ans. Anil would get +2 +5 cake. + + +Page no. 168 + Section 7.6 +Figure it out +Q1. Find the missing numbers: +a. 5 glasses of juice shared equally among 4 friends is the same as ____ glasses of +juice shared equally among 8 friends. +So, +𝟓 +𝟒 = 𝟖 +Ans. 10 +b. 4 kg of potatoes divided equally in 3 bags is the same as 12 kgs of potatoes divided +equally in __ bags. +So, +𝟒 +𝟑 = +𝟏𝟐 +Ans. 9 +c. 7 rotis divided among 5 children is the same as____rotis divided among _____ +children. +So, +𝟕 +𝟓 = +Ans. One of the choices is 14, 10 +(Try for other options also.) + +Page no. 170 + Section 7.6 +Q. Suppose the number of children is kept the same, but the number of units that are +being shared is increased? What can you say about each child’s share now? Why? +Discuss how your reasoning explains +𝟏 +𝟓 < +𝟐 +𝟓, +𝟑 +𝟕 < +𝟒 +𝟕 and +𝟏 +𝟐 < +𝟓 +𝟖 . +Ans. Each child will have larger share now. +If the number of units to be shared increase with same number of children then each child will +have larger share. +Q. Now, decide in which of the two groups will each child get a larger share: +1. Group 1: 3 glasses of sugarcane juice divided equally among 4 children. +Group 2: 7 glasses of sugarcane juice divided equally among 10 children. +2. Group 1: 4 glasses of sugarcane juice divided equally among 7 children. +Group 2: 5 glasses of sugarcane juice divided equally among 7 children. + +Which groups were easier to compare? Why? +Ans. +Each child’s share in group 1 = +4 +7 +Each child’s share in group 2 = +5 +7 +And +5 +7 > +4 +7 +The groups in second part were easier to compare because the number of children is +same. + +Page no. 172 + Section 7.6 +Q. Find equivalent fractions for the given pairs of fractions such that the fractional units +are the same. +a. +𝟕 +𝟐 and +𝟑 +𝟓 +b. +𝟖 +𝟑 and +𝟓 +𝟔 +c. +𝟑 +𝟒 and +𝟑 +𝟓 +d. +𝟔 +𝟕 and +𝟖 +𝟓 + +e. +𝟗 +𝟒 and +𝟓 +𝟐 +f. +𝟏 +𝟏𝟎 and +𝟐 +𝟗 +g. +𝟖 +𝟑 and +𝟏𝟏 +𝟒 +h. +𝟏𝟑 +𝟔 and +𝟏 +𝟗 +Ans. +a. +35 +10 and +6 +10 +b. +16 +6 and +5 +6 +𝑐. +15 +20 and +12 +20 +𝑑. +30 +35 and +56 +35 +𝑒. +9 +4 and +10 +4 +𝑓. +9 +90 and +20 +90 +𝑔. +32 +12 and +33 +12 +ℎ. +39 +18 and +2 +18 + + + + + + + + +Page no. 173 + Section 7.6 +Figure it out +Q. Express the following fractions in lowest terms: +a. +𝟏𝟕 +𝟓𝟏 +b. +𝟔𝟒 +𝟏𝟒𝟒 +c. +𝟏𝟐𝟔 +𝟏𝟒𝟕 +d. +𝟓𝟐𝟓 +𝟏𝟏𝟐 +Ans. a. +1 +3 + 𝑏. +4 +9 +c. +6 +7 +d. +75 +16 + +Page no. 174 + Section 7.7 +Figure it out +Q1. Compare the following fractions and justify your answers: +a. +𝟖 +𝟑, +𝟓 +𝟐 + +b. +𝟒 +𝟗 , +𝟑 +𝟕 + +c. +𝟕 +𝟏𝟎 , +𝟗 +𝟏𝟒 + +d. +𝟏𝟐 +𝟓 , +𝟖 +𝟓 +e. +𝟗 +𝟒 , +𝟓 +𝟐 +Ans. + a. +8 +3 > +5 +2 +b. +4 +9 > +3 +7 +𝑐. +7 +10 > +9 +14 +d. +12 +5 > +8 +5 +𝑒. +9 +4 < +5 +2 + + + + +Q2. Write the following fractions in ascending order. +a. +𝟕 +𝟏𝟎, +𝟏𝟏 +𝟏𝟓, +𝟐 +𝟓 + +b. +𝟏𝟗 +𝟐𝟒, +𝟓 +𝟔, +𝟕 +𝟏𝟐 +Ans. a. +2 +5 < +7 +10 < +11 +15 +b. +7 +12 < +19 +24 < +5 +6 +Q3. Write the following fractions in descending order. +a. +𝟐𝟓 +𝟏𝟔, +𝟕 +𝟖, +𝟏𝟑 +𝟒, +𝟏𝟕 +𝟑𝟐 + +b. +𝟑 +𝟒, +𝟏𝟐 +𝟓, +𝟕 +𝟏𝟐, +𝟓 +𝟒 +Ans. a. +13 +4 > +25 +16 > +7 +8 > +17 +32 +b. +12 +5 > +5 +4 > +3 +4 > +7 +12 + + Section 7.8 +Page no. 177 +Q. Try adding +𝟒 +𝟕 + +𝟔 +𝟕 using a number line. Do you get the same answer? +Ans. +4 +7 + +6 +7 + +4 +7 + +6 +7 = +10 +7 = 1 + +3 +7 +Yes, using a number line, the answer is same. + +Page no. 179 + Section 7.8 +Figure it out +Q1. Add the following fractions using Brahmagupta’s method: + + + + + + + +Ans. +a. +13 +7 +b. +13 +12 +c. +9 +6 = +93 +6 3 = +3 +2 +d. +20 +21 +e. +77 +60 +f. +22 +15 +g. +22 +15 +h. +49 +40 +i. +23 +4 +j. +62 +21 +k. +77 +60 +l. +199 +105 +m. +83 +12 +Q2. Rahim mixes +𝟐 +𝟑 litres of yellow paint with +𝟑 +𝟒 litres of blue paint to make green paint. +What is the volume of green paint he has made? +Ans. 1 +5 +12 litres +Q3. Geeta bought +𝟐 +𝟓 meter of lace and Shamim bought +𝟑 +𝟒 meter of the same lace to put a +complete border on a table cloth whose perimeter is 1 meter long. Find the total +length of the lace they both have bought. Will the lace be sufficient to cover the +whole border? +Ans. Total length of the lace = 1 +3 +20 m +Yes. + + + Section 7.8 +Figure it out (Page no. 181) +Q1. + +5 +8 - +3 +8 +Ans. +2 +8 (= +1 +4, in lowest terms) +Q2. + +𝟕 +𝟗 - +𝟓 +𝟗 +Ans. +2 +9 +Q3. + +𝟏𝟎 +𝟐𝟕 - +𝟏 +𝟐𝟕 +Ans. +9 +27 (= +1 +3, in lowest terms) + +Page no. 181 + Section 7.8 +Figure it out +Ans. 1. +1 +4 2. +2 +9 3 +1 +3 + +Figure it out +Q1. Carry out the following subtractions using Brahmagupta’s method: +a. +8 +15 - +3 +15 +b. +2 +5 - +4 +15 +c. +5 +6 - +4 +9 + +d. +2 +3 - +1 +2 + +Ans. a. +1 +3 +b. +2 +15 +c. +7 +18 +d. +1 +6 + + + + + +Q2. Subtract as indicated: +a. +𝟏𝟑 +𝟒 from +𝟏𝟎 +𝟑 +b. +𝟏𝟖 +𝟓 from +𝟐𝟑 +𝟑 + +c. +𝟐𝟗 +𝟕 from +𝟒𝟓 +𝟕 +Ans. a. +1 +12 +b. +61 +15 +c. +16 +7 +Q3. Solve the following problems: +a. Jaya’s school is +𝟕 +𝟏𝟎 km from her home. She takes an auto for +𝟏 +𝟐 km from her home +daily, and then walks the remaining distance to reach her school. How much does +she walk daily to reach the school? +b. Jeevika takes +𝟏𝟎 +𝟑 minutes to take a complete round of the park and her friend +Namit takes +𝟏𝟑 +𝟒 minutes to do the same. Who takes less time and by how much? +Ans. a. +1 +5 Km +b. Namit takes less time than Jeevika, by +1 +12 minutes" +class_6,8,Playing with Constructions,ncert_books/class_6/Ganita_Prakash/fegp108.pdf,"Playing with +Constructions +8 +8.1 Artwork +Observe the following figures and try drawing them freehand. +Fig. 8.1 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +188 +Now, arm yourself with a ruler and a compass. Let us explore if +we can draw these figures with these tools and get familiar with a +compass. +Observe the way a compass is made. What can one draw with the +compass? Explore! +Do you know what curves are? They are any shapes that can be +drawn on paper with a pencil, and include straight lines, circles and +other figures as shown below: +Mark a point ‘P’ in your notebook. Then, mark as many points as +possible, in different directions, that are 4 cm away from P. + +Think: Imagine marking all the points of 4 cm distance from the +point P. How would they look? + +Try to draw it and verify if it is correct by taking some points on +the curve and checking if their distances from P are indeed 4 cm. +Explore, if you have not already done so, +and see if a compass can be used for this +purpose. +You can start by marking a few points +of distance 4 cm from P using the compass. +How can this be done? +R +4 cm +4 cm +P +Q +Reprint 2025-26 + +Playing with Constructions +189 +You will have to open up the compass against a ruler (see Fig. 8.2) +such that the distance between the tip of the compass and the pencil +is 4 cm. + +Now, try to get the full curve. +Hint: Keep the point of the compass fixed moving only the pencil. +What is the shape of the curve? It is a circle! +Take a point on the circle. What will be its distance from P—equal +to 4 cm, less than 4 cm or greater than 4 cm? Similarly, what will be +the distance between P and another point on the circle? +As shown in the figure, the point P is called the centre of the circle +and the distance between the centre and any point on the circle is +called the radius of the circle. +radius +1 +0 +2 +3 +4 +5 +6 +7 +centre +Fig. 8.2 +P +Having explored the use of a compass, go ahead and recreate the +images in Fig. 8.1. +Can you make the figures look as good as the figures shown there? +Try again if you want to! +Also, has the use of instruments made the construction easier? +Now try constructing the following figures. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +190 + Construct +1. A Person + +How will you draw this? + +This figure has two components. + +You might have figured out a way of drawing the first part. For +drawing the second part, see this. + +The challenge here is to find out where to place the tip of the +compass and the radius to be taken for drawing this curve. You +can fix a radius in the compass and try placing the tip of the +Reprint 2025-26 + +Playing with Constructions +191 +compass in different locations to see which point works for +getting the curve. Use your estimate where to keep the tip. +2. Wavy Wave + +Construct this. + +As the length of the central line is not specified, we can take it to +be of any length. + +Let us take AB to be the central line such that the length of AB is +8 cm. We write this as AB = 8 cm. + +Here, the first wave is drawn as a half circle. +A +X +? +8 cm +B +? + Figure it Out +1. What radius should be taken in the compass to get this half circle? +What should be the length of AX? +2. Take a central line of a different length and try to draw the wave +on it. +3. Try to recreate the figure where the waves are smaller than a +half circle (as appearing in the neck of the figure, ‘A Person’). +The challenge here is to get both the waves to be identical. +This may be tricky! +Try +This +Reprint 2025-26 + +Ganita Prakash | Grade 6 +192 +3. Eyes + +How do you draw these eyes with a compass? + +For a hint, go to the end of the chapter. +  Make other artwork of your choice with a ruler and a compass. +8.2 Squares and Rectangles +Now, let us look at some basic figures having straight lines in their +boundary. +Fig. 8.3 +What shapes are these? Yes, these are our familiar squares and +rectangles. But what makes them squares and rectangles? +Consider this rectangle ABCD. +The points A, B, C and D are the corners +of the rectangle. Lines AB, BC, CD and DA +are its sides. Its angles are ∠A, ∠B, ∠C +and ∠D. +The blue sides AB and CD are called +opposite sides, as they lie opposite to +each other. Likewise, AD and BC is the +other pair of opposite sides. +A +D +B +C +Fig. 8.4 +Reprint 2025-26 + +Playing with Constructions +193 +Recall that, in a rectangle: +R1) The opposite sides are equal in length, and +R2) All the angles are 90°. +As in the case of rectangles, the corners and sides are defined for +a square in the same manner. +A square satisfies the following two properties: +S1) All the sides are equal, and +S2) All the angles are 90 +o. +See the rectangle in Fig. 8.4 and the name given to it: ABCD. This +rectangle can also be named in other ways — BCDA, CDAB, DABC, +ADCB, DCBA, CBAD and BADC. So, can a rectangle be named using +any combination of the labels around its corners? No! For example, +it cannot be named ABDC or ACBD. Can you see what names are +allowed and what names are not? +In a valid name, the corners occur in an order of travel around +the rectangle, starting from any corner. + Which of the following is not a name for this square? + +1. +PQSR + +2. +SPQR + +3. +RSPQ + +4. +QRSP +Rotated Squares and Rectangles +Here is a square piece of paper +having all its sides equal in length +and all angles equal to 90°. It is +rotated as shown in the figure. Is it +still a square? +Let us check if the rotated paper +still satisfies the properties of a square. +• +Are all the sides still equal? Yes. +• +Are all the angles still 90°? Yes. +S +R +P +Q +Reprint 2025-26 + +Ganita Prakash | Grade 6 +194 +Rotating a square does not change its +lengths and angles. +Therefore, this rotated figure satisfies both +the properties of a square and so, it is a square. +By the same reasoning, a rotated rectangle +is still a rectangle. + Figure it Out +1. Draw the rectangle and four squares configuration (shown in +Fig. 8.3) on a dot paper. + +What did you do to recreate this figure so that the four squares +are placed symmetrically around the rectangle? Discuss with +your classmates. +2. Identify +if +there +are +any +squares +in +this +collection. +Use measurements if needed. +A +D +C +B + + + Think: Is it possible to reason out if the sides are equal or +not, and if the angles are right or not without using any +measuring instruments in the above figure? Can we do this +by only looking at the position of corners in the dot grid? +3. Draw at least 3 rotated squares and rectangles on a dot grid. Draw +them such that their corners are on the dots. Verify if the squares +and rectangles that you have drawn satisfy their respective +properties. +Reprint 2025-26 + +Playing with Constructions +195 +8.3 Constructing Squares and Rectangles +Now, let us start constructing squares and rectangles. How would +you construct a square with a side of 6 cm? +For help, you can see the following figures. A square PQRS of side +length 6 cm is constructed. +P +Q +6 cm +P +Q +6 cm +Mark a point to draw a +perpendicular to PQ through P. +Mark S on the perpendicular such that +PS = 6 cm using a ruler. +Method 2 +This can also be done using a compass. +Step 1 +Step 2 +6 cm +S +P +Q +6 cm +90° +Step 3 +Method 1 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +196 +Can you see why PS should be +6 cm long? +6 cm +S +P +Q +6 cm +Step 4 +Draw a perpendicular to line +segment PQ through Q. +R +P +Q +6 cm +Step 5 +If we had used the compass, then the next +point can easily be marked using it! +R +S +P +Q +How long is the side RS and what +are the measures of ∠R and ∠S? +Step 6 +R +S +P +Q +6 cm +6 cm +6 cm +90º +90º +Reprint 2025-26 + +Playing with Constructions +197 + Construct +1. Draw a rectangle with sides of length 4 cm and 6 cm. After +drawing, check if it satisfies both the rectangle properties. +2. Draw a rectangle of sides 2 cm and 10 cm. After drawing, check if +it satisfies both the rectangle properties. +3. Is it possible to construct a 4-sided figure in which— +•• +all the angles are equal to 90º but +•• +opposite sides are not equal? +8.4 An Exploration in Rectangles +Construct a rectangle ABCD with AB = 7 cm and BC = 4 cm. +Imagine X to be a point that can be moved anywhere along +the side AD. Similarly, imagine Y to be a point that can be moved +anywhere along the side BC. Note that X can also be placed +on the end point A or D. Similarly, Y can also be placed on the +end point B or C. +A +A +B +B = Y +B +B = Y +D +D = X +C +C +Y +1 cm +5 mm +4 cm +X +A +A = X +D +D +C +C +Y +2 cm +1 cm +X +Try +This +Reprint 2025-26 + +Ganita Prakash | Grade 6 +198 + At which positions will the points X and Y be at their closest? +When do you think they will be the farthest? What does your +intuition say? Discuss with your classmates. +Now, verify your guesses by placing the points X and Y on the +sides and measure how near or far they are. +The distance between X and Y can be obtained by measuring the +length of the line XY. +How does the minimum distance between the points X and Y +compare to the length of AB? +Change the positions of X and Y to check if there are other +positions where they are at their nearest or farthest. You could +construct multiple copies of the rectangle and try out various +positions of X and Y. +How will you keep track of the lengths XY for different positions +of X and Y? +Here is one way of doing it. Suppose here are some of the positions +of X and Y that you have considered: +• +When X is 5 mm away from A and Y is 3 cm away from B, +XY = ___ cm __ mm +• +When X is 1 cm away from A and Y is 1 cm away from B, +XY = ___ cm ___ mm +• +When X is 2 cm away from A and Y is 4 cm away from B, +XY = ___ cm ___ mm and so on. +  Is there a shorthand way of writing it down? In all the sentences, +only the position of X, Y and the length XY changes. So we could +write this as: +Distance of X from A +Distance of Y from B +Length of XY +Math +Talk +Reprint 2025-26 + +Playing with Constructions +199 +  Have you checked what happens to the length XY when X and Y +are placed at the same distance away from A and B, respectively? +For example, as in the cases like these: +Distance of X +from A +Distance of Y +from B +Length of XY +5 mm +5 mm +1 cm +1 cm +1 cm 5 mm +1 cm 5 mm +and so on. + In each of these cases, observe + +1. how the length XY compares to that of AB and + +2. the shape of the 4-sided figure ABYX. +  How does the farthest distance between X and Y compare with +the length of AC? BD? + Construct + +Breaking Rectangles + +Construct +a +rectangle +that can be divided into 3 +identical squares as shown +in the figure. + +Solution + +If this seem difficult, let us simplify the problem. + Explore + +What about constructing a rectangle that can be divided into +two identical squares? Can you try it? + +It is wise to first plan and then construct. But how do we plan? +Can you think of a way? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +200 + +One way is to visualise the final figure by drawing a rough +diagram of it. +A +B +C +F +E +D + +What can we infer from this figure? + +Can you identify the equal sides? + +Since, the two squares are identical, + +AB = BC and FE = ED + +Since ABEF and BCDE are squares, all the sides in each of the +squares are equal. This is written as — + +AF = AB = BE = FE + +BE = BC = CD = ED + +So, all the shorter lines are equal! + +A convention is followed to represent equal sides. It is done by +putting a ‘|’ on the line. Refer to the rough figure. + +Using this analysis, can you try constructing it? Remember, all +that was asked for is a rectangle that can be divided into two +identical squares and with no measurements imposed. + +To draw the rectangle ACDF, one could assign any length to AF. For +example, if we assign AF = 4 cm, then what must the length of AC be? + + Explore: Can the rectangle now be completed? + +In fact, one could proceed by drawing AF without even +measuring its length using a ruler. We could then construct a +line perpendicular to AF that is long enough to contain the other +side. As, AB = AF, we need to somehow transfer the length of AF +Reprint 2025-26 + +Playing with Constructions +201 +to get the point B. How do we do it without a ruler? Can it be +done using a compass? + Observe, how the length of AF is measured using a compass. +A +F +A +F +B + +Use it to mark out the points B and C, and complete the rectangle. +  With this idea, try constructing a rectangle that can be divided +into three identical squares. + +  Give the lengths of the sides of a rectangle that cannot be +divided into — +• +two identical squares; +• +three identical squares. + Construct + +1. +A Square within a Rectangle + +Construct a rectangle of sides 8 cm and 4 cm. How will you +construct a square inside, as shown in the figure, such that the +centre of the square is the same as the centre of the rectangle? +8 cm +4 cm +Reprint 2025-26 + +Ganita Prakash | Grade 6 +202 + + +Hint: Draw a rough figure. What will be the sidelength of the +square? What will be the distance between the corners of +the square and the outer rectangle? + +2. +Falling Squares + + + + +Now, try this. +Square of +side 5 cm +Square of +side 7 cm +Square of +side 3 cm + +3. Shadings + +Construct this. Choose measurements of your +choice. Note that the larger 4-sided figure is +a square and so are the smaller ones. +Make sure that the +squares are aligned +the way they are +shown. +4 cm +4 cm +4 cm +Each is a +Square of +side 4 cm +Reprint 2025-26 + +Playing with Constructions +203 + +4. +Square with a Hole + + + + + + +Observe that the circular +hole is the same as the +centre of the square. + + +Hint: Think where the centre of the circle should be. + +5. +Square with more Holes + +6. Square with Curves + +This is a square with 8 cm sidelengths. + +Hint: Think where the tip of the compass +can be placed to get all the 4 arcs +to bulge uniformly from each of +the sides. Try it out! +8.5 Exploring Diagonals of Rectangles and Squares +Consider a rectangle PQRS. Join PR and QS. +These two lines are called the diagonals +of the rectangle. +Compare the lengths of the diagonals. +First predict the answer. Then construct a +Try +This +P +S +Q +R +a +b +c +d +e +f +g +h +Reprint 2025-26 + +Ganita Prakash | Grade 6 +204 +rectangle marking the points as shown and measure the diagonals. +In rectangle PQRS, the right angles at P and R are referred to as +opposite angles. The other pair of opposite angles are the right +angles at Q and S. +Observe that a diagonal divides each of the pair of opposite +angles into two smaller angles. In the figure, the diagonal PR divides +angle R into two smaller angles which we simply call g and h. The +diagonal also divides angle P into c and d. Are g and h equal? Are +c and d equal? +First predict the answers, and then measure the angles. What do +you observe? Identify pairs of angles that are equal. + Explore +How should the rectangle be constructed so that the diagonal divides +the opposite angles into equal parts? +How will you record your observations? First, identify the +parameters that need to be tracked. They are the sides of the rectangle +and the 8 angles formed by the two diagonals. Are there any other +measurements that you would want to keep track of? +Sides +A +B +C +D +E +F +G +H +In your experimentation, did you consider the case when all four +sides of the rectangle are equal? That is, did you consider the case of +a square? See what happens in this special case! + +  What general laws did you observe with respect to the +angles and sides? Try to frame and discuss them with +your classmates. + +How can one be sure if the laws that you have observed +will always be true? +Math +Talk +Reprint 2025-26 + +Playing with Constructions +205 + Construct +1. Construct a rectangle in which one of the diagonals divides +the opposite angles into 60° and 30°. + +Solution + +Let us start with a rough diagram. +D +A +B +C + +In what order should its parts be drawn? + +We will briefly sketch a possible order of construction. + +Step 1 +A +B +A +B + +AB is drawn with an arbitrary length. What is the next point that +can be located? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +206 + +Step 2 +A +60o +B +C + +Step 3 + +We know the line on which D lies. Draw a line through A +perpendicular to AB. +A +60o +B +C + + +Now ∠A is divided into two angles. One measures 60°. Check +what the other angle is. + +There are at least two ways of finding the point D — +• One uses the fact that all the angles of a rectangle are right +angles. +• The other uses the fact that opposite sides are equal. +Reprint 2025-26 + +Playing with Constructions +207 + +Step 4   + +Method 1 +A +D +60o +30o +B +C +Draw a line perpendicular to BC at C to get the point D. + +Method 2 +A +A +D +60o +60o +B +B +C +C + +Using a compass, mark the point D such that AD = BC. + +Join CD to get the required rectangle. +We have seen how to construct rectangles when their sides are +given. But what do we do if a side and a diagonal is given? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +208 +2. Construct a rectangle where one of its sides is 5 cm and the +length of a diagonal is 7 cm. + +Solution +Let us draw a rough diagram. +Let us decide the steps of construction. +Which line can be drawn first? + +Step 1 +The base CD measuring length 5 cm can be easily constructed. +D +C +5 cm +Next? + +Step 2 +Draw a perpendicular to line DC at the point C. Let us call this line l. +5 cm +l +D +C + +This is easy as we know that this line is perpendicular to the +base. The point B should be somewhere on this line l. +  How do we spot it? What else do we know about the position of B? + +We know that it is at a distance of 7 cm from the point D. +A +B +C +D +5 cm +7 cm +Reprint 2025-26 + +Playing with Constructions +209 + + +One of the ways of marking B is by taking a ruler and trying +to move it around to get a point on line l that is 7 cm from point D. +However, this requires trial and error. There is another efficient +method which doesn’t involve trial and error. + + +For this, instead of trying to get that one required point of distance +7 cm from D, let us explore a way of getting all the points of +distance 7 cm from D. + + +We know what this shape is! + +Step 3 + +Method 1 +D +C +l + +Construct a circle of radius 7 cm with point D as the centre. + + +Can you spot the point B here? Remember that it is 7 cm +away from point D and on the line l. + + +Consider the point at which the circle and the line intersect. +What is its distance from point D? If needed, check your figure. +What do you observe? + + +The point where the circle intersects the line l is the required +point B. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +210 + +Method 2 + +To locate the point B, was it necessary to draw the + +entire circle? We can see that only the arc near the line l is needed. +So, the third step can also be done as shown in the figure below. +D +C +l + + +Having marked the three points of the rectangle, we only need +to complete it. Recall that we were in a similar situation in the +previous problem also. We saw two methods of completing the +rectangle from here. We could follow any one of those methods. + +Step 4 +D +A +C +B +90° +90° + +Construct perpendiculars to DC and BC passing through D and B, +respectively. The point where these lines intersect is the fourth +point A. + + +Check if ABCD is indeed a rectangle satisfying properties R1 +and R2. +Reprint 2025-26 + +Playing with Constructions +211 + Construct +1. Construct a rectangle in which one of the diagonals divides the +opposite angles into 50° and 40°. +2. Construct a rectangle in which one of the diagonals divides the +opposite angles into 45° and 45°. What do you observe about the +sides? +3. Construct a rectangle one of whose sides is 4 cm and the diagonal +is of length 8 cm. +4. Construct a rectangle one of whose sides is 3 cm and the diagonal +is of length 7 cm. +8.6 Points Equidistant from Two Given Points + Construct + +House + +Recreate this figure. + + +Note that all the lines forming the +border of the house are of length 5 cm. + +Solution + +The first task is to identify in what +sequence the lines and curve will +have to be drawn. + +Step 1 +5 cm +5 cm +5 cm +B +D +C +E +1 cm +2 cm +5 cm +5 cm +5 cm +5 cm +5 cm +A +B +D +C +E +1 cm +2 cm +Reprint 2025-26 + +Ganita Prakash | Grade 6 +212 + +Can you complete the figure? Try! + + +We need to locate the point A that is of distance 5 cm from +the points B and C. You might have realised that this can be done +using a ruler. However, this leads to a lot of trial and error. This +construction can be further simplified. How? + + +If you have guessed that this can be done by the use of +compass, you are right! Go ahead and explore how the point A +can be located without trial and error. + + +There is a similarity between the problem of finding point +A in this problem and point B at step 3 of the second solved +example of the previous section (see page 209). + +Step 2 +5 cm +5 cm +5 cm +B +D +C +E +1 cm +2 cm + +Draw a curve that has all its points of 5 cm from the point B; the +circle centred at B should be with 5 cm radius. + + +Does this help in spotting the point A? Construct and explore +in the figure. + + +The point A can be located by finding the correct point on the +circle that is of distance 5 cm from the point C. Again, this can be +done using a ruler. But can we use a compass for this? +Reprint 2025-26 + +Playing with Constructions +213 + +Step 3 + +Method 1 + +Take a radius of 5 cm in the compass and with C as the centre, +draw a circle. +5 cm +5 cm +5 cm +B +D +C +E +1 cm +2 cm + + +Are you able to spot the point A? Check the figure in your +notebook. What do you observe? + + +See the point at which both the circles intersect. How far is +it from the point B? + +How far is it from C? + +Thus, this is the point A! + Think + +Was it necessary to draw two full circles to get the point A? We +only needed part of both the circles. + +Method 2 + +So the point A could have been obtained just by drawing arcs of +radius 5 cm from points B and C. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +214 +5 cm +5 cm +5 cm +B +D +C +E +1 cm +2 cm +A + +Join A to B and A to C by straight lines. + + +Having obtained point A, what remains is the construction +of the remaining arc. How do we do it? + +Can we use the fact that A is of distance 5 cm from both B and C? + +Step 4 + +Take 5 cm radius in the compass and from A, draw the arc +touching B and C as shown in the figure. +5 cm +5 cm +5 cm +5 cm +5 cm +A +B +D +C +E +1 cm +2 cm + +The house is ready! +Reprint 2025-26 + +Playing with Constructions +215 + Construct +1. Construct a bigger house in which all the sides are of length 7 cm. +2. Try to recreate ‘A Person’, ‘Wavy Wave’, and ‘Eyes’ from the section +‘Artworkʼ, using ideas involved in the ‘House’ construction. +3. Is there a 4-sided figure in which all the sides are equal in length +but is not a square? If such a figure exists, can you construct it? +Hints +A) Eyes (from 8.1 Artwork and Construct above (page no. 215). +Part of the construction is shown earlier. Observe it carefully. You will +see two horizontal lines drawn lightly. In geometric constructions, one +often constructs supporting curves or figures that are not part of the given +figure but help in constructing it. +A +B + +The technique to draw the upper and the lower curves of the eye is the +same as that used in the figure, ‘A Personʼ. Points A and B are the locations +where the tip of the compass is placed when drawing the curves of the +eye. Note that the upper curve and the lower curve should together form +a symmetrical figure. For this to happen, where should these points A and +B be placed? Make a good estimate. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +216 + +Try to get the eyes as symmetrical and identical as possible. This might +need many trials. +B) (From Construct above (page no. 211). +For the purpose of construction, let us take the side lengths to be of 5 cm. +Consider this figure. +A +5 cm +5 cm +C +B +We need to identify only one more point to make this a 4-sided +figure. That point, let us call it D, should be 5 cm from both B and C. +How can such a point be found? +Can any of the ideas used in the ‘House’ problem be used here? +S u m m a r y + + +All the points of a circle are at the same distance from its centre. This +distance is called the radius of the circle. + + +A compass can be used to construct circles and their parts. + + +A rough diagram can be useful in planning how to construct a given +figure. + + +A rectangle can be constructed given the lengths of its sides or that of +one of its sides and a diagonal. +Reprint 2025-26 + +[1] + +CHAPTER 8 — SOLUTIONS +Playing with Constructions + +Page no. 188 +Section 8.1 + Think: Imagine marking all the points of 4 cm distance from the point P. How would +they look? + +Ans. + +On joining, they are forming a circle. + + + + +Page no. 191 +Section 8.1 +Figure it out +Q.1. What radius should be taken in the compass to get this half circle? What should be +the length of AX? +Ans. 2 cm. AX = 4 cm. + +Q.2. Take a central line of a different length and try to draw the wave on it. +Ans. + + +Q.3. Try to recreate the figure where the waves are smaller than a half circle (as +appearing in the neck of the figure ‘A Person’). The challenge here is to get both the +waves to be identical. This may be tricky! +Ans. + + + +Page no. 193 +Section 8.2 + Q. Which of the following is not a name for this square? +1. PQSR +2. SPQR +4 cm + +[2] + +3. RSPQ +4. QRSP +Ans. PQSR is not a name for the given square. + +Page no. 194 +Section 8.2 +Figure it out +Draw a rectangle & then leave one dot distance diagonally to place the four smaller squares. +Q.2. Identify if there are any squares in this collection. Use measurements if needed. +Ans. A is a square. + + Think: Is it possible to reason out if the sides are equal or not, and if the angles are +right or not without using any measuring instruments in the above figure? Can we +do this by only looking at the position of corners in the dot grid? +Ans. Yes, In the above figure without measuring one can reason out for equal sides & right +angles. +Yes, the position of the points on the grid paper makes it possible +Q.3. Draw at least 3 rotated squares and rectangles on a dot grid. Draw them such that +their corners are on the dots. Verify if the squares and rectangles that you have +drawn satisfy their respective properties. +Ans. + + + + + +[3] + +Page no. 197 +Section 8.3 + Construct +Q.1. Draw a rectangle with sides of length 4 cm and 6 cm. After drawing, check if it +satisfies both the rectangle properties. +Ans. A = B = C = D = 90 +AB = CD = 4 cm & AD = BC = 6 cm. + + + +Q.2. Draw a rectangle of sides 2 cm and 10 cm. After drawing, check if it satisfies both +the rectangle properties. +Ans. +P = Q = R = S = 90 +PQ = SR = 10 cm & PS = QR = 2 cm + +Q.3. Is it possible to construct a 4-sided figure in which— +• all the angles are equal to 90º but +• opposite sides are not equal? +Ans. No. + +Page no. 198 +Section 8.4 + + Q. Is there a shorthand way of writing it down? In all the sentences, only the +position of X, Y and the length XY changes. So we could write this as: +Distance of X from A +Distance of Y from B +Length of XY + + + +Ans. +Distance of X from A +Distance of Y from B +Length of XY +5 mm +1 cm +2 cm +3 cm +1 cm +4 cm +7.4 cm +7 cm +7.3 cm + + + +[4] + +Page No. 199 + Q. Have you checked what happens to the length XY when X and Y are placed at +the same distance away from A and B, respectively? For example, as in the cases +like these: +Distance of X from A +Distance of Y from B +Length of XY +5 mm +1 cm +1 cm 5 mm +5 cm +1 cm +4 cm + +And so on. +Ans. +Distance of X from A +Distance of Y from B +Length of XY +5 mm +1 cm +1 cm 5 mm +5 mm +1 cm +1 cm 5 mm +7 cm +7 cm +7 cm + +Page no. 199 + Q. In each of these cases, observe +i. how the length XY compares to that of AB and +ii. the shape of the 4-sided figure ABYX. +Ans. i. +XY = AB +ii. +ABYX is a rectangle + + Q. How does the farthest distance between X and Y compare with the length of AC? +BD? +Ans. The farthest distance between X and Y is equal to AC or BD. + +Construct +Q. Construct a rectangle that can be divided into 3 identical squares as shown in the +figure. +Ans. Hint: Take the length of the rectangle three times its breadth. + +Page no. 201 + Give the lengths of the sides of a rectangle that cannot be divided into — +• two identical squares; +• three identical squares +Ans. +• Lengths of sides of a rectangle that cannot be divided into two identical squares: +Length = 4 cm and Breadth = 2.5cm (Try for more!) +• Length of sides of a rectangle that cannot be divided into three identical squares: +Length = 7cm and Breadth = 2cm +(Try for other possibilities.) + + + +[5] + +Page no. 201 +Q.4. Square with a Hole +Ans. The centre of the circle should be at the meeting point of two line segments connecting +opposite vertices. + +Page no. 204 +Section 8.5 + Explore +Q. How should the rectangle be constructed so that the diagonal divides the opposite +angles into equal parts? +Ans. The rectangle should be constructed in such a manner that the two adjacent sides are +equal i.e. the rectangle becomes a square. + +Page no. 211 +Construct +Q.1. Construct a rectangle in which one of the diagonals divides the opposite angles into +50°and 40°. +Ans. +Q.2. Construct a rectangle in which one of the diagonals divides the opposite angles into +45° and 45°. What do you observe about the sides? +Ans. + + + +[6] + +Q.3. Construct a rectangle one of whose sides is 4 cm and the diagonal is of length 8 cm. +Ans. +Q.4. Construct a rectangle one of whose sides is 3 cm and the diagonal is of length 7 cm. +Ans. + + + + +[7] + +Page no. 215 +Construct +Q.1. Construct a bigger house in which all the sides are of length 7 cm. +Ans. + + + +[8] + +Q.2. Try to recreate ‘A Person’, ‘Wavy Wave’ and ‘Eyes’ from the section Artwork, +using ideas involved in the ‘House’ construction. +Ans. +Q.3. Is there a 4-sided figure in which all the sides are equal in length but is not a +square? If such a figure exists, can you construct it? +Ans." +class_6,9,Symmetry,ncert_books/class_6/Ganita_Prakash/fegp109.pdf,"SYMMETRY +9 +Look around you — you may find many objects that catch your +attention. Some such things are shown below: +Butterfly +Flower +Pinwheel +Rangoli +There is something beautiful about the pictures above. +The flower looks the same from many different angles. What +about the butterfly? No doubt, the colours are very attractive. But +what else about the butterfly appeals to you? +In these pictures, it appears that some parts of the figure are +repeated and these repetitions seem to occur in a definite pattern. +Can you see what repeats in the beautiful rangoli figure? In the +Reprint 2025-26 + +Ganita Prakash | Grade 6 +218 +rangoli, the red petals come back onto themselves when the flower +is rotated by 90˚ around the centre and so do the other parts of the +rangoli. +What about the pinwheel? Can you spot which pattern is repeating? +Hint: Look at the hexagon first. +Now, can you say what figure repeats +along each side of the hexagon? What +is the shape of the figure that is stuck to +each side? Do you recognise it? How do +these shapes move as you move along the +boundary of the hexagon? What about +the other pictures — what is it about +those structures that appeals to you and +what are the patterns in those structures +that repeat? +On the other hand, look at this picture +of clouds. There is no such repetitive pattern. +We can say that the first four figures are symmetrical and the last one is +not symmetrical. A symmetry refers to a part or parts of a figure that +are repeated in some definite pattern. +Taj Mahal +Gopuram +What are the symmetries that you see in these beautiful structures? +Clouds +Reprint 2025-26 + +Symmetry +219 +9.1 Line of Symmetry +Figure (a) shows the picture of a blue triangle with a dotted line. +What if you fold the triangle along the dotted line? Yes, one half +of the triangle covers the other half completely. These are called +mirror halves! +(a) +(b) +What about Figure (b) with the four puzzle pieces and a dotted line +passing through the middle? Are they mirror halves? No, when we fold +along the line, the left half does not exactly fit over the right half. +A line that cuts a figure into two parts that exactly overlap when +folded along that line is called a line of symmetry of the figure. + Figure it Out +1. Do you see any line of symmetry in the figures at the start of the +chapter? What about in the picture of the cloud? +2. For each of the following figures, identify the line(s) of symmetry +if it exists. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +220 +Figures with more than one line of symmetry +Does a square have only one line of symmetry? +Take a square piece of paper. By folding, find all its lines of symmetry. +Fold 1 +Fold 2 +Fold 4 +Fold 3 +Here are the different folds giving different lines of symmetry. +• Fold the paper into half vertically. +• Fold it again into half horizontally (i.e., you have folded it +twice). Now open out the folds. +Vertical Fold +Horizontal Fold +Again fold the square into half (for a third time now), but this +time along a diagonal, as shown in the figure. Again, open it. +Reprint 2025-26 + +Symmetry +221 +Fold it into half (for the fourth time), but this time along the other +diagonal, as shown in the figure. Open out the fold. +  Is there any other way to fold the square so that the two halves +overlap? How many lines of symmetry does the square shape have? + +Thus, figures can have multiple lines of symmetry. The figures +below also have multiple lines of symmetry. Can you find them all? +  We saw that the diagonal of a square is also a line of symmetry. +Let us take a rectangle that is not a square. Is its diagonal a line of +symmetry? + First, see the rectangle and answer this +question. Then, take a rectangular piece of +paper and check if the two parts overlap by +folding it along its diagonal. What do you +observe? +Reflection +So far we have been saying that when we fold a figure along a line of +symmetry, the two parts overlap completely. We could also say that +the part of the figure on one side of the line of symmetry is reflected +by the line to the other side; similarly, the part of the figure on the +other side of the line of symmetry is reflected to the first side! Let us +understand this by labeling some points on the figure. +The figure shows a square with its corners labeled A, B, C and D. +Let us first consider the vertical line of symmetry. When we reflect +Reprint 2025-26 + +Ganita Prakash | Grade 6 +222 +the square along this line, the points B, C on the right get reflected to +the left side and occupy the positions occupied earlier by A, D. What +happens to the points A, D? A occupies the position occupied by B and +D that of C! +A +D +B +C +  What if we reflect along the diagonal from A to C? Where do +points A, B, C and D go? What if we reflect along the horizontal line of +symmetry? +A figure that has a line or lines of symmetry is thus, also said to have +reflection symmetry. +Generating shapes having lines of symmetry +So far we have seen symmetrical figures and asymmetrical +figures. How does one generate such symmetrical figures? Let us +explore this. +Ink Blot Devils +You enjoyed doing this earlier in Class 5. Take a piece of paper. Fold +it in half. Open the paper and spill a few drops of ink (or paint) on +one half. +Now press the halves together and then open the paper again. +• What do you see? +• Is the resulting figure symmetric? +• If yes, where is the line of symmetry? +• Is there any other line along which it can be folded to produce +two identical parts? +• Try making more such patterns. +Reprint 2025-26 + +Symmetry +223 +Paper Folding and Cutting +Here is another way of making symmetric shapes! +In these two figures, a sheet of paper is +folded and a cut is made along the dotted +line shown. Draw a sketch of how the paper +will look when unfolded. +Do you see a line of symmetry in this +figure? What is it? +Make different symmetric shapes by folding and cutting. +There are more ways of folding and cutting pieces of paper to get +symmetric shapes! +Use thin rectangular coloured +paper. Fold it several times and +create some intricate patterns by +cutting the paper, like the one shown +here. Identify the lines of symmetry +in the repeating design. Use such +decorative paper cut-outs for festive +occasions. + +  Figure it Out + +Punching Game + +The fold is a line of symmetry. Punch holes at different locations of +a folded square sheet of paper using a punching machine and create +different symmetric patterns. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +224 +1. In each of the following figures, a hole was punched in a folded +square sheet of paper and then the paper was unfolded. Identify +the line along which the paper was folded. + +Figure (d) was created by punching a single hole. How was the +paper folded? + + +a. +b. +c. +d. +2. Given the line(s) of symmetry, find the other hole(s): +a. +b. +c. +d. +e. +3. Here are some questions on paper cutting. + +Consider a vertical fold. We represent it this way: + +Vertical Fold +Reprint 2025-26 + +Symmetry +225 + +Similarly, a horizontal fold is represented as follows: +Horizontal Fold +4. After each of the following cuts, predict the shape of the +hole when the paper is opened. After you have made your +prediction, make the cutouts and verify your answer. +a. +b. +d. +c. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +226 +5 + Suppose you have to get each of these shapes with some folds and +a single straight cut. How will you do it? + +a. The hole in the centre is a square. + +b. The hole in the centre is a square. + + Note: For the above two questions, check if the 4-sided figures in +the centre satisfy both the properties of a square. +6. How many lines of symmetry do these shapes have? + +a. + +b. +A triangle with equal sides and equal angles. +Reprint 2025-26 + +Symmetry +227 + +c. +A hexagon with equal sides and equal angles. +7. Trace each figure and draw the lines of symmetry, if any: + + +Reprint 2025-26 + +Ganita Prakash | Grade 6 +228 +8. Find the lines of symmetry for the kolam below. +9. Draw the following. + a. +A triangle with exactly one line of symmetry. + b. +A triangle with exactly three lines of symmetry. + c. +A triangle with no line of symmetry. +Is it possible to draw a triangle with exactly two lines of symmetry? +10. Draw the following. In each case, the figure should contain at least +one curved boundary. + a. +A figure with exactly one line of symmetry. + b. +A figure with exactly two lines of symmetry. + c. +A figure with exactly four lines of symmetry. +11. +Copy the following on squared paper. Complete them so that the +blue line is a line of symmetry. Problem (a) has been done for you. +(a) +(b) +(c) +Reprint 2025-26 + +Symmetry +229 +(d) +(e) +(f) + +Hint: For (c) and (f), see if rotating the book helps! +12. Copy the following drawing on squared paper. Complete each one +of them so that the resulting figure has the two blue lines as lines +of symmetry. +(a) +(b) +(c) +(d) +(e) +(f) +Reprint 2025-26 + +Ganita Prakash | Grade 6 +230 +13. Copy the following on a dot grid. For each figure draw two more +lines to make a shape that has a line of symmetry. +9.2 Rotational Symmetry +The paper windmill in the picture looks +symmetrical but there is no line of symmetry! +However, if you fold it, the two halves will not +exactly overlap. On the other hand, if you rotate +it by 90° about the red point at the centre, the +windmill looks exactly the same. +We say that the windmill has rotational +symmetry. +When talking of rotational symmetry, there is always a fixed +point about which the object is rotated. This fixed point is called the +centre of rotation. +Will the windmill above look exactly the same when rotated +through an angle of less than 90°? +No! +An angle through which a figure can be rotated to look exactly the +same is called an angle of rotational symmetry, or just an angle of +symmetry, for short. +Reprint 2025-26 + +Symmetry +231 +For the windmill, the angles of symmetry are 90° (quarter turn), +180° (half turn), 270° (three-quarter turn) and 360° (full turn). +Observe that when any figure is rotated by 360°, it comes back to its +original position, so 360° is always an angle of symmetry. +Thus, we see that the windmill has 4 angles of symmetry. +Do you know of any other shape that has exactly four angles of +symmetry? +How many angles of symmetry does a square have? How much +rotation does it require to get the initial square? +We get back a square overlapping with itself after 90° of rotation. +This takes point A to the position of point B, point B to the position +of point C, point C to the position of point D, and point D back to the +position of point A. Do you know where to mark the centre of rotation? +A +B +C +D +B +C +D +A +Imaginary +reference line +Line after +rotation +Square after +rotation +Initial +position +What are the other angles of symmetry? +Line after +rotation +Initial +position +C +D +A +B +180o +Line after +rotation +Initial +position +A +B +C +D +270o +360o +A +B +D +C +Reprint 2025-26 + +Ganita Prakash | Grade 6 +232 +Example: Find the angles of symmetry of the following strip. +Solution: Let us rotate the strip in a clockwise direction about its centre. +A rotation of 180° results in the figure above. Does this overlap +with the original figure. +No. Why? +Another rotation through 180° from this position gives the original +shape. +This figure comes back to its original shape only after one +complete rotation through 360°. So, we say that this figure does not +have rotational symmetry. +Rotational Symmetry of Figures with +Radial Arms +Consider this figure, a picture with +4 radial arms. How many angles of +symmetry does it have? What are they? +Note that the angle between adjacent +central dotted lines is 90°. +Can you change the angles between +the radial arms so that the figure still has +4 angles of symmetry? Try drawing it. +To check if the figure drawn indeed has 4 angles of symmetry, +you could draw the figure on two different pieces of paper. Cut out +the radial arms from one of the papers. Keep the figure on the paper +fixed and rotate the cutout to check for rotational symmetry. +How will you modify the figure above so that it has only two +angles of symmetry? +90O +Reprint 2025-26 + +Symmetry +233 +Here is one way: +We have seen figures having 4 and 2 angles of symmetry. Can we +get a figure having exactly 3 angles of symmetry? Can you use radial +arms for this? +Let us try with 3 radial arms as in the figure below. How many +angles of symmetry does it have and what are they? +Here is a figure with three radial arms. +Trace and cut out a copy of this figure. By rotating the cutout over +this figure determine its angles of rotation. +We see that only a full turn or a rotation of 360° will bring the figure +back into itself. So this figure does not have rotational symmetry as +360 degrees is its only angle of symmetry. +However, can anything in the figure be changed to make it have +3 angles of symmetry? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +234 +Can it be done by changing the angles between the dotted lines? +If a figure with three radial arms should have rotational symmetry, +then a rotated version of it should overlap with the original. Here +are rough diagrams of both of them. +If these two figures must overlap, what can you tell about the +angles? +C +B +A +B +A +C +Observe that ∠A must overlap ∠B, ∠B must overlap ∠C and ∠C +must overlap ∠A. +So, ∠A = ∠B = ∠C. What must this angle be? +We know that a full turn has 360 degrees. This is equally distributed +amongst these three angles. So each angle must be 360° +3 = 120°. +So, the radial arms figure with 3 arms shows rotational symmetry +when the angle between the adjacent dotted lines is 120°. Use paper +cutouts to verify this observation. +Now how many angles of rotation does the figure have and what +are they? +Initial +position +After 120° +rotation +After 240° +(120° + 120°) +rotation +After 360° +(120° + 120°+ 120°) +rotation +Note: The colours have been added to show the rotations. +Reprint 2025-26 + +Symmetry +235 +Let us explore more figures. + Can you draw a figure with radial arms that has a) exactly 5 +angles of symmetry, b) 6 angles of symmetry? Also find the angles of +symmetry in each case. +Hint: Use 5 radial arms for the first case. What should the angle +between two adjacent radial arms be? + Consider a figure with radial arms having exactly 7 angles +of symmetry. What will be its smallest angle of symmetry? Is the +number of degrees a whole number in this case? If not, express it as +a mixed fraction. +Let us find the angles of symmetry for other kinds of figures. + Figure it Out +1. Find the angles of symmetry for the given figures about the point +marked •. +(a) +(b) +(c) +2. Which of the following figures have more than one angle of +symmetry? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +236 +3. Give the order of rotational symmetry for each figure: +Let us list down the angles of symmetry for all the cases above. +• Angles of symmetry when there are exactly 2 of them: 180°, 360°. +• Angles of symmetry when there are exactly 3 of them: 120°, +240°, 360°. +• Angles of symmetry when there are exactly 4 of them: 90°, +180°, 270°, 360°. +Do you observe something common about the angles of symmetries +in these cases? The first set of numbers are all multiples of 180. The +second are all multiples of 120. The third are all multiples of 90. + In each case, the angles are the multiples of the smallest angle. +You may wonder and ask if this will always happen. What do you +think? + True or False +• Every figure will have 360 degrees as an angle of symmetry. +Reprint 2025-26 + +Symmetry +237 +• If the smallest angle of symmetry of a figure is a natural number +in degrees, then it is a factor of 360. +Is there a smallest angle of symmetry for all figures? It turns out +that this is the case for most figures, except for the most symmetric +shapes like the circle, whose symmetries we now discuss. +Symmetries of a circle +The circle is a fascinating figure. What happens when you rotate a +circle clockwise about its centre? It coincides with itself. It does not +matter what angle you rotate it by! So, for a circle, every angle is an +angle of symmetry. +Now take a point on the rim of the circle and join it to the centre. +Extend the segment to a diameter of the circle. Is that diameter +a line  of reflection symmetry? It is. Every diameter is a line of +symmetry! +Like wheels, we can find other objects around us having rotational +symmetry. Find them. Some of them are shown below: +Flower +Wheel +Fan +Reprint 2025-26 + +Ganita Prakash | Grade 6 +238 + Figure it Out +1. Colour the sectors of the circle below so that the figure has i) +3 angles of symmetry, ii) 4 angles of symmetry, iii) what are +the possible numbers of angles of symmetry you can obtain +by colouring the sectors in different ways? +2. Draw two figures other than a circle and a square that have both +reflection symmetry and rotational symmetry. +3. Draw, wherever possible, a rough sketch of: +a. A triangle with at least two lines of symmetry and at least two +angles of symmetry. +b. A triangle with only one line of symmetry but not having +rotational symmetry. +c. A quadrilateral with rotational symmetry but no reflection +symmetry. +d. A quadrilateral with reflection symmetry but not having +rotational symmetry. +4. In a figure, 60° is the smallest angle of symmetry. What are +the other angles of symmetry of this figure? +5. In a figure, 60° is an angle of symmetry. The figure has two angles +of symmetry less than 60°. What is its smallest angle of symmetry? +6. Can we have a figure with rotational symmetry whose smallest +angle of symmetry is: + +a. 45°? + +b. 17°? +Try +This +Reprint 2025-26 + +Symmetry +239 +7. This is a picture of the new Parliament Building in Delhi. +a. Does the outer boundary of the picture have reflection +symmetry? If so, draw the lines of symmetries. How many are +they? +b. Does it have rotational symmetry around its centre? If so, find +the angles of rotational symmetry. + 8. How many lines of symmetry do the shapes in the first shape +sequence in Chapter 1, Table 3, the Regular Polygons, have? What +number sequence do you get? + 9. How many angles of symmetry do the shapes in the first shape +sequence in Chapter 1, Table 3, the Regular Polygons, have? What +number sequence do you get? +10. How many lines of symmetry do the shapes in the last shape +sequence in Chapter 1, Table 3, the Koch Snowflake sequence, +have? How many angles of symmetry? +11. How many lines of symmetry and angles of +symmetry does Ashoka Chakra have? +Playing with Tiles +a. Use the colour tiles + given at the end of the +book to complete the following figure so that it has exactly 2 lines +of symmetry. +b. Use 16 such tiles to make figures that have exactly: + +1 line of symmetry + +2 lines of symmetry +c. Use these tiles in making creative symmetric designs. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +240 +Reprint 2025-26 + +Symmetry +241 +  Game +Draw a 6 by 6 grid. Two players +take +turns +covering +two +adjacent squares by drawing +a line. The line can be placed +either way: horizontally or +vertically. The lines cannot +overlap. The game goes on till +a player is not able to place any +more lines. The player who is +not able to place a line loses. +With what strategy can one play to win this game? +Sum m a r y + + +When a figure is made up of parts that repeat in a definite pattern, +we say that the figure has symmetry. We say that such a figure is +symmetrical. + + +A line that cuts a plane figure into two parts that exactly overlap +when folded along that line is called a line of symmetry or axis of +symmetry of the figure. + + +A figure may have multiple lines of symmetry. + + +Sometimes a figure looks exactly the same when it is rotated by an +angle about a fixed point. Such an angle is called an angle of symmetry +of the figure. A figure that has an angle of symmetry strictly between +0 and 360 degrees is said to have rotational symmetry. The point +of the figure about which the rotation occurs is called the centre of +rotation. + + +A figure may have multiple angles of symmetry. + + +Some figures may have a line of symmetry but no angle of symmetry, +while others may have angles of symmetry but no lines of symmetry. Some +figures may have both lines of symmetry as well as angles of symmetry. +Not allowed +Reprint 2025-26 + +[1] + +CHAPTER 9 — SOLUTIONS +Symmetry +Section 9.1 +Page No. 219 + Figure it Out +Q1. Do you see any line of symmetry in the figures at the start of the chapter? What +about in the picture of the cloud? +Ans. Yes, there are 6, 4 and 1 lines of symmetry in the figures of flower, rangoli and butterfly +respectively. There is no line of symmetry in the figures of pinwheel and cloud. +Q2. For each of the following figures, identify the line(s) of symmetry if it exists. +Ans. + +Section 9.1 +Page no. 221 +Q. Is there any other way to fold the square so that the two halves overlap? How many +lines of symmetry does the square shape have? +Ans. No, there is no other way to fold the square. + The square shape has 4 lines of symmetry. +Q. We saw that the diagonal of a square is also a line of symmetry. Let us take a rectangle +that is not a square. Is its diagonal a line of symmetry? +Ans. No, Rectangle’s diagonal is not a line of symmetry. + +Page no. 222 +Q. What if we reflect along the diagonal from A to C? Where do points A, B, C and D go? +What if we reflect along the horizontal line of symmetry? +Ans. If we reflect along the diagonal from A to C, D occupies the position occupied by B +earlier. A and C remain at the same place. +If we reflect along the horizontal line of symmetry, D and C occupies the position earlier +occupied by A and B respectively. + +[2] + + +Section 9.1 +Page no. 223 +Q1 In each of the following figures, a hole was punched in a folded square sheet of paper +and then the paper was unfolded. Identify the line along which the paper was folded. +Figure (d) was created by punching a single hole. How was the paper folded? +Ans. + + +For figure (d), paper was folded vertically and then horizontally or vice versa. +Q2. Given the line(s) of symmetry, find the other hole(s): +Ans. + + +Q4. After each of the following cuts, predict the shape of the hole when the paper is +opened. After you have made your prediction, make the cutouts and verify your +answer. + + +[3] + + +Q5. Suppose you have to get each of these shapes with some folds and a single straight +cut. How will you do it? +a. The hole in the centre is square. +b. The hole in the centre is a square. +Note : For the above two questions, check if the 4-sided figures in the centre satisfy both the +properties of a square. +Ans. +a. At first fold the paper horizontally then vertically. + Now cut a small square at the center (all sides closed corner). + + + +b. At first fold the paper horizontally then vertically. + +Now at closed corner, cut along with a slanting line. + + + + + +[4] + +Q6. How many lines of symmetry do these shapes have? +Ans. a. + b. A triangle with equal sides and equal angles. + + c. A hexagon with equal sides and equal angles. + +Q7. Trace each figure and draw the lines of symmetry, if any: + + +[5] + +Ans. + + +Q8. Find the lines of symmetry for the kolam below. +Ans. + +Q9. Draw the following. +a. A triangle with exactly one line of symmetry +b. A triangle with exactly three lines of symmetry +c. A triangle with no line of symmetry +Is it possible to draw a triangle with exactly two lines of symmetry? + +[6] + +Ans. +a. + +b. + +c. + +No, it is not possible to draw a triangle with exactly two lines of symmetry. + +Q10. Draw the following. In each case, the figure should contain at least one curved +boundary. +a. A figure with exactly one line of Symmetry +b. A figure with exactly two lines of symmetry +c. A figure with exactly four lines of symmetry +Ans. + + + + + +[7] + +Q11. Copy the following on squared paper. Complete them so that the blue line is a line +of symmetry. Problem (a) has been done for you. +Hint: For (c) and (f), see if rotating the book helps! +Ans. + + + + + + + +[8] + +Q12. Copy the following drawing on squared paper. Complete each one of them so that +the resulting figure has the two blue lines as lines of symmetry. +Ans. + + +Q13. Copy the following on a dot grid. For each figure draw two more lines to make a +shape that has a line of symmetry. +Ans. + +[9] + +Section 9.2 +Page no. 235 + Can you draw a figure with radial arms that has a) exactly 5 angles of symmetry, b) +6 angles of symmetry? Also find the angles of symmetry in each case. +Hint: Use 5 radial arms for the first case. What should the angle between two adjacent radial +arms be? +Ans. + + +Angles of symmetry = 72°, 144°, 216°, 288°, 360°. + +Angles of symmetry = 60°, 120°, 180°, 240°, 300°, 360°. + + + +[10] + + Consider a figure with radial arms having exactly 7 angles of symmetry. What will be +its smallest angle of symmetry? Is the number of degrees a whole number in this +case? If not, express it as a mixed faction. +Let us find the angles of symmetry for other kinds of figures. +Ans. +360° ÷ 7 = 51 +3 +7 +No, its smallest angle of symmetry is not a whole number. + +Page no. 235 +Figure it out +Q1. Find the angles of symmetry for the given figures about the point marked. +Ans. +a. Angles of symmetry = 90°, 180°, 270°, 360° + +b. Angle of symmetry = 360° + +c. Angles of symmetry = 180°, 360° + +Q2. Which of the following figures have more than one angle of symmetry? +Ans. +The following figures have more than one angle of symmetry– + + + + + + +[11] + +Q3. Give the order of rotational symmetry for each figure: +Ans. +Orders of rotational symmetry + + +Page no. 236 + In each case, the angles are the multiples of the smallest angle. You may wonder and +ask if this will always happen. What do you think? +Ans. Yes, the angles of symmetry are always the multiples of the smallest angle. For example, +the second angle is twice the amount of rotation of the first rotation. + + True or False +• Every figure will have 360 degrees as an angle of symmetry. +• If the smallest angle of symmetry of a figure is a natural number in degrees, then +it is a factor of 360. +Ans. +True + + +True + + + +[12] + +Section 9.2 +Page no. 238 +Figure it out +Q1. Color the sectors of the circle below so that the figure has i) 3 angles of symmetry, ii) +4 angles of symmetry, iii) what are the possible numbers of angles of symmetry you +can obtain by coloring the sectors in different ways? +Ans. + +(i) +3 Angles of symmetry +(ii) +4 angles of symmetry +(iii) +12 angles of symmetry are possible to obtain. + +Q2. Draw two figures other than a circle and a square that have both reflection symmetry +and rotational symmetry. +Ans. + + +Numbers of line symmetry = 4 + +Order of rotational symmetry = 4 + + + + + + +Numbers of line symmetry = 4 +Order of rotational symmetry = 4 + + + + + +[13] + +Q3. Draw, wherever possible, a rough sketch of +a. A triangle with at least two lines of symmetry and at least two angles of symmetry. +b. A triangle with only one line of symmetry but not having rotational symmetry. +c. A quadrilateral with rotational symmetry but no reflection symmetry. +d. A quadrilateral with reflection symmetry but not having rotational symmetry. +Ans. +a. + +3 lines of symmetry + +3 angles of symmetry + +b. + 1 line of symmetry + No rotational symmetry + +c. + + No line of symmetry + 2 angles (180°, 360°) + + + + + +[14] + +d. A quadrilateral with reflection symmetry but not having rotational symmetry + + +Q4. In a figure, 60° is the smallest angle of symmetry. What are the other angles of +symmetry of the figure? +Ans. +Other angles of symmetry = 120°, 180°, 240°, 300°, 360°. + +Q5. In the figure, 60° in an angle of symmetry. The figure has two angles of symmetry +less then 60°. What is its smallest angle of symmetry? +Ans. +The smallest angle of symmetry = 20° + +Q6. Can we have a figure with rotational symmetry whose smallest angle of symmetry is +a. 45°? +b. 17°? +Ans. +a. yes, as 360° is a multiple of 45°. + +b. No, as 360° is not a multiple of 17°. + +Q7. This is a picture of the new Parliament Building in Delhi. +a. Does the outer boundary of the picture have reflection symmetry? If so, draw the +lines of symmetries. How many are they? +b. Does it have rotational symmetry around its centre? If so, find the angles of +rotational symmetry. +Ans. +a. Yes, the outer boundary of the picture has 3 lines of symmetry. + +b. Yes, the outer boundary has rotational symmetry. The angles of rotational symmetry + +are 120°, 240°, 360°. + + +[15] + +Q8. How many lines of symmetry do the shapes in the first shape sequence in Chapter 1, +Table 3, the Regular Polygons, have? What number sequence do you get? +Ans. + + + + + + + +It is a counting number sequence. +Q10. How many lines of symmetry do the shapes in the last shape sequence in Chapter 1, +Table 3, the Koch Snowflake sequence, have? How many angles of symmetry? +Ans. +Number of lines of symmetry : 3, 6, 6,6,6 + + Angles of symmetry : 3, 6, 6,6,6 + +Q11. How many lines of symmetry and angles of symmetry does Ashoka Chakra have? +Ans. +Number of lines of symmetry = 24 + +Number of angles of symmetry = 24 + +Regular Polygon +No. of line symmetry +Triangle +Quadrilateral +Pentagon +Hexagon +Heptagon +Octagon +Nonagon +Decagon +3 +4 +5 +6 +7 +8 +9 +10" +class_6,10,The Other Side of Zero,ncert_books/class_6/Ganita_Prakash/fegp110.pdf,"Integers +THE OTHER SIDE OF +ZERO +10 +More and more numbers! +Recall that the very first numbers we learned about in the study of +mathematics were the counting numbers 1, 2, 3, 4, … +Then we learned that there are even more numbers! For example, +there is the number 0 (zero), representing nothing, which comes before +1. The number 0 has a very important history in India and now in the +world. For example, around the world we learn to write numbers in +the Indian number system using the digits 0 to 9, allowing us to write +numbers however large or however small using just these 10 digits. +We then learned about more numbers that exist between the +numbers 0, 1, 2, 3, 4, … , such as +1 +2 , +3 +2 , and 13 +6 . These are called +fractions. +But are there still more numbers? Well, 0 is an additional number +that we didn’t know about earlier, and it comes before 1 and is less +than 1. Are there perhaps more numbers that come before 0 and are +less than 0? +Phrased another way, we have seen the number line: +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +However, this is actually only a number ‘ray’, in the language we +learned earlier in geometry; this ray starts at 0 and goes forever to +the right. Do there exist numbers to the left of 0, so that this number +ray can be completed to a true number line? +That is what we will investigate in this chapter! +Reprint 2025-26 + +The Other Side of Zero +243 + Can there be a number less than 0? Can you think of any ways to +have less than 0 of something? +10.1 Bela’s Building of Fun +Children flock to Bela’s ice cream factory to see +and taste her tasty ice cream. To make it even +more fun for them, Bela purchased a multi- +storied building and filled it with attractions. +She named it Bela’s Building of Fun. +But this was no ordinary building! +Observe that some of the floors in the +‘Building of Fun’ are below the ground. What are the shops that you +find on these floors? What is there on the ground floor? +A lift is used to go up and down between the floors. It has two +buttons: ‘+’ to go up and ‘–’ to go down. Can you spot the lift? +To go to the Art Centre from the ‘Welcome Hall’, you +must press the ‘+’ button twice. +We say that the button press is + + or + 2. +To go down two floors, you must press the ‘–’ +button twice, which we write as – – or – 2. +So if you press + 1 (i.e., if you press the ‘+’ +button once), then you will go up one floor and if +you press – 1 (i.e., if you press the ‘–’ button +once), then you will go down 1 floor. +Lift button presses and numbers: ++++ is written as + 3 + – – – – is written as –  4 + What do you press to go four floors up? +What do you press to go three floors down? +Reprint 2025-26 + +Ganita Prakash | Grade 6 +244 +Numbering the floors in the building of fun +Entry to the ‘Building of Fun’ is at the ground floor level and is called +the ‘Welcome Hall’. Starting from the ground floor, you can reach the +Food Court by pressing + 1 and can reach the Art Centre by pressing ++ 2. So, we can say that the Food Court is on Floor + 1 and that the Art +Centre is on Floor + 2. +Starting from the ground floor, you must press – 1 to reach the Toy +Store. So, the Toy Store is on Floor – 1 similarly starting from the ground +floor, you must press – 2 to reach the Video Games shop. So, the Video +Games shop is on Floor – 2. +The ground floor is called Floor 0. Can you see why? +  Number all the floors in the Building of Fun. +Did you notice that + 3 is the floor number of the Book +Store, but it is also the number of floors you move +when you press + 3? Similarly, – 3 is the floor number +but it is also the number of floors you go down when +you press – 3, i.e., when you press – – – . +A number with a ‘+’ sign in front is called a positive +number. A number with a ‘–’ sign in front is called a +negative number. +In the ‘Building of Fun’, the floors are numbered +using the ground floor, Floor 0, as a reference or starting +point. The floors above the ground floor are numbered +with positive numbers. To get to them from the ground +floor, one must press the ‘+’ button some number of +times. The floors below the ground are numbered with +negative numbers. To get to them from the ground floor, +one must press the ‘–’ button some number of times. +Zero is neither a positive nor a negative number. +We do not put a ‘+’ or ‘–’ sign in front of it. +Reprint 2025-26 + +The Other Side of Zero +245 +Addition to keep track of movement +Start from the Food Court and press + 2 in the lift. Where will you +reach? ____________ +We can describe this using an expression: +Starting Floor + Movement = Target Floor. +The starting floor is + 1 (Food Court) and the number of button +presses is + 2. Therefore, you reach the target floor (+ 1) + (+ 2) = + 3 +(Book Store). + Figure it Out +1. You start from Floor + 2 and press – 3 in the lift. Where will you +reach? Write an expression for this movement. +2. Evaluate these expressions (you may think of them as Starting +Floor + Movement by referring to the Building of Fun). +a. (+ 1) + (+ 4) = _______ +b. (+ 4) + (+ 1) = _______ +c. (+ 4) + (–  3) = _______ +d. (– 1) + (+ 2) = _______ +e. (– 1) + (+ 1) = _______ +f. 0 + (+ 2) = _________ +g. 0 + (– 2) = _________ +3. +Starting from different floors, find the movements required to +reach Floor – 5. For example, if I start at Floor + 2, I must press – 7 +to reach Floor – 5. The expression is (+ 2) + (– 7) = – 5. + +Find more such starting positions and the movements needed to +reach Floor – 5 and write the expressions. +Combining button presses is also addition +Gurmit was in the Toy Store and wanted to go down two floors. +But by mistake he pressed the ‘+’ button two times. He realised his +mistake and quickly pressed the ‘–’ button three times. How many +floors below or above the Toy Store will Gurmit reach? +Gurmit will go one floor down. We can show the movement +resulting from combining button presses as an expression: +(+ 2) + (– 3) = – 1. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +246 + Figure it out +Evaluate these expressions by thinking of them as the resulting movement +of combining button presses: +a. (+ 1) + (+ 4) = _____________ +b. (+ 4) + (+ 1) = _____________ +c. (+ 4) + (–  3) + (– 2) = _______ +d. (– 1) + (+ 2) + (– 3) = _______ +Back to zero! +On the ground floor, Basant is in a great hurry and by +mistake he presses +3. What can he do to cancel it and +stay on the ground floor? He can cancel it by pressing +– 3. That is, (+3) + (– 3) = 0. +We call – 3 the inverse of +3. Similarly, the inverse of +– 3 is +3. +If Basant now presses +4 and then presses – 4 in the +lift, where will he reach? +Here is another way to think of the concept of +inverse. If you are at Floor +4 and you press its inverse +– 4, then you are back to zero, the ground floor! If you +are at Floor – 2 and press its inverse +2, then you go to +(– 2) + (+2) = 0, again the ground floor! +  Write the inverses of these numbers: + + + ++4, –4, –3, 0, +2, –1. +  Connect the inverses by drawing lines. ++9 ++7 +–8 +–5 +–7 ++8 ++5 +–9 +Comparing numbers using floors + Who is on the lowest floor? +1. Jay is in the Art Centre. So, he is on Floor +2. +2. Asin is in the Sports Centre. So, she is on Floor ___. +3. Binnu is in the Cinema Centre. So, she is on Floor ____. +4. Aman is in the Toys Store. So, he is on Floor ____. +Reprint 2025-26 + +The Other Side of Zero +247 +Floor +3 is lower than Floor +4. So, we write +3 < +4. We +also write +4 > +3. +  Should we write –3 < – 4 or – 4 < – 3? +Floor – 4 is lower than Floor – 3. So, – 4 < – 3. It is also +correct to write – 3 > – 4 + Figure it Out +1. Compare the following numbers using the Building of +Fun and fill in the boxes with < or >. +a. – 2 + +5 +b. – 5 + + 4 +c. – 5 + – 3 +d. + 6 + – 6 +e. 0 + – 4 +f. 0 + + 4 +Notice that all negative number floors are below +Floor 0. So, all negative numbers are less than 0. All +the positive number floors are above Floor 0. So, all +positive numbers are greater than 0. +2. Imagine the Building of Fun with more floors. Compare +the numbers and fill in the boxes with < or >: +a. – 10 + – 12 +b. + 17 + – 10 +c. 0 + – 20 +d. + 9 + – 9 +e. – 25 + – 7 +f. + 15 + – 17 +3. If Floor A = – 12, Floor D = – 1 and Floor E = + 1 in the +building shown on the right as a line, find the numbers +of Floors B, C, F, G, and H. +4. Mark the following floors of the building shown on +the right. +a. – 7 +b. –  4 +c. + 3 +d. –  10 +Subtraction to find which button to press +In earlier classes, we understood the meaning of subtraction as ‘take +away’. For example, there are 10 books on the shelf. I take away 4 +books. How many are left on the shelf? +We can express the answer using subtraction: 10 – 4 = 6 or ‘Ten +take away four is six.’ +0 ++1 +E +F +G +H +D +C +B +A +–1 +–12 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +248 +You may also be familiar with another meaning of subtraction which +is related to comparison or making quantities equal. For example, +consider this situation: I have `10 with me and my sister has `6. +Now, I can ask the question: ʻHow much more money should my +sister get in order to have the same amount as me?ʼ +We can write this in two ways: 6 + ? = 10 Or 10 – 6 = ? +Here, we see the connection between ‘finding the missing number +to be added’ and subtraction. +For subtraction of positive and negative numbers, we will use +this meaning of subtraction as ‘making equal’ or ‘finding the missing +number to be added’. +  Evaluate 15 – 5, 100 – 10 and 74 – 34 from this perspective. +In general, when there are two unequal quantities, subtraction can +indicate the change needed to make the quantities equal. Subtraction +shows how much the starting quantity should change in order to +become the target quantity. In the context of different floor levels, +what is the change required to reach the Target Floor from the Starting +Floor? Notice that the change needed may be positive (for an increase) +or negative (for a decrease). +Teacher’s Note +Your starting floor is the Art Centre and your target floor is the +Sports Centre. What should be your button press? +You need to go three floors up, so you should press + 3. We can +write this as an expression using subtraction: +Target Floor – Starting Floor = Movement needed. +In the above example, the starting floor is + 2 (Art Centre) and +the target floor is + 5. The button press to get to + 5 from + 2 is + 3. +Therefore, +(+ 5) – (+ 2) = + 3 +Reprint 2025-26 + +The Other Side of Zero +249 +Explanation +Recall the connection between addition and subtraction. For 3 + ? = 5, +we can find the missing number using subtraction: 5 – 3 = 2. That is, +subtraction is the same as finding the missing number to be added. +We know that — +Starting Floor + Movement needed = Target Floor +If the movement needed is to be found, then, +Starting Floor + ? = Target Floor +So, +Target Floor – Starting Floor = ? = Movement needed +More examples: +a. If the Target Floor is – 1 and Starting Floor is – 2, what button +should you press? + +You need to go one floor up, so, you should press + 1. + +Expression: (– 1) – (– 2) = (+1) +b. If the Target Floor is – 1 and Starting floor is +3, what button +should you press? + +You need to go four floors down, so, you should press – 4. + +Expression: (– 1) – (+ 3) = (– 4) +c. If the Target Floor is +2 and Starting Floor is – 2, what button +should you press? + +You need to go four floors up, so, you should press +4. + +Expression: (+ 2) – (– 2) = (+ 4) + Figure it Out + +Complete these expressions. You may think of them as finding the +movement needed to reach the Target Floor from the Starting Floor. +a. (+ 1) – (+ 4) = _______ +b. (0) – (+ 2) = _________ +c. (+ 4) – (+ 1) = _______ +d. (0) – (– 2) = _________ +e. (+ 4) – (– 3) = _______ +f. (– 4) – (– 3) = ________ +g. (– 1) – (+ 2) = _______ +h. (– 2) – (– 2) = ________ +i. +(– 1) – (+1) = _______ +j. (+ 3) – (– 3) = ________ +Reprint 2025-26 + +Ganita Prakash | Grade 6 +250 +Adding and subtracting larger numbers +The picture shows a mine, a place where +minerals are extracted by digging into the +rock. The truck is at the ground level, but +the minerals are present both above and +below the ground level. There is a fast +moving lift which moves up and down in +a mineshaft carrying people and ore. +Some of the levels are marked in the +picture. The ground level is marked 0. +Levels above the ground are marked by +positive numbers and levels below the +ground are marked by negative numbers. +The number indicates how many metres +above or below the ground level it is. +In the mine, just like in the Building of Fun: +Starting Level + Movement = Target Level +For example, +(+ 40) + (+ 60) = + 100 +(– 90) + (– 55) = – 145 +Target Level – Starting Level = Movement needed +For example, +(+ 40) – (– 50) = + 90 + +(– 90) – (+ 40) = – 130 +How many negative numbers are there? +Bela’s Building of Fun had only six floors above and five floors below. +That is numbers – 5 to + 6. In the mine above, we have numbers from +– 200 to + 180. But we can imagine larger buildings or mineshafts. +Just as positive numbers + 1, + 2, + 3, ... keep going up without an end, +similarly, negative numbers – 1, – 2, – 3, ... keep going down. Positive +and negative numbers, with zero, are called integers. They go both +ways from 0: … – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, … +Reprint 2025-26 + +The Other Side of Zero +251 + Figure it Out +Complete these expressions. +a. (+ 40) + ______ = + 200 +b. (+ 40) + _______ = – 200 +c. (– 50) + ______ = + 200 +d. (– 50) + _______ = – 200 +e. (– 200) – (– 40) = _______ +f. (+ 200) – (+ 40) = _______ +g. (– 200) – (+ 40) = _______ +Check your answers by thinking about the movement in the mineshaft. +Adding, subtracting, and comparing any numbers +To add and subtract even larger integers, we can imagine even larger +lifts! In fact, we can imagine a lift that can extend forever upwards +and forever downwards, starting from Level 0. There does not even +have to be any building or mine around — just an ‘infinite lift’! +We can use this imagination to add and subtract any integers we like. +For example, suppose we want to carry out the subtraction + 2000 +– (– 200). We can imagine a lift with 2000 levels above the ground +and 200 below the ground. Recall that, +Target Level – Starting Level = Movement needed +To go from the Starting Floor – 200 to the Target Floor + 2000, we +must press + 2200 (+ 200 to get to zero, and then + 2000 more after +that to get to + 2200). Therefore, (+ 2000) – (– 200) = + 2200. +Notice that (+ 2000) + (+ 200) is also + 2200. + Try evaluating the following expressions by similarly drawing or +imagining a suitable lift: +a. – 125 + (– 30) +b. + 105 – (– 55) +c. + 105 + (+ 55) +d. + 80 – (– 150) +e. + 80 + (+ 150) +f. – 99 – (– 200) +g. – 99 + (+ 200) +h. + 1500 – (– 1500) ++1000 ++900 ++800 ++700 ++600 ++500 ++400 ++300 ++200 ++100 +– 100 +– 200 +– 300 +– 400 +– 500 +– 600 +– 700 +– 800 +– 900 +– 1000 +0 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +252 +In the above example, we saw that + 2000 – (– 200) = + 2000 + (+ 200) += + 2200. In other words, subtracting a negative number is the same +as adding the corresponding positive number. That is, we can +replace subtraction of a negative number by addition of a positive +number! + In the other exercises that you did above, did you notice +that subtracting a negative number was the same as adding +the corresponding positive number? +Take a look at the ‘infinite lift’ above. Does it remind you +of a number line? In what ways? +Back to the number line +The ‘infinite lift’ we saw above looked very much like a number line, +didn’t it? In fact, if we rotate it by 90°, it basically becomes a number +line. It also tells us how to complete the number ray to a number line, +answering the question that we had asked at the beginning of the +chapter. To the left of 0 are the negative numbers – 1, – 2, – 3, … +Usually we drop the ‘+’ sign on positive numbers and simply write +them as 1, 2, 3, … +– 10 – 9 +– 8 +– 7 +– 6 +– 5 +– 4 +– 3 +– 2 +– 1 +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +Instead of traveling along the number line using a lift, we can +simply imagine walking on it. To the right is the positive (forward) +direction, and to the left is the negative (backward) direction. +Smaller numbers are now to the left of bigger numbers and +bigger numbers are to the right of smaller numbers. So, 2 < 5; +– 3 < 2; and – 5 < – 3. + If, from 5 you wish to go over to 9, how far must you travel along +the number line? +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +0 +Math +Talk +Reprint 2025-26 + +The Other Side of Zero +253 +You must travel 4 steps. That is why 5 + 4 = 9. +(Remember: Starting Number + Movement = Target Number) +The corresponding subtraction statement is 9 – 5 = 4. +(Remember: Target Number – Starting Number = Movement +needed) + Now, from 9, if you wish to go to 3, how much must you travel +along the number line? +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +You must move 6 steps backward, i.e., you must move –6. Hence, +we write 9 + (–6) = 3. +(Remember again : Starting Number + Movement = Target +Number) +The corresponding subtraction statement is 3 – 9 = – 6. +(Remember again: Target Number – Starting Number = Movement +needed) + Now, from 3, if you wish to go to – 2, how far must you travel? +– 10 – 9 +– 8 +– 7 +– 6 +– 5 +– 4 +– 3 +– 2 +– 1 +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +You must travel – 5 steps, i.e., 5 steps backward. Thus, 3 + (– 5) = – 2. +The corresponding subtraction statement is: – 2 – 3 = – 5. + Figure it Out +– 10 – 9 +– 8 +– 7 +– 6 +– 5 +– 4 +– 3 +– 2 +– 1 +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +1. Mark 3 positive numbers and 3 negative numbers on the number +line above. +2. Write down the above 3 marked negative numbers in the following +boxes: + + +Reprint 2025-26 + +Ganita Prakash | Grade 6 +254 +3. Is 2 > – 3? Why? Is – 2 < 3? Why? +4. What are a.  – 5 + 0 b.  7 + (– 7) c. – 10 + 20 d. 10 – 20 e. 7 – (– 7) +f.  – 8 – (– 10)? +Using the unmarked number line to add and subtract +Just as you can do additions, subtractions and comparisons with small +numbers using the number line above, you can also do them with +large numbers by imagining an ‘infinite number line’ or drawing an +‘unmarked number line’ as follows: +0 +This line shows only the position of zero. Other numbers are not +marked. It can be convenient to use this unmarked number line to +add and subtract integers. You can show, or simply imagine, the scale +of the number line and the positions of numbers on it. +For example, this unmarked number line (UNL) shows the +addition problem: 85 + (– 60) = ? + ++85 ++25 +– 60 +0 +We then can visualise that 85 + (– 60) = 25 +The following UNL shows a subtraction problem which can also be +written as a missing addend problem: (– 100) – (+ 250) = ? or 250 + ? = – 100. ++ 250 +– 100 +0 +? +We can then visualise that ? = –350 in this problem. +In this way, you can carry out addition and subtraction problems, +with positive and negative numbers, on paper or in your head using an +unmarked number line. +Reprint 2025-26 + +The Other Side of Zero +255 +  Use unmarked number lines to evaluate these expressions: +a. – 125 + (– 30) = _______ +b. + 105 –  (– 55) = _______ +c. + 80 – (– 150) = _______ +d. – 99 – (– 200) = _______ +Converting subtraction to addition and addition to subtraction +Recall that Target Floor – Starting Floor = Movement needed +or +Target Floor = Starting Floor + Movement needed +If we start at 2 and wish to go to – 3, what is the movement needed? +First method: Looking at the number line, we see we need to +move – 5 (i.e., 5 in the backward direction). Therefore, – 3 – 2 = – 5. +The movement needed is –5. +Second method: Break the journey from 2 to –3 into two parts. +a. From 2 to 0, the movement is 0 – 2 = – 2. +b. From 0 to –3, the movement is – 3 – 0 = – 3. +The total movement is the sum of the two movements: – 3 + (– 2) = – 5. +Look at the two coloured expressions. There is no subtraction in +the second one! +In this way, we can always convert subtraction to addition. The +number that is being subtracted can be replaced by its inverse +and then added instead. +Similarly, a number that is being added can be replaced by +its inverse and then subtracted. In this way, we can also always +convert addition to subtraction. +Examples: +a. (+ 7) – (+ 5) = (+ 7) + (– 5) +b. (– 3) – (+ 8) = (– 3) + (– 8) +c. (+ 8) – (– 2) = (+ 8) + (+ 2) +d. (+ 6) – (– 9) = (+ 6) + (+ 9) +– 3 +0 +2 +– 5 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +256 +10.2 The Token Model +Using tokens for addition +In Bela’s Building of Fun, the lift attendant is bored. To entertain +himself, he keeps a box containing lots of positive (red) and negative +(green) tokens. Each time he presses the ‘+’ button, he takes a +positive token from the box and puts it in his pocket. Similarly, +each time he presses the ‘–’ button, he takes a negative token and +puts it in his pocket. +He starts on the ground floor (Floor 0) with an empty pocket. After +one hour, he checks his pocket and finds 5 positive and 3 negative +tokens. On which floor is he now? +He must have pressed ‘+’ five times and ‘–’ 3 times and (+ 5)+(– 3)= + 2. +So, he is at Floor + 2 now. +Here is another way to do the calculation. +A positive token and a negative token cancel each other, because +the value of this pair of tokens together is zero. These two tokens in +his pocket meant that he pressed ‘+’ once and ‘–’ once, respectively, +and these cancel each other. We say that a positive and a negative +token make a ʻzero pairʼ. When you remove all the zero pairs, you are +left with two positive tokens, so (+5) + (–3) = +2. +We can perform any such addition using tokens! +Example: Add + 5 and – 8. +Reprint 2025-26 + +The Other Side of Zero +257 +From the picture, we see that we can remove five zero pairs, and we +are then left with – 3. Therefore (+ 5) + (– 8) = – 3. + Figure it Out +1. +Complete the additions using tokens. + +a. (+ 6) + (+ 4) +b. (– 3) + (– 2) + +c. (+ 5) + (– 7) +d. (– 2) + (+ 6) +2. Cancel the zero pairs in the following two sets of tokens. On what +floor is the lift attendant in each case? What is the corresponding +addition statement in each case? +a. +b. +Using tokens for subtraction +We have seen how to perform addition of integers with positive +tokens and negative tokens. We can also perform subtraction using +tokens! +Example: Let us subtract: +(+5) – (+4). +This is easy to do. From 5 positives +take away 4 positives to see the result. +Example: Let us subtract: +(–7) – (–5). +Is (– 7) – (– 5) the same as (– 7) + (+ 5)? +Example: Let us subtract: (+ 5) – (+ 6). +Put down 5 positives. +But there are not enough tokens to take out 6 positives! +(+ 5) – (+ 4) = + 1 +(–7) – (–5) = –2 +Reprint 2025-26 + +Ganita Prakash | Grade 6 +258 +To get around this issue, we can put out an extra zero pair (a +positive and a negative), knowing that this does not change the value +of the set of tokens. +Now, we can take out 6 positives! +See what is left: +We conclude that (+5) – (+6) = – 1. + Figure it Out +1. +Evaluate the following differences using tokens. Check that you +get the same result as with other methods you now know: +a. (+ 10) – (+ 7) +b. (– 8) – (– 4) +c. (– 9) – (– 4) +d. (+ 9) – (+ 12) +e. (– 5) – (– 7) +f. (– 2) – (– 6) +2. Complete the subtractions: +a. (– 5) – (– 7) +b. (+ 10) – (+ 13) +c. (– 7) – (– 9) +d. (+ 3) – (+ 8) +e. (– 2) – (– 7) +f. (+ 3) – (+ 15) +Example: + 4 –  (– 6). +Start with 4 positives. +We have to take out 6 negatives from these. But there are not +enough negatives. +This is not a problem. We add some zero pairs as this does not +change the value of the set of tokens. +But how many zero pairs? We have to take away 6 negatives so +we put down 6 zero pairs: +Now we can take away 6 negatives: + +Therefore, + 4 – (– 6) = + 10. +Reprint 2025-26 + +The Other Side of Zero +259 + Figure it Out +1. +Try to subtract: – 3 – (+ 5). + +How many zero pairs will you have to put in? What is the result? +2. Evaluate the following using tokens. +a. (– 3) – (+ 10) +b. (+ 8) – (– 7) +c. (– 5) – (+ 9) +d. (– 9) – (+ 10) +e. (+ 6) – (– 4) +f. (– 2) – (+ 7) +10.3 Integers in Other Places +Credits and debits +Suppose you open a bank account at your local bank with the `100 +that you had been saving over the last month. Your bank balance +therefore, starts at `100. +Then you make `60 at your job the next day and you deposit it in +your account. This is shown in your bank passbook as a ‘credit’. + Your new bank balance is _______. + +The next day you pay your electric bill of `30 using your bank +account. This is shown in your bank passbook as a ‘debit’. + Your bank balance is now ______. + +The next day you make a major purchase for your business of +`150. Again this is shown as a debit. + What is your bank balance now? ______ + +Is this possible? +Yes, some banks do allow your account balance to become negative, +temporarily! Some banks also charge you an additional amount if +your balance becomes negative, in the form of ‘interest’ or a ‘fee’. +Your strategic large purchase the previous day allows you to +make, `200 at your business the next day. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +260 + What is your balance now? ______ +You can think of ‘credits’ as positive numbers and ‘debits’ as negative +numbers. The total of all your credits (positive numbers) and debits +(negative numbers) is your total bank account balance. This can be +positive or negative! +In general, it is better to try to keep a positive balance in your +bank account! + Figure it Out +1. +Suppose you start with `0 in your bank account, and then you +have credits of `30, `40, and `50, and debits of `40, `50, and `60. +What is your bank account balance now? +2. Suppose you start with `0 in your bank account, and then you have +debits of `1, 2, 4, 8, 16, 32, 64, and 128, and then a single credit of +`256. What is your bank account balance now? +3. +Why is it generally better to try and maintain a positive balance in +your bank account? What are circumstances under which it may +be worthwhile to temporarily have a negative balance? +As you can see, positive and negative numbers along with zero are +extremely useful in the world of banking and accounting. +Geographical cross sections +We measure the height of geographical features like mountains, +plateaus, and deserts from ‘sea level’. The height at sea level is 0m. +Heights above sea level are represented using positive numbers and +heights below sea level are represented using negative numbers. +Reprint 2025-26 + +The Other Side of Zero +261 + Figure it Out +1. +Looking at the geographical cross section, fill in the respective +heights: + +a. + b. + c. + +d. + + +e. + f. + g. + +-1500 +-1000 +-500 +0 +500 +1000 +1500 +Height (m) +A +B +C +D +E +F +G +Sea +level +Ask what a geographical cross section is by showing the figure in this +page. It is like imagining a vertical slice taken out at some location on +the Earth. This is what would be seen from a side view. Discuss the +notion of ‘sea level’ for measuring heights and depths in geography. +Teacher’s Note +2. Which is the highest point in this geographical cross section? +Which is the lowest point? +3. +Can you write the points A, B, …, G in a sequence of decreasing order +of heights? Can you write the points in a sequence of increasing +order of heights? +4. What is the highest point above sea level on Earth? What is its +height? +5. +What is the lowest point with respect to sea level on land or on the +ocean floor? What is its height? (This height should be negative). +Reprint 2025-26 + +Ganita Prakash | Grade 6 +262 +Temperature +During summer time you would have heard in the news that there +is a ‘heat wave’. What do you think will be the temperature during +the summer when you feel very hot? In the winter we have cooler +or colder temperatures. +What has been the maximum temperature during the summer +and the minimum temperature during the winter last year in your +area? Find out. +When we measure temperature, we use Celsius as the unit of +measure (°C). The thermometers below are showing 40°C and 15°C +temperatures. + Figure it Out +1. +Do you know that there are some places +in India where temperatures can go +below 0°C? Find out the places in India +where +temperatures +sometimes +go +below 0°C. What is common among +these places? Why does it become colder +there and not in other places? +2. Leh in Ladakh gets very cold during +the winter. The following is a table of +temperature readings taken during +different times of the day and night in +Leh on a day in November. Match the +temperature with the appropriate time +of the day and night. +Temperature +Time +14°C +02:00 a.m. +8°C +11:00 p.m. +–2°C +02:00 p.m. +–4°C +11:00 a.m. +Math +Talk +0 +10 +–10 +–20 +20 +30 +40 +50 +60 +70 +80 +90 +˚C +100 +0 +10 +–10 +–20 +20 +30 +40 +50 +60 +70 +80 +90 +˚C +100 +Reprint 2025-26 + +The Other Side of Zero +263 +Talk about thermometers and how they are used to measure +temperature. Bring a laboratory thermometer to the class and +measure the temperature of hot water and cold water. Point out to +children that there are markings in the thermometer that are below +0°C. Have a discussion on what 0°C indicates, namely, the freezing +point of water. +Teacher’s Note +10.4 Explorations with Integers +A hollow integer grid +4 +– 1 +– 3 +– 3 +1 +– 1 +– 1 +2 +5 +– 3 +– 5 +0 +– 5 +–��8 +– 2 +7 +There is something special about the numbers in these two grids. +Let us explore what that is. +Top row: +Bottom row: +Left column: +Right column: + +4 + (– 1) + (– 3)  = 0 +(– 1) + (– 1) + 2    = 0 + +4 + (– 3) + (– 1)  = 0 +(– 3) + 1   + 2   = 0 + +5 + (– 3) + (– 5)  = ____ +(– 8) + (– 2) + 7   = ____ + +5 + 0   + (– 8)  = ____ +(– 5) + (– 5)  + 7   = ____ +In each grid, the numbers in each of the two rows (the top row +and the bottom row) and the numbers in each of the two columns +(the leftmost column and the rightmost column) add up to give the +same number. We shall call this sum as the ‘border sum’. The border +sum of the first grid is ‘0’. + Figure it Out +1. +Do the calculations for the second grid above and find the +border sum. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +264 +2. Complete the grids to make the required border sum: +– 10 +– 5 +9 +6 +8 +– 5 +– 2 +7 +– 5 +Border sum is +4 +Border sum is – 2 +Border sum is – 4 +3. For the last grid above, find more than one way of filling the +numbers to get border sum – 4. +4. Which other grids can be filled in multiple ways? What could be +the reason? +5. Make a border integer square puzzle and challenge your classmates. +An amazing grid of numbers! +Below is a grid having some numbers. Follow the steps as shown +until no number is left. +3 +4 +0 +9 +– 2 +– 1 +– 5 +4 +1 +2 +– 2 +7 +– 7 +– 6 +– 10 +– 1 +Strike out the row and column of the +chosen number +Circle any number +Circle any unstruck number +When there are no more unstruck numbers, STOP. Add the circled +numbers. +In the example below, the circled numbers are – 1, 9, – 7, – 2. If you +add them, you get – 1. +3 +4 +0 +9 +– 2 +– 1 +– 5 +4 +1 +2 +– 2 +7 +– 7 +– 6 +– 10 +– 1 +3 +4 +0 +9 +– 2 +– 1 +– 5 +4 +1 +2 +– 2 +7 +– 7 +– 6 +– 10 +– 1 +3 +4 +0 +9 +– 2 +– 1 +– 5 +4 +1 +2 +– 2 +7 +– 7 +– 6 +– 10 +– 1 +3 +4 +0 +9 +– 2 +– 1 +– 5 +4 +1 +2 +–  2 +7 +– 7 +– 6 +– 10 +– 1 +Reprint 2025-26 + +The Other Side of Zero +265 + Figure it Out +1. +Try afresh, choose different numbers this time. What sum did you +get? Was it different from the first time? Try a few more times! +2. Play the same game with the grids below. What answer did +you get? +7 +10 +13 +16 +– 2 +1 +4 +7 +– 11 +– 8 +– 5 +– 2 +– 20 +– 7 +– 14 +– 11 +– 11 +– 10 +– 9 +– 8 +– 7 +– 6 +– 5 +– 4 +– 3 +– 2 +– 1 +0 +1 +2 +3 +4 +3. +What could be so special about these grids? Is the +magic in the numbers or the way they are arranged or +both? Can you make more such grids? + Figure it Out + +1. +Write all the integers between the given pairs, in increasing order. +a. 0 and – 7 +b. – 4 and 4 +c. – 8 and – 15 +d. – 30 and – 23 +2.  Give three numbers such that their sum is – 8. +3.  There are two dice whose faces have these numbers: – 1, 2, – 3, +4, – 5, 6. The smallest possible sum upon rolling these dice is – 10 += (– 5) + (– 5) and the largest possible sum is 12 = (6) + (6). Some +numbers between (– 10) and (+ 12) are not possible to get by adding +numbers on these two dice. Find those numbers. +4. Solve these: +8 – 13 +(– 8) – (13) +(– 13) – (– 8) +(– 13) + (– 8) +8 + (– 13) +(– 8) – (– 13) +(13) – 8 +13 – (– 8) +5.  Find the years below. +a. From the present year, which year was it 150 years ago? ________ +b. From the present year, which year was it 2200 years ago? _______ +Hint: Recall that there was no year 0. +Try +This +Reprint 2025-26 + +Ganita Prakash | Grade 6 +266 +c. What will be the year 320 years after 680 BCE? ________ +6. +Complete the following sequences: +a. (– 40), (– 34), (– 28), (– 22), _____, ______, ______ +b. 3, 4, 2, 5, 1, 6, 0, 7, _____, _____, _____ +c. _____, ______, 12, 6, 1, (– 3), (– 6), _____, ______, ______ +7. +Here are six integer cards: (+ 1), (+ 7), (+ 18), (– 5), (– 2), (– 9). + +You can pick any of these and make an expression using addition(s) +and subtraction(s). + +Here is an expression: (+ 18) + (+ 1) – (+ 7) – (– 2) which gives a value +(+ 14). Now, pick cards and make an expression such that its value +is closer to (–  30). +8. The sum of two positive integers is always positive but a (positive +integer) – (positive integer) can be positive or negative. What about +a. (positive) – (negative) +b. (positive) + (negative) +c. (negative) + (negative) +d. (negative) – (negative) +e. (negative) – (positive) +f. (negative) + (positive) +9. +This string has a total of 100 tokens arranged in a particular +pattern. What is the value of the string? +10.5 A Pinch of History +Like general fractions, general integers (including zero and the +negative numbers) were first conceived of and used in Asia, thousands +of years ago, before they eventually spread across the world in more +modern times. +The first known instances of the use of negative numbers +occurred in the context of accounting. In one of China’s most +important mathematical works, The Nine Chapters on Mathematical +Art (Jiuzhang Suanshu)—which was completed by the first or second +century CE—positive and negative numbers were represented using +red and black rods, much like the way we represented them using +green and red tokens! +Reprint 2025-26 + +The Other Side of Zero +267 +There was a strong culture of accountancy also in India in ancient +times. The concept of credit and debit was written about extensively +by Kautilya in his Arthaśhāstra (c. 300 BCE), including the recognition +that an account balance could be negative. The explicit use of negative +numbers in the context of accounting is seen in a number of ancient +Indian works, including in the Bakśhālī manuscript from around the +year 300 CE, where a negative number was written using a special +symbol placed after the number (rather than before the number as +we do today). +The first general treatment of positive numbers, negative numbers, +and zero — all on an equal footing as equally-valid numbers on +which one can perform the basic operations of addition, subtraction, +multiplication, and even division — was given by Brahmagupta in +his Brāhma-sphuṭa-siddhānta in the year 628 CE. Brahmagupta gave +clear and explicit rules for operations on all numbers — positive, +negative, and zero — that essentially formed the modern way of +understanding these numbers that we still use today! +Some of Brahmagupta’s key rules for addition and subtraction of +positive numbers, negative numbers and zero are given below: +Brahmagupta’s Rules for Addition (Brāhma-sphuṭa-siddhānta +18.30, 628 CE) +1. +The sum of two positives is positive (e.g., 2 + 3 = 5). +2. The sum of two negatives is negative. To add two negatives, add +the numbers (without the signs), and then place a minus sign to +obtain the result (e.g., (– 2) + (– 3) = – 5). +3. +To add a positive number and a negative number, subtract the +smaller number (without the sign) from the greater number +(without the sign), and place the sign of the greater number to +obtain the result (e.g., –  5 + 3 = – 2, 2 + (– 3) = – 1 and – 3 + 5 = 2). +4. The sum of a number and its inverse is zero (e.g., 2 + (– 2) = 0). +5. +The sum of any number and zero is the same number (e.g., – 2 + 0 += – 2 and 0 + 0 = 0). +Reprint 2025-26 + +Ganita Prakash | Grade 6 +268 +Brahmagupta’s Rules for Subtraction (Brāhma-sphuṭa-siddhānta +18.31-18.32) +1. +If a smaller positive is subtracted from a larger positive, the result +is positive (e.g., 3 – 2 = 1). +2. If a larger positive is subtracted from a smaller positive, the result +is negative (e.g., 2– 3 = – 1). +3. +Subtracting a negative number is the same as adding the +corresponding positive number (e.g., 2 – (– 3) = 2 + 3). +4. Subtracting a number from itself gives zero (e.g., 2 – 2 = 0 and – 2 – +(– 2) = 0). +5. +Subtracting zero from a number gives the same number (e.g., – 2 +– 0 = – 2 and 0 –  0 = 0). Subtracting a number from zero gives the +number’s inverse (e.g., 0 –  (– 2) = 2). +Once you understand Brahmagupta’s rules, you can do addition +and subtraction with any numbers whatsoever — positive, negative, +and zero! + Figure it Out +1. +Can you explain each of Brahmagupta’s rules in terms of Bela’s +Building of Fun, or in terms of a number line? +2. Give your own examples of each rule. +Brahmagupta was the first to describe zero as a number on +an equal  footing with positive numbers as well as with negative +numbers, and the first to give explicit rules for performing arithmetic +operations on all such numbers, positive, negative, and zero—forming +what is now called a ring. It would change the way the world does +mathematics. +However, it took many centuries for the rest of the world to +adopt zero and negative numbers as numbers. These numbers +were transmitted to, accepted by, and further studied by the Arab +world by the 9th century, before making their way to Europe by the +13th century. +Reprint 2025-26 + +The Other Side of Zero +269 +Surprisingly, negative numbers were still not accepted by many +European mathematicians even in the 18th century. Lazare Carnot, a +French mathematician in the 18th century, called negative numbers +‘absurd’. But over time, zero as well as negative numbers proved +to be indispensable in global mathematics and science, and are +now considered to be critical numbers on an equal footing with +and as important as positive numbers — just as Brahmagupta had +recommended and explicitly described way back in the year 628 CE! +This abstraction of arithmetic rules on all numbers paved the way +for the modern development of algebra, which we will learn about +in future classes. +Su m m a r y + + +There are numbers that are less than zero. They are written with a ‘–’ +sign in front of them (e.g., – 2), and are called negative numbers. They +lie to the left of zero on the number line. + + +The numbers ..., – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, ... are called integers. The +numbers 1, 2, 3, 4, ... are called positive integers and the numbers …, +– 4, – 3, – 2, – 1 are called negative integers. Zero (0) is neither positive +nor negative. + + +Every given number has another number associated to it which when +added to the given number gives zero. This is called the additive +inverse of the number. For example, the additive inverse of 7 is – 7 and +the additive inverse of – 543 is 543. + + +Addition can be interpreted as Starting Position + Movement = Target +Position. + + +Addition can also be interpreted as the combination of movements or +increases/decreases: Movement 1 + Movement 2 = Total Movement. + + +Subtraction can be interpreted as Target Position – Starting Position += Movement. +Reprint 2025-26 + +Ganita Prakash | Grade 6 +270 + + +In general, we can add two numbers by following Brahmagupta’s Rules +for Addition: +a. If both numbers are positive, add the numbers and the result is a +positive number (e.g., 2 + 3 = 5). +b. If both numbers are negative, add the numbers (without the signs), +and then place a minus sign to obtain the result (– 2 + (– 3) = – 5). +c. If one number is positive and the other is negative, subtract the +smaller number (without the sign) from the greater number +(without the sign), and place the sign of the greater number to +obtain the result (e.g., – 5 + 3 = – 2). +d. A number plus its additive inverse is zero (e.g., 2 + (– 2) = 0). +e. A number plus zero gives back the same number (e.g., – 2 + 0 = – 2). + + +We can subtract two integers by converting the problem into an addition +problem and then following the rules of addition. Subtraction of an +integer is the same as the addition of its additive inverse. + + +Integers can be compared: … – 3 < – 2 < – 1 < 0 < +1 < +2 < +3 < ... Smaller +numbers are to the left of larger numbers on the number line. + + +We can give meaning to positive and negative numbers by interpreting +them as credits and debits. We can also interpret positive numbers +as distances above a reference point like the ground level. Similarly, +negative numbers can be interpreted as distances below the ground +level. When measuring temperatures in degrees Celsius, positive +temperatures are those above the freezing point of water, and negative +temperatures are those below the freezing point of water. +Reprint 2025-26 + +The Other Side of Zero +271 +Integers: Snakes and Ladders +Rules +• This is a two player game. Each player has 1 pawn. Both +players start at 0. Players can reach either – 50 or + 50 to win +but need not decide or fix this before or during play. +• Each player rolls two dice at a time. One dice has numbers +from + 1 to + 6 and the other dice has numbers from – 1 to – 6. +• After each roll of the two dice, the player can add or subtract +them in any order and then move the steps that indicate the +result. A positive result means moving towards + 50 and a +negative result means moving towards – 50. +Reprint 2025-26 + +Learning Material Sheets +Reprint 2025-26 + +Reprint 2025-26 + +Reprint 2025-26 + +Reprint 2025-26 + +Reprint 2025-26 + +A +B +C +D +E +F +G +Tangram +Note: Cut each shape along the white border. +Reprint 2025-26 + +Reprint 2025-26 + +1 UNIT +1 +2 +2 +2 +1 +3 +2 +3 +3 +3 +1 +4 +2 +4 +3 +4 +4 +4 +1 +5 +2 +5 +3 +5 +4 +5 +5 +5 +1 +6 +2 +6 +3 +6 +4 +6 +5 +6 +6 +6 +1 +7 +2 +7 +3 +7 +4 +7 +5 +7 +6 +7 +7 +7 +1 +8 +2 +8 +3 +8 +4 +8 +5 +8 +6 +8 +7 +8 +8 +8 +1 +9 +2 +9 +3 +9 +4 +9 +5 +9 +6 +9 +7 +9 +8 +9 +9 +9 + 1 +10 + 2 +10 + 3 +10 + 4 +10 + 5 +10 + 6 +10 + 7 +10 + 8 +10 + 9 +10 +10 +10 +Fraction Wall +Note: Cut each shape along the white border. +Reprint 2025-26 + +Reprint 2025-26 + +Note: Cut the tiles along the white border. +Reprint 2025-26 + +Reprint 2025-26 + +Notes +Reprint 2025-26 + +Notes +Reprint 2025-26 + +[1] + +CHAPTER 10 — SOLUTIONS +The Other Side of Zero + +Page no. 243 +Q. What do you press to go four floors up? What do you press to go three floors down? +Ans. +• + + + + or + 4 +• - - - or – 3 + +Page No. 244 +Q. Number all the floors in the Building of Fun. +Ans. ++3: Book Store + ++2: Art Centre + ++1: Food Court + +0: Welcome Hall + +-1: Toy Store + +-2: Video Game. + +Page No. 245 + Figure it out +Q.1. You start from Floor + 2 and press – 3 in the lift. Where will you reach? Write an +expression for this movement. +Ans. +(+2) + (-3) = -1; Toy store. +Q.2. Evaluate these expressions (you may think of them as Starting Floor + Movement by +referring to the Building of Fun). +a. (+ 1) + (+ 4) = _______ +b. (+ 4) + (+ 1) = _______ +c. (+ 4) + (– 3) = _______ + d. (– 1) + (+ 2) = _______ +e. (– 1) + (+ 1) = _______ + f. 0 + (+ 2) = _________ +g. 0 + (– 2) = _________ +Ans. +a. (+5) +b. (+5) +c. (+1) +d. (+1) + +e. 0 + +f. (+2) +g. (-2) + + +[2] + +Q.3. Starting from different floors, find the movements required to reach Floor – 5. For +example, if I start at Floor + 2, I must press – 7 to reach Floor – 5. The expression +is (+ 2) + (– 7) = – 5. +Find more such starting positions and the movements needed to reach Floor – 5 and +write the expressions. +Ans. +Starting floor + Movement = Target Floor + ++1 + (-6) = (-5) + +-2 + (-3) = (-5) + +0 + (-5) = (-5) + +(Try more possibilities) + +Page No. 246 +Figure it out +Q. Evaluate these expressions by thinking of them as the resulting movement of +combining button presses: +a. (+ 1) + (+ 4) = _____________ +b. (+ 4) + (+ 1) = _____________ +c. (+ 4) + (– 3) + (– 2) = _______ +d. (– 1) + (+ 2) + (– 3) = _______ +Ans. +a. (+5) b. (+5) c. (-1) +d. (-2) + +Q. Write the inverses of these numbers: ++4, –4, –3, 0, +2, –1. +Ans. +Inverse of +4 = -4 + +Inverse of -4 = +4 + +Inverse of -3 = +3 + +Inverse of 0 = 0 + +Inverse of +2 = (-2) + +Inverse of -1 = +1 + +Q. Connect the inverses by drawing lines. +(+5) (-7) +(–8) +(+9) + +(–9) +(+8) +(-5) +(+7) + +[3] + + +Ans. (+5) +(-7) +(–8) +(+9) + +(–9) +(+8) +(-5) +(+7) + + Q. Who is on the lowest floor? +1. Jay is in the Art Centre. So, he is on Floor +2. +2. Asin is in the Sports Centre. So, she is on Floor ___. +3. Binnu is in the Cinema Centre. So, she is on Floor ____. +4. Aman is in the Toys Shop. So, he is on Floor ____. +Ans. Binnu is on the lowest floor. + +1. ++2 + +2. ++5 + +3. +-3 + +4. +-1 + +Page No. 247 +Figure it out +Q.1. Compare the following numbers using the Building of Fun and fill in the boxes +with < or >. +Ans. a. (–2) + +5 +b. (–5) + + 4 + +c. (–5) + – 3 +d. + 6 + – 6 +e. 0 + – 4 +f. 0 + + 4 +Ans. +a. < + +b. < + +c. < + +d. > + +e. > + +f. < + +Q.2. Imagine the Building of Fun with more floors. Compare the numbers and fill in the +boxes with < or >: +a. (–10) + (–12) + +b. + 17 + (–10) +c. 0 + (–20) + +d. + 9 + (–9) +e. (–25) + (–7) + +f. + 15 + (–17) +Ans. +a. > + +b. > + +c. > + +d. > + +e. < + +f. > + +a. + +[4] + +Q.3. If Floor A = – 12, Floor D = – 1 and Floor E = + 1 in the building shown on the right +as a line, find the numbers of Floors B, C, F, G and H. +Ans. +B = -9 +C = -6 +F = +2 G = +6 H = +11 +Q.4. Mark the following floors of the building shown on the right. +a. -7 + +b. -4 + +c. +3 +d. -10 +Ans. +a. P = -7 +b. Q = - 4 +c. R = +3 +d. S = -10 + +Page 248 +Q. Evaluate 15 – 5, 100 – 10 and 74 – 34 from this perspective. +• 15 -5 can be expressed as 5 + ? = 15 . So, ? = 10 +• 100 – 10 can be expressed as 10 + ? = 100. So, ? = 90 +• 74 – 34 can be expressed as 34 + ? = 74. So, ? = 40 + +Page No. 249 +Figure it out +Q. Complete these expressions. You may think of them as finding the movement needed +to reach the Target Floor from the Starting Floor. +a. (+ 1) – (+ 4) = _______ + +b. (0) – (+ 2) = _________ +c. (+ 4) – (+ 1) = _______ + +d. (0) – (– 2) = _________ +e. (+ 4) – (– 3) = _______ + +f. (– 4) – (– 3) = ________ +g. (– 1) – (+ 2) = _______ + +h. (– 2) – (– 2) += ________ +i. (– 1) – (+1) = _______ + +j. (+ 3) – (– 3) = ________ + +Ans. +a. (-3) + +b. (-2) + +c. (+3) d. (+2) + +e. (+7) f. (-1) + +g. (-3) + +h. (0) + +i. (-2) + +j. (+6) + +E +P +C +S +A +Q +R +B +D +-12 -11 -10 -9 +-8 +-7 -6 +-5 +-4 +-3 ++8 +-2 +-1 0 ++1 ++7 ++2 +3 +4 ++6 ++5 +G + +[5] + +Page No. 251 + Figure it out +Q. Complete these expressions. +a. (+ 40) + ______ = + 200 + +b. (+ 40) + _______ = – 200 +c. (– 50) + ______ = + 200 + +d. (– 50) + _______ = – 200 +e. (– 200) – (– 40) = _______ +f. (+ 200) – (+ 40) = _______ +g. (– 200) – (+ 40) = _______ +Ans. +a. (+160) +b. (-240) +c. (+250) +d. (-150) + +e. (-160) +f. (+160) +g. (-240) + +Page No. 252 +Q. In the other exercises that you did above, did you notice that subtracting a negative +number was the same as adding the corresponding positive number? +Ans. +Yes. The infinite lift reminds us of a number line. +Q. If, from 5 you wish to go over to 9, how far must you travel along the number line? +Ans. +4 steps. +Q. Now, from 9, if you wish to go to 3, how much must you travel along the number line? +Ans. +One must travel 6 steps backward or move (-6) from 9 to go to 3. + +Q. Now, from 3, if you wish to go to – 2, how far must you travel? +Ans. +One must travel 5 steps backward or move (-5) from 3 to go to -2. + +Page No. 253 +Figure it out + + +Q1. Mark 3 positive numbers and 3 negative numbers on the number line above. +Ans. +Mark A = 2, B = 5, C = 8 + + +P = -1, Q = -3, R = -7 + + +(Try other possibilities) + +B +-10 -9 +-8 +-7 +-6 +-5 -4 +-3 +-2 +-1 +10 +0 +1 +2 +3 +9 +4 +5 +6 +8 +7 +A +C +P +Q +R + +[6] + +Q2. Write down the above 3 marked negative numbers in the following boxes: + + + + +Ans. +-7, +-3, +-1 +Q3. Is 2 > – 3? Why? Is – 2 < 3? Why? +Ans. +Yes, 2 > -3. On the number line 2 is on the right side of (-3). +Yes, -2 < 3, as 3 is on the right of -2. +Q4. What are (i) – 5 + 0 (ii) 7 + (– 7) (iii) – 10 + 20 (iv) 10 – 20 (v) 7 – (– 7) +(vi) – 8 – (– 10)? +Ans. +(i) – 5 + 0 = -5 (ii) 7 + (– 7) = 0 +(iii) – 10 + 20 =10 + + +(iv) 10 – 20 = -10 + (v) 7 – (– 7) = 14 +(vi) – 8 – (– 10) = 2 + +Page No. 255 +Q. Use unmarked number lines to evaluate these expressions: +a. – 125 + (– 30) = _______ +b. + 105 – (– 55) = _______ +c. + 80 – (– 150) = _______ +d. – 99 – (– 200) = _______ +Ans. +a. – 125 + (– 30) = -155 +b. + 105 – (– 55) = 160 +c. + 80 – (– 150) = 230 +d. – 99 – (– 200) = 101 + +Page No. 257 +Figure it out +Q.1. Complete the additions using tokens. +a. (+ 6) + (+ 4) +b. (– 3) + (– 2) +c. (+ 5) + (– 7) +d. (– 2) + (+ 6) +Ans. +a. +10 + +b. -5 + +[7] + + +c. (-2) + +d. (+4) +Q.2. Cancel the zero pairs in the following two sets of tokens. On what floor is the lift +attendant in each case? What is the corresponding addition statement in each case? +a. + + + + +b. + + + + + + + + + + + + + + + + +Ans. +a. (+3) + (-5) = (-2) + + +b. +(+6) + (-3) = (+3) + +Attendant is at (-2) Floor + + +Attendant is at (+3) Floor + +Page No. 258 +Figure it out +Q.1. Evaluate the following differences using tokens. Check that you get the same result +as with other methods you now know: +a. (+ 10) – (+ 7) +b. (– 8) – (– 4) +c. (– 9) – (– 4) +d. (+ 9) – (+ 12) +e. (– 5) – (– 7) +f. (– 2) – (– 6) +Ans. +a. + + + + + + + + + + + +(+10) – (+7) = (+3) + +b. + + + + + + + + + +(-8) – (-4) = (-8) + (+4) = (-4) + +c. + + + + + + + + + + +(-9) – (-4) = (-9) + (+4) = (-5) + +d. + + + + + + + + + + + + + + + + + + +(+9) – (+12) = (-3) + +e. + + + + + + + + +[8] + + + + + + +(-5) - (-7) = (+2) + +f. + + + + + + + + + + + + + +(-2) – (-6) = +4 +Try for other methods. +Q.2. Complete the subtractions: +a. (– 5) – (– 7) +b. (+ 10) – (+ 13) + +c. (– 7) – (– 9) +d. (+ 3) – (+ 8) +e. (– 2) – (– 7) + +f. (+ 3) – (+ 15) +Ans. +a. +2 + +b. -3 + +c. +2 + +d. (-5) + +e. +5 + +f. (-12) + +Section 10.2 +Page No. 259 + Figure it out +Q.1. Try to subtract: – 3 – (+ 5). +How many zero pairs will you have to put in? What is the result? +Ans. + -8 + +We have to put five zero pairs. +Q.2. Evaluate the following using tokens. +a. (– 3) – (+ 10) +b. (+ 8) – (– 7) +c. (– 5) – (+ 9) +d. (– 9) – (+ 10) +e. (+ 6) – (– 4) +f. (– 2) – (+ 7) +Ans. +a. -13 +b. +15 +c. -14 +d. -19 +e. +10 +f. -9 + + + +[9] + +Section 10.3 +Page No. 259 +Q. Your new bank balance is _______. +Ans. +₹ 100 + ₹ 60 = ₹ 160 +Q. Your bank balance is now ______. +Ans. +₹ 160 - ₹ 30 = ₹ 130 +Q. What is your bank balance now? ______ + Is this possible? +Ans. +₹ 130 - ₹ 150 = - ₹ 20 + +Page No. 260 +Q. What is your balance now? ______ +Ans. +₹ 180 + +Section 10.3 +Page 260 + Figure it Out +Q1. Suppose you start with 0 rupees in your bank account, and then you have credits of +₹30, ₹40, and ₹50, and debits of ₹40, ₹50, and ₹60. What is your bank account +balance now?] +Ans. (+30) + (+40) + (+50) – (40) – (50) – (60) +Balance amount in Bank Account = ₹ (-30) + +Q2. Suppose you start with 0 rupees in your bank account, and then you have debits of +₹1, 2, 4, 8, 16, 32, 64, and 128, and then a single credit of ₹256. What is your bank +account balance now? +Ans. (1) – (2) – (4) – (8) – (16) – (32) – (64) – (128) + (256) +Balance amount in the Bank Account = ₹ 1 + + + +[10] + + Figure it Out +Page 261 +Q.1. Looking at the geographical cross section fill in the respective heights: +A + + +B + +C + + +D +E. + + +F. + +G. +Ans. +Approximate heights are: + A. (+1500) m + +B. (-500) m + +C. (+300) m + +D. (-1200) m + +E. (+1200) m + +F. (-200) m + +G. (+100) m + +Q.2. Which is the highest point in this geographical cross-section? Which is the lowest +point? +Ans. +Highest point is A & the lowest point is D. +Q.3. Can you write the points A, B, …., G in a sequence of decreasing order of heights? +Can you write the point in a sequence of increasing order of heights? +Ans. +A, E, C, G, F, B, D. (Decreasing order) + +D, B, F, G, C, E, A. (Increasing order) +Q.4. What is the highest point above sea level on Earth? What is its height? +Ans. +Mount-Everest. Its height above sea level is 8848m. +Q.5. What is the lowest point with respect to sea level on land or on the ocean floor? What +is its height? (This height should be negative). +Ans. The lowest know point on the Earth is the challenger deep, located in the Mariana trench +in the Pacific Ocean. Its depth is approximately -10994 m. + + + +[11] + +Page 262 + Figure it Out +Q.2 Leh in Ladakh gets very cold during the winter. The following is a table of +temperature readings taken during different times of the day and night in Leh on a +day in November. Match the temperature with the appropriate time of the day and +night. +Ans. +Temperature +Time +14°C +02:00 p.m. +8°C +11:00 a.m +–2°C +11:00 p.m. +–4°C +02:00 a.m. + +Section 10.4 +Page 263 + Figure it Out +Q.1 Do the calculations for the second grid above and find the border sum. +Ans. -3, -3, -3, -3 +Q.2 Complete the grids to make the required border sum: +Ans. One of the ways: +-10 +10 +4 + 5 + +-5 + 9 +-10 +5 + + + Border sum +4 + +6 +8 +-16 +11 + +-5 +-19 +-2 +19 + Border sum -2 + +7 +-2 +-9 +-3 + +-5 +-8 +-6 +10 + Border sum -4 +Think of other ways. + + +[12] + +Page 265 + Figure it Out +Q.2 + + + + + + + + + +-8 + + + + + +-14 + + +Figure it Out + + +Q.1 Write all the integers between the given pairs, in increasing order. +a. 0 and –7 +b. –4 and 4 +c. –8 and –15 +d. –30 and –23 +Ans. a. -6,-5,-4,-3,-2,-1 + + +b. -3,-2,-1,0,1,2,3 + c. -14,-13,-12,-11,-10,-9 +d. -29,-28,-27,-26,-25,-24 +Q.2 Give three numbers such that their sum is –8. +Ans. One of the combinations is -5,7,-10. +Think of other combinations. +Q.3 There are two dice whose faces have these numbers: –1, 2, –3, 4, –5, 6. The smallest +possible sum upon rolling these dice is –10 = (–5) + (–5) and the largest possible +sum is 12 = (6)+(6). Some numbers between (–10) and (+12) are not possible to get +by adding numbers on these two dice. Find those numbers. +Ans. -9,-7,-5,0,2,7,9,11 + +Q.4 +Solve these: + + + +Ans. +-5 +-21 +-5 +-21 +-5 +5 +5 +21 + + + + +[13] + +Q.5 Find the years below. +a. From the present year, which year was it 150 years ago? ________ +b. From the present year, which year was it 2200 years ago? _______ +Hint: Recall that there was no year 0. +c. What will be the year 320 years after 680 BCE? ________ +Ans. c. -360 BCE +Q.6 Complete the following sequences: +a. (–40), (–34), (–28), (–22), _____, ______, ______ +b. 3, 4, 2, 5, 1, 6, 0, 7, _____, _____, _____ +c. _____, ______, 12, 6, 1, (–3), (–6), _____, ______, ______ +Ans. a. -16, -10, -4 + +b. -1,8, -2 + +c. 27, 19, … -8,-9,-9 +Q.7 Here are six integer cards: (+1), (+7), (+18), (–5), (–2), (–9). You can pick any of these +and make an expression using addition(s) and subtraction(s). Here is an expression: +(+18)+(+1)–(+7) – (–2) which gives a value (+14). Now, pick cards and make an +expression such that its value is closer to (– 30). +Ans. One of the ways is (-2) + (-9) – (+18) – (+1) = -30 +Q.8 The sum of two positive integers is always positive but a (positive integer) – (positive +integer) can be positive or negative. What about +a. (positive) – (negative) + +b. (positive) + (negative) + + + +c. (negative) + (negative) + +d. (negative) – (negative) + + +e. (negative) – (positive) + +f. (negative) + (positive) + +Ans. a. positive + +b. can be positive or negative + + c. negative + +d. can be positive or negative + + e. negative + +f. can be positive or negative +Q.9 This string has a total of 100 tokens arranged in a particular pattern. What is the +value of the string? + +Ans. 20" +class_7,1,Large Numbers Around Us,ncert_books/class_7/gegp1dd/gegp101.pdf,"1.1 A Lakh Varieties! +Eshwarappa is a farmer in Chintamani, +a town in Karnataka. He visits the +market regularly to buy seeds for his +rice field. During one such visit he +overheard a conversation between +Ramanna and Lakshmamma. Ramanna +said, “Earlier our country had about a +lakh varieties of rice. Farmers used to +preserve different varieties of seeds +and use them to grow rice. Now, we +only have a handful of varieties. Also, +farmers have to come to the market to buy seeds”. +Lakshmamma said, “There is a seed bank near my house. So far, they +have collected about a hundred indigenous varieties of rice seeds from +different places. You can also buy seeds from there.” +You may have heard the word ‘lakh�� +before. Do you know how big one lakh is? Let +us find out. +Eshwarappa shared this incident with his +daughter Roxie and son Estu . +Estu was surprised to know that there +were about one lakh varieties of rice in this +country. He wondered “One lakh! So far I +have only tasted 3 varieties. If we tried a new +variety each day, would we even come close +to tasting all the varieties in a lifetime of 100 +years?” +What do you think? Guess. +LARGE NUMBERS +AROUND US +1 + +Ganita Prakash | Grade 7 +But how much is one lakh? Observe the pattern and fill in the boxes +given below. +999 +The largest 3-digit number is +The smallest 4-digit number is +The largest 4-digit number is +The smallest 5-digit number is +The largest 5-digit number is +1,00,000 +The smallest 6-digit number is ++1 ++1 ++1 +1,00,000 is read as “One Lakh” +99,995 +99,996 +99,998 +Roxie and Estu found that if they ate +one variety of rice a day, they would come +nowhere close to a lakh in a lifetime! Roxie +suggests, “What if we ate 2 varieties of +rice every day? Would we then be able to +eat 1 lakh varieties of rice in 100 years?” +What if a person ate 3 varieties of rice every day? Will they be able to +taste all the lakh varieties in a 100 year lifetime? Find out. +Estu said, “We know how many days there are in a year — 365, if +we ignore leap years. If we live for y years, the number of days in our +lifetime will be 365 × y.” +2 + +Large Numbers Around Us +Choose a number for y. How close to one lakh is the number of days in +y years, for the y of your choice? +Figure it Out +1. According to the 2011 Census, the population of the town of +Chintamani was about 75,000. How much less than one lakh is 75,000? +2. The estimated population of Chintamani in the year 2024 is 1,06,000. +How much more than one lakh is 1,06,000? +3. By how much did the population of Chintamani increase from 2011 +to 2024? +Getting a Feel of Large Numbers +You may have come across interesting facts like these: +• The world’s tallest statue is the ‘Statue of Unity’ +in Gujarat depicting Sardar Vallabhbhai Patel. Its +height is about 180 metres. +• Kunchikal waterfall in Karnataka is said to drop +from a height of about 450 metres. +It is not always easy to get a sense of how big these measurements +are. But, we can get a better sense of their size when we compare them +with something familiar. Let us see an example. +Look at the picture on the right. Somu is 1 metre tall. If each floor +is about four times his height, what is the approximate height of the +building? +Which is taller — The Statue of Unity or this building? How much taller? +____________m. +We can see that the +height of the Statue of Unity +is close to 4 times the height +of Somu’s building. +How much taller is the +Kunchikal +waterfall +than +Somu's +building? +____________m. +How many floors should +Somu’s building have to be +as high as the waterfall? +____________ . +500 +400 +300 +200 +100 +meters +0 +3 + +Ganita Prakash | Grade 7 +Is One Lakh a Very Large Number? +Eshwarappa asked Roxie and Estu, “Is a lakh big or +small?” +Roxie feels that 1 lakh is a large number: +• “We had one lakh varieties of rice — that +is a lot.” +• “Living 1 lakh days would mean living for 274 +years — that is a really long time!” +• “If 1 lakh people stood shoulder to shoulder in a line they could +stretch as far as 38 kilometres.” +Estu, however, thinks it is not that big: +• “Do you know that the cricket stadium in +Ahmedabad has a seating capacity of more +than 1 lakh? One lakh people in such a small +area!” +• “Most humans have 80,000 to 1,20,000 hairs on +their heads. Imagine, 1 lakh hairs fit in such a +tiny space!” +• “I heard that there are some species of fish +where a female fish can lay almost one lakh +eggs at once very comfortably. Some even lay tens of lakhs of eggs +at a time.” +How do you view a lakh — is a lakh big or small? +Reading and Writing Numbers +We have already been using commas for 5-digit numbers like 45,830 in +the Indian place value system. As numbers grow bigger, using commas +helps in reading the numbers easily. We use a comma in between the +digits representing the “ten thousands” place and the “one lakh” place +as you have seen just before (1,00,000). +The number name of 12,78,830 is twelve lakh seventy eight thousand +eight hundred thirty. +Similarly, the number 15,75,000 in words is fifteen lakh seventy five +thousand. +Write each of the numbers given below in words: +(a) 3,00,600 +(b) 5,04,085 +(c) 27,30,000 +(d) 70,53,138 +4 + +Large Numbers Around Us +Write the corresponding number in the Indian place value system for +each of the following: +(a) One lakh twenty three thousand four hundred and fifty six +(b) Four lakh seven thousand seven hundred and four +(c) Fifty lakhs five thousand and fifty +(d) Ten lakhs two hundred and thirty five +1.2 Land of Tens +In the Land of Tens, there are special calculators with special buttons. +1. The Thoughtful Thousands only has a +1000 button. How many +times should it be pressed to show: +(a) Three thousand? 3 times +(b) 10,000? ____________ +(c) Fifty three thousand? ___________ +(d) 90,000? ______________ +(e) One Lakh? ________________ +(f) ____________? 153 times +(g) How many thousands are required to make one lakh? +2. The Tedious Tens only has a +10 button. How +many times should it be pressed to show: +(a) Five hundred? _____________ +(b) 780? _________ +(c) 1000? _________ +(d) 3700? ________ +(e) 10,000? ___________ +(f) One lakh? _____________ +(g) ____________? 435 times +3. The Handy Hundreds only has a +100 button. How many +times should it be pressed to show: +(a) Four hundred? ___________times +(b) 3,700? __________ +Note to the Teacher: Encourage students to make connections between these facts. +For example, can the whole population of Chintamani fit in the stadium? How +can we imagine the line of 38 km, having a lakh people, sitting next to each +other in the stadium? +5 + +Ganita Prakash | Grade 7 +(c) 10,000? __________ +(d) Fifty three thousand? __________ +(e) 90,000? __________ +(f) 97,600? __________ +(g) 1,00,000? __________ +(h) _________? 582 times +(i) + How +many +hundreds +are +required +to +make +ten +thousand? +(j) + How many hundreds are required to make one lakh? +(k) Handy Hundreds says, “There are some numbers which +Tedious Tens and Thoughtful Thousands can’t show but I +can.” Is this statement true? Think and explore. +4. Creative Chitti is a different kind of +calculator. It has the following buttons: ++1, +10, +100, +1000, +10000, +100000 and ++1000000. It always has multiple ways of +doing things. “How so?”, you might ask. +To get the number 321, it presses +10 +thirty two times and +1 once. Will it get +321? Alternatively, it can press +100 two +times and +10 twelve times and +1 once. +5. Two of the many different ways to get +5072 are shown below: +These two ways can be expressed as: +(a) (50 × 100) + (7 × 10) + (2 × 1) = 5072 +(b) (3×1000) + (20×100) + (72 ×1) = 5072 +Find a different way to get 5072 and write +an expression for the same. +Figure it Out +For each number given below, write +expressions for at least two different ways +to obtain the number through button clicks. +Think like Chitti and be creative. +(a) 8300 +(b) 40629 +(c) 56354 +Buttons +5072 ++10,00,000 ++1,00,000 ++10,000 ++1,000 +3 ++100 +50 +20 ++10 +7 ++1 +2 +72 +6 + +Large Numbers Around Us +(d) 66666 +(e) 367813 +Creative Chitti has some questions for you — +(a) You have to make exactly 30 button presses. What is the largest +3-digit number you can make? What is the smallest 3-digit +number you can make? +(b) 997 can be made using 25 clicks. +Can you make 997 with a different +number of clicks? +Create questions like these and challenge +your classmates. +Systematic Sippy is a different kind of +calculator. It has the following buttons: +1, ++10, +100, +1000, +10000, +100000. It wants +to be used as minimally as possible. +How can we get the numbers (a) 5072, (b) 8300 using as few button +clicks as possible? +Find out which buttons should be +clicked and how many times to get the +desired numbers given in the table. +The aim is to click as few buttons as +possible. +Here is one way to get the number +5072. This method uses 23 button +clicks in total. +Is there another way to get 5072 +using less than 23 button clicks? +Write the expression for the same. +Figure it Out +1. For the numbers in the previous +exercise, find out how to get each +number by making the smallest +number of button clicks and write +the expression. +2. Do you see any connection between each number and the +corresponding smallest number of button clicks? +3. If you notice, the expressions for the least button clicks also +give the Indian place value notation of the numbers. Think +about why this is so. +Buttons +5072 ++10,00,000 ++1,00,000 ++10,000 ++1,000 +5 ++100 +0 ++10 +6 ++1 +12 +Math +Talk +7 + +Ganita Prakash | Grade 7 +What if we press the +10,00,000 button ten times? What number +will come up? How many zeroes will it have? What should we call it? +The number will be 100 lakhs, which is also called a crore. 1 crore is +written as 1,00,00,000 — it is 1 followed by seven zeroes. +1.3 Of Crores and Crores! +The table below shows some numbers according to both the Indian +system and the American system (also called the International system) +of naming numerals and placing commas. Observe the placement of +commas in both systems. +Indian System +American System +1,000 +One thousand +1,000 +One thousand +10,000 +Ten thousand +10,000 +Ten thousand +1,00,000 +One lakh +100,000 +Hundred +thousand +10,00,000 +Ten lakhs +1,000,000 +One million +1,00,00,000 +One crore +10,000,000 +Ten million +10,00,00,000 +Ten crores +100,000,000 +Hundred +million +1,00,00,00,000 +One arab or One +hundred crores +1,000,000,000 +One billion +Notice that in the Indian system, commas are placed to group the +digits in a 3-2-2-2... pattern from right to left (thousands, lakhs, crores, +etc.). In the American system, the digits are grouped uniformly in a +3-3-3-3… pattern from right to left (thousands, millions, billions, etc.). +The Indian system of naming numbers is also followed in Bhutan, +Nepal, Sri Lanka, Pakistan, Bangladesh, Maldives, Afghanistan, and +Myanmar. The words lakh and crore originate from the Sanskrit words +lakṣha (लक्ष) and koṭi (कोटि‍). The American system is also used in many +countries. +Observe the number of zeroes in 1 lakh and 1 crore. +1 lakh, written in numbers would be 1 followed by 5 zeroes. +1 crore, written in numbers would be 1 followed by 7 zeroes. +A lakh is a hundred times a thousand, a crore is a hundred times a +lakh and an arab is a hundred times a crore (i.e., a hundred thousand +is 1 lakh, 100 lakhs is 1 crore, and 100 crores is 1 arab). +How many zeros does a thousand lakh have? _____ +8 + +Large Numbers Around Us +How many zeros does a hundred thousand have? ____ +The number 9876501234 can be easily read by placing commas first: +(a) 9,87,65,01,234 → 9 arab 87 crore 65 lakhs 1 thousand and 234 +or 987 crore 65 lakh 1 thousand 234 (in the Indian system). +(b) 9,876,501,234 → 9 billion 876 million 501 thousand 234 (in the +American system). +Figure it Out +1. Read the following numbers in Indian place value notation and +write their number names in both the Indian and American systems: +(a) 4050678 +(b) +48121620 +(c) 20022002 +(d) +246813579 +(e) 345000543 +(f) +1020304050 +2. Write the following numbers in Indian place value notation: +(a) One crore one lakh one thousand ten +(b) One billion one million one thousand one +(c) Ten crore twenty lakh thirty thousand forty +(d) Nine billion eighty million seven hundred thousand six hundred +3. Compare and write ‘<’, ‘>’ or ‘=’: +(a) 30 thousand ____ 3 lakhs +(b) 500 lakhs ______ 5 million +(c) 800 thousand ____ 8 million +(d) 640 crore ______ 60 billion +We shall come across even bigger numbers in later grades. +1.4 Exact and Approximate Values +1 lakh people +visited the +book fair. +Ohh, so +many +people! +If I had not gone, +they would have +written “99,999 +people visited the +book fair last week”. +9 + +Ganita Prakash | Grade 7 +What do you think of this conversation? Have you read or heard +such headlines or statements? +Very often, exact numbers are not required and just an approximation +is sufficient. For example, according to the 2011 census, the population +of Chintamani town is 76,068. Instead, saying that the population is +about 75,000 is enough to give an idea of how big the quantity is. +Bappi, call and +check if Toofan +Express is on time. +What +number +did you +call? +I called the +rounded off +number, of course! +My teacher told +us that efficient +people deal with +rounded off +numbers. +There are situations where it makes sense to round up a number +(rounding up is when the approximated number is more than the actual +number). For example, if a school has 732 people including students, +teachers and staff: the principal might order 750 sweets instead of +700 sweets. +There are situations where it is better to round down (rounding down +is when the approximated number is less than the actual number). For +example, if the cost of an item is ₹470, the shopkeeper may say that the +cost is around ₹450 instead of saying it is around ₹500. +Think and share situations where it is appropriate to (a) round up, +(b) round down, (c) either rounding up or rounding down is okay and +(d) when exact numbers are needed. +10 + +Large Numbers Around Us +Nearest Neighbours +With large numbers it is useful to know the nearest thousand, lakh or +crore. For example, the nearest neighbours of the number 6,72,85,183 +are shown in the table below. +Nearest thousand +6,72,85,000 +Nearest ten thousand +6,72,90,000 +Nearest lakh +6,73,00,000 +Nearest ten lakh +6,70,00,000 +Nearest crore +7,00,00,000 +Similarly, write the five nearest neighbours for these numbers: +(a) 3,87,69,957 +(b) 29,05,32,481 +I have a number for which all five nearest neighbours are +5,00,00,000. What could the number be? How many such numbers +are there? +Roxie and Estu are estimating the values of simple expressions. +1. 4,63,128 + 4,19,682, +Roxie: “The sum is near 8,00,000 and is more than 8,00,000.” +Estu: “The sum is near 9,00,000 and is less than 9,00,000.” +(a) Are these estimates correct? Whose estimate is closer to the +sum? +(b) Will the sum be greater than 8,50,000 or less than 8,50,000? +Why do you think so? +(c) Will the sum be greater than 8,83,128 or less than 8,83,128? +Why do you think so? +(d) Exact value of 4,63,128 + 4,19,682 = ___________ +2. 14,63,128 – 4,90,020 +Roxie: “The difference is near 10,00,000 and is less than 10,00,000.” +Estu: “The difference is near 9,00,000 and is more than 9,00,000.” +(a) Are these estimates correct? Whose estimate is closer to the +difference? +(b) Will the difference be greater than 9,50,000 or less than +9,50,000? Why do you think so? +Math +Talk +11 + +Ganita Prakash | Grade 7 +(c) Will the difference be greater than 9,63,128 or less than +9,63,128? Why do you think so? +(d) Exact value of 14,63,128 – 4,90,020 = __________ +Populations of Cities +Observe the populations of some Indian cities in the table below. +Rank +City +Population (2011) +Population (2001) +1 +Mumbai +1,24,42,373 +1,19,78,450 +2 +New Delhi +1,10,07,835 +98,79,172 +3 +Bengaluru +84,25,970 +43,01,326 +4 +Hyderabad +68,09,970 +36,37,483 +5 +Ahmedabad +55,70,585 +35,20,085 +6 +Chennai +46,81,087 +43,43,645 +7 +Kolkata +44,86,679 +45,72,876 +8 +Surat +44,67,797 +24,33,835 +9 +Vadodara +35,52,371 +16,90,000 +10 +Pune +31,15,431 +25,38,473 +11 +Jaipur +30,46,163 +23,22,575 +12 +Lucknow +28,15,601 +21,85,927 +13 +Kanpur +27,67,031 +25,51,337 +14 +Nagpur +24,05,665 +20,52,066 +Note to the Teacher: Ask students questions like — “what could the numbers +be if the sum had to be less than 8,50,000."" +12 + +Large Numbers Around Us +15 +Indore +19,60,631 +14,74,968 +16 +Thane +18,18,872 +12,62,551 +17 +Bhopal +17,98,218 +14,37,354 +18 +Visakhapatnam +17,28,128 +13,45,938 +19 +Pimpri-Chinchwad +17,27,692 +10,12,472 +20 +Patna +16,84,222 +13,66,444 +From the information given in the table, answer the following questions +by approximation: +1. What is your general observation about this data? Share it with +the class. +2. What is an appropriate title for the above table? +3. How much is the population of Pune in 2011? Approximately, by +how much has it increased compared to 2001? +4. Which city’s population increased the most between 2001 and 2011? +5. Are there cities whose population has almost doubled? Which +are they? +6. By what number should we multiply Patna’s population to get a +number/population close to that of Mumbai? +1.5 Patterns in Products +Roxie and Estu are playing with multiplication. They encounter an +interesting technique for multiplying a number by 10, 100, 1000, and +so on. +A Multiplication Shortcut +Roxie evaluated 116 × 5 as follows: +116 × 5 = 116 × 10 +2 + + += 58 × 10 + + += 580. +58 +13 + +Ganita Prakash | Grade 7 +Estu evaluated 824 × 25 as follows: +206 +824 × 25 += 824 × 100 +4 + + + += 20600. +Using the meaning of multiplication and division, can you explain +why multiplying by 5 is the same as dividing by 2 and multiplying +by 10? +Figure it Out +1. Find quick ways to calculate these products: +(a) 2 × 1768 × 50 +(b) 72 × 125 [Hint: 125 = 1000 +8 + ] +(c) 125 × 40 × 8 × 25 +2. Calculate these products quickly. +(a) 25 × 12 = _____________ +(b) 25 × 240 = _____________ +(c) 250 × 120 = _____________ +(d) 2500 × 12 =_____________ +(e) ______×______= 120000000 +How Long is the Product? +In each of the following boxes, the multiplications produce interesting +patterns. Evaluate them to find the pattern. Extend the multiplications +based on the observed pattern. +11 × 11 = +111 × 111 = +1111 × 1111 = +66 × 61 = +666 × 661 = +6666 × 6661 = +3 × 5 = +33 × 35 = +333 × 335 = +101 × 101 = +102 × 102 = +103 × 103 = +Math +Talk +14 + +Large Numbers Around Us +Observe the number of digits in the two numbers being multiplied +and their product in each case. Is there any connection between the +numbers being multiplied and the number of digits in their product? +Roxie says that the product of two 2-digit numbers can only be a 3- or +a 4-digit number. Is she correct? +Should we try all possible multiplications with 2-digit numbers to tell +whether Roxie’s claim is true? Or is there a better way to find out? +She explains her reasoning: “We want to +know about the number of digits in the product +of two 2-digit numbers. To know the smallest +such product I took 10 × 10, so all other products +will be greater than 100. +To know the greatest such product I +multiplied the smallest 3-digit numbers (100 +× 100) to get 10,000; so the product of all the +2-digit multiplications will be less than 10,000.” +Can multiplying a 3-digit number with another 3-digit number give a +4-digit number? +Can multiplying a 4-digit number with a 2-digit number give a 5-digit +number? +Observe the multiplication statements below. Do you notice any +patterns? See if this pattern extends for other numbers as well. +1-digit +× +1-digit += +1-digit +or +2-digit +2-digit +× +1-digit += +2-digit +or +3-digit +2-digit +× +2-digit += +3-digit +or +4-digit +3-digit +× +3-digit += +5-digit +or +6-digit +5-digit +× +5-digit += +or +8-digit +× +3-digit += +or +12-digit +× +13-digit += +or +15 + +Ganita Prakash | Grade 7 +Fascinating Facts about Large Numbers +Some interesting facts about large numbers are hidden below. Calculate +the product or quotient to uncover the facts. Once you find the product +or quotient, read the number in both Indian and American naming +systems. Share your thoughts and questions about the fact with the +class after you discover each number. +1250 × 380 +______________ +is the number of kīrtanas composed by Purandaradāsa +according to legends. +Purandaradāsa was a composer and singer in the 15th +century. His kīrtanas spanned social reform, bhakti +and spirituality. He systematised methods for teaching +Carnatic music which is followed to the present day. +How many years did he live to compose +so many songs? At what age did he start +composing songs? +If he composed 4,75,000 songs, how +many songs per year did he have to +compose? +2100 × 70,000 +_______________ +is the approximate distance +in kilometers, between the Earth +and the Sun. +This distance keeps varying throughout the year. +The farthest distance is about 152 million kilometers. +16 + +Large Numbers Around Us +How did they measure +the distance between the +Earth and the Sun? +6400 × 62,500 +_________________ +is the average number of litres of water the Amazon river +discharges into the Atlantic Ocean every second. +The river's flow into the Atlantic is so much that +drinkable freshwater is found even 160 kilometers into +the open sea. +As you did before, divide the given numbers to uncover interesting +facts about division. Share your thoughts and questions with the class +after you uncover each number. +13,95,000 ÷ 150 +_________________ +is the distance (in kms) of the longest single-train journey +in the world. The train runs in Russia between Moscow +and Vladivostok. The duration of this journey is about 7 +days. The longest train route in India is from Dibrugarh +in Assam to Kanyakumari in Tamil Nadu; it covers 4219 +kms in about 76 hours. + +17 + +Ganita Prakash | Grade 7 +Adult blue whales can weigh more than +10,50,00,000 ÷ 700 +_________________ +kilograms. A newborn blue whale weighs around 2,700 kg, +which is similar to the weight of an adult hippopotamus. The +heart of a blue whale was recorded to be nearly 700 kg. The +tongue of a blue whale weighs as much as an elephant. Blue +whales can eat up to 3500 kg of krill every day. The largest +known land animal, Argentinosauras, is estimated to weigh +90,000 kgs. +52,00,00,00,000 ÷ 130 +_________________ +was the weight, in tonnes, of global plastic waste +generated in the year 2021. +Large Number Fact +In a single gram of healthy soil there can be 100 million to 1 billion bacteria and 1 lakh to +1 million fungi, which can support plants’ growth and health. +Share such large-number facts you know / come across with your class. +1.6 Did You Ever Wonder…? +Estu is amused by all these interesting facts about large numbers. While +thinking about these, he came up with an unusual question, “Could the +entire population of Mumbai fit into 1 lakh buses?” +What do you think? +How can we find out? +Let us assume a bus can accommodate 50 people. Then 1 lakh buses +can accommodate 1 lakh × 50 = 50 lakh people. +18 + +Large Numbers Around Us +The population of Mumbai is more than 1 crore 24 lakhs (we saw +this in an earlier table). So, the entire population of Mumbai cannot fit +in 1 lakh buses. +The RMS Titanic ship carried about 2500 passengers. Can the population +of Mumbai fit into 5000 such ships? +Inspired by this strange question, Roxie wondered, “If I could travel +100 kilometers every day, could I reach the Moon in 10 years?” (The +distance between the Earth and the Moon is 3,84,400 km.) +How far would she have travelled in a year? +How far would she have travelled in 10 years? +Is it not easier to perform these calculations in stages? You can use +this method for all large calculations. +Find out if you can reach the Sun in a lifetime, if you travel 1000 +kilometers every day. (You had written down the distance between the +Earth and the Sun in a previous exercise.) +Make necessary reasonable assumptions and answer the questions +below: +(a) If a single sheet of paper weighs 5 grams, could you lift one +lakh sheets of paper together at the same time? +(b) If 250 babies are born every minute across the world, will a +million babies be born in a day? +(c) Can you count 1 million coins in a day? Assume you can count +1 coin every second. +Think and create more such fun questions and share them with your +class. +Figure it Out +1. Using all digits from 0 – 9 exactly once (the first digit cannot be 0) to +create a 10-digit number, write the — +19 + +Ganita Prakash | Grade 7 +(a) Largest multiple of 5 +(b) Smallest even number +2. The number 10,30,285 in words is Ten lakhs thirty thousand two +hundred eighty five, which has 43 letters. Give a 7-digit number +name which has the maximum number of letters. +3. Write a 9-digit number where exchanging any two digits results in +a bigger number. How many such numbers exist? +4. Strike out 10 digits from the number 12345123451234512345 so +that the remaining number is as large as possible. +5. The words ‘zero’ and ‘one’ share letters ‘e’ and ‘o’. The words ‘one’ +and ‘two’ share a letter ‘o’, and the words ‘two’ and ‘three’ also +share a letter ‘t’. How far do you have to count to find two consecutive +numbers which do not share an English letter in common? +6. Suppose you write down all the numbers 1, 2, 3, 4, …, 9, 10, 11, ... +The tenth digit you write is ‘1’ and the eleventh digit is ‘0’, as +part of the number 10. +(a) What would the 1000th digit be? At which number would it +occur? +(b) What number would contain the millionth digit? +(c) When would you have written the digit ‘5’ for the 5000th +time? +7. A calculator has only ‘+10,000’ and ‘+100’ buttons. Write an +expression describing the number of button clicks to be made for +the following numbers: +(a) 20,800 +(b) 92,100 +(c) 1,20,500 +(d) 65,30,000 +(e) 70,25,700 +8. How many lakhs make a billion? +9. You are given two sets of number cards numbered from 1 – 9. Place +a number card in each box below to get the (a) largest possible sum +(b) smallest possible difference of the two resulting numbers. +Try +This +Try +This +20 + +10. You are given some number cards; 4000, 13000, 300, 70000, +150000, 20, 5. Using the cards get as close as you can to the +numbers below using any operation you want. Each card +can be used only once for making a particular number. +(a) 1,10,000: Closest I could make is 4000 × (20 + 5) + 13000 += 1,13,000 +(b) 2,00,000: +(c) 5,80,000: +(d) 12,45,000: +(e) 20,90,800: +11. Find out how many coins should be stacked to match the height of +the Statue of Unity. Assume each coin is 1 mm thick. +12. Grey-headed albatrosses have a roughly 7-feet wide wingspan. +They are known to migrate across several oceans. Albatrosses can +cover about 900 – 1000 km in a day. One of the longest single trips +recorded is about 12,000 km. How many days would such a trip +take to cross the Pacific Ocean approximately? +13. A bar-tailed godwit holds the record for the longest recorded +non-stop flight. It travelled 13,560 km from Alaska to Australia +without stopping. Its journey started on 13 October 2022 +and continued for about 11 days. Find out the approximate +distance it covered every day. Find out the approximate distance it +covered every hour. +14. Bald eagles are known to fly as high as 4500 – 6000 m above the +ground level. Mount Everest is about 8850 m high. Aeroplanes can +fly as high as 10,000 – 12,800 m. How many times bigger are these +heights compared to Somu’s building? + +Ganita Prakash | Grade 7 +• We came across large numbers — lakhs, crores and arabs; millions +and billions. We learnt how to read and write these numbers in the +Indian and American/International naming systems. +(a) 1 lakh is 1 followed by 5 zeroes: 1,00,000 +(b) 1 crore is 1 followed by 7 zeroes: 1,00,00,000 +(c) 1 million is 1 followed by 6 zeroes: 1,000,000 (which is also ten +lakhs) +(d) 1 arab is 1 followed by 9 zeroes: 1,000,000,000 (which is also 100 +crore or 1 billion) +• We generally round up or round down large numbers. Many times it +is enough just to know roughly how big or small something is. +• To get a sense of large numbers or quantities, we can check how +many times bigger they are compared to numbers or quantities that +are more familiar. +• We saw how to factorise numbers and regroup them to make +multiplications simpler. +• We carried out interesting thought experiments such as  — “Would +one be able to watch 1000 movies in a year?” +SUMMARY +22 + +Toothpick Digits +We can write digits as shown in the image below: +0123456789 +You can either use toothpicks or matchsticks, or just write the +digits in this way, using lines to represent sticks. +To make the digit 7, three sticks are needed. +Write or make the number 5108. How many sticks are required? +1. Make or write the number 42,019. It would require exactly +23 sticks. +2. Starting with 42,019, add or write two more sticks, and +make a bigger number. One example is 42,078. What other +numbers bigger than 42,019 can you make in this way? +3. Preetham wants to insert the digit ‘1’ somewhere among +the digits ‘4’, ‘2’, ‘0’, ‘1’ and ‘9’. Where should he place the +digit ‘1’ to get the biggest possible number? +4. What other numbers can he make by placing the digit ‘1’? +1. Make or write the number 63,890. +2. Starting with 63,890, rearrange exactly four sticks and +make a bigger number. One example is 88,078. What other +numbers bigger than 63,890 can you make in this way? +1. Make any number using exactly 24 sticks or lines. +2. What is the biggest number that can be made using 24 +sticks or lines? +3. What is the smallest number that can be made using 24 +sticks or lines? +Make your own questions and challenge each other. + +1 + + +Page No. 3 + Figure it out +1. According to the 2011 Census, the population of the town of +Chintamani was about 75,000. How much less than one lakh is +75,000? +Ans: 75,000 is 25,000 less than 1 lakh. +2. The estimated population of Chintamani in the year 2024 is +1,06,000. +How much more than one lakh is 1,06,000? +Ans: 1,06,000 is 6,000 more than 1 lakh. +3. By how much did the population of Chintamani increase from +2011 to 2024? +Ans: Increase in population from 2011 to 2024 is 31,000. +Look at the picture on the right. Somu is 1 metre tall. If each floor +is about four times his height, what is the approximate height of +the building? (Refer textbook for the picture) +Ans: 40 m + Which is taller — The Statue of Unity or this building? How much +taller? ____________m. +Ans: The Statue of Unity is 140 m taller than the building. + How tall is the Kunchikal waterfall compared to the building? 410 m. + How many floors should Somu’s building have to be as high as the +waterfall? approx 113 floors . +Page No. 4 +Write each of the numbers given below in words: +(a) +3,00,600 +Ans: Three lakh six hundred +(b) +5,04,085 +Ans: Five lakh four thousand eighty five +Chapter – 1 +Large Numbers Around Us + +2 + +(c) +27,30,000 +Ans: Twenty seven lakh thirty thousand +(d) + 70,53,138 +Ans: Seventy lakh fifty three thousand one hundred thirty eight +Page No. 5 +Write the corresponding number for each of the following: +(a) One lakh twenty-three thousand four hundred and fifty-six. +Ans: 1,23,456 +(b) Four lakh seven thousand seven hundred and four. +Ans: 4,07,704 +(c) Fifty lakhs five thousand and fifty. +Ans: 50,05,050 +(d) Ten lakhs two hundred and thirty-five. +Ans: 10,00,235 +Thoughtful thousand + +(a) Three thousand? 3 times +(b) 10,000? 10 times + +(c) Fifty three thousand? 53 times +(d) 90,000? 90 times + +(e) One Lakh? 100 times +(f) 1,53,000 ? 153 times +(g) How many thousands are required to make one lakh? 100 +Tedious tens + +(a) Five hundred? 50 times +(b) 780? + 78 times +(c) 1000? + 100 times +(d) 3700? + 370 times +(e) 10,000? 1000 times + +(f) One lakh? 10,000 times + +3 + +(g) 4350 ? 435 times +Handy Hundreds +(a) Four hundred? 4 times +(b) 3,700? 37 times +(c) 10,000? 100 times + +(d) Fifty three thousand? 530 times +(e) 90,000? 900 times + +(f) 97,600? 976 times + +(g) 1,00,000? 1000 times + +(h) 58200 ? 582 times +(i) How many hundreds are required to make ten thousand ? 100 +(j) How many hundreds are required to make one lakh? 1000 +(k) Handy Hundreds says, “There are some numbers which Tedious +Tens and Thoughtful Thousands can’t show but I can”. Is this +statement true? Think and explore. +Ans: No +Page No. 6 + Find a different way to get 5072 and write an expression for the same. + + + + + + + + + + +Buttons +5072 + + ++10,00,000 + + + ++1,00,000 + + + ++10,000 + + + ++1,000 +5 +4 +1 ++100 + +10 +40 ++10 +7 +7 +6 ++1 +2 +2 +12 + +4 + + Figure it out +1. For each number given below, write expressions for at least two +different ways to obtain the number through button clicks. Think +like Chitti and be creative. +Ans: Some ways are given below. Try more ways. +(a) 8300 +Ans: (8 × 1000) + (3 × 100) + +or, (5 × 1000) + (33 × 100) +(b) 40629 +Ans: (40 × 1000) + (6 × 100) + (2 × 10) + (9 × 1) + +or, (30 × 1000) + (106 × 100) +(29 × 1) +(c) 56354 +Ans: (56 × 1000) + (3 × 100) + (5 × 10) + (4 × 1) + or, (46 × 1000) + (103 × 100) + (54 × 1) +(d) 66666 +Ans: (66 × 1000) + (6 × 100) + (66 × 1) + +or, (6 × 10000) + (6 × 1000) + (6 × 100) + (6 × 10) + (6 × 1) +(e) 367813 +Ans: +(3 × 100000) + (6 × 10000) + (7 × 1000) ) + (8 × 100) + (1 × 10) + (3 × 1) +or, (36 × 10000) + (78 × 1000) + (13 × 1) +Page No. 7 +Creative Chitti has some questions for you — +(a) You have to make exactly 30 button presses. What is the largest 3- +digit number you can make? What is the smallest 3-digit number +you can make? +Ans: 993 is the largest 3-digit number that can be made using 30 button +presses. + + +102 is the smallest 3-digit number that can be made using 30 +button presses. + +5 + +(b) 997 can be made using 25 clicks. Can you make 997 with a different +number of clicks? +Ans: 997 can also be made as – + + +(8 × 100) + (19 × 10) + (7 × 1) + + +Number of clicks = 8 + 19 + 7 + = 34 + How can we get the numbers (a) 5072, (b) 8300, using as few button +clicks as possible? +Ans: +(a) 5072 = (5 × 1000) + (7 × 10) + (2 × 1) + + +Number of clicks = 14 +(b) 8300 = (8 × 1000) + (3 × 100) + + +Number of clicks = 11 + Figure it out +1. For the numbers in the previous exercise, find out how to get each +number by making the smallest number of button clicks and write +the expression. +Ans: +(a) 8300 = (8 × 1000) + (3 × 100) + + +Number of clicks = 11 +(b) 40629 = (4 × 10000) + (6 × 100) + (2 × 10) + (9 × 1) + + +Number of clicks = 4 + 6 + 2 + 9 = 21 +(c) 56354 = (5 × 10000) + (6 × 1000) + (3 × 100) + (5 × 10) + (4 × 1) + + +Number of clicks = 5 + 6 + 3 + 5 + 4 = 23 +(d) 66666 = (6 × 10000) + (6 × 1000) + (6 × 100) + (6 × 10) + (6 × 1) + + +Number of clicks = 6 + 6 + 6 + 6 + 6 = 30 +(f) 367813 = +(3 × 100000) + (6 × 10000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1) + + +Number of clicks = 3 + 6 + 7 + 8 + 1 + 3 = 28 + +6 + +2. Do you see any connection between each number and the +corresponding smallest number of button clicks? +Ans: The smallest number of buttons click for each number is the sum of +its digit. +Page No. 8 + How many zeros does a thousand lakh have? 8 zeros + How many zeros does a hundred thousand have? 5 zeros +Page No. 9 + Figure it out +1. Read the numbers and write their number names in both the Indian +and American systems: +Sl. No. +Indian System +American System +(a) 4050678 +40,50,678 +Forty Lakh Fifty Thousand Six +Hundred Seventy-Eight +4,050,678 +Four +Millions +Fifty +Thousand Six Hundred +Seventy-Eight +(b) 48121620 +4,81,21,620 +Four +Crore +Eighty-One +Lakh +Twenty-One +Thousand +Six +Hundred Twenty +48,121,620 +Forty Eight Millions One +Hundred Twenty One +Thousand Six Hundred +Twenty +(c) 20022002 +2,00,22,002 +Two Crore Twenty Two Thousand +Two +20,022,002 +Twenty millions twenty +two thousand and two +(d) 246813579 +24,68,13,579 +Twenty Four Crore Sixty Eight +Lakh Thirteen Thousand Five +Hundred Seventy Nine +246,813,579 +Two Hundred Forty Six +Millions Eight Hundred +Thirteen Thousand Five +Hundred Seventy Nine +(e) 345000543 +34,50,00,543 +345,000,543 + +7 + +Thirty four crore fifty lakh five +hundred forty three +Three hundred forty +five millions five +hundred forty three +(f) 1020304050 +1,02,03,04,050 +One Arab two crore three lakh +four thousand fifty +1,020,304,050 +One +billion +twenty +millions three hundred +four thousand fifty +2. Write the following numbers in Indian place value notation: +(a) One crore one lakh one thousand Ten +Ans: 1,01,01,010 +(b) One billion one million one thousand one +Ans: 1,00,10,01,001 +(c) Ten crore twenty lakh thirty thousand forty +Ans: 10,20,30,040 +(d) Nine billion eighty million seven hundred thousand six hundred +Ans: 9,08,07,00,600 +3. Compare and write ‘<’, ‘>’ or ‘=’: +(a) 30 thousand __<__ 3 lakhs +(b) 500 lakhs ___>___ 5 million +(c) 800 thousand __<__ 8 million +(d) 640 crore ___<___ 60 billion +Page No. 10 + Think and share situations where it is appropriate to (a) round up, +(b) round down, (c) either rounding up or rounding down is okay +and (d) when exact numbers are needed. + +Ans: +(a) Buying food/fruits for a group. +(b) Estimating time remaining to leave for the school (catching the +school bus). +(c) Distance estimation between places, especially far off places. +(d) Handling money in banks. +Find more such situations. + + + +8 + +Page No. 11 + Similarly, write the five nearest neighbours for these numbers: +(a) 3,87,69,957 +Nearest thousand +3,87,70,000 +Nearest ten thousand +3,87,70,000 +Nearest lakh +3,88,00,000 +Nearest ten lakh +3,90,00,000 +Nearest crore +4,00,00,000 + +(b) 29,05,32,481 +Nearest thousand +29,05,32,000 +Nearest ten thousand +29,05,30,000 +Nearest lakh +29,05,00,000 +Nearest ten lakh +29,10,00,000 +Nearest crore +29,00,00,000 + +Page No. 13 + From the information given in the table answer the following +questions by approximation: (Refer table on P. 12–13 +1. What is your general observation about this data? Share with the +class. +Ans: One of the observations is – +Most cities shown in the table have seen a significant rise in population +from 2001 to 2011. +Think of more observations. +2. What is an appropriate title for the above table? +Ans: Population of some Indian cities in 2001 and 2011. +3. How much is the population of Pune in 2011? Approximately, by how +much has it increased compared to 2001? +Ans: Population of Pune in 2011 is 31,15,431. +The population of Pune has approximately increased by 6 lakh. + +4. Which city’s population increased the most between 2001 and -2011? +Ans: Bengaluru has shown maximum increase in population which is +41,24,644. + + +9 + +5. Are there cities whose population has almost doubled? Which are +they? +Ans: Yes, they are Bengaluru, Hyderabad, Surat and Vadodara +6. By what number should we multiply Patna’s population to get a +number/population close to that of Mumbai? +Ans: We need to multiply Patna’s population by 7.39 to get a number close to +the population of Mumbai. +Page No. 14 + Using the meaning of multiplication and division, can you explain why +multiplying by 5 is the same as dividing by 2 and multiplying by 10? +Ans: 10 ÷ 2 is the same as 5. + + Figure it out +1. Find quick ways to calculate these products: +(a) 2 × 1768 × 50 += 2 × 50 × 1768 += 100 × 1768 + +(b) 72 × 125 [Hint: 125 = +1000 +8 += 72 × +1000 +8 += +72 +8 × 1000 = 9 × 1000 + +(c) 125 × 40 × 8 × 25 += 125 × 8 �� 40 × 25 += 1000 × 1000 + +2. Calculate these products quickly. +(a) 25 × 12 = _____________ +25 × 12 = +100 +4 × 12 + = 100 × +12 +4 + + = 100 × 3 + +(b) 25 × 240 = _____________ +25 × 240 = +100 +4 × 240 += 100 × +240 +4 + = 100 × 60 + +10 + + +(c) 250 × 120 = _____________ +250 × 120 = +1000 +4 × 120 += 1000 × +120 +4 + = 1000 × 30 +(d) 2500 × 12 =_____________ += 2500 × 12 = +10,000 +4 + × 12 += 10,000 × +12 +4 + = 10,000 × 3 + +(e) __1200____×_1,00,000_____= 120000000 + +Page No. 15 + Observe the number of digits in the two numbers being multiplied +and their product in each case. Is there any connection between the +numbers being multiplied and the number of digits in their +product? +Ans: +The maximum digits in their product will be the sum of the digits of +the two numbers. +The minimum digit in their product will be one less than their sum. + + Roxie says that the product of two 2-digit numbers can only be a 3- or +a 4-digit number. Is she correct? +Ans: Yes, Roxie is correct. + + Should we try all possible multiplications with 2-digit numbers to +tell whether Roxie’s claim is true? Or is there a better way to find +out? +Ans: She can just check the minimum and maximum number of digits +possible. + Can multiplying a 3-digit number with another 3-digit number give a +4-digit number? +Ans: No. + Can multiplying a 4-digit number with a 2-digit number give a 5-digit +number? +Ans: Yes. + +11 + +Observe the multiplication statements below. Do you notice any patterns? See if +this pattern extends for other numbers as well. +Page No. 16 +5-digit +× +5-digit += +9 +or +10 + + + + + + + +8-digit +× +3-digit += +10 +or +11 + + + + + + + +12-digit +× +13-digit += +24 +or +25 +Page No. 19 + Find out if you can reach the Sun in a lifetime, if you travel 1000 +kilometers every day. (You had written down the distance between +the Earth and the Sun in a previous exercise.) +Ans: We cannot reach the sun in a lifetime because it takes approx. 403 +years. + Make necessary reasonable assumptions and answer the questions +below: +(a) If a single sheet of paper weighs 5 grams, could you lift one lakh +sheets of paper together at the same time? + +Ans: Weight of one lakh (1,00,000) sheets = 500 kg. +Since a person cannot lift 500 kg so, we cannot lift one lakh sheets +of paper. +(b) If 250 babies are born every minute across the world, will a +million babies be born in a day? +Ans: Babies born in one day = 3,60,000 babies. +3,60,000 is less than one million (10 lakh) +Hence, a million babies cannot born in a day. +(c) Can you count 1 million coins in a day? Assume you can count 1 +coin every second. +Ans: No. Because, Coins counted per day = 86,400 + + + + + + +12 + + Figure it out +Page No. 19 +1. Using all digits from 0 – 9 exactly once (the first digit cannot be 0) to +create a 10-digit number, write the — +(a) Largest multiple of 5 +Ans: largest multiple of 5 = 9876543210 + +(b) Smallest even number +Ans: The smallest such even number = 1023456798 + +2. The number 10,30,285 in words is ten lakhs thirty thousand two +hundred eighty-five, which has 42 letters. Give a 7-digit number name +which has the maximum number of letters. +Ans: One such number is 77,77,777, which has 60 letters. + +3. Write a 9-digit number where exchanging any two digits results in a +bigger number. How many such numbers exist? +Ans: One such number is 987654312. +Exchanging last two digits, we get 987654321, which is bigger number +than the initial number. +Try more! + +4. Strike out 10 digits from the number 12345123451234512345 so that +the remaining number is as large as possible. +Ans: 5534512345 + +6. Suppose you write down all the numbers 1, 2, 3, 4, …, 9, 10, 11, and so +on. The tenth digit you write is ‘1’ and the eleventh digit is ‘0’, as part +of the number 10. (This question can be solved in different ways. One +of the way is —) +(a) What would the 1000th digit be? At which number would it +occur? +Ans: Digits from 1 to 9 = 9 digits +Digits from 10 to 99 = 99-10+1 = 90 numbers × 2 = 180 digits +Total digits from 1 to 99 = 9 + 180 = 189 digits +Remaining digits to reach 1000th digit = 1000 – 189 = 811 +Number of 3-digit numbers = +811 +3 = 270 full numbers + 1 digit left +over. +The first 3-digit number is 100 +270th 3-digit number is 100 + 270 – 1 = 369 + +13 + +The next number is 370. +The first digit of 370, which is 3, is the 1000th digit. + +(b) What number would contain the millionth digit? +Ans: The millionth digit occurs in the number 185184+1=1,85,185. +(c) When would you have written the digit ‘5’ for the 5000th time? +Ans: 13995 +7. A calculator has only ‘+10,000’ and ‘+100’ buttons. Write an +expression describing the number of button clicks to be made for the +following numbers: +(a) 20,800 +Ans: 20,800 = (2 × 10,000) + (8 × 100) +Total clicks = 2 + 8 = 10 button clicks. + +(b) 92,100 +Ans: 92,100 = (9 × 10,000) + (21 × 100) +Total clicks = 9 + 21 = 30 button clicks. + +(c) 1,20,500 +Ans: 1,20,500 = (12 × 10,000) + (5 × 100) +Total clicks = 12 + 5 = 17 button clicks. + +(d) 65,30,000 +Ans: 65,30,000 = (653 × 10,000) +Total clicks = 653 button clicks. + +(e) 70,25,700 +Ans: 70,25,700 = (702 × 10,000) + (57 × 100) +Total clicks = 759 button clicks. + +8. How many lakhs make a billion? +Ans: 10,000 lakh make a billion. +9. You are given two sets of number cards numbered from 1 – 9. Place a +number card in each box below to get the (a) largest possible sum (b) +smallest possible difference of two resulting numbers. +(a) Ans: +Sum = 1,00,54,320 +9 +9 +8 +8 +7 +7 +6 +6 +5 +5 +4 +4 + +14 + + +(b) Ans: +Difference = 10,22,447 + + +10. You are given some number cards; 4000, 13000, 300, 70000, 150000, 20, +5. Using the cards get as close as you can to the numbers below using +any operation you want. Each card can be used only once for making +a particular number. +(a) 1,10,000: 4000 × (20 + 5) + 13000 = 1,13,000 +(b) 2,00,000: +Ans: closest estimate = 1,50,000 + 70,000 – (4000 × 5) = 2,00,000. + +(c) 5,80,000: +Ans: closest estimate = (1,50,000 × 4) – (4000 × 5) = 5,80,000. + +(d) 12,45,000: +Ans: +closest estimate = (70,000 × 20) – 1,50,000 – 4,000 – (300 × 5) = +12,44,500. + +(e) 20,90,800: +Ans: closest estimate = (1,50,000 × 14) + 4,000 – 13,000 = 20,91,000. + +11. Find out how many coins should be stacked to match the height of the +Statue of Unity. Assume each coin is 1 mm thick. +Ans: +Height of each coin = 1 mm +Height of Statue of Unity = 180 m = 180 × 100 × 10 mm = 1, 80,000 mm. +Number of coins to be stacked = 1,80,000 mm ÷ 1 mm = 1,80,000 coins. + +12. Grey-headed Albatrosses have a roughly 7-feet wide wingspan. They +are known to migrate across several oceans. Albatrosses can cover +about 900 – 1000 km in a day. One of the longest single trips recorded +is about 12,000 km. How many days would such a trip take to cross an +ocean approximately? + + +1 +1 +2 +2 +3 +3 +4 +9 +9 +8 +8 +7 + +15 + +Ans: +Estimated number of days if it flies 900 km/day = 12,000 ÷ 900 + = 13.3 days +Estimated number of days if it flies 1000 km/day = 12,000 ÷ 1000 + = 12 days +Hence it would take approximately 12 to 14 days for a grey headed +albatross to complete a 12,000 km trip to cross an ocean. + +13. A bar-tailed godwit holds the record for the longest recorded non- +stop flight. It travelled 13,560 km from Alaska to Australia without +stopping. Its journey started on 13 October 2022 and continued for +about 11 days. Find out the approximate distance it covered every +day. Find out the approximate distance it covered every hour. +Ans: +Distance covered per day = 13,560 km ÷ 11 days = 1232.73 km or 1233 km. +Distance covered per hour = 1233 ÷ 24 = 51.36 km or 51 km. +14. Bald eagles are known to fly as high as 4500 – 6000 m above the ground +level. Mount Everest is about 8850 m high. Aeroplanes can fly as high +as 10,000 – 12,800 m. How many times bigger are these heights +compared to Somu’s building? + +Ans: +Height of Somu’s building = 40 m +(i) Bald eagles’ flight height = 4500 – 6000 m +Lower estimate = 4500 ÷ 40 = 112.5 times +Upper estimate = 6000 ÷ 40 = 150 times +Bald eagles fly about 112 to 150 times higher than Somu’s building. + +(ii) Mount Everest height = 8850 m +Now, 8850 ÷ 40 = 221.25 +Mount Everest is approximately 221 times taller than Somu’s +building. + +(iii) Aeroplanes flight height = 10,000 – 12,000 m +Lower estimate = 10,000 ÷ 40 = 250 +Upper estimate = 12800 ÷ 40 = 320. +Airplanes fly about 250 to 320 times as high as Somu’s building." +class_7,2,Arithmetic Expressions,ncert_books/class_7/gegp1dd/gegp102.pdf,"2.1 Simple Expressions +You may have seen mathematical phrases like 13 + 2, 20 – 4, 12 × 5, and +18 ÷ 3. Such phrases are called arithmetic expressions. +Every arithmetic expression has a value which is the number it +evaluates to. For example, the value of the expression 13 + 2 is 15. This +expression can be read as ‘13 plus 2’ or ‘the sum of 13 and 2’. +We use the equality sign ‘=’ to denote the relationship between an +arithmetic expression and its value. For example: +13 + 2 = 15. +Example 1: Mallika spends ₹25 every day for lunch at school. Write +the expression for the total amount she spends on lunch in a week +from Monday to Friday. +The expression for the total amount is 5 × 25. +5 × 25 is “5 times 25” or “the product of 5 and 25”. +Different expressions can have the same value. Here are multiple +ways to express the number 12, using two numbers and any of the four +operations +, – , × and ÷: +10 + 2, 15 – 3, 3 × 4, 24 ÷ 2. +Choose your favourite number and write as many expressions as you +can having that value. +Comparing Expressions +As we compare numbers using ‘=’, ‘<’ and ‘>’ signs, we can also compare +expressions. We compare expressions based on their values and write +the ‘equal to’, ‘greater than’ or ‘less than’ sign accordingly. For example, +10 + 2 > 7 + 1 +ARITHMETIC +EXPRESSIONS +2 + +Arithmetic Expressions +25 +because the value of 10 + 2 = 12 is greater than the value of 7 + 1 =  8. +Similarly, +13 – 2 < 4 × 3. +Figure it Out +1. Fill in the blanks to make the expressions equal on both sides of +the = sign: +(a) 13 + 4 = ____ + 6 +(b) 22 + ____ = 6 × 5 +(c) 8 × ____ = 64 ÷ 2 +(d) 34 – ____ = 25 +2. Arrange the following expressions in ascending (increasing) order +of their values. +(a) 67 – 19 +(b) 67 – 20 +(c) 35 + 25 +(d) 5 × 11 +(e) 120 ÷ 3 +Example 2: Which is greater? 1023 + 125 or 1022 + 128? +Imagining a situation could help us answer this +without finding the values. Raja had 1023 marbles +and got 125 more today. Now he has 1023 + 125 +marbles. Joy had 1022 marbles and got 128 +more today. Now he has 1022 + 128 marbles. Who +has more? +This situation can be represented as shown in +the picture on the right. To begin with, Raja had 1 +more marble than Joy. But Joy got 3 more marbles +than Raja today. We can see that Joy has (two) +more marbles than Raja now. +That is, + + + +1023 + 125 < 1022 + 128. +Example 3: Which is greater? 113 – 25 or 112 – 24? +Imagine a situation, Raja had 113 marbles and lost +25 of them. He has 113 – 25 marbles. Joy had 112 +marbles and lost 24 today. He has 112 – 24 marbles. +Who has more marbles left with them? +Raja had 1 marble more than Joy. But he also +lost 1 marble more than Joy did. Therefore, they +have an equal number of marbles now. +That is, +113 – 25 = 112 – 24. +Raja (1023 + 125) +Joy (1022 + 128) +1022 +1022 +125 +125 +1 +1 +1 +1 +Raja (113 – 25) +remove +112 +24 +1 +Joy (112 – 24) +remove +112 +24 + +Ganita Prakash | Grade 7 +26 +Use ‘>’ or ‘<’ or ‘=’ in each of the following expressions to compare +them. Can you do it without complicated calculations? Explain your +thinking in each case. +(a) 245 + 289 + 246 + 285 +(b) 273 – 145 + 272 – 144 +(c) 364 + 587 + 363 + 589 +(d) 124 + 245 + 129 + 245 +(e) 213 – 77 + 214 – 76 +2.2 Reading and Evaluating Complex Expressions +Sometimes, when an expression is not accompanied by a context, there +can be more than one way of evaluating its value. In such cases, we +need some tools and rules to specify how exactly the expression has to +be evaluated. +To give an example with language, look +at the following sentences: +(a) Sentence: “Shalini sat next to a +friend with toys”. + +Meaning: The friend has toys and +Shalini sat next to her. +(b) Sentence: “Shalini sat next to a +friend, with toys”. + +Meaning: Shalini has the toys +and she sat with them next to her +friend. +This sentence without the punctuation could have been interpreted +in two different ways. The appropriate use of a comma specifies how +the sentence has to be understood. +Let us see an expression that can be evaluated in more than one way. +Example 4: Mallesh brought 30 marbles to the playground. Arun +brought 5 bags of marbles with 4 marbles in each bag. How many +marbles did Mallesh and Arun bring to the playground? +Mallesh summarized this by writing the mathematical expression — +30 + 5 × 4. + +Arithmetic Expressions +27 +Without knowing the context behind this expression, Purna found +the value of this expression to be 140. He added 30 and 5 first, to get 35, +and then multiplied 35 by 4 to get 140. +Mallesh found the value of this expression to be 50. He multiplied 5 +and 4 first to get 20 and added 20 to 30 to get 50. +In this case, Mallesh is right. But why did Purna get it wrong? +Just looking at the expression 30 + 5 × 4, it is not clear whether we +should do the addition first or multiplication. +Just as punctuation marks are used to resolve confusions in language, +brackets and the notion of terms are used in mathematics to resolve +confusions in evaluating expressions. +Brackets in Expressions +In the expression to find the number of marbles — 30 + 5 × 4 — we had +to first multiply 5 and 4, and then add this product to 30. This order of +operations is clarified by the use of brackets as follows: +30 + (5 × 4). +When evaluating an expression having brackets, we need to first find +the values of the expressions inside the brackets before performing +other operations. So, in the above expression, we first find the value +of 5 × 4, and then do the addition. Thus, this expression describes the +number of marbles: +30 + (5 × 4 ) = 30 + 20 = 50. +Example 5: Irfan bought a pack of biscuits for ₹15 and a packet of toor +dal for ₹56. He gave the shopkeeper ₹100. Write an expression that can +help us calculate the change Irfan will get back from the shopkeeper. +Irfan spent ₹15 on a biscuit packet and ₹56 on toor dal. So, the total +cost in rupees is 15 + 56. He gave ₹100 to the shopkeeper. So, he should +get back 100 minus the total cost. Can we write that expression as— +100 – 15 + 56 ? +Can we first subtract 15 from 100 and then add 56 to the result? We +will get 141. It is absurd that he gets more money than he paid the +shopkeeper! +We can use brackets in this case: +100 – (15 + 56). +Evaluating the expression within the brackets first, we get 100 minus +71, which is 29. So, Irfan will get back ₹29. + +Ganita Prakash | Grade 7 +28 +Terms in Expressions +Suppose we have the expression 30 + 5 × 4 without any brackets. Does +it have no meaning? +When there are expressions having multiple operations, and the +order of operations is not specified by the brackets, we use the notion +of terms to determine the order. +Terms are the parts of an expression separated by a ‘+’ sign. For +example, in 12+7, the terms are 12 and 7, as marked below. +12 +7 ++ +12 + 7 = +We will keep marking each term of an expression as above. Note +that this way of marking the terms is not a usual practice. This will be +done until you become familiar with this concept. +Now, what are the terms in 83 – 14? We know that subtracting a +number is the same as adding the inverse of the number. Recall that +the inverse of a given number has the sign opposite to it. For example, +the inverse of 14 is –14, and the inverse of –14 is 14. Thus, subtracting +14 from 83 is the same as adding –14 to 83. That is, +83 +– 14 ++ +83 – 14 = + +Thus, the terms of the expression 83 – 14 are 83 and –14. +Check if replacing subtraction by addition in this way does not change +the value of the expression, by taking different examples. +Can you explain why subtracting a number is the same as adding its +inverse, using the Token Model of integers that we saw in the Class +6 textbook of mathematics? +All subtractions in an expression are converted to additions in this +manner to identify the terms. +Here are some more examples of expressions and their terms: +– 18 +– 3 ++ +–18 – 3 = +6 × 5 +3 ++ +6 × 5 + 3 = +2 +– 10 ++ +2 – 10 + 4 × 6 = +4 × 6 ++ +Note that 6 × 5, 4 × 6 are single terms as they do not have any ‘+’ sign. +In the following table, some expressions are given. Complete the table. +Try +This + +Arithmetic Expressions +29 +Expression +Expression as the sum of its terms +Terms +13 – 2 + 6 +13 +– 2 ++ +6 ++ +13, – 2, 6 +5 + 6 × 3 +5 +6 × 3 ++ +4 + 15 – 9 ++ ++ +23 – 2 × 4 + 16 ++ ++ +28 + 19 – 8 ++ ++ +Now we will see how terms are used to determine the order of +operations to find the value of an expression. +We will start with expressions having only additions (with all the +subtractions suitably converted into additions). +Does changing the order in which the terms are added give different +values? +Swapping and Grouping +Let us consider a simple expression having only two terms. +Example 6: Madhu is flying a drone from a terrace. The drone goes 6 m +up and then 4 m down. Write an expression to show how high the final +position of the drone is from the terrace. +The drone is 6 – 4 = 2 m above the terrace. Writing it as sum of terms: +6 +– 4 ++ +2 += +Will the sum change if we swap the terms? +– 4 + 6 ++ +2 += +It doesn’t in this case. +We already know that swapping the terms does not change the sum +when both the terms are positive numbers. +Will this also hold when there are terms having negative numbers as +well? Take some more expressions and check. + +Ganita Prakash | Grade 7 +30 +Can you explain why this is happening using the Token Model of +integers that we saw in the Class 6 textbook of mathematics? +Thus, in an expression having two terms, swapping them does not +change the value. +Term 1 +Term 2 +Term 2 ++ +Term 1 += ++ +Now consider an expression having three terms: (–7) + 10 + (–11). Let +us add these terms in the following two different orders: +– 7 + 10 ++ +– 11 ++ +(adding the first two terms and then adding their sum to the third +term) +– 7 + 10 ++ +– 11 ++ +(adding the last two terms and then adding their sum to the first +term) +What do you see? The sums are the same in both cases. +Again, we know that while adding positive numbers, grouping them +in any of the above two ways gives the same sum. +Will this also hold when there are terms having negative numbers as +well? Take some more expressions and check. +Can you explain why this is happening using the Token Model of +integers that we saw in the Class 6 textbook of mathematics? +Thus, grouping the terms of an expression in either of the following +ways gives the same value. +Term 1 +Term 2 +Term 1 +Term 3 +Term 2 ++ +Term 3 ++ ++ += ++ +Let us consider the expression (–7) + 10 + (–11) again. What happens +when we change the order and add –7 and –11 first, and then add this +sum to 10? Will we get the same sum as before? +We see that adding the terms of the expression (–7) + 10 + (–11) in +any order gives the same sum of –8. +Try +This +Try +This + +Arithmetic Expressions +31 +Does adding the terms of an expression in any order give the same +value? Take some more expressions and check. Consider expressions +with more than 3 terms also. +Can you explain why this is happening using the Token Model of +integers that we saw in the Class 6 textbook of mathematics? +Thus, the addition of terms in any order gives the same value. +Therefore, in an expression having only additions, it does not matter +in what order the terms are added: they all give the same value. +Now let us consider expressions having multiplication and division +also, without the order of operations specified by the brackets. The +values of such expressions are found by first evaluating the terms. +Once all the terms are evaluated, they are added. +For example, the expression 30 + 5 × 4 is evaluated as follows: +30 +30 +5 × 4 +20 ++ ++ +30 + 5 × 4 = +50 += += +The expression 5 × (3 + 2) + 78 + 3 is evaluated as follows: +5 × (3 + 2) +3 ++ +5 × (3 + 2) + 78 + 3 = ++ +7 × 8 +Where (3+2) is first evaluated and this sum is multiplied by 5 (= 25). +The expression 7 × 8 is evaluated (= 56). This simplifies to 25 + 56 + 3 = +84. +Manasa is adding a long list of numbers. It took her five +minutes to add them all and she got the answer 11749. Then +she realised that she had forgotten to include the fourth +number 9055. Does she have to start all over again? +In mathematics we use the phrase commutative property +of addition instead of saying “swapping terms does not +change the sum”. Similarly, “grouping does not change the sum” is +called the associative property of addition. +Swapping the Order of Things in Everyday Life +Manasa is going outside to play. Her mother says, “Wear +your hat and shoes!” Which one should she +wear first? She can wear her hat first and then +her shoes. Or she can wear her shoes first and +then her hat. +Manasa will look exactly the same in both cases. +Imagine a different situation: Manasa’s mother +says “Wear your socks and shoes!” Now the +Try +This +1342 +774 +8611 +9055 +1022 + +Ganita Prakash | Grade 7 +32 +order matters. She should wear socks and then shoes. If she wears +shoes and then socks, Manasa will feel very uncomfortable and look +very different. +More Expressions and Their Terms +Example 7: Amu, Charan, Madhu, and John went to a hotel and ordered +four dosas. Each dosa cost ₹23, and they wish to thank the waiter by +tipping ₹5. Write an expression describing the total cost. +Cost of 4 dosas = 4 × 23 +Can the total amount with tip be written as 4 × 23 + 5? Evaluating it, +we get +4 × 23 +92 +5 +5 ++ ++ +4 × 23 + 5 = +97 += += +Thus, 4 × 23 + 5 is a correct way of writing the expression. +If the total number of friends goes up to 7 and the tip remains the +same, how much will they have to pay? Write an expression for this +situation and identify its terms. +Example 8: Children in a class are playing “Fire in the mountain, run, +run, run!”. Whenever the teacher calls out a number, students are +supposed to arrange themselves in groups of that number. Whoever is +not part of the announced group size, is out. +Ruby +wanted +to rest and sat on +one side. The other +33 +students +were +playing the game in +the class. +The teacher called +out ‘5’. Once +children settled, +Ruby wrote +6 × 5 + 3 +(understood as 3 more than 6 × 5) +Think and discuss why she wrote this. +The expression written as a sum of terms is— +6 × 5 +3 . ++ ++ ++ ++ + +Arithmetic Expressions +33 +For each of the cases below, write the expression and identify its terms: +If the teacher had called out ‘4’, Ruby would write ____________ +If the teacher had called out ‘7’, Ruby would write ____________ +Write expressions like the above for your class size. +Example 9: Raghu bought 100 kg of rice from the wholesale market +and packed them into 2 kg packets. He already had four 2 kg packets. +Write an expression for the number of 2 kg packets of rice he has now +and identify the terms. +He had 4 packets. The number of new 2 kg packets of rice is 100 ÷ 2, +which we also write as 100 +2 . +The number of 2 kg packets he has now is 4 + 100 +2 . The terms are— +4 +100 +2 +. ++ +Example 10: Kannan has to pay ₹432 to a shopkeeper using coins of ₹1 +and ₹5, and notes of ₹10, ₹20, ₹50 and ₹100. How can he do it? +There is more than one possibility. For example, +432 = 4 × 100 + 1 × 20 + 1 × 10 + 2 × 1 +Meaning: 4 notes of ₹100, 1 note of ₹20, 1 note of ₹10 and 2 notes of ₹1 +432 = 8 × 50 + 1 × 10 + 4 × 5 + 2 × 1 +Meaning: 8 notes of ₹50, 1 note of ₹10, 4 notes of ₹5 and 2 notes of ₹1 +Identify the terms in the two expressions above. +Can you think of some more ways of giving ₹432 to someone? +Example 11: Here are two pictures. Which of these two arrangements +matches with the expression 5 × 2 + 3? + +Ganita Prakash | Grade 7 +34 +Let us write this expression as a sum of terms. +5 × 2 +10 +3 +3 ++ ++ +13 += += +This expression 5 × 2 + 3 can be understood as 3 more than 5 × 2, +which describes the arrangement on the left. +What is the expression for the arrangement in the right making use of +the number of yellow and blue squares? +Do you recall the use of brackets? We need to use brackets for this. +2 × (5 + 3) +Notice that this arrangement can also be described using— +5 + 3 + 5 + 3 +OR +5 × 2 + 3 × 2 +Figure it Out +1. Find the values of the following expressions by writing the terms +in each case. +(a) 28 – 7 + 8 +(b) 39 – 2 × 6 + 11 +(c) 40 – 10 + 10 + 10 +(d) 48 – 10 × 2 + 16 ÷ 2 +(e) 6 × 3 – 4 × 8 × 5 +2. Write a story/situation for each of the following expressions and +find their values. +(a) 89 + 21 – 10 +(b) 5 × 12 – 6 +(c) 4 × 9 + 2 × 6 +3. For each of the following situations, write the expression describing +the situation, identify its terms and find the value of the expression. +(a) Queen Alia gave 100 gold coins to Princess Elsa and 100 gold +coins to Princess Anna last year. Princess Elsa used the coins +to start a business and doubled her coins. Princess Anna +bought jewellery and has only half of the coins left. Write +an expression describing how many gold coins Princess Elsa +and Princess Anna together have. +(b) A metro train ticket between two stations is ₹40 for an adult +and ₹20 for a child. What is the total cost of tickets: +(i) + for four adults and three children? +(ii) +for two groups having three adults each? + +Arithmetic Expressions +35 +(c) Find the total height of the window + +by +writing +an +expression +describing the relationship among +the measurements shown in the +picture. +Removing Brackets — I +Let us find the value of this expression, +200 – (40 + 3). +We first evaluate the expression inside the +bracket to 43 and then subtract it from 200. +But it is simpler to first subtract 40 from 200: +200 – 40 = 160. +And then subtract 3 from 160: +160 – 3 = 157. +What we did here was 200 – 40 – 3. Notice, that we did not do +200 – 40 + 3. +So, +200 – (40 + 3) = 200 – 40 – 3. +Example 12: We also saw this earlier in the case of Irfan purchasing a +biscuit packet (₹15) and a toor dal packet (₹56). When he paid ₹100, the +change he gets in rupees is: +100 – (15 + 56) = 29. +The change could also have been calculated as follows: +(a) First subtract the cost of the biscuit packet (15) from 100: +100 – 15 = 85. + +This is the amount the shopkeeper owes Irfan if he had +purchased only the biscuits. As he has purchased toor dal +also, its cost is taken from this remaining amount of 85. +(b) So, to find the change, we need to subtract the cost of toor +dal from 85. +85 – 56 = 29. + +What we have done here is 100 – 15 – 56. So, +100 – (15 + 56) = 100 – 15 – 56. +Notice how upon removing the brackets preceded by a negative +sign, the signs of the terms inside the brackets change. Observe +3 cm +2 cm +5 cm +Border +Grill +Gap +Total +Height + +Ganita Prakash | Grade 7 +36 +the signs of 40 and 3 in the first example, and that of 15 and 56 in +the second. +Example 13: Consider the expression 500 – (250 – 100). Is it possible to +write this expression without the brackets? +To evaluate this expression, we need to subtract 250 – 100 = 150 +from 500: +500 – (250 – 100) = 500 – 150 = 350. +If we were to directly subtract 250 from 500, then we would have +subtracted 100 more than what we needed to. So, we should add back +that 100 to 500 – 250 to make the expression take the same value as 500 +– (250 – 100). This sequence of operations is 500 – 250+100. Thus, +500 – (250 – 100) = 500 – 250 + 100. +Check that 500 – (250 – 100) is not equal to 500 – 250 – 100. +Notice again that when the brackets preceded by a negative sign +are removed, the signs of the terms inside the brackets change. In this +case, the signs of 250 and – 100 change to – 250 and 100. +Example 14: Hira has a rare coin collection. She has 28 coins in one bag +and 35 coins in another. She gifts her friend 10 coins from the second +bag. Write an expression for the number of coins left with Hira. +This can be expressed by 28 + (35 – 10). +We know that this is the same as 28 + (35 + (–10)). Since the terms +can be added in any order, this expression can simply be written as +28 + 35 + (–10), or 28 + 35 – 10. Thus, +28 + (35 – 10) = 28 + 35 – 10 = 53. +When the brackets are NOT preceded by a negative sign, the terms +within them do not change their signs upon removing the brackets. +Notice the sign of the terms 35 and – 10 in the above expression. +Rather than simply remembering rules for when to change the sign and +when not to, you can figure it out for yourself by thinking about the +meanings of the expressions. +Tinker the Terms I +What happens to the value of an expression if we increase or decrease +the value of one of its terms? +Some expressions are given in following three columns. In each +column, one or more terms are changed from the first expression. Go +through the example (in the first column) and fill the blanks, doing as +little computation as possible. + +Arithmetic Expressions +37 +53 +– 16 ++ +37 += +53 +– 16 ++ +37 += +– 87 +– 16 ++ += +54 +– 16 ++ +38 += +54 is one more than +53, so the value will +be 1 more than 37. +53 +– 15 ++ += +Is –15 one more or +one less than –16? +52 +– 16 ++ += +52 is one less than +53, so the value will +be 1 less than 37. +53 +– 17 ++ += +Is –17 one more or +one less than –16? +– 88 +– 15 ++ += +– 86 +– 18 ++ += +– 97 +– 26 ++ += +Figure it Out +1. Fill in the blanks with numbers, and boxes with operation signs +such that the expressions on both sides are equal. +(a) 24 + (6 – 4) = 24 + 6 + _____ +(b) 38 + (_____ + _____) = 38 + 9 – 4 +(c) 24 – (6 +4) = 24 + 6 – 4 +(d) 24 – 6 – 4 = 24 – 6 + _____ +(e) 27 – (8 + 3) = 27 8 3 +(f) 27– (_____ + _____) = 27 – 8 + 3 +2. Remove the brackets and write the expression having the same +value. +(a) 14 + (12 + 10) +(b) 14 – (12 + 10) +(c) 14 + (12 – 10) +(d) 14 – (12 – 10) +(e) –14 + 12 – 10 +(f) +14 – (–12 – 10) +3. Find the values of the following expressions. For each pair, first +try to guess whether they have the same value. When are the two +expressions equal? +(a) (6 + 10) – 2 and 6 + (10 – 2) +(b) 16 – (8 – 3) and (16 – 8) – 3 +(c) 27 – (18 + 4) and 27 + (–18 – 4) +4. In each of the sets of expressions below, identify those that +have the same value. Do not evaluate them, but rather use your +understanding of terms. + +Ganita Prakash | Grade 7 +38 +(a) 319 + 537, 319 – 537, – 537 + 319, 537 – 319 +(b) 87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109, 87 – 46 + 109, 87 +– (46 + 109), (87 – 46) + 109 +5. Add brackets at appropriate places in the expressions such that +they lead to the values indicated. +(a) 34 – 9 + 12 = 13 +(b) 56 – 14 – 8 = 34 +(c) –22 – 12 + 10 + 22 = – 22 +6. Using only reasoning of how terms change their values, fill the +blanks to make the expressions on either side of the equality (=) +equal. +(a) 423 + ______= 419 + ______ +(b) 207 – 68 = 210 – ______ +7. Using the numbers 2, 3 and 5, and the operators ‘+’ and ‘–’, and     +brackets, as necessary, generate expressions to give as many +different values as possible. For example, 2 – 3 + 5 = 4 and 3 – (5 – 2) += 0. +8. Whenever Jasoda has to subtract 9 from a number, she subtracts 10 +and adds 1 to it. For example, 36 – 9 = 26 + 1. +(a) Do you think she always gets the correct answer? Why? +(b) Can you think of other similar strategies? Give some +examples. +9. Consider the two expressions: a) 73 – 14 + 1, b) 73 – 14 – 1. For each +of these expressions, identify the expressions from the following +collection that are equal to it. +(a) 73 – (14 + 1) +b) +73 – (14 – 1) +(c) 73 + (– 14 + 1) +d) +73 + (– 14 – 1) +Removing Brackets — II +Example 15: Lhamo and Norbu went to a hotel. Each of them ordered +a vegetable cutlet and a rasgulla. A vegetable cutlet costs ₹43 and a +rasgulla costs ₹24. Write an expression for the amount they will have +to pay. +As each of them had one vegetable cutlet and one rasagulla, each of +their shares can be represented by 43 + 24. +What about the total amount they have to pay? Can it be described by +the expression: 2 × 43 + 24? +Math +Talk + +Arithmetic Expressions +39 +Writing it as sum of terms gives: +2 × 43 +24 ++ +This expression means 24 more than +2 × 43. But, we want an expression +which means twice or double of 43 + 24. +We can make use of brackets to write +such an expression: +2 × (43 + 24). +So, we can say that together they have +to pay 2 × (43 + 24). This is also the same +as paying for two vegetable cutlets and +two rasgullas: +2 × 43 + 2 × 24. +Therefore, +2 × (43 + 24) = 2 × 43 + 2 × 24. +If another friend, Sangmu, joins them and orders the same items, what +will be the expression for the total amount to be paid? +Example 16: In the Republic Day parade, there are boy scouts and girl +guides marching together. The scouts march in 4 rows with 5 scouts in +each row. The guides march in 3 rows with 5 guides in each row (see +the figure below). How many scouts and guides are marching in this +parade? +The number of boy scouts marching is 4 × 5. The number of girl +guides marching is 3 × 5. +The total number of +scouts and guides will be +4 × 5 + 3 × 5. +This can also be found +by first finding the total +number of rows, i.e., 4 + +3, and then multiplying +their sum by the number +of children in each row. +Thus, the number of boys +and girls can be found by +(4 + 3) × 5. +Therefore, 4 × 5 + 3 × 5 += (4 + 3) × 5. +Computing these expressions, we get +4 × 5 +20 +3 × 5 +15 ++ ++ +4 × 5 + 3 × 5 = +35 += += + +2 × 43 + 2 × 24 +2 × (43 + 24) +₹ 43 +₹ 43 +₹ 24 +₹ 24 +3 × 5 +(4+3) × 5 +4 × 5 + +Ganita Prakash | Grade 7 +40 +(4 + 3) × 5 = 7 × 5 = 35 +5 × 4 + 3 ≠ 5 × (4 + 3). Can you explain why? +Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5? +The observations that we have made in the previous two examples +can be seen in a general way as follows. +Consider 10 × 98 + 3 × 98. This means taking the sum of 10 times 98 +and 3 times 98. +98 + 98 + 98 + 98 + 98 + 98 + 98 + 98 + 98 + 98 + 98 + 98 + 98 +10 times +3 times +Clearly, this is the same as 10 + 3 = 13 times 98. Thus, +10 × 98 +3 × 98 = (10 + 3) × 98. +Writing this equality the other way, we get +(10 + 3) 98 = 10 × 98 + 3 × 98. +Swapping the numbers in the products above, this property can be +seen in the following form: +98 × 10 + 9 × 83 = 98 (10 + 3), and +98 (10 + 3) = 98 × 10 + 98 × 3. +Similarly, let us consider the expression 14 × 10 – 6 × 10. This means +subtracting 6 times 10 from 14 times 10. +10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 +14 times +Clearly, this is 14 – 6 = 8 times 10. Thus, +14 × 10 – 6 × 10 = (14 – 6) × 10, +or +(14 – 6) × 10 = 14 × 10 – 6 × 10 +This property can be nicely summed up as follows: +The multiple of a sum (difference) is the same as the sum (difference) +of the multiples. +Tinker the Terms II +Let us understand what happens when we change the numbers +occurring in a product. + +Arithmetic Expressions +41 +Example 17: Given 53 × 18 = 954. Find out 63 × 18. +As 63 × 18 means 63 times 18, +63 × 18 = (53 + 10) × 18 + + += 53 ×18 + 10×18 + + += 954 + 180 + + += 1134. +Example 18: Find an effective way of evaluating 97 × 25. +97 × 25 means 97 times 25. +We can write it as (100 – 3) × 25 +We know that this is the same as the difference of 100 times 25 and +3 times 25: +97 × 25 = 100 × 25 – 3 × 25 +Find this value. +Use this method to find the following products: +(a) 95 × 8 +(b) 104 × 15 +(c) 49 × 50 + +Is this quicker than the multiplication procedure you use generally? +Which other products might be quicker to find like the ones above? +Figure it Out +1. Fill in the blanks with numbers, and boxes by signs, so that the +expressions on both sides are equal. +(a) 3 × (6 + 7) = 3 × 6 + 3 × 7 +(b) (8 + 3) × 4 = 8 × 4 + 3 × 4 +(c) 3 × (5 + 8) = 3 × 5 + 3 × ____ +(d) (9 + 2) × 4 = 9 × 4 + 2 ×____ +(e) 3 × (____ + 4) = 3 ____+____ +(f) (____+ 6) × 4 = 13 × 4 + ____ +(g) 3 × (____+____) = 3 × 5 + 3 × 2 +(h) (____+____)×____= 2 × 4 + 3 × 4 +(i) +5 × (9 – 2) = 5 × 9 – 5 × ____ +(j) +(5 – 2) × 7 = 5 × 7 – 2 × ____ +(k) 5 × (8 – 3) = 5 × 8 + 5 × ____ +(l) +(8 – 3) × 7 = 8 × 7 + 3 × 7 +Math +Talk + +Ganita Prakash | Grade 7 +42 +(m) 5 × (12 –____) =____ + 5 ×____ +(n) (15 –____) × 7 =____ + 6 × 7 +(o) 5 × (____–____) = 5 × 9 – 5 × 4 +(p) (____–____) × ____= 17 × 7 – 9 × 7 +2. In the boxes below, fill ‘<’, ‘>’ or ‘=’ after analysing the expressions +on the LHS and RHS. Use reasoning and understanding of terms and +brackets to figure this out and not by evaluating the expressions. +(a) (8 – 3) × 29 + (3 – 8) × 29 +(b) 15 + 9 × 18 + (15 + 9) × 18 +(c) 23 × (17 – 9) + 23 × 17 + 23 × 9 +(d) (34 – 28) × 42 + 34 × 42 – 28 × 42 +3. Here is one way to make 14: _2_ × ( _1_ + _6_ ) = 14. Are there other +ways of getting 14? Fill them out below: +(a) _____× (_____+_____) = 14 +(b) _____× (_____+_____) = 14 +(c) _____× (_____+_____) = 14 +(d) _____× (_____+_____) = 14 +4. Find out the sum of the numbers given in each picture below in +at least two different ways. Describe how you solved it through +expressions. +Figure it Out +1. Read the situations given below. Write appropriate expressions for +each of them and find their values. +(a) The district market in Begur operates on all seven days of +a week. Rahim supplies 9 kg of mangoes each day from his +orchard and Shyam supplies 11 kg of mangoes each day from +his orchard to this market. Find the amount of mangoes +supplied by them in a week to the local district market. + +Arithmetic Expressions +43 +(b) Binu earns ₹20,000 per month. She spends ₹5,000 on rent, +₹5,000 on food, and ₹2,000 on other expenses every month. +What is the amount Binu will save by the end of a year? +(c) During the daytime a snail climbs 3 cm up a post, and during +the night while asleep, accidentally slips down by 2 cm. The +post is 10 cm high, and a delicious treat is on its top. In how +many days will the snail get the treat? +2. Melvin reads a two-page story every day except on Tuesdays and +Saturdays. How many stories would he complete reading in 8 +weeks? Which of the expressions below describes this scenario? +(a) 5 × 2 × 8 +(b) (7 – 2) × 8 +(c) 8 × 7 +(d) 7 × 2 × 8 +(e) 7 × 5 – 2 +(f) (7 + 2) × 8 +(g) 7 × 8 – 2 × 8 +(h) (7 – 5) × 8 +3. Find different ways of evaluating the following expressions: +(a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 +(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 +4. Compare the following pairs of expressions using ‘<’, ‘>’ or ‘=’ or by +reasoning. +(a) 49 – 7 + 8 + 49 – 7 + 8 +(b) 83 × 42 – 18 + 83 × 40 – 18 +(c) 145 – 17 × 8 + 145 – 17 × 6 +(d) 23 × 48 – 35 + 23 × (48 – 35) +(e) (16 – 11) × 12 + –11 × 12 + 16 × 12 +(f) (76 – 53) × 88 + 88 × (53 – 76) +(g) 25 × (42 + 16) + 25 × (43 + 15) +(h) 36 × (28 – 16) + 35 × (27 – 15) + +Ganita Prakash | Grade 7 +44 +5. Identify which of the following expressions are equal to the given +expression without computation. You may rewrite the expressions +using terms or removing brackets. There can be more than one +expression which is equal to the given expression. +(a) 83 – 37 – 12 +(i) + 84 – 38 – 12 +(ii) +84 – (37 + 12) +(iii) 83 – 38 – 13 +(iv) – 37 + 83 –12 +(b) 93 + 37 × 44 + 76 +(i) +37 + 93 × 44 + 76 +(ii) +93 + 37 × 76 + 44 +(iii) (93 + 37) × (44 + 76) +(iv) 37 × 44 + 93 + 76 +5. Choose a number and create ten different expressions having +that value. +• We have been reading and evaluating simple expressions for quite +some time now. Here we started by revising the meaning of some +simple expressions and their values. +• We learnt how to compare certain expressions through reasoning +instead of bluntly evaluating them. +• To help read and evaluate complex expressions without confusion, +we use terms and brackets. +• When an expression is written as a sum of terms, changing the order +of the terms or grouping the terms does not change the value of the +expression. This is because the “commutative property of addition” +and the “associative property of addition”, respectively. +• To evaluate expressions within brackets, we saw that when we +remove brackets preceded by a negative sign, the terms within the +bracket change their sign. +• We also learnt about the “distributive property” — multiplying a +number with an expression inside brackets is equal to the multiplying +the number with each term in the bracket. +SUMMARY + +Arithmetic Expressions +45 +  +Expression Engineer! +Using three 3’s along with the four operations (addition, subtraction, +multiplication, and division) and brackets as needed we can create +several expressions. For example, (3 + 3)/3 = 2, 3 + 3 – 3 = 3, 3 × +3 + 3 = 12, and so on. +Using four 4’s, create expressions to get all values from 1 to 20. +Using the numbers 1, 2, 3, 4, and 5 exactly once in any order get +as many values as possible between  – 10 and +10. +Using the numbers 0 to 9 exactly once in any order, make an +expression with a value 100. +What other similar interesting questions can you ask? + +1 + + +Page No. 24 + Choose your favourite number and write as many expressions as you +can having that value. +Ans: Let us choose the number 24. +The arithmetic expressions for the number can be as follows: +12 + 12 = 24; 4 × 6 = 24; 48 ÷ 2 = 24, 34 – 10 = 24, 20 + 4 = 24 etc. +Now you choose your number and form arithmetic expressions for that +number. +Page No. 25 + Figure it Out +1. Fill in the blanks to make the expressions equal on both sides of the = +sign: +(a) 13 + 4 = _____ + 6 + +(b) 22 + ______ = 6 × 5 +(c) 8 × ______ = 64 ÷ 2 + +(d) 34 – ______ = 25 +Ans: 1. (a) 11 (b) 8 (c) 4 (d) 9 +2. Arrange the following expressions in ascending (increasing) order of +their values. +(a) 67 – 19 + +(b) 67 – 20 +(c) 35 + 25 + + +(d) 5 × 11 + +(e) 120 ÷ 3 +Ans: 2. 120 ÷ 3 < 67 – 20 < 67 – 19 < 5 × 11 < 35 + 25 +Page No. 26 + Use ‘>’ or ‘<’ ‘=’ in each of the following expressions to compare them. +Can you do it without complicated calculations? Explain your thinking +in each case. +Ans: (a) > (b) = (c) < (d) < (e) < + +Chapter – 2 +Arithmetic Expressions + +2 + +Page No. 28 – 29 + Check if replacing subtraction by addition in this way does not change +the value of the expression, by taking different examples. +Ans: Let us take the numbers. 18 and 10 +18 – 10 = 8 and 18 + (– 10) = 8 + + Can you explain why subtracting a number is the same as adding its +inverse, using the Token Model of integers that we saw in the Class 6 +textbook of mathematics? +Ans: Refer chapter 10 of class VI Mathematics Textbook. + Does changing the order in which the terms are added give different +values? +Ans: No, +For example, 14 + 10 +( – 5) = 19 or (– 5) + 10 + 14 = 19 +Page No. 29 + Will this also hold when there are terms having negative numbers as +well? Take some more expressions and check. +Ans: +Yes, +For example: (– 4) + (– 2) = – 6 or (– 2) + (– 4) = – 6 +(– 6) + (– 8) = – 14 or (– 8) + (– 6) = – 14 +Write some more examples. +Page No. 30 + Can you explain why this is happening using the Token Model of +integers that we saw in the Class 6 textbook of mathematics? +Ans: + + + + +3 + + Will this also hold when there are terms having negative numbers as +well? Take some more expressions and check. +Ans: Yes +For example, + + + + +Other way + + Can you explain why this is happening using the Token Model of +integers that we saw in the Class 6 textbook of mathematics? +Ans: Yes. +Consider the expression: − 6 + (− 7) + (– 13) + +Page No. 31 + Does adding the terms of an expression in any order give the same +value? Take some more expressions and check. Consider expressions +with more than 3 terms also. +Ans: Yes, adding the terms of an expression in any order gives the same +value Consider an expression with 3 terms +9 + 4 + 6 +Order 1: 9 + 4 + 6 + = (9 + 4) + 6 = 13 + 6 = 19 +Order 2: 9 + 4 + 6 + = 9 + (4 + 6) = 9 + 10 = 19 +An expression with 4 terms (including a negative number) +Consider the expression: 15 + (– 8) + 5 + 12 +Order 1: (15 + (– 8)) + 5 + 12 = (7 + 5) + 12 = 12 + 12 = 24 +Order 2: 15 + 5 + 12 + (– 8) + = (15 + 5 + 12) + (– 8) + +4 + + = ((15 + 5) + 12) + (– 8) = (20 + 12) + (– 8) = 32 + (– 8) = 24 + + Manasa is adding a long list of numbers. It took her five minutes to +add them all and she got the answer 11749. Then she realised that she +had forgotten to include the fourth number 9055. Does she have to +start all over again? + +Ans: No, she can add fourth number, 9055 to the sum she got (11749) to get +the correct sum of the list of given numbers. That is 11749 + 9055 = +20804 (Using Associative Property) +Page No. 32 + If the total number of friends goes up to 7 and the tip remains the same, +how much will they have to pay? Write an expression for this situation +and identify its terms. +Ans: +The expression for the total cost = 7 × 23 + 5 = 161 + 5 = ₹166. +The terms in the expression 7 × 23 + 5 are 7 × 23, 5. +Page No. 33 + For each of the cases below, write the expression and identify its terms: +(a) If the teacher had called out ‘4’, Ruby would write _________ +(b) If the teacher had called out ‘7’, Ruby would write _________ +Write an expression like the above for your class size. +Ans: +(a) 8 × 4 + 1 +Terms: 8 × 4, 1 +(b) 4 × 7 + 5 +Terms: 4 × 7, 5 + +5 + + Identify the terms in the two expressions above. +Ans: +Way 1: 432 = 4 × 100 + 1 × 20 + 1 × 10 + 2 × 1 +Terms: 4 × 100, 1 × 20, 1 × 10, and 2 × 1 +Way 2: 432 = 8 × 50 + 1 × 10 + 4 × 5 + 2 × 1 +Terms: 8 × 50, 1 × 10, 4 × 5, and 2 × 1 + Can you think of some more ways of giving ₹ 432 to someone? +Ans: +40 × 10 + 3 × 10 + 2 × 1 +Terms: 40 × 10, 3 × 10 and 2×1 +Try for some more expressions. +Page No. 34 – 35 + Figure it Out +1. Find the values of the following expressions by writing the +terms in each case. +(a) 28 – 7 + 8 +(b) 39 – 2 × 6 + 11 +(c) 40 – 10 + 10 + 10 +(d) 48 – 10 × 2 + 16 ÷ 2 +(e) 6 × 3 – 4 × 8 × 5 +Ans: + +(a) 28 – 7 + 8 +Terms: 28, – 7, and 8 + + + + + +Value: 29 +(b) 39 – 2 × 6 + 11 +Terms: 39, – 2 × 6 and 11 + + + + +Value: 38 +(c) 40 – 10 + 10 + 10 = 40 + (– 10) + 10 + 10 +Terms: 40, – 10, 10 and 10 + + + + +Value: 50 +(d) 48 – 10 × 2 + 16 ÷ 2 = 48 + (– 10 × 2) + (16 ÷ 2) + Terms: 48, – 10 × 2, 16 ÷ 2 + + + + +Value: 36 +(e) 6 × 3 – 4 × 8 × 5 = (6 × 3) + (– 4 × 8 × 5) + Terms: 6 × 3, 4 × 8 × 5 + + + + +Value: – 142 + + +6 + +2. Write a story/situation for each of the following expressions and +find their values. +(а) 89 + 21 – 10 +(b) 5 × 12 – 6 +(c) 4 × 9 + 2 × 6 +Ans: +(a) 89 + 21 – 10; + +One of the stories could be; +A library had 89 books on its shelves. The librarian bought 21 new +books and added them to the collection. Later, 10 books were +borrowed by students. How many books remain on the shelves? +Value of the expression: 100 +(b) 5 × 12 – 6 +Make a story of your own for the given expression. +Value of the expression: 54 +(c) 4 × 9 + 2 × 6 +Make a story of your own for the given expression. +Value of the expression: 48 +3. For each of the following situations, write the expression describing +the situation, identify its terms, and find the value of the expression. +(а) Queen Alia gave 100 gold coins to Princess Elsa and 100 gold coins +to Princess Anna last year. Princess Elsa used the coins to start a +business and doubled her coins. Princess Anna bought jewellery +and has only half of the coins left. Write an expression describing +how many gold coins Princess Elsa and Princess Anna together +have. + + + + + +7 + +(b) A metro train ticket between two stations is ₹40 for an adult and +₹20 for a child. What is the total cost of the tickets? +(i) for four adults and three +children? +(ii) for two groups having three +adults each? + (c) Find the total height of the window +by +writing +an +expression +describing the relationship among +the measurements shown in the +picture. +Ans: +(a) +Expression describing the situation = 2 × 100 + +100 +2 +Terms: 2 × 100, +100 +2 + + +Value: 250 gold coins. +(b) +(i) Expression = 4 × 40 + 3 × 20 +Terms: 4 × 40, 3 × 20 + +Value: ₹220 +(ii) Expression = 2 × (3 × 40). +Terms: 2 × (3 × 40) +Value: ₹240 +(c) +Expression = 7 × 5 + 6 × 2 + 2 × 3 +Terms: 7 × 5, 6 × 2, 2 × 3 +Total height: 53 cm + +Page No. 37 +Some expressions are given in the following three columns. In each +column, one or more terms are changed from the first expression. Go +through the example (in the first column) and fill in the blanks, doing +as little computation as possible. + +8 + + +Ans: + + + +9 + +Page No. 37 – 38 + Figure it Out +1. Fill in the blanks with numbers, and boxes with operation signs such +that the expressions on both sides are equal. +(a) 24 + (6 – 4) = 24 + 6 + _________ +(b) 38 + (_________ + ________) = 38 + 9 – 4 +(c) 24 – (6 + 4) = 24 + 6 – 4 +(d) 24 – 6 – 4 = 24 – 6 + _________ +(e) 27 – (8 + 3) = 27 _________ 8 _________ 3 +(f) 27 – (_________ + ________) = 27 – 8 + 3 +Ans: +(a) 24 + (6 – 4) = 24 + 6 + 4 +(b) 38 + (9 + 4) = 38 + 9 – 4 +(c) 24 – (6 + 4) = 24 + 6 – 4 +(d) 24 – 6 – 4 = 24 – 6 + 4 +(e) 27 – (8 + 3) = 27 – 8 – 3 +(f) 27 – (8 + 3) = 27 – 8 + 3 +2. Remove the brackets and write the expression having the same +value. +(a) 14 + (12 + 10) +(b) 14 – (12 + 10) +(c) 14 + (12 – 10) +(d) 14 – (12 – 10) +(e) – 14 + 12 – 10 +(f) 14 – ( – 12 – 10) +Ans: (a) and (f) +3. Find the values of the following expressions. For each pair, first try +to guess whether they have the same value. When are the two +expressions equal? +(a) (6 + 10) – 2 and 6 + (10 – 2) +(b) 16 – (8 – 3) and (16 – 8) – 3 +(c) 27 – (18 + 4) and 27 + (– 18 – 4) + + + + +10 + +Ans: + +(a) (6 + 10) – 2 = 6 + 10 – 2 = 14 +6 + (10 – 2) = 6 + 10 – 2 = 14 +So, (6 + 10) – 2 = 6 + (10 – 2) +(b) 16 – (8 – 3) and (16 – 8) – 3 +16 – (8 – 3) = 16 – 8 + 3 = 11 +(16 – 8) – 3 = 16 – 8 – 3 = 5 +So, 16 – (8 – 3) ≠ (16 – 8) – 3 +(c) 27 – (18 + 4) and 27 + ( – 18 – 4) +27 – (18 + 4) = 27 – 18 – 4 = 5 +27 + ( – 18 – 4) = 27 – 18 – 4 = 5 +So, 27 – (18 + 4) = 27 + ( – 18 – 4) +4. In each of the sets of expressions below, identify those that have the +same value. Do not evaluate them, but rather use your +understanding of terms. +(a) 319 + 537, 319 – 537, – 537 + 319, 537 – 319 +(b) 87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109, 87 – 46 + 109, 87 – (46 + +109), (87 – 46) + 109 +Ans: +(a) 319 – 537 and – 537 + 319 have the same terms So, 319 – 537 and – +537 + 319 have the same value. +(b) 87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109 and also, 87 – 46 + 109 and +(87 – 46) + 109 have the same terms and so have equal values. + +5. Add brackets at appropriate places in the expressions such that they +lead to the values indicated. +(a) 34 – 9 + 12 = 13 +(b) 56 – 14 – 8 = 34 +(c) – 22 – 12 – 10 + 22 = – 22 +Ans: +(a) 34 – (9 + 12) = 34 – 21 = 13 +(b) (56 – 14) – 8 = 42 – 8 = 34 +(c) – 22 – (12 + 10) + 22 = – 22 – 22 + 22 = – 22 + + + +11 + +6. Using only reasoning of how terms change their values, fill the +blanks to make the expressions on either side of the equality (=) +equal. +(a) 423 + ________ = 419 + ________ +(b) 207 – 68 = 210 – ________ +Ans: +(a) 423 + 419 = 419 + 423 (Commutative Property) +(b) 207 – 68 = 207 + 3 – 3 – 68 = 207 + 3 – (3 + 68) = 210 – 71 (Associative +Property) +7. Using the numbers 2, 3, and 5, and the operators ‘+’ and ‘ – ‘, and +brackets, as necessary, generate expressions to give as many +different values as possible. +For example, 2 – 3 + 5 = 4 and 3 – (5 – 2) = 0 +Ans: Some of the expressions are +1. 3 – 2 – 5 = – 4 +2. 5 + 2 + 3 = 10 +3. 5 + 2 – 3 = 4 +4. 2 – (3 + 5) = – 6 +Form more expressions. +8. Whenever Jasoda has to subtract 9 from a number, she subtracts 10 +and adds 1 to it. +For example, 36 – 9 = 26 + 1 +(a) Do you think she always gets the correct answer? Why? +(b) Can you think of other similar strategies? Give some examples. +Ans: + +(a) Yes +36 – 9 = 36 – (10 – 1) = 36 – 10 + 1 = 26 + 1 +(b) Yes +To subtract 8: Subtract 10 and add 2 (since – 8 = – 10 + 2). +Ex: 55 – 8 = 55 – (10 – 2) = 55 – 10 + 2 +More such strategies can be thought of. + + + + +12 + +9. Consider the two expressions: +(a) 73 – 14 + 1 +(b) 73 – 14 – 1 +For each of these expressions, identify the expressions from the +following collection that are equal to it. +(a) 73 – (14 + 1) +(b) 73 – (14 – 1) +(c) 73 + (– 14 + 1) +(d) 73 + (– 14 – 1) +Ans: Expressions (b) and (c) are equal to the expression 73 – 14 + 1, and +expressions (a) and (d) are equal to the expression 73 – 14 – 1. +Page No. 39 + If another friend, Sangmu, joins them and orders the same items, +what will be the expression for the total amount to be paid? +Ans: Expression: 3 × (43 + 24). +Page No. 40 + 5 × 4 + 3 ≠ 5 × (4 + 3). Can you explain why? +Ans: +5 × 4 + 3 = (5 × 4) + 3 = 23. +5 × (4 + 3) = 5 × 7 = 35 +So, 5 × 4 + 3 ≠ 5 × (4 + 3) + Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5? +Ans: Yes, all three expressions are equal. +Page No. 41 + Use this method to find the following products: +(a) 95 × 8 +(b) 104 × 15 +(c) +49 × 50 +Is this quicker than the multiplication procedure you use generally? + + +13 + +Ans: +(a) + +95 × 8 = (100 – 5) × 8 += 100 × 8 – 5 × 8 += 800 – 40 += 760 +(b) + +104 × 15 = (100 + 4) × 15 += 100 × 15 + 4 × 15 +Try further. +(c) + +49 × 50 = (50 – 1) × 50 += 50 × 50 – 50 × 1 +Try further. +Yes, this procedure is quicker than the multiplication procedure we +generally use. + Which other products might be quicker to find, like the ones above? +Ans: +This method is useful when one of the numbers is close to a multiple +of 10, 50, 100, 1000, etc. +For example: +98 × 7 = (100 – 2) × 7 +; +103 × 9 = (100 + 3) × 9 +49 × 5 = (50 – 1) × 5 +Find some more! +Page No. 41 – 42 + Figure it Out +1. Fill in the blanks with numbers and boxes by signs, so that the +expressions on both sides are equal. +(а) 3 × (6 + 7) = 3 × 6 + 3 × 7 +(b) (8 + 3) × 4 = 8 × 4 + 3 × 4 +(c) 3 × (5 + 8) = 3 × 5 + 3 × ________ +(d) (9 + 2) × 4 = 9 × 4 + 2 × ________ +(e) 3 × ( ________ + 4) = 3 ________ + ________ +(f) (________ + 6) × 4 = 13 × 4 + ________ +(g) 3 × (________ + ________) = 3 × 5 + 3 × 2 +(h) (________ + ________) × ________ = 2 × 4 + 3 × 4 + +14 + +(i) 5 × (9 – 2) = 5 × 9 – 5 × ________ +(j) (5 – 2) × 7 = 5 × 7 – 2 × ________ +(k) 5 × (8 – 3) = 5 × 8 + 5 × ________ +(l) (8 – 3) × 7 = 8 × 7 + 3 × 7 +(m) 5 × (12 – ________) = ________ + 5 × ________ +(n) (15 – ________) × 7 = ________ + 6 × 7 +(o) 5 × (________ – ________) = 5 × 9 – 5 × 4 +(p) (________ – ________) × ________ = 17 × 7 – 9 × 7 +Ans: +(c) 3 × (5 + 8) = 3 × 5 +3 × 8 +(d) (9 + 2) × 4 = 9 × 4 +2 × 4 +(e) 3 × (10 + 4) = 3×10 + 3 × 4 +(f) (13 + 6) × 4 = 13 × 4 + 6 × 4 +(g) 3 × (5 + 2) = 3 × 5 + 3 × 2 +(h) (2 + 3) × 4 = 2 × 4 + 3 × 4 +(i) 5 × (9 – 2) = 5 × 9 – 5 × 2 +(j) (5 – 2) × 7 = 5 × 7 – 2 × 7 +(k) 5 × (8 – 3) = 5 × 8 +5 × 3 +(l) (8 – 3) × 7 = 8 × 7 +3 × 7 +(m) 5 × (12 – 3) = 5 × 12 +5 × 3 +(n) (15 – 6) × 7 = 15 × 7 +6 × 7 +(o) 5 × (9 – 4) = 5 × 9 – 5 × 4 +(p) (17 – 9) × 7 = 17 × 7 – 9 × 7 +[In (e), (m) you may put any number other than 10 and 3 respectively.] + + + ++ ++ +– +– +– +– + +15 + +2. In the boxes below, fill in ‘<’, ‘>’ or ‘=’ after analysing the expressions +on the LHS and RHS. Use reasoning and understanding of terms and +brackets to figure this out, and not by evaluating the expressions. +(a) (8 – 3) × 29 +(3 – 8) × 29 +(b) 15 + 9 × 18 +(15 + 9) × 18 +(c) 23 × (17 – 9) +23 × 17 + 23 × 9 +(d) (34 – 28) × 42 +34 × 42 – 28 × 42 +Ans: +(a) > : since 8 – 3 > 3 – 8 + +(b) < +(c) < +(d) = +[Try to reason in other parts] +3. Here is one way to make 14: 2 × (1 + 6) = 14. Are there other ways of +getting 14? Fill them out below: +(a) ________ × (________ + ________) = 14 +(b) ________ × (________ + ________) = 14 +(c) ________ × (________ + ________) = 14 +(d) ________ × (________ + ________) = 14 +Ans: +(a) +2 × (5 + 2) = 14 +(b) +2 × (3 + 4) = 14 +(c) +7 × (1 + 1) = 14 +(d) +2 × (6 + 1) = 14 +4. Find out the sum of the numbers given in each picture below in at +least two different ways. Describe how you solved it through +expressions. + + + + +(I) +(II) + +16 + +Ans: +For I: Way1: +(5 × 4) + (4 × 8) = 52 +Way 2: +2 × (4 + 8 + 4) + (8 + 4 + 8) = 52 +For II: Way 1: +(8 × 5) + (8 × 6) = 88 +Way 2: +8 × (5 + 6) = 88 +Find more ways. +Page No. 42 – 44 + Figure it Out +1. Read the situations given below. Write appropriate expressions for +each of them and find their values. +(a) The district market in Begur operates on all seven days of the +week. Rahim supplies 9 kg of mangoes each day from his +orchard, and Shyam supplies 11 kg of mangoes each day from his +orchard to this market. Find the number of mangoes supplied by +them in a week to the local district market. +(b) Binu earns ₹ 20,000 per month. She spends ₹ 5,000 on rent, ₹ 5,000 +on food, and ₹ 2,000 on other expenses every month. What is the +amount Binu will save by the end of the year? +(c) During the daytime, a snail climbs 3 cm up a post, and during the +night, while asleep, accidentally slips down by 2 cm. The post is +10 cm high, and a delicious treat is on top. In how many days will +the snail get the treat? +Ans: +(a) Expression: 7 × (9 + 11) kg +Value: 140 kg +(b) Expression: 12 × 20,000 – 12 × (5000 + 5000 + 2000) +Amount Binu will save by the end of the year = ₹96,000 +(c) In the daytime the snail climbs 3 cm up the post and slips down by 2 +cm at night. +So, the distance climbed by the snail on the post in full day cycle += 3 – 2 = 1 cm +∴ The distance climbed by the snail in 7 days = 7 cm The height of the +post is 10 cm. +On the 8th day the snail will climb i.e. 7 × (3 – 2) + 3 × 1 = 10 cm + +17 + +2. Melvin reads a two – page story every day except on Tuesdays and +Saturdays. How many stories would he complete reading in 8 +weeks? Which of the expressions below describes this scenario? +(a) 5 × 2 × 8 +(b) (7 – 2) × 8 +(c) 8 × 7 +(d) 7 × 2 × 8 +(e) 7 × 5 – 2 +(f) +(7 + 2) × 8 +(g) 7 × 8 – 2 × 8 +(h) (7 – 5) × 8 +Ans: Expressions describing the scenario (total stories) = (b) and (g). +3. Find different ways of evaluating the following expressions: +(a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 +(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 +Ans: + +(a) Way 1 +: 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 + += (1 + 3 + 5 + 7 + 9) + (– 2 – 4 – 6 – 8 – 10) +Way 2: 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 + + + = (1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) +(9 – 10) + + Try more ways. +(b) Way 1: 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + + = (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) +Way 2: 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + += (1 + 1 + 1 + 1 + 1) + (– 1 – 1 – 1 – 1 – 1) +Try more ways. +4. Compare the following pairs of expressions using ‘<’, ‘>’, or ‘=,’ or by +reasoning. +(a) 49 – 7 + 8 + 49 – 7 + 8 +(b) 83 × 42 – 18 + 83 × 40 – 18 +(c) 145 – 17 × 8 + 145 – 17 × 6 +(d) 23 × 48 – 35 + 23 × (48 – 35) +(e) (16 – 11) × 12 + – 11 × 12 + 16 × 12 +(f) (76 – 53) × 88 + 88 × (53 – 76) + +18 + +(g) 25 × (42 + 16) + 25 × (43 + 15) +(h) 36 × (28 – 16) + 35 × (27 – 15) +Ans: +(a) = + +(b) > +(c) < +(d) > +(e) = + +(f) > + +(g) = +(h) > +5. Identify which of the following expressions are equal to the given +expression without computation. You may rewrite the expressions +using terms or removing brackets. There can be more than one +expression that is equal to the given expression. +(a) 83 – 37 – 12 +(i) 84 – 38 – 12 +(ii) 84 – (37 + 12) +(iii) 83 – 38 – 13 +(iv) – 37 + 83 – 12 + +(b) 93 + 37 × 44 + 76 +(i) 37 + 93 × 44 + 76 +(ii) 93 + 37 × 76 + 44 +(iii) (93 + 37) × (44 + 76) +(iv) 37 × 44 + 93 + 76 +Ans: +(a) (i) and (iv) +(b) (iv) +6. Choose a number and create ten different expressions having that +value. +Ans: Let’s choose the number 26. Here are ten different expressions with +the value 26: +1. 10 + 16 +2. 30 – 4 +3. 13 × 2 +4. +52 +2 +5. 5 + (3 × 7) +6. (5 × 5) + 1 +Try for some other numbers." +class_7,3,A Peek Beyond the Point,ncert_books/class_7/gegp1dd/gegp103.pdf,"3.1 The Need for Smaller Units +Sonu’s mother was fixing a +toy. She was trying to join +two pieces with the help of +a +screw. +Sonu +was +watching his mother with +great curiosity. His mother +was unable to join the +pieces. Sonu asked why. +His mother said that the +screw was not of the +right size. +She brought another screw from the box and was able to fix the toy. +The two screws looked the same to Sonu. But when he observed them +closely, he saw they were of slightly different lengths. +Sonu was fascinated +by how such a small +difference in lengths +could matter so much. +He was curious to +know the difference +in lengths. He was also +curious to know how +little +the +difference +was +because +the +screws looked nearly +the same. +In the following figure, screws are placed above a scale. Measure +them and write their length in the space provided. +A PEEK BEYOND +THE POINT +3 + +A Peek Beyond the Point +Which scale helped you measure the length of the screws accurately? +Why? +What is the meaning of 2 7 +10 cm (the length of the first screw)? +As seen on the ruler, the unit length between two consecutive +numbers is divided into 10 equal parts. To get the length 2 7 +10 cm, we +go from 0 to 2 and then take seven parts of 1 +10. The length of the screw +is 2 cm and 7 +10 cm. Similarly, we can make sense of the length 3 2 +10 cm. +We read 2 7 +10 cm as two and seven-tenth centimeters, and 3 2 +10 cm +as three and two-tenth centimeters. +Can you explain why the unit was divided into smaller parts to measure +the screws? +Measure the following objects using a scale and write their +measurements in centimeters (as shown earlier for the lengths of the +screws): pen, sharpener, and any other object of your choice. +Write the measurements of the objects shown in the picture: +Between 2 cm + and 3 cm +_______________ +_______________ +_______________ +More than 2 1 +2 + +cm but less than +3 cm +2 7 +10 + cm +47 + +Ganita Prakash | Grade 7 +As seen here, when exact measures are required we can make use of +smaller units of measurement. +3.2 A Tenth Part +The length of the pencil shown in the figure below is 3 4 +10 units, which can +also be read as 3 units and four one-tenths, i.e., (3 × 1) + (4 × 1 +10) units. += 10 times 1 +10 = 10 × 1 +10 = 1 unit +This length is the same as 34 one-tenths units because 10 one-tenths +units make one unit. +1 +10 + 1 +10 + 1 +10 + 1 +10 + 1 +10 + 1 +10 + 1 +10 + 1 +10 + 1 +10 + 1 +10 +48 + +A Peek Beyond the Point +34 × 1 +10 = 34 +10 = 10 +10 + 10 +10 + 10 +10 + 4 +10 (34 one-tenths) + + = 1 + 1 + 1 + 4 +10 (3 and 4 one-tenths) +A few numbers with fractional units are shown below along with +how to read them. + 4 1 +10 + ‘four and one-tenth’ + 4 +10 + ‘four one-tenths’ or ‘four-tenths’ + 41 +10 + ‘forty-one one-tenths’ or ‘forty-one tenths’ +41 1 +10 + ‘forty-one and one-tenth’ +For the objects shown below, write their lengths in two ways and +read them aloud. An example is given for the USB cable. (Note that the +unit length used in each diagram is not the same). +The length of the USB cable is 4 and 8 +10 units or 48 +10 units. +Arrange these lengths in increasing order: +(a) 9 +10 +(b) 1 7 +10 +(c) 130 +10 +(d) 13 1 +10 +(e) 10 5 +10 +(f) 7 6 +10 +(g) 6 7 +10 +(h) +4 +10 +49 + +Ganita Prakash | Grade 7 +Arrange the following lengths in increasing order: 4 1 +10, 4 +10, 41 +10, 41 1 +10. +Sonu is measuring some of his body parts. The length of Sonu’s lower +arm is 2 7 +10 units, and that of his upper arm is 3 6 +10 units. What is the +total length of his arm? +To get the total length, let us see the lower and upper arm length as +2 units and 7 one-tenths, and 3 units and 6 one-tenths, respectively. +So, there are (2 + 3) units and (7 + 6) one-tenths. Together, they make +5 units and 13 one-tenths. But 13 one-tenths is 1 unit and 3 one-tenths. +So, the total length is 6 units and 3 one-tenths. +(a) + + +(2 + 3) + ( +7 +10 + 6 +10) + + + + += (2 + 3) + ( +13 +10) + + + + += 5 + 13 +10 + + + + += 5 + 10 +10 + 3 +10 = 5 + 1 + 3 +10 + + + + += 6 + 3 +10 + + + + += 6 3 +10  +Or, both the lengths can be converted to tenths and then added: +(c) 27 one-tenths and 35 one-tenths is 62 one-tenths +27 +10 + 35 +10 = 62 +10 +62 +10 is the same as 60 one-tenths ( +60 +10) and 2 one-tenths ( +2 +10), which is +(b) + + +2 7 +10 + + + + + 3 6 +10 + + + + = 5 13 +10 + + + + = 6 3 +10  +50 + +A Peek Beyond the Point +equal to 6 units and 2 one-tenths, i.e., 6 2 +10. +The lengths of the body parts of a honeybee are given. Find its +total length. +Head: 2 3 +10 units +Thorax: 5 4 +10 units +Abdomen: 7 5 +10 units +The length of Shylaja’s hand is 12 4 +10 units, +and her palm is 6 7 +10 units, as shown in the picture. What is the length +of the longest (middle) finger? +The length of the finger can be found by +evaluating (12 + 4 +10) – (6 + 7 +10). This can be +done in different ways. For example, +(a) 12 + 4 +10 – 6 – 7 +10 + += (12 – 6) + ( +4 +10 – 7 +10) + += 6 – 3 +10 + += 5 + 1 – 3 +10 + += 5 + 10 +10 – 3 +10 + += 5 + 7 +10 = 5 7 +10  +As in the case of counting numbers, it is convenient to start +subtraction from the tenths. We cannot remove 7 one-tenths from 4 +Head Thorax +abdomen +12 4 +10 units +6 7 +10 units +Discuss what is +being done here +and why. +(b) 12 4 +10 + +11 14 +10 + +– 6 7 +10 + +– 6 7 +10 + + + + += 5 7 +10 +51 + +Ganita Prakash | Grade 7 +one-tenths. So we split a unit from 12 and convert it to 10 one-tenths. +Now, the number has 11 units and 14 one-tenths. We subtract 7 one- +tenths from 14 one-tenths and then subtract 6 units from 11 units. +Try computing the difference by converting both lengths to tenths. +A Celestial Pearl Danio’s length is 2 4 +10 cm, and the length of a Philippine +Goby is 9 +10 cm. What is the difference in their lengths? +How big are these fish compared to your finger? +Observe the given sequences of numbers. Identify the change after +each term and extend the pattern: +(a) 4, 4 3 +10 , 4 6 +10 , +_________, _________, _________, _________ +(b) 8 2 +10 , 8 7 +10 , 9 2 +10 , +_________, _________, _________, _________ +(c) 7 6 +10 , 8 7 +10 , +_________, _________, _________, _________ +(d) 5 7 +10 , 5 3 +10 , +_________, _________, _________, _________ +(e) 13 5 +10 , 13, 12 5 +10 , +_________, _________, _________, _________ +(f) 11 5 +10 , 10 4 +10 , 9 3 +10 , _________, _________, _________, _________ +3.3 A Hundredth Part +The length of a sheet of paper was 8 +9 +10 units, which can also be said as +8 units and 9 one-tenths. It is folded +in half along its length. What is its +length now? +Celestial Pearl Danio +Philippine Goby +52 + +A Peek Beyond the Point +We can say that its +length is between 4 4 +10 +units and 4 5 +10 units. But +we cannot state its exact +measurement, since there +are no markings. Earlier, +we split a unit into 10 +one-tenths +to +measure +smaller lengths. We can +do something similar and +split each one-tenth into +10 parts. +What is the length of this +smaller part? How many +such smaller parts make +a unit length? +As shown in the figure +below, each one-tenth has +10 smaller parts, and there are 10 one-tenths in a unit; therefore, there +will be 100 smaller parts in a unit. Therefore, one part’s length will +be 1 +100 of a unit. +Returning to our question, what is the length of the folded paper? +We can see that it ends at 4 4 +10 5 +100, read as 4 units and 4 one-tenths +and 5 one-hundredths. +How many one-hundredths make one-tenth? Can we also say that +the length is 4 units and 45 one-hundredths? +Math +Talk +53 + +Ganita Prakash | Grade 7 +Observe the figure below. Notice the markings and the corresponding +lengths written in the boxes when measured from 0. Fill the lengths in +the empty boxes. +The length of the wire in the first picture is given in three different +ways. Can you see how they denote the same length? +1 1 +10 4 +100 +One and one-tenth and four-hundredths +1 14 +100 +One and fourteen-hundredths +114 +100 +One Hundred and Fourteen-hundredths +For the lengths shown below write the measurements and read out the +measures in words. +1 3 +10 +1 +10 +2 +10 +20 +100 +1 +100 +99 +100 +130 +100 +0 +1 +2 +54 + +A Peek Beyond the Point +In each group, identify the longest and the shortest lengths. Mark each +length on the scale. +(a) +3 +10 , 3 +100 , 33 +100 + +(b) 3 1 +10 , 30 +10 , 1 3 +10 + +(c) +45 +100 , 54 +100 , 5 +10 , 4 +10 + +(d) 3 6 +10 , 3 6 +100 , 3 6 +10 6 +100 + +55 + +Ganita Prakash | Grade 7 +(e) +8 +10 2 +100 , 9 +100 , 1 8 +100 +(f) 7 3 +10 5 +100 , 7 5 +10 , 7 41 +100 +(g) 65 +10 15 +100 , 5 87 +100 , 5 7 +100 +What will be the sum of 15 3 +10 4 +100 and 2 6 +10 8 +100 ? +This can be solved in different ways. Some are shown below. +(a) Method 1 + +(15 + 2) + ( +3 +10 + 6 +10) + ( +4 +100 + 8 +100) + += 17 + 9 +10 + 12 +100 + += 17 + 9 +10 + 1 +10 + 2 +100 + += 17 + 10 +10 + 2 +100 + += 18 2 +100 . + +15 3 +10 4 +100 ++ + 2 6 +10 8 +100 += +17 9 +10 12 +100 += +17 10 +10 2 +100 += +18 + 2 +100 +(b) Method 2 +10 hundredths is +the same as 1 tenth. +56 + +A Peek Beyond the Point +Are both these methods different? +Observe the addition done below for 483 + 268. Do you see any +similarities between the methods shown above? + + + (400 + 80 + 3) + (200 + 60 + 8) + + += (400 + 200) + (80 + 60) + (3 + 8) + + += 600 + 140 + 11 + + += 600 + 150 + 1 + + += 700 + 50 + 1 + + += 751 + One can also find the sum 15 3 +10 4 +100 + 2 6 +10 8 +100 by converting to +hundredths, as follows. +(c) (15 + 2) + ( +34 +100 + 68 +100) + + + += 17 + 102 +100 + += 17 + 1 + 2 +100 + += 18 2 +100 +(d) ( +1534 +100 ) + ( +268 +100) + += 1802 +100 + += 1802 +100 + 2 +100 + += 18 2 +100 +What is the difference: 25 9 +10 – 6 4 +10 7 +100 ? +One way to solve this is as follows: +Math +Talk +100 hundredths is +same as 1 unit. +15 is the same as +1500 hundredths +and 2 is the same as +200 hundredths. +25 9 +10 +– 6 4 +10 +25 8 +10 10 +100 +– 6 4 +10 7 +100 +25 8 +10 10 +100 +– 6 4 +10 7 +100 += 19 4 +10 3 +100 +57 + +Ganita Prakash | Grade 7 +Solve this by converting to hundredths. +What is the difference 15 3 +10 4 +100 –2 6 +10 8 +100 ? +One way to solve this is as follows: +15 3 +10 4 +100 +15 2 +10 14 +100 +14 12 +10 14 +100 +– 2 6 +10 8 +100 +– 2 6 +10 8 +100 +– 2 6 +10 8 +100 += 2 6 +10 6 +100 +Observe the subtraction done below for 653 – 268. Do you see any +similarities with the methods shown above? + + (600 + 50 + 3) – (200 + 60 + 8) + += (600 – 200) + (50 – 60) + (3 – 8) + += (600 – 200) + (40 – 60) + (13 – 8) + += (600 – 200) + (40 – 60) + 5 + += (500 – 200) + (140 – 60) + 5 + += 300 + 80 + 5 + += 385 +Figure it Out +Find the sums and differences: +(a) +3 +10 + 3 4 +100 +(b) 9 5 +10 7 +100 + 2 1 +10 3 +100 +(c) 15 6 +10 4 +100 + 14 3 +10 6 +100 +(d) 7 7 +100 – 4 4 +100 +(e) 8 6 +100 – 5 3 +100 +(f) 12 6 +100 2 +100 – 9 +10 9 +100 +3.4 Decimal Place Value +You may have noticed that whenever we need to measure something +more accurately, we split a part into 10 (smaller) equal parts ― we split +a unit into 10 one-tenths and then split each one-tenth into 10, one- +hundredths and then we use these smaller parts to measure. +Math +Talk +Math +Talk +58 + +A Peek Beyond the Point +Can we not split a unit into 4 equal parts, 5 equal parts, 8 equal parts, +or any other number of equal parts instead? +Yes, we can. The example below +compares how the same length is +represented when the unit is split +into 10 equal parts and when the +unit is split into 4 equal parts. +If an even more precise measure +is needed, each quarter can further +be split into four equal parts. Each +part then measures 1 +16 of a unit, i.e., +16 such parts make 1 unit. +Then why split a unit into 10 parts every time? +The reason is the special role that 10 plays in the Indian place value +system. For a whole number written in the Indian place value system — +for example, 281 — the place value of 2 is hundreds (100), that of 8 is +tens (10), and that of 4 is one (1). Each place value is 10 times bigger +than the one immediately to its right. Equivalently, each place value is +10 times smaller than the one immediately to its left: +10 ones make 1 ten, +10 tens make 1 hundred, +10 hundreds make 1 thousand, and so on. +10,000 +1000 +100 +10 +1 +× 10 +÷ 10 +× 10 +÷ 10 +× 10 +÷ 10 +× 10 +÷ 10 +In order to extend this system of writing numbers to quantities +smaller than one, we divide one into 10 equal parts. What does +this give? It gives one-tenth. Further dividing it into 10 parts gives +one-hundredth, and so on. +59 + +Ganita Prakash | Grade 7 +Can we extend this further? +What will the fraction be when 1 +100 is split into 10 equal parts? +It will be +1 +1000 , i.e., a thousand such parts make up a unit. +Just as when we extend to the left of 10,000, we get bigger place +values at each step, we can also extend to the right of +1 +1000 , getting +smaller place values at each step. +10,000 +1000 +100 +10 +1 +1 +10 +1 +100 +× 10 +÷ 10 +× 10 +÷ 10 +× 10 +÷ 10 +× 10 +÷ 10 +× 10 +÷ 10 +× 10 +÷ 10 +60 + +A Peek Beyond the Point +This way of writing numbers is called the “decimal system” since it +is based on the number 10; “decem” means ten in Latin, which in turn +is cognate to the Sanskrit daśha meaning 10, with similar words for +10 occurring across many Indian languages including Odia, Konkani, +Marathi, Gujarati, Hindi, Kashmiri, Bodo, and Assamese. We shall learn +about other ways of writing numbers in later grades. +How Big? +We already know that a hundred 10s make 1000, and a hundred 100s +make 10000. +We can ask similar questions about fractional parts: +(a) How many thousandths make one unit? +(b) How many thousandths make one tenth? +(c) How many thousandths make one hundredth? +(d) How many tenths make one ten? +(e) How many hundredths make one ten? +Make a few more questions of this kind and answer them. +Notation, Writing and Reading of Numbers +We have been writing numbers in a particular way, say 456, instead of +writing them as 4 × 100 (4 hundreds) + 5 × 10 (5 tens) + 6 × 1 (6 ones). +Similarly, can we skip writing tenths and hundredths? +Can the quantity 4 2 +10 be written as 42 (skipping the 1 +10 in 2 × 1 +10) ? +If yes, how would we know if 42 means 4 tens and 2 units or it +means 4 units and 2 tenths? +Similarly, 705 could mean: +(a) 7 hundreds, and 0 tens and 5 ones (700 + 0 + 5) +(b) 7 tens and 0 units and 5 tenths (70 + 0 + 5 +10) +(c) 7 units and 0 tenths and 5 hundredths (7 + 0 +10 + 5 +100) +1,00,000 +10,000 +1000 +100 +10 +1 +1 +10 +1 +100 +1 +1000 +1 +10,000 +× 10 +× 10 +× 10 +× 10 +× 10 +× 10 +× 10 +× 10 +× 10 +÷ 10 +÷ 10 +÷ 10 +÷ 10 +÷ 10 +÷ 10 +÷ 10 +÷ 10 +÷ 10 +Math +Talk +61 + +Ganita Prakash | Grade 7 +Since these are different quantities, we need to have distinct ways +of writing them. +To identify the place value where integers end and the fractional +parts start, we use a point or period (‘.’) as a separator, called a +decimal point. +The above quantities in decimal notation are then: +Quantity +Decimal Notation +7 hundreds and 5 ones +(700 + 0 + 5) +705 +7 tens and 5 tenths (70 + 0 + 5 +10) +70.5 +7 units and 5 hundredths +(7 + 0 + 5 +100) +7.05 +These numbers, when shown through place value, are as follows: +Decimal +number +Hundreds +Tens +Units +Tenths +Hundredths +705 +7 × 100 +0 × 10 +5 × 1 +70.5 +7 × 10 +0 × 1 +5 × 1 +10 +7.05 +7 × 1 +0 × 1 +10 +5 × 1 +100 +× 10 × 1 +×  1 +10 ×  1 +100 +number of +tens +number of +units +number of +one-tents +number of one-hundredths +62 + +A Peek Beyond the Point +Thus decimal notation is a natural extension of the Indian place +value system to numbers also having fractional parts. Just as 705 means +7 × 100 + 5 × 1, the number 70.5 means 7 × 10 + 5 × 1 +10 , and 7.05 means +7 × 1 +5 × 1 +100. +We have seen how to write numbers using the decimal point (‘.’). But +how do we read/say these numbers? +We know that 705 is read as seven hundred and five. +70.5 is read as seventy point five, short for seventy and five-tenths. +7.05 is read as seven point zero five, short for seven and five +hundredths. +0.274 is read as zero point two seven four. We don’t read it as zero +point two hundred and seventy four as 0.274 means 2 one-tenths +and 7 one-hundredths and 4 one-thousandths. +Make a place value table similar to the one above. Write each quantity +in decimal form and in terms of place value, and read the number: +(a) 2 ones, 3 tenths and 5 hundredths +(b) 1 ten and 5 tenths +(c) 4 ones and 6 hundredths +(d) 1 hundred, 1 one and 1 hundredth +(e) 8 +100 and 9 +10 +(f) 5 +100 +(g) +1 +10 +(h) 2 1 +100 , 4 1 +10 and 7 +7 +1000 +In the chapter on large numbers, we learned how to write 23 +hundreds. +23 hundreds = 23 × 100 = 2000 + 300 = 2300. +Thousands +Hundreds +Tens +Units +23 +2 +3 +0 +0 +Similarly, 23 tens would be: +23 tens = 23 × 10 = 200 + 30 = 230. +63 + +Ganita Prakash | Grade 7 +Thousands +Hundreds +Tens +Units +23 +2 +3 +0 +How can we write 234 tenths in decimal form? +234 tenths = 234 +10 += 200 +10 + 30 +10 + 4 +10 += 20 + 3 + 4 +10 += 23.4. +Hundreds +Tens +Units +Tenths +Hundredths +234 +2 +3 +4 +Write these quantities in decimal form: (a) 234 hundredths, (b) 105 tenths. +3.5 Units of Measurement +Length Conversion +We have been using a scale to measure length for a few years. We +already know that 1 cm = 10 mm (millimeters). +How many cm is 1 mm? +1 mm = 1 +10 cm = 0.1 cm (i.e., one-tenth of a cm). +How many cm is (a) 5 mm? (b) 12 mm? +5 mm = 5 +10 cm = 0.5 cm +12 mm = 10 mm + 2 mm + + = 1 cm + 2 +10 cm + + = 1.2 cm. +64 + +A Peek Beyond the Point +How many mm is 5.6 cm? Since each cm has 10 mm, 5.6 cm (5 cm + +0.6 cm) is 56 mm. +Fill in the blanks below (mm <–> cm) +12 mm = 1.2 cm +56 mm = 5.6 cm +70 mm = _______ +________ = 0.9 cm +134 mm =__________ +________ = 203.6 cm +The illustration below shows how small some things are! Try taking +an approximate measurement of each. +• The three blue stripes represent the typical relative sizes of pen +strokes: fine stroke, medium stroke, and bold stroke. +• A human hair is about 0.1 mm in thickness. +• The thickness of a newspaper can range from 0.05 to 0.08 mm. +• Mustard seeds have a thickness of 1 – 2 mm. +• The smallest ant species discovered so far, Carabera Bruni, has a +total length of 0.8 – 1 mm. They are found in Sri Lanka and China. +• The smallest land snail species discovered so far, Acmella Nana, +has a shell diameter of 0.7 mm. They are found in Malaysia. +We also know that 1 m = 100 cm. Based on this, we can say that +1 cm = 1 +100 m = 0.01 m. +1.5 mm +1 mm +0.5 mm +65 + +Ganita Prakash | Grade 7 +How many m is (a) 10 cm? (b) 15 cm? +10 cm = 1 +10 m = 0.1 m +Since each cm is one-hundredth of a meter, 15 cm can be written as +15 cm = 15 +100 m += 10 +100 m + 5 +100 m += 1 +10 m + 5 +100 m += 0.15 m. +Fill in the blanks below (cm <–> m): +36 cm = _______ +50 cm = _______ +_______ = 0.89 m +4 cm = _______ +325 cm = _______ +________ = 2.07 m +How many mm does 1 meter have? +Can we write 1 mm = +1 +1000 m? +Here, we have some more interesting facts about small things in +nature! +1 cm +1 cm +0 +66 + +A Peek Beyond the Point +• The egg of a hummingbird typically is 1.3 cm long and 0.9 cm wide. +• The Philippine Goby is about 0.9 cm long. It can be found in the +Philippines and other Southeast Asian countries. +• The smallest known jellyfish, Irukandji, has a bell size of 0.5 – 2.5 +cm. Its tentacles can be as long as 1 m. They are found in Australia. +Its venom can be fatal to humans. +• The Wolfi octopus, also known as the Star-sucker Pygmy Octopus, +is the smallest known octopus in the world. Their typical size is +around 1 – 2.5 cm and they weigh less than 1 gm. They are found +in the Pacific Ocean. +Weight Conversion +Let us look at kilograms (kg). We know that 1 kg = 1000 gram (g). We +can say that +1 g = +1 +1000 kg = 0.001 kg. +How many kilograms is 5 g? +5 g = +5 +1000 kg = 0.005 kg. +How many kilograms is 10 g? +10 g = 10 +1000 kg = 1 +100 kg = 0.010 kg. +As each gram is one-thousandth of a kg, 254 g can be written as +254 g = 254 +1000 kg += ( +200 +1000 + 50 +1000 + +4 +1000) kg += ( +2 +10 + 5 +100 + +4 +1000) kg += 0.254 kg. +Fill in the blanks below (g <–> kg) +67 + +Ganita Prakash | Grade 7 +465 g = _______ +68 g = _________ +1560 g = ________ +704 g = _______ +________ = 0.56 kg +_______ = 2.5 kg +Look at the picture below showing different quantities of rice. +Starting from the 1g heap, subsequent heaps can be found that are 10 +times heavier than the previous heap/packets. The combined weight of +rice in this picture is 11.111 kg. +Also, +1 gram = 1000 milligrams (mg). So, 1 mg = +1 +1000 g = 0.001 g. +Rupee ─ Paise conversion +You may have heard of ‘paisa’. 100 paise is equal to 1 rupee. As we have +coins and notes for rupees, coins for paise were also used commonly +until recently. There were coins for 1 paisa, 2 paise, 3 paise, 5 paise, 10 +paise, 20 paise, 25 paise, and 50 paise. All denominations of 25 paise +and less were removed from use in the year 2011. But we still see paise +in bills, account statements, etc. +1 rupee = 100 paise +1 paisa = 1 +100 rupee = 0.01 rupee +As each paisa is one-hundredth of a rupee, +75 paise = 75 +100 rupee +68 + +A Peek Beyond the Point += ( +70 +100 + 5 +100) rupee += ( +7 +100 + 5 +100) rupee += 0.75 rupee. +Fill in the blanks below (rupee <–> paise) +10 p = ___________ +_______p = ₹ 0.05 +________p = ₹ 0.36 +_________ = ₹ 0.50 +99 p = _________ +250 p = _________ +During the 1970s, a masala dosa +cost just 50 paise, one could +buy a banana for 20-25 paise, +a handful of peppermints were +available for 2 paise or 3 paise, +and a kg of rice cost ₹2.45. +Discuss with adults at home/school the prices of different products +and services during their childhood. Try to find old coins and stamps. +3.6 Locating and Comparing Decimals +Let us consider the decimal number 1.4. It is equal to 1 unit and 4 +tenths. This means that the unit between 1 and 2 is divided into 10 +Try +This +69 + +Ganita Prakash | Grade 7 +equal parts, and 4 such parts are taken. Hence, 1.4 lies between 1 and +2. Draw the number line and divide the unit between 1 and 2 into 10 +equal parts. Take the fourth part, and we have 1.4 on the number line. +1 +1.1 +1.2 +1.3 +1.4 +1.5 +1.6 +1.7 +1.8 +1.9 +2 +Name all the divisions between 1 and 1.1 on the number line. +1 +1.04 +1.1 +1.2 +1.3 +Identify and write the decimal numbers against the letters. +A +5 +5.1 +5.3 +5.4 +B +C +D +There is Zero Dilemma! +Sonu says that 0.2 can also be written as 0.20, 0.200; Zara thinks that +putting zeros on the right side may alter the value of the decimal +number. What do you think? +We can figure this out by looking at the quantities these numbers +represent using place value. +Decimal +number +Units +Tenths +Hundredths +Thousandths +0.2 +0 +2 +0.20 +0 +2 +0 +0.200 +0 +2 +0 +0 +0.02 +0 +0 +2 +0.002 +0 +0 +0 +2 +We can see that 0.2, 0.20, and 0.200 are all equal as they represent +the same quantity, i.e., 2 tenths. But 0.2, 0.02, and 0.002 are different. +70 + +A Peek Beyond the Point +Can you tell which of these is the smallest and which is the largest? +Which of these are the same: 4.5, 4.05, 0.405, 4.050, 4.50, 4.005, 04.50? +Observe the number lines in Figure (a) below. At each level, a +particular segment of the number line is magnified to locate the +number 4.185. +Identify the decimal number in the last number line in Figure (b) +denoted by ‘?’. +0 +4 +5 +4.1 +4.18 +4.185 +4.19 +10 +4.2 +0 +10 +? +(a) +(b) +Make such number lines for the decimal numbers: (a) 9.876 (b) 0.407. +In the number line shown below, what decimal numbers do the boxes +labelled ‘a’, ‘b’, and ‘c’ denote? +5 +b +a +c +10 +The box with ‘b’ corresponds to the decimal number 7.5; are you +able to see how? There are 5 units between 5 and 10, divided into 10 +equal parts. Hence, every 2 divisions make a unit, and so every division +is 1 +2 unit. What numbers do ‘a’ and ‘c’ denote? +71 + +Ganita Prakash | Grade 7 +Using similar reasoning find out the decimal numbers in the boxes +below. +Which is larger: 6.456 or +6.465? +To answer this, we +can use the number line +to locate both decimal +numbers +and +show +which is larger. +This +can +also +be +done by comparing the +corresponding digits at +each place value, as we +do with whole numbers. +This +comparison +is +visualised step by step below. Note that the visualisation below is not to scale. +6.456 +6.465 +6 +Both numbers have 6 units. +6.456 +6.465 +6 +6.4 +Both numbers have 6 units +and 4 tenths. +6.456 +6.465 +6 +6.4 +6.45 +6.46 +Both numbers have 6 units +and 4 tenths, but the first +number has only 5 +hundredths, whereas the +second number has 6 +hundredths. +8 +4.3 +d +f +g +h +e +8.1 +4.8 +72 + +A Peek Beyond the Point +We start by comparing the most significant digits (digits with the +highest place value) of the two numbers. If the digits are the same, +we compare the next smaller place value. We keep going till we find +a position where the digits are not equal. The number with the larger +digit at this position is the greater of the two. +Why can we stop comparing at this point? Can we be sure that +whatever digits are there after this will not affect our +conclusion? +Which decimal number is greater? +(a) 1.23 or 1.32 +(b) 3.81 or 13.800 +(c) 1.009 or 1.090 +Closest Decimals +Consider the decimal numbers 0.9, 1.1, 1.01, and 1.11. Identify the +decimal number that is closest to 1. +Let us compare the decimal numbers. Arranging these in ascending +order, we get 0.9 < 1 < 1.01 < 1.1 < 1.11. Among the neighbours of 1, 1.01 +is 1/100 away from 1 whereas 0.9 is 10/100 away from 1. Therefore, +1.01 is closest to 1. +Which of the above is closest to 1.09? +Which among these is closest to 4: 3.56, 3.65, 3.099? +Which among these is closest to 1: 0.8, 0.69, 1.08? +In each case below use the digits 4, 1, 8, 2, and 5 exactly once and +try to make a decimal number as close as possible to 25. + +Math +Talk +Math +Talk +73 + +Ganita Prakash | Grade 7 +3.7 Addition and Subtraction of Decimals +Priya requires 2.7 m of cloth for her skirt, and Shylaja requires 3.5m +for her kurti. What is the total quantity of cloth needed? +We have to find the sum of 2.7m + 3.5m. +Earlier, we saw how to add 2 7 +10 + 3 5 +10 (also shown below). Can you +carry out the same addition using decimal notation? It is shown below. +Share your observations. +The total quantity of cloth needed is 6.2 m. +How much longer is Shylaja’s cloth compared to Priya’s? +We have to find the difference of 3.5m – 2.7m. Again, observe how +the differences 3 5 +10 – 2 7 +10 and 3.5m – 2.7m are computed. +As you can see, the standard procedure for adding and subtracting +whole numbers can be used to add and subtract decimals. +A detailed view of the underlying place value calculation is shown +below for the sum 75.345 + 86.691. Its compact form is shown next to it. + 2 7 +10 + + 3 5 +10 += 5 12 +10 += 6 2 +10 + 2.7 + + 3.5 + = 6.2 +1 + 3 5 +10 + – 2 7 +10 +2 15 +10 + – 2 7 +10 + 3.5 + – 2.7 + 3.5 + – 2.7 += 0 8 +10 + = 0.8 +2 1 +74 + +A Peek Beyond the Point +Write the detailed place value computation for 84.691 – 77.345, and +its compact form. +Figure it Out +1. Find the sums +(a) 5.3 + 2.6 +(b) 18 + 8.8 +(c) +2.15 + 5.26 +(d) 9.01 + 9.10 +(e) +29.19 + 9.91 +(f) +0.934 + 0.6 +(g) +0.75 + 0.03 +(h) 6.236 + 0.487 +2. Find the differences +(a) 5.6 – 2.3 +(b) 18 – 8.8 +(c) +10.4 – 4.5 +(d) 17 – 16.198 +(e) +17 – 0.05 +(f) +34.505 – 18.1 +(g) +9.9 – 9.09 +(h) 6.236 – 0.487 +Decimal Sequences +Observe this sequence of decimal numbers and identify the change +after each term. +4.4, 4.8. 5.2, 5.6, 6.0, … +We can see that 0.4 is being added to a term to get the next term. +Continue this sequence and write the next 3 terms. +Try +This ++ += +7 × 10 +5 × 1 +3 × 1 +10 +4 × 1 +100 +5 × +1 +1000 +8 × 10 +6 × 1 +6 × 1 +10 +9 × 1 +100 +1 × +1 +1000 +16 × 10 +12 × 1 +10 × 1 +10 +13 × 1 +100 +6 × +1 +1000 +1 × 1 +1 × 10 +1 × 100 +1 × 100 +1 × 1 +10 + 11 1 + 75.345 + +86.691 +=162.036 +75 + +Ganita Prakash | Grade 7 +Similarly, identify the change and write the next 3 terms for each +sequence given below. Try to do this computation mentally. +(a) 4.4, 4.45, 4.5, … +(b) 25.75, 26.25, 26.75, … +(c) +10.56, 10.67, 10.78, … +(d) 13.5, 16, 18.5, … +(e) +8.5, 9.4, 10.3, … +(f) +5, 4.95, 4.90, … +(g) +12.45, 11.95, 11.45, … +(h) 36.5, 33, 29.5, … +Make your own sequences and challenge your classmates to extend +the pattern. +Estimating Sums and Differences +Sonu has observed sums and differences of decimal numbers and says, +“If we add two decimal numbers, then the sum will always be greater +than the sum of their whole number parts. Also, the sum will always be +less than 2 more than the sum of their whole number parts.” +Let us use an example to understand what his claim means: +If the two numbers to be added are 25.936 and 8.202, the claim is +that their sum will be greater than 25 + 8 (whole number parts) and +will be less than 25 + 1 + 8 + 1. +What do you think about this claim? Verify if this is true for these +numbers. Will it work for any 2 decimal numbers? +What about for the sum of 25.93603259 and 8.202? +Similarly, come up with a way to narrow down the range of whole +numbers within which the difference of two decimal numbers +will lie. +Note to the Teacher: Estimating the result before computing may help in +identifying if a mistake happens with the calculation. +3.8 More on the Decimal System +Decimal and Measurement Disasters +Decimal point and unit conversion mistakes may seem minor sometimes +but they can lead to serious problems. Here are some actual incidents +in which such errors caused major issues. +Math +Talk +Try +This +76 + +A Peek Beyond the Point +• In 2013, the finance office of Amsterdam City Council (Netherlands) +mistakenly sent out €188 million in housing benefits instead of the +intended €1.8 million due to a programming error that processed +payments in euro cents instead of euros. (1 euro-cent = 1/100 +euro). +• In 1983, a decimal error nearly caused a disaster for an Air Canada +Boeing 767. The ground staff miscalculated the fuel, loading 22,300 +pounds instead of kilograms—about half of what was needed (1 +pound ~ 0.453 kg). The plane ran out of fuel mid-air, forcing the +pilots to make an emergency landing at an abandoned airfield. +Fortunately, everyone survived. +Several incidents have occurred due to incorrect reading of decimal +numbers while giving medication. For example, reading 0.05 mg as 0.5 +mg can lead to using a medicine 10 times more than the prescribed +quantity. It is therefore important to pay attention to units and the +location of the decimal point. +Deceptive Decimal Notation +Sarayu gets a message: “The bus will reach the station 4.5 hours post +noon.” When will the bus reach the station: 4:05 p.m., 4:50 p.m., 4:25 p.m.? +None of these! Here, 0.5 hours means splitting an hour into 10 +equal parts and taking 5 parts out of it. Each part will be 6 minutes (60 +minutes/10) long. 5 such parts make 30 minutes. So, the bus will reach +the station at 4:30. +Here is a short-story of a decimal mishap: A girl measures the width +of an opening as 2 ft 5 inches but conveys to the carpenter to make a +door 2.5 ft wide. The carpenter makes a door of width 2 ft 6 inches +(since 1 ft = 12 inches, 0.5 ft = 6 inches), and it wouldn’t close fully. +I said the door’s +width should be +2.5 ft. +It is 2.5 ft! You +can verify. +77 + +Ganita Prakash | Grade 7 +If you watch cricket, you might have noticed decimal-looking +numbers like ‘Overs left: 5.5’. Does this mean 5 overs and 5 balls or 5 +overs and 3 balls? Here, 5.5 overs means 5 5 +6 overs (as 1 over = 6 balls), +i.e., 5 overs and 5 balls. +Where else can we see such ‘non-decimals’ with a decimal-like +notation? +A Pinch of History – Decimal Notation Over Time +Decimal fractions (i.e., fractions with denominators like 1 +10 , +1 +100 , +1 +1000 , and so on) are used in the works of a number of ancient Indian +astronomers and mathematicians, including in the important 8th +century works of Śhrīdharāchārya on arithmetic and algebra. Decimal +notation, in essentially its modern form, was described in detail in +Kitāb al-Fuṣūl fī al-Ḥisāb al Hindī (The Book of Chapters on Indian +Arithmetic) by Abūl Ḥassan al-Uqlīdisī, an Arab mathematician, in +around 950 CE. He represented the number 0.059375 as 0ˈ 059375. +In the 15th century, to separate whole numbers from fractional +parts, a number of different notations were used: +• a vertical mark on the last digit of the whole number part (as +shown above), +• use of different colours and +• a numerical superscript giving the number of fractional decimal +places (0.36 would be written as 36 +2 +). +In the 16th century, John Napier, a Scottish mathematician, and +Christopher Clavius, a German mathematician, used the point/period +(‘.’) to separate the whole number and the fractional parts, while +François Viète, a French mathematician, used the comma (‘,’) instead. +Currently, several countries use the comma to separate the integer +part and the fractional part. In these countries, the number 1,000.5 +is written as 1 000,5 (space as a thousand separator). But the decimal +point has endured as the most popular notation for writing numbers +having fractional parts in the Indian place value system. +Figure it Out +1. Convert the following fractions into decimals: +(a) +5 +100 +(b) +16 +1000 +(c) +12 +10 +(d) +254 +1000 +Math +Talk +78 + +A Peek Beyond the Point +2. Convert the following decimals into a sum of tenths, hundredths +and thousandths: +(a) 0.34 +(b) 1.02 +(c) +0.8 +(d) 0.362 +3. What decimal number does each letter represent in the number +line below? +b +a +c +6.6 +6.5 +6.4 +4. Arrange the following quantities in descending order: +(a) 11.01, 1.011, 1.101, 11.10, 1.01 +(b) 2.567, 2.675, 2.768, 2.499, 2.698 +(c) 4.678 g, 4.595 g, 4.600 g, 4.656 g, 4.666 g +(d) 33.13 m, 33.31 m, 33.133 m, 33.331 m, 33.313 m +5. Using the digits 1, 4, 0, 8, and 6 make: +(a) the decimal number closest to 30 +(b) the smallest possible decimal number between 100 and 1000. +6. Will a decimal number with more digits be greater than a decimal +number with fewer digits? +7. Mahi purchases 0.25 kg of beans, 0.3 kg of carrots, 0.5 kg of potatoes, +0.2 kg of capsicums, and 0.05 kg of ginger. Calculate the total weight +of the items she bought. +8. Pinto supplies 3.79 L, 4.2 L, and 4.25 L of milk to a milk dairy in +the first three days. In 6 days, he supplies 25 litres of milk. Find the +total quantity of milk supplied to the dairy in the last three days. +9. Tinku weighed 35.75 kg in January and 34.50 kg in February. Has +he gained or lost weight? How much is the change? +10. Extend the pattern: 5.5, 6.4, 6.39, 7.29, 7.28, 6.18, 6.17, ____, _____ +11. How many millimeters make 1 kilometer? +12. Indian Railways offers optional travel insurance for passengers +who book e-tickets. It costs 45 paise per passenger. If 1 lakh people +opt for insurance in a day, what is the total insurance fee paid? +13. Which is greater? +(a) 10 +1000 or 1 +10 ? +79 + +Ganita Prakash | Grade 7 +(b) One-hundredth or 90 thousandths? +(c) One-thousandth or 90 hundredths? +14. Write the decimal forms of the quantities mentioned (an example +is given): +(a) 87 ones, 5 tenths and 60 hundredths = 88.10 +(b) 12 tens and 12 tenths +(c) 10 tens, 10 ones, 10 tenths, and 10 hundredths +(d) 25 tens, 25 ones, 25 tenths, and 25 hundredths +15. Using each digit 0 – 9 not more than once, fill the boxes below so +that the sum is closest to 10.5: + +16. Write the following fractions in decimal form: +(a) 1 +2 +(b) 3 +2 +(c) 1 +4 +(d) 3 +4 +(e) 1 +5 +(f) 4 +5 +• We can split a unit into smaller parts to get more exact/accurate +measurements. +• We extended the Indian place value system and saw that + » +1 unit = 10 one-tenths, + » +1 tenth = 10 one-hundredths, + » +1 hundredth = 10 one-thousandths, + » +10 one-hundredths = 1 tenth, + » +100 one-hundredths = 1 unit. +• A decimal point (‘.’) is used in the Indian place value system to separate +the whole number part of a number from its fractional part. +• We also learnt how to compare decimal numbers, locate them on the +number line, and perform addition and subtraction on them. +SUMMARY +Try +This +80 + +1 + +Page No. 47 + Write the measurements of the objects shown in the picture. + +Ans: +Eraser → 2 4 +10 cm, Pencil → 4 5 +10 cm → 4 1 +2 cm, Chalk → 1 4 +10 𝑐𝑚 +Page No. 49 + Arrange the lengths in increasing order: +(a) +𝟗 +𝟏𝟎 + +(b) 1 +𝟕 +𝟏𝟎 + +(c) +𝟏𝟑𝟎 +𝟏𝟎 + + +(d) 13 +𝟏 +𝟏𝟎 +(e) 10 +𝟓 +𝟏𝟎 + +(f) 7 +𝟔 +𝟏𝟎 + +(g) 6 +𝟕 +𝟏𝟎 + + +(h) +𝟒 +𝟏𝟎 +Ans: + +4 +10, +9 +10 , 1 +7 +10 , 6 +7 +10 , 7 +6 +10 , 10 +5 +10 , +130 +10 , 13 +1 +10 + +Chapter – 3 +A Peek Beyond the Point + +2 + +Page No. 50 + Arrange the following lengths in increasing order: 4 +1 +10 , +4 +10 , +41 +10 , 41 +1 +10 +Ans: +Increasing order: +4 +10 , +41 +10 = 4 +1 +10 , 41 +1 +10 +Page No. 51 + The lengths of the body parts of a honeybee are given. Find its total +length. + +Head = 2 3 +10 units +Thorax = 5 4 +10 units +Abdomen = 7 5 +10 units + + Ans: 15 +2 +10 units. +Page No. 52 + A Celestial Pearl Danio’s length is 𝟐 +𝟒 +𝟏𝟎 cm, and the length of a +Philippine Goby is +𝟗 +𝟏𝟎 cm. What is the difference in their lengths? +Ans: 1 +5 +10 cm + Observe the given sequences of numbers. Identify the change after +each term and extend the pattern: +(a) 4, 4 +𝟑 +𝟏𝟎, 4 +𝟔 +𝟏𝟎, +_______, _______, _______, _______ +(b) 8 +𝟐 +𝟏𝟎, 8 +𝟕 +𝟏𝟎, 9 +𝟐 +𝟏𝟎, +_______, _______, _______, _______ +(c) 7 +𝟔 +𝟏𝟎, 8 +𝟕 +𝟏𝟎, +_______, _______, _______, _______ + + +3 + +(d) 5 +𝟕 +𝟏𝟎, 5 +𝟑 +𝟏𝟎, +_______, _______, _______, _______ +(e) 13 +𝟓 +𝟏𝟎, 13, 12 +𝟓 +𝟏𝟎, _______, _______, _______, _______ +(f) +11 +𝟓 +𝟏𝟎, 10 +𝟒 +𝟏𝟎, 9 +𝟑 +𝟏𝟎, _______, _______, _______, _______ +Ans: +(a) 4, 4 3 +10 , 4 6 +10 , 4 9 +10 , 5 2 +10 , 5 5 +10 , 5 8 +10 → increment of 3 +10 each time. +(b) 8 2 +10 , 8 7 +10 , 9 2 +10 , 9 7 +10 , 10 2 +10 , 10 7 +10 , 11 2 +10 →increment of 5 +10 each time. +(c) 7 6 +10 , 8 7 +10 , 9 8 +10 , 10 9 +10 , 12, 13 1 +10 →increment of 1 + 1 +10 +(d) 5 7 +10 , 5 3 +10 , 4 9 +10 , 4 5 +10 , 4 1 +10 , 3 7 +10 →subtracting: 4 +10 each time +(e) 13 5 +10 , 13 , 12 5 +10 , 12, 11 5 +10 , 11, 10 5 +10 →decreasing by 5 +10 +(f) 11 5 +10 , 10 4 +10 , 9 3 +10 , 8 2 +10 , 7 1 +10 , 6 → decreasing by 1 1 +10 +Page No. 53 + How many one‑hundredths make one‑tenth? Can we also say that +the length is 4 units and 45 one‑hundredths? +Ans: +1 +10 = +10 +100 +So, 10 one‑hundredths make one‑tenth. +Yes, we can write it 4 + +45 +100 = 4 units and 45 one‑hundredths. + + + + + +4 + +Page No. 54 + Observe +the +figure +below. +Notice +the +markings +and +the +corresponding lengths written in the boxes when measured from 0. +Fill the lengths in the empty boxes. + Ans: + + For the lengths shown below write the measurements and read out +the measures in words. +Ans: + +Measurement −5 37 +100 +In words: Five and thirtyseven-hundredths. + +Measurement −15 3 +100 +In words: Fifteen and three-hundredths. + + + +5 + +Page No. 55 + +Measurement −7 52 +100 +In words: Seven and fiftytwo-hundredths. + +Measurement −9 80 +100 +In words: Nine and eighty-hundredths. + In each group, identify the longest and the shortest lengths. Mark +each length on the scale. +Ans: + + +6 + + +(a) Longest = +33 +100 + +Shortest = +3 +100 +(b) Longest = 3 +1 +10 +Shortest = 1 +3 +10 +(c) Longest = +54 +100 + +Shortest = +4 +10 +(d) Longest = 3 +6 +10 +6 +100 Shortest = 3 6 +100 + + + + +7 + +Page No. 56 + +(e) Longest = 1 +8 +100 +Shortest = +9 +100 +(f) Longest = 7 +5 +100 + +Shortest = 7 +3 +10 +5 +100 +(g) Longest = +65 +10 +15 +100 +Shortest = 5 +7 +100 + + + + +8 + +Page No. 58 + Figure it Out +Find the sums and differences +(a) +3 +10 + 3 +4 +100 +Ans: 3 +34 +100 +(b) 9 +5 +10 +7 +100 + 2 +1 +10 +3 +100 +Ans: 11 +7 +10 +(c) 15 +6 +10 +4 +100 + 14 +3 +10 +6 +100 +Ans: 30 +(d) 7 +7 +100 – 4 +4 +100 +Ans: 3 +3 +100 +(e) 8 +6 +100 – 5 +3 +100 +Ans: 3 +3 +100 +(f) 12 +6 +10 +2 +100 – +9 +10 +9 +100 +Ans: 11 +63 +100 +Page No. 63 + Make a place value table similar to the one above. Write each +quantity in decimal form and in terms of place value, and read the +number: +Ans: + +9 + +Question +Hundreds +Tens +Ones +Tenths +Hundredths +Thousandth +Decimal +Form +Read as +a +2 ones, 3 +tenths and 5 +hundredths +– +– +2 +3 +5 +– +2 + +3 +10 + +5 +100 += 2.35 +Two point +three five (or +two and +thirty-five +hundredths) +b + 1 ten and 5 +tenths +– +1 +– +5 +– +– +10 + +5 +10 = 10.5 +Ten point +five (or ten +and five +tenths) +c + 4 ones and +6 +hundredths +– +– +4 +– +6 +– +4 + 0 + +6 +100 += 4.06 +Four point +zero six (or +four and six +hundredths) +d + 1 hundred, +1 one and 1 +hundredth +1 +– +1 +– +1 +– +100 + 0 + 1 + 0 + +1 +100 +=101.01 +One hundred +one point +zero one (or +one hundred +one and one +hundredth) +e + +8 +100 and +9 +10 +– +– +– +9 +8 +– +9 +10 + +8 +100 += 0.98 +Zero point +nine eight +(or ninety- +eight +hundredths) + +10 + + +Page No. 64 + Write these quantities in decimal form: + (a) 234 hundredths +Ans: 2.34 + (b) 105 tenths. +Ans: 10.5 + +f +5 +100 +– +– +– + +5 +– +5 +100 = 0.05 +Zero point +zero five (or +five +hundredths) +g +1 +10 +– +– +– +1 +– +– +1 +10 = 0.1 +Zero point +one (or one +tenth) +h +2 +1 +10, 4 +1 +10, +and 7 +7 +1000 + +– +2 +4 +7 +0 +1 +- +1 +– +- +- +- +7 +2.01 +4 + 1 +10 = 4.1 +7 + 0 +10 + 0 +100 + +7 +1000 += 7.007 +Two point +zero one +and +Four point +one +and +Seven point +zero zero +seven + +13.117 +Thirteen +point one +one seven. + +11 + + Fill in the blanks below (mm < – > cm) + +Ans: + +12 mm= 1.2 cm +56 mm = 5.6 cm +70 mm = 7.0 cm +9 mm = 0.9 cm +134 mm = 13.4 cm +2036 mm = 203.6 cm +Page No. 66 + Fill in blanks (cm <-> m): +36 cm = 0.36 m +50 cm = 0.5 m +89 cm = 0.89 m +4 cm = 0.04 m +325 cm = 3.25 m +207 cm = 2.07 m + How many mm does 1 meter have? +Ans: 1m = 1000 mm + Can we write 1 mm = +𝟏 +𝟏𝟎𝟎𝟎 m? +Ans: Yes +Page No. 68 + Fill in the blanks below (g < – > kg) +Ans: +465 g = 0.465 kg +68 g = 0.068 kg +1560 g = 1.56 kg +704 g = 0.704 kg +560 g = 0.56 kg +2500 g = 2.5 kg + + + +12 + + +Page No. 69 + Fill in the blanks below (rupee < – > paise) +10 p = ₹0.10 +5 p = ₹0.05 +36 p = ₹0.36 +50 p = ₹0.50 +99 p = ₹0.99 +250 p = ₹2.50 +Page No. 70 + Name all the divisions between 1 and 1.1 on the number line. +Ans: The small divisions between 1 and 1.1 on the number line (if the sub- +division is of 10 parts) are: 1.01, 1.02, 1.03, 1.04, 1.05, 1.06, 1.07, 1.08, 1.09 + Identify and write the decimal numbers against the letters. +Ans: + +Page No. 71 + Can you tell which of these is the smallest and which is the largest? +Ans: +Decimal Number +Units +Tenths +Hundredths +Thousandths +0.2 +0 +2 +0 +0 +0.20 +0 +2 +0 +0 +0.200 +0 +2 +0 +0 +0.02 +0 +0 +2 +0 +0.002 +0 +0 +0 +2 +0.2 = 0.20 = 0.200 is the largest. + +13 + +0.002 is the smallest. + Which of these are the same: 4.5, 4.05, 0.405, 4.050, 4.50, 4.005, 04.50? +Ans: 4.5, 4.50, and 04.50 are equal. +4.05 and 4.050 are equal. + Identify the decimal number in the last number line in Figure (b) +denoted by ‘?’ +Ans: + + Make such number lines for the decimal numbers: (a) 9.876 (b) 0.407. +Ans: + +(a) + + + + + + (b) + + + + + +14 + + In the number line shown below, what decimal numbers do the boxes +labelled ‘a’, ‘b’, and ‘c’ denote? +Ans: + +Page No. 72 + Using similar reasoning find out the decimal numbers in the boxes +below. +Ans: + + +15 + +Page No. 73 + Which decimal number is greater? + (a) 1.23 or 1.32 (b) 3.81 or 13.800 (c) 1.009 or 1.090 +Ans: +(a) 1.32 is larger. +(b) 13.800 is larger. +(c) 1.090 is larger. + Consider the decimal numbers 0.9, 1.1, 1.01 and 1.11 +Which of the above is closest to 1.09? +Ans: 1.1 is closest to 1.09. + Which among these is closest to 4: 3.56, 3.65, 3.099? +Ans: 3.65 is closest to 4. + Which among these is closest to 1: 0.8, 0.69, 1.08? +Ans: 1.08 is closest to 1. + In each case below use the digits 4, 1, 8, 2, and 5 exactly once and try +to make a decimal number as close as possible to 25. +Ans: + + +Page No. 75 + Figure it Out +1. Find the sums +(a) 5.3 + 2.6 + + +(b) 18 + 8.8 + +(c) 2.15 + 5.26 + +(d) 9.01 + 9.10 +(e) 29.19 + 9.91 + +(f) 0.934 + 0.6 + +(g) 0.75 + 0.03 + + +(h) 6.236 + 0.487 + +16 + +Ans: +(a) 7.9 + + + +(b) 26.8 + + +(c) 7.41 + + + +(d) 18.11 +(e) 39.10 + + +(f) 1.534 + + +(g) 0.78 + + + +(h) 6.723 +2. Find the differences +(a) 5.6 – 2.3 + + +(b) 18 – 8.8 + +(c) 10.4 – 4.5 + + +(d) 17 – 16.198 +(e) 17 – 0.05 + + +(f) 34.505 – 18.1 +(g) 9.9 – 9.09 + + +(h) 6.236 – 0.487 +Ans: +(a) 3.3 + + +(b) 9.2 + + +(c) 5.9 + + +(d) 0.802 +(e) 16.95 + +(f) 16.405 + + +(g) 0.81 + + +(h) 5.749 + Continue this sequence 4.4, 4.8. 5.2, 5.6, 6.0, … and write the next 3 +terms. +Ans: 4.4, 4.8. 5.2, 5.6, 6.0, 6.4, 6.8, 7.2 +Page No. 76 + Similarly, identify the change and write the next 3 terms for each +sequence given below. +Try to do this computation mentally. +(a) 4.4, 4.45, 4.5, … + +(b) 25.75, 26.25, 26.75, … +(c) 10.56, 10.67, 10.78, … +(d) 13.5, 16, 18.5, … +(e) 8.5, 9.4, 10.3, … + +(f) 5, 4.95, 4.90, … +(g) 12.45, 11.95, 11.45, … +(h) 36.5, 33, 29.5, … +Ans: +(a) 4.4, 4.45, 4.5,4.55, 4.6, 4.65 → increment of + 0.05 +(b) 25.75, 26.25, 26.75, 27.25, 27.75, 28.25 → add 0.5 +(c) 10.56, 10.67, 10.78, 10.89, 11.0, 11.11→ Add 0.11 +(d) 13.5, 16, 18.5, 21.0, 23.5, 26.0 → Add 2.5 + +17 + +(e) 8.5, 9.4, 10.3, 11.2, 12.1, 13.0 → Add 0.9 +(f) 5, 4.95, 4.90, 4.85, 4.80, 4.75 → Subtract 0.05 +(g) 12.45, 11.95, 11.45, 10.95, 10.45, 9.95 → Subtract 0.50 +(h) 36.5, 33, 29.5, 26.0, 22.5, 19.0 → Subtract 3.5 +Page No. 78 + Figure it out +1. Convert the following fractions into decimals: +(a) +𝟓 +𝟏𝟎𝟎 + +(b) +𝟏𝟔 +𝟏𝟎𝟎��� + +(c) +𝟏𝟐 +𝟏𝟎 +(d) +𝟐𝟓𝟒 +𝟏𝟎𝟎𝟎 +Ans: + (a) +5 +100 = 0.05 +(b) +16 +1000 = 0.016 +(c) +12 +10 = 1.2 +(d) +254 +1000 = 0.254 +Page No. 79 +2. Convert the following decimals into a sum of tenths, hundredths +and thousandths: +(a) 0.34 +(b) 1.02 +(c) 0.8 +(d) 0.362 +Ans: + (a) +3 +10 + +4 +100 + (b) +100 +100 + +2 +100 = 1 + +2 +100 + (c) +8 +10 + + + (d) +3 +10 + +6 +100 + +2 +1000 +3. What decimal number does each letter represent in the number +line below? +Ans: + +18 + + +4. Arrange the following quantities in descending order: +(a) 11.01, 1.011, 1.101, 11.10, 1.01 +→ Descending order: 11.10, 11.01, 1.101, 1.011, 1.01. +(b) 2.567, 2.675, 2.768, 2.499, 2.698 +→ Descending order: 2.768, 2.698, 2.675, 2.567, 2.499. +(c) 4.678 g, 4.595 g, 4.600 g, 4.656 g, 4.666 g +→Descending order: 4.678, 4.666, 4.656, 4.600, 4.595. +(d) 33.13 m, 33.31 m, 33.133 m, 33.331 m, 33.313 m +→ Descending order: 33.331, 33.313, 33.31, 33.133, 33.13. +5. Using the digits 1, 4, 0, 8, and 6 make: (a) the decimal number +closest to 30 (b) the smallest possible decimal number between +100 and 1000. +Ans: Using the digits 1, 4, 0, 8, and 6, we can make: +(a) 40.168 +(b) 104.68 + 6. Will a decimal number with more digits be greater than a +decimal number with fewer digits? +Ans: No. It is not necessary. For example, 2.05 (2 digits after decimal) vs +2.5 (1 digit after decimal). 2.5 >2.05 +7. Mahi purchases 0.25 kg of beans, 0.3 kg of carrots, 0.5 kg of potatoes, +0.2 kg of capsicums, and 0.05 kg of ginger. Calculate the total +weight of the items she bought. +Ans: 1.3 kg + 8. Pinto supplies 3.79 L, 4.2 L, and 4.25 L of milk to a milk dairy in the +first three days. In 6 days, he supplies 25 liters of milk. Find the +total quantity of milk supplied to the dairy in the last three days. + +19 + +Ans: 12.76 L +9. Tinku weighed 35.75 kg in January and 34.50 kg in February. Has +he gained or lost weight? How much is the change? +Ans: He has lost weight. The change in the weight = 1.25 kg + +10. Extend the pattern: 5.5, 6.4, 6.39, 7.29, 7.28, 8.18, 8.17, ____, _____ +Ans: 9.07, 9.06 +11. How many millimeters make 1 kilometer? +Ans: 1000000 mm +12. Indian Railways offers optional travel insurance for passengers +who book e-tickets. It costs 45 paise per passenger. If 1 lakh +people opt for insurance in a day, what is the total insurance fee +paid? +Ans: The total insurance fee paid for 1 lakh passengers is ₹45,000. +13. Which is greater? +(a) +𝟏𝟎 +𝟏𝟎𝟎𝟎 or +𝟏 +𝟏𝟎? +(b) One-hundredth or 90 thousandths? +(c) One-thousandth or 90 hundredths? + Ans: +(a) +1 +10 > +1 +100 +(b) 90 thousandths > One-hundredth +(c) 90 hundredths > One-thousandth +Page No. 80 +14. Write the decimal forms of the quantities mentioned (an +example is given): +(a) 87 ones, 5 tenths and 60 hundredths +(b) 12 tens and 12 tenths +(c) 10 tens, 10 ones, 10 tenths, and 10 hundredths +(d) 25 tens, 25 ones, 25 tenths, and 25 hundredths +Ans: + +20 + +(a) 88.10 + +121.2 +(b) 111.1 + +277.75 + + +15. Using each digit 0 – 9 not more than once, fill the boxes below so +that the sum is closest to 10.5: +Ans: + +16. Write the following fractions in decimal form: +(a) +1 +2 + + +(b) +3 +2 + +(c) +1 +4 +(d) +3 +4 + + +(e) +1 +5 + +(f) +4 +5 +Ans: +(a) 0.5 + + +(b) 1.5 + +(c) 0.25 +(d) 0.75 + +(e) 0.2 + +(f) 0.8" +class_7,4,Expressions using Letter-Numbers,ncert_books/class_7/gegp1dd/gegp104.pdf,"4.1 The Notion of Letter-Numbers +In this chapter we shall look at a concise way of expressing mathematical +relations and patterns. We shall see how this helps us in thinking about +these relationships and patterns, and in explaining why they may hold +true. +Example 1: Shabnam is 3 years older than Aftab. When Aftab’s age +10 years, Shabnam’s age will be 13 years. Now Aftab’s age is 18 years, +what will Shabnam’s age be? _______ +Given Aftab’s age, how will you find out Shabnam’s age? +Easy: We add 3 to Aftab’s age to get Shabnam’s age. +Can we write this as an expression? +Shabnam’s age is 3 years more than Aftab’s. In short, this can be +written as: +Shabnam’s age = Aftab’s age + 3. +Such +mathematical +relations +are +generally represented in a shorthand +form. In the relation above, instead +of writing the phrase ‘Aftab’s Age’, +the convention is to use a convenient +symbol. Usually, letters or short phrases +are used for this purpose. +Let us say we use the letter a to +denote Aftab’s age (we could have +used any other letter), and s to denote +Shabnam’s age. Then the expression to +find Shabnam’s age will be a + 3, which can be written as +s = a + 3. +If a is 23 (Aftab’s age in years), then what is Shabnam’s age? +Expression for +Shabnam’s age +4 + 3 +10 + 3 +23 + 3 +? + 3 +a + 3 +Aftab’s +age +4 +10 +23 +? +a +Fig. 4.1 +EXPRESSIONS +USING LETTER- +NUMBERS +4 + +Ganita Prakash | Grade 7 +Replacing a by 23 in the expression a + 3, we get, s = 23 + 3 = 26 years. +Letters such as a and s that are used to represent numbers are called +letter-numbers. Mathematical expressions containing letter-numbers, +such as the expression a + 3, are called algebraic expressions. +Given the age of Shabnam, write an expression to find Aftab’s age. +We know that Aftab is 3 years younger than Shabnam. So, Aftab’s +age will be 3 less than Shabnam’s. This can be described as + + +Aftab’s age = Shabnam’s age – 3. +If we again use the letter a to denote Aftab’s age and the letter s +to denote Shabnam’s age, then the algebraic expression would be: +a = s – 3, meaning 3 less than s. +Use this expression to find Aftab’s age if Shabnam’s age is 20. +Example 2: Parthiv is making matchstick patterns. He repeatedly +places Ls next to each other. Each L has two matchsticks as shown in +Figure 4.2. +Fig. 4.2 +How many matchsticks are needed to make 5 Ls? It will be 5 × 2. +How many matchsticks are needed to make 7 Ls? It will be 7 × 2. +How many matchsticks are needed to make 45 Ls? It will be 45 × 2. +Now, what is the relation between the number of Ls and the number +of sticks? +First, let us describe the relationship or the pattern here. Every L +needs 2 matchsticks. So the number of matchsticks needed will be 2 +times the number of L’s. This can be written as: +Number of matchsticks = 2 × Number of L’s +Now, we can use any letter to denote the number of L’s. Let’s use n. +The algebraic expression for the number of matchsticks will be: +2 × n. +This expression tells us how many matchsticks are needed to make +n L’s. To find the number of matchsticks, we just replace n by the +number of L. +Example 3: Ketaki prepares and supplies coconut-jaggery laddus. The +price of a coconut is ₹35 and the price of 1 kg jaggery is ₹60. +82 + +Expressions Using Letter-Numbers +How much should she pay if she buys 10 coconuts and 5 kg jaggery? +Cost of 10 coconuts = 10 × ₹35 +Cost of 5 kg jaggery = 5 × ₹60 +Total cost = 10 × ₹35 + 5 × ₹60 = ₹350 + ₹300 = ₹650. +How much should she pay if she buys 8 coconuts and 9 kg jaggery? +Write an algebraic expression to find the total amount to be paid for a +given number of coconuts and quantity of jaggery. +Let us identify the relationships and then write the expressions. +Quantity needed +Relationship +Expression +Cost of coconuts +Number of +coconuts × 35 +c × 35 +Cost of jaggery +Number of kgs of +jaggery × 60 +j × 60 +Here, ‘c’ represents the number of coconuts and ‘j’ represents the +number of kgs of jaggery. The total amount to be paid will be: +Cost of coconuts + Cost of jaggery. +The corresponding algebraic expression can be written as: +c × 35 + j × 60 +Use this expression (or formula) to find the total amount to be paid for +7 coconuts and 4 kg jaggery. +Notice that for different values of ‘c’ and ‘j’, the value of the expression +also changes. +Writing this expression as a sum of terms we get: ++ +c × 35 +j × 60 +Example 4: We are familiar with calculating the perimeters of simple +shapes. Write expressions for perimeters. +The perimeter of a square is 4 times the length of its side. This can +be written as the expression: 4 × q, where q stands for the sidelength. +What is the perimeter of a square with sidelength 7 cm? Use the +expression to find out. +You must have realised how the use of letter-numbers and algebraic +expressions allows us to express general mathematical relations in +83 + +Ganita Prakash | Grade 7 +a concise way. Mathematical relations expressed this way are often +called formulas. +Figure it Out +1. Write formulas for the perimeter of: +(a) triangle with all sides equal. +(b) a regular pentagon (as we have learnt last year, we use the +word ‘regular’ to say that all sidelengths and angle measures +are equal) +(c) a regular hexagon +2. Munirathna has a 20 m long pipe. However, he wants a longer +watering pipe for his garden. He joins another pipe of some length to +this one. Give the expression for the combined length of the pipe. Use +the letter-number ‘k’ to denote the length in meters of the other pipe. +3. What is the total amount Krithika has, if she has the following +numbers of notes of ₹100, ₹20 and ₹5? Complete the following table: +No. of +₹100 notes +No. of +₹20 notes +No. of +₹5 notes +Expression and total +amount +3 +5 +6 +6 × 100 + 4 × 20 + 3 × 5 += 695 +8 +4 +z +x +y +z +4. Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller +mill to start running. Once it is running, each kg of grain takes +8 seconds to grind into powder. Which of the expressions below +describes the time taken to complete grind ‘y’ kg of grain, assuming +the machine is off initially? +(a) 10 + 8 + y +(b) +(10 + 8) × y +(c) 10 × 8 × y +(d) +10 + 8 × y +(e) 10 × y + 8 +5. Write algebraic expressions using letters of your choice. +(a) 5 more than a number +(b) 4 less than a number +84 + +Expressions Using Letter-Numbers +(c) 2 less than 13 times a number +(d) 13 less than 2 times a number +6. Describe situations corresponding to the following algebraic +expressions: +(a) 8 × x + 3 × y +(b) 15 × j – 2 × k +7. In a calendar month, if any 2 × 3 grid full of dates is chosen as +shown in the picture, write expressions for the dates in the blank +cells if the bottom middle cell has date ‘w’. +4.2 Revisiting Arithmetic Expressions +We learnt to write expressions as sums of terms and it became easy +for us to read arithmetic expressions. Many times they could have +been read in multiple ways and it was confusing. We used swapping +(adding two numbers in any order) and grouping (adding numbers +by grouping them conveniently) to find easy ways of evaluating +expressions. Swapping and grouping terms does not change the value of +the expression. We also learnt to use brackets in expressions, including +brackets with a negative sign outside. We learnt the distributive +property (multiple of a sum is the same as sum of multiples). +Let us revise these concepts and find the values of the following +expressions: +1. +23 – 10 × 2 +2. +83 + 28 – 13 + 32 +3. +34 – 14 + 20 +4. +42 + 15 – (8 – 7) +5. +68 – (18 + 13) +6. +7 × 4 + 9 × 6 +7. +20 + 8 × (16 – 6) +Let us evaluate the first expression, 23 – 10 × 2. First we shall write +the terms of the expression. Notice that one of the terms is a number, +while the other one has to be converted to a number before adding the +two terms. +23 – 10 × 2 = 23 + –10 × 2 = 23 + –20 = 3 +w – 1 +w +85 + +Ganita Prakash | Grade 7 +Let us now evaluate the second one. All the terms of this expression +are numbers. If we notice the terms, we find that it will be easier to +evaluate if we swap and group the terms. +83 + 28 – 13 + 32 = +Let us now look at the fifth expression. It has brackets with a negative +sign outside. This can be evaluated in two ways — by solving the bracket +first (like the solution on the left side) or by removing the brackets +appropriately (as on the right side). +Now, find the values of the other arithmetic expressions. +Algebraic expressions also take number values when the letter- +numbers they contain are replaced by numbers. In Example 1, for +finding Shabnam’s age when Aftab is 23 years old, we replaced the +letter-number a in the expression a + 3 by 23, and it took the value 26. +4.3 Omission of the Multiplication Symbol in +Algebraic Expressions +Look at this number sequence: +4, 8, 12, 16, 20, 24, 28, ... +How can we describe this sequence or pattern? Easy: These are the +numbers appearing in the multiplication table of 4 (multiples of 4 in +an increasing order). +What is the third term of this sequence? It is 4 × 3. +What is the 29th term of this sequence? It is 4 × 29. +Find an algebraic expression to get the nth term of this sequence. +Note that here ‘n’ is a letter-number that denotes a position in the +sequence. +83 +28 +–13 +32 ++ ++ ++ +70 +60 +130 += ++ += +37 += +68 +–(18 + 13) += ++ +68 +–31 += ++ +68 +–(18 + 13) += ++ +50 +–13 +37 += ++ += +68 +–18 +–13 += ++ ++ +OR +86 + +Expressions Using Letter-Numbers +As it is the sequence of multiples of 4, it can be seen that the nth term +will be 4 times n: +4 × n +As a standard practice, we shorten 4 × n to 4n by skipping the +multiplication sign. We write the number first, followed by the letter(s). +Find the value of the expression 7k when k = 4. The value is 7 × 4 = 28. +Find the value that the expression 5m + 3 takes when m = 2. +As 5m stands for 5 × m, the value of the expression when m = 2 is +5 × 2 + 3 = 13. +Mind the Mistake, Mend the Mistake +Some simplifications are shown below where the letter-numbers are +replaced by numbers and the value of the expression is obtained. +1. Observe each of them and identify if there is a mistake. +2. If you think there is a mistake, try to explain what might have gone +wrong. +3. Then, correct it and give the value of the expression. +4.4 Simplification of Algebraic Expressions +Earlier we found expressions to find perimeters of different regular +figures in terms of their sides. Let us now find an expression to find the +perimeter of a rectangle. +1 +If a = – 4, +then 10 – a = 6. +2 +If d = 6, +then 3d = 36. +3 +If s = 7, +then 3s – 2 = 15. +4 +If r = 8, +then 2r + 1 = 29. +5 +If j = 5, +then 2j = 10. +6 +If m = –6, +then 3 (m + 1) = 19. +7 +If f = 3, g = 1 +then 2f – 2g = 2. +8 +If t = 4, b = 3 +then 2t + b = 24. +9 +If h = 5, n = 6 +then h – (3 – n) = 4. +87 + +Ganita Prakash | Grade 7 +As in the previous cases, we will first describe how to get the perimeter +when the length and the breadth of the rectangle are known: +Find the sum of length + breadth + length + breadth. +Let us use the letter-numbers l and b in place of length and breadth +respectively. Let p denote the perimeter of the rectangle. Then we have +p = l + b + l + b +As we know, these represent numbers, and so the terms of an +expression can be added in any order. Hence the above expression can +be written as: += l + l + b + b +As l + l = 2 × l = 2l, and b + b = 2 × b = 2b, we have +p = 2l + 2b. +Notice that the initial expression (l + b + l + b) and the final expression +(2l + 2b) that we got for the perimeter look different. However, they +are equal since the expression was obtained from the initial one by +applying the same rules and operations we do for numbers; they are +equal in the sense that they both take the same values when letter- +numbers are replaced by numbers. +For example, if we assign l = 3, b = 4, we get +l + b + l + b = 3 + 4 + 3 + 4 = 14, and +2l + 2b = 2 × 3 + 2 × 4 = 14. +We call the expression 2l + 2b the simplified form of l + b + l + b. +Let us see some more examples of simplification. +Example 5: Here is a table showing the number of pencils and erasers +sold in a shop. The price per pencil is c, and the price per eraser is d. Find +the total money earned by the shopkeeper during these three days. +Day 1 +Day 2 +Day 3 +Pencils +(Price ‘c’) +5 +3 +10 +Erasers +(Price ‘d’) +4 +6 +1 +Let us first find the money earned by the sale of pencils. +The money earned by selling pencils on Day 1 is 5c. Similarly, the +money earned by selling pencils on Day 2 is _____, and Day 3 is ______. +The total money earned by the sale of pencils is 5c + 3c + 10c. Can we +simplify this expression further and reduce the number of terms? +88 + +Expressions Using Letter-Numbers +The expression means 5 times c is added to 3 times c is added to 10 +times c. So in total, the letter-number c is added (5 + 3 + 10) times. This +is what we have seen as the distributive property of numbers. Thus, +5 × c + 3 × c + 10 × c = (5 + 3 + 10) × c +(5 + 3 + 10) × c can be simplified to 18 × c = 18c. +If c = ₹50, find the total amount earned by the scale of pencils. +Write the expression for the total money earned by selling erasers. +Then, simplify the expression. +The expression for the total money earned by selling pencils and +erasers during these three days is 18c + 11d. +Can the expression 18c + 11d be simplified further? +There is no way of further simplifying this expression as it contains +different letter-numbers. It is in its simplest form. +In this problem, we saw the expression 5c + 3c + 10c getting simplified +to the expression 18c. +Check that both expressions take the same value when c is replaced by +different numbers. +Example 6: A big rectangle is split into two smaller rectangles as +shown. Write an expression describing the area of the bigger rectangle. +The areas of the smaller rectangles are 4v sq. units and 3v sq. units. +The area of the bigger rectangle can be found in two ways: (i) by +directly using its side lengths v and (4 + 3), or (ii) by adding the areas +of the smaller rectangles. +The first way gives 7v, and the second way gives 4v + 3v. We know +that they are equal: 4v + 3v = 7v, and this is the required expression for +the area of the bigger rectangle. +As earlier, a big rectangle is split into two smaller rectangles as shown +below. Write an expression to find the area of the rectangle AEFD. +Another way of seeing +the distributive +property. +3 +4 +v +89 + +Ganita Prakash | Grade 7 +Even in this case, the area of rectangle AEFD can be found in two +ways: (i) by directly using the side lengths n and (12 – 4), or (ii) +subtracting the area of the rectangle EBCF from that of ABCD. +The first method gives us 8n, and the second method gives us +12n – 4n, and they are equal, since 12n – 4n = 8n. This is the expression +for the area of the rectangle AEFD. +Sets of terms such as (5c, c, 10c), (12n, – 4n) that involve the same +letter-numbers are called like terms. Sets of terms such as {18c, 11d} +are called unlike terms as they have different letter-numbers. +As we have seen, like terms can be added together and simplified +into a single term. +Example 7: A shop rents out chairs +and tables for a day’s use. To rent +them, one has to first pay the following +amount per piece. +When the furniture is returned, the +shopkeeper pays back some amount +as follows. +Write an expression for the total +number of rupees paid if x chairs and +y tables are rented. +For x chairs and y tables, let us find the total amount paid at the +beginning and the amount one gets back after returning the furniture. +Describe the procedure to get these amounts. +The total amount in rupees paid at the beginning is 40x + 75y, and +the total amount returned is 6x + 10y. +So, the total amount paid = (40x + 75y) – (6x + 10y). +Can we simplify this expression? If yes, how? If not, why not? +4 +C +B +F +E +D +A +12 +n +Item +Amount +Chair +₹40 +Table +₹75 +Amount +returned +Chair +₹6 +Table +₹10 +Math +Talk +90 + +Expressions Using Letter-Numbers +Recalling how we open brackets in an arithmetic expression, we get +(40x + 75y) – (6x + 10y) = (40x + 75y) – 6x – 10y +Since the terms can be added in any order, the remaining bracket +can be opened and the expression becomes 40x + 75y + – 6x + – 10y +We can group the like terms together, This results in +40x + – 6x + 75y + – 10y += (40 – 6)x + (75 – 10)y += 34x + 65y. +The expression (40x + 75y) – (6x + 10y) is simplified to 34x + 65y, +which is the total amount paid in rupees. +Could we have written the initial expression as +(40x + 75y) + (– 6x – 10y)? +Example 8: Charu has been through three rounds of a quiz. Her scores +in the three rounds are 7p – 3q, 8p – 4q, and 6p – 2q. Here, p represents +the score for a correct answer and q represents the penalty for an +incorrect answer. +What do each of the expressions mean? +If the score for a correct answer is 4 (p = 4) and the penalty for a wrong +answer is 1 (q = 1), find Charu’s score in the first round. +Charu’s score is 7 × 4 – 3 × 1. We can evaluate this expression by +writing it as a sum of terms. +7 × 4 – 3 × 1 = 7 × 4 + – 3 × 1 = 28 + – 3 = 25 +What are her scores in the second and third rounds? +What if there is no penalty? What will be the value of q in that +situation? +What is her final score after the three rounds? +Her +final +score +will +be +the +sum +of +the +three +scores: +(7p – 3q) + (8p – 4q) + (6p – 2q). +Since the terms can be added in any order, we can remove the +brackets and write +7p + – 3q + 8p + – 4q + 6p + –2q += 7p + 8p + 6p + – (3q) + – (4q) + – (2q) (by swapping and grouping) += (7 + 8 + 6)p + – (3 + 4 + 2)q += 21p + – 9q += 21p – 9q. +Charu’s total score after three rounds is 21p – 9q. Her friend Krishita’s +score after three rounds is 23p – 7q. +Math +Talk +91 + +Ganita Prakash | Grade 7 +Give some possible scores for Krishita in the three rounds so that they +add up to give 23p – 7q. +Can we say who scored more? Can you explain why? +How much more has Krishita scored than Charu? This can be found by +finding the difference between the two scores. +23p – 7q – (21p – 9q) +Simplify this expression further. +Example 9: Simplify the expression 4 (x + y) – y +Using the distributive property, this expression can be simplified to +4 (x + y) – y = 4x + 4y – y + += 4x + 4y + – y + += 4x + (4 – 1)y + += 4x + 3y. +Example 10: Are the expressions 5u and 5 + u equal to each other? +The expression 5u means 5 times the number u, and the expression +5 + u means 5 more than the number u. These two being different +operations, they should give different values for most values of u. +Let us check this. +Fill the blanks below by replacing the letter-numbers by numbers; an +example is shown. Then compare the values that 5u and 5 + u take. +If the expressions 5u and 5 + u are equal, then they should take the +10 +5u +u = 11 +u = 8 +u = 2 +u = 2 +u = 5 +u = 5 +u = 8 +u = 11 +5 + u +7 +92 + +Expressions Using Letter-Numbers +same values for any given value of u. But we can see that they do not. +So, these two expressions are not equal. +Are the expressions 10y – 3 and 10(y – 3) equal? +10y – 3, short for 10 × y – 3, means 3 less than 10 times y, +10(y – 3), short for 10 × (y – 3), means 10 times (3 less than y). +Let us compare the values that these expressions take for different +values of y. + +After filling in the two diagrams, do you think the two expressions are +equal? +Example 11: What is the sum of the numbers +in the picture (unknown values are denoted by +letter-numbers)? + There are many ways to go about it. Here, we +show some of them. +1. +Adding row wise gives: +(4 × 3) + (r + s) + (r + s) + (4 × 3) +2. +Adding like terms together gives: +(8 × 3) + (r + r) + (s + s) +3. +Adding the upper half and doubling gives: +2 × (4 × 3 + r + s) +The three expressions might seem different. We can simplify each +one and see that they all are the same: 2r + 2s + 24. +Figure it Out +1. Add the numbers in each picture below. Write their +corresponding expressions and simplify them. Try adding the +numbers in each picture in a couple different ways and see +Math +Talk +17 +10y – 3 +y = 2 +y = 10 +y = 0 +y = 0 +y = 7 +y = 7 +y = 10 +y = 2 +10 (y – 3) +–10 +3 +r +s +r +s +3 +3 +3 +3 +3 +3 +3 +Math +Talk +93 + +Ganita Prakash | Grade 7 +that you get the same thing. + +2. Simplify each of the following expressions: +(a) p + p + p + p, p + p + p + q, + +p + q + p – q, +(b) p – q + p – q, + +p + q – p + q, +(c) p + q – (p + q), + +p – q – p – q +(d) 2d – d – d – d, + +2d – d – d – c, +(e) 2d – d – (d – c), + +2d – (d – d) – c, +(f) 2d – d – c – c +Mind the Mistake, Mend the Mistake +Some simplifications of algebraic expressions are done below. The +expression on the right-hand side should be in its simplest form. +• Observe each of them and see if there is a mistake. +• If you think there is a mistake, try to explain what might have +gone wrong. +• Then, simplify it correctly. + +Expression +Simplest Form +Correct Simplest + + +Form +1.  3a + 2b + +5 +2.  3b – 2b – b + +0 +3.  6 (p + 2) + +6p + 8 +4.  (4x + 3y) – (3x + 4y) +x + y +5.  5 – (2 – 6z) + +3 – 6z +6.  2 + (x + 3) + +2x – 6 +7.  2y + (3y – 6) + +– y + 6 +8.  7p – p + 5q – 2q +7p + 3q +9.  5 (2w + 3x + 4w) +10w + 15x + 20w +–5g +5k +5k +–5g +5k +5k +5k +5k +5k +5k +5k +5k +–5g +5k +5k +–5g +2p +3q +3q +2p +2p +3q +3q +2p +–2 +3 +3 +–2 +5y +x +x +5y +–6 +2 +94 + +Expressions Using Letter-Numbers +10.  3j + 6k + 9h + 12 +3 (j + 2k + 3h + 4) +11.  4 (2r + 3s + 5) +– 20 – 8r – 12s +Take a look at all the corrected simplest forms (i.e. brackets are removed, +like terms are added, and terms with only numbers are also added). Is +there any relation between the number of terms and the number of +letter-numbers these expressions have? +4.5 Pick Patterns and Reveal Relationships +In the first section we got a glimpse of algebraic expressions and +how to use them to describe simple patterns and relationships in a +concise and elegant manner. Here, we continue to look for general +relationships between quantities in different scenarios, find patterns +and, interestingly, even explain why these patterns occur. + Remember the importance of describing in simple language, or +visualising mathematical relationships, before trying to write them as +expressions. +Formula Detective +Look at the picture given. In each case, the number machine takes in +the 2 numbers at the top of the ‘Y’ as inputs, performs some operations +and produces the result at the bottom. The machine performs the same +operations on its inputs in each case. +Find out the formula of this number machine. +The formula for the number machine above is “two times the first +number minus the second number”. When written as an algebraic +expression, the formula is 2a – b. The expression for the first set of +inputs is 2 × 5 – 2 = 8. Check that the formula holds true for each set +of inputs. +5 +8 +15 +7 +10 +expression: +expression: +expression: +expression: +expression: +8 +9 +10 +6 +2 +1 +11 +10 +4 +95 + +Ganita Prakash | Grade 7 +Find the formulas of the number machines below and write the +expression for each set of inputs. +Now, make a formula on your own. Write a few number machines as +examples using that formula. Challenge your classmates to figure it out! +Note to the Teacher: Not just solving problems but creating new questions is +also very much a part of learning and doing mathematics! + +Algebraic Expressions to Describe Patterns +Example 12: Somjit noticed a repeating pattern along the border +of a saree. +A +B +C +D +E +F +Somjit wonders if there is a way to describe all the positions where the +(i) Design A occurs, (ii) Design B occurs, and (iii) Design C occurs. +5 +5 +7 +18 +18 +expression: +expression: +expression: +expression: +expression: +8 +9 +10 +a +2 +1 +11 +10 +b +4 +5 +1 +7 +expression: +expression: +expression: +expression: +expression: +6 +3 +10 +a +1 +0 +2 +3 +b +96 + +Expressions Using Letter-Numbers +Let us start with design C. It appears for the first time at position 3, +the second time at position 6. +Where would design C appear for the nth time? +We can see that this design appears in positions that are multiples of +3. So the nth occurrence of Design C will be at position 3n. +Similarly, find the formula that gives the position where the other +Designs appear for the nth time. +The positions where B occurs are 2, 5, 8, 11, 14, and so on. +We can see that the position of the nth appearance of Design B is one +less than the position at which Design C appears for the nth time. Thus, +the nth occurrence of Design B is at position: +3n – 1 +Similarly, the expression describing the position at which the design +A appears for the nth time is: 3n – 2. +Given a position number can we find out the design that appears there? +Which Design appears at Position 122? +If the position is a multiple of 3, then clearly we have Design C. As +seen earlier, if the position is one less than a multiple of 3, it has Design +B, and if it is 2 less than a multiple of 3, then it has Design A. +Can the remainder obtained by dividing the position number by 3 be +used for this? Observe the table below. +Position no. +Quotient on division by 3 +Remainder +99 +33 +0 +122 +40 +2 +148 +49 +1 +Use this to find what design appears at positions 99, 122, and 148. +Patterns in a Calendar +Here is the calendar of November 2024. Consider 2 × 2 squares, +as marked in the calendar. The numbers in this square show an +interesting property. +97 + +Ganita Prakash | Grade 7 +Let us take the marked 2 × 2 square, and +consider the numbers lying on the diagonals; +12 and 20; 13 and 19. Find their sums; 12 + 20, +13 + 19. What do you observe? +They are equal. +Let us extend the numbers in the calendar beyond +30, creating endless rows. +Will the diagonal sums be equal in every 2 × 2 square in this endless +grid? How can we be sure? +To be sure of this we cannot check with all 2 × 2 squares as there are +an unlimited number of them. +Let us consider a 2 × 2 square. Its top left number can be any number. +Let us call it ‘a’. +Given that we know the top left number, how do we +find the other numbers in this 2 × 2 square? +As we have been doing, first let us describe the other +numbers in words. +a +? +? +? +12 +13 +19 +20 +98 + +Expressions Using Letter-Numbers +• the number to the right of ‘a’ will be 1 more than it. +• the number below ‘a’ will be 7 more than it. +• the number diagonal to ‘a’ will be 8 more than it. +So the other numbers in the 2 × 2 square can be +represented as shown in the grid. Let us find the +diagonal sums; a + (a + 8), and (a + 1) + (a + 7). +Let us simplify them. +Since the terms can be added in any order, the +brackets can be opened. +a + (a + 8) = a + a + 8 = 2a + 8 +(a + 1) + (a + 7) = a + 1 + a + 7 = a + a + 1 + 7 = 2a + 8 +We see that both diagonal sums are equal to 2a + 8 (8 more than +2 times a). +Verify this expression for diagonal sums by considering any 2 × 2 +square and taking its top left number to be ‘a’. +Thus, we have shown that diagonal sums are equal for any value of +a, i.e., for any 2 × 2 square! + +Consider a set of numbers from the calendar (having endless rows) +forming under the following shape: +Find the sum of all the numbers. Compare it with the number in the +centre: 15. Repeat this for another set of numbers that forms this shape. +What do you observe? +We see that the total sum is always 5 times the number in the centre. +Will this always happen? How do you show this? +[Hint: Consider a general set of numbers that forms this shape. +Take the number at the centre to be ‘a’. Express the other numbers in +terms of ‘a’.] +Find other shapes for which the sum of the numbers within the +figure is always a multiple of one of the numbers. +a +a + 1 +a + 7 +a + 8 +This problem is an example that shows the power of algebraic modelling +in verifying whether a pattern will always hold. +8 +14 +15 +16 +22 +Math +Talk +Math +Talk +99 + +Ganita Prakash | Grade 7 +Matchstick Patterns +Look at the picture below. It is a pattern using matchsticks. Can you +identify what the pattern is? +We can see that Step 1 has 1 triangle, Step 2 has 2 triangles, Step 3 +has 3 triangles, and so on. +Can you tell how many matchsticks there will be in the next step, +Step 5? It is 11. You can also draw this and see. +How many matchsticks will there be in Step 33, Step 84, and Step 108? +Of course, we can draw and count, but is there a quicker way to find +the answers using the pattern present here? +What is the general rule to find the number of matchsticks in the +next step? We can see that at each step 2 matchsticks are placed to get +the next one, i.e., the number of matchsticks increases by 2 every time. +Step Number +1 +2 +3 +4 +5 +6 +No. of +Matchsticks +3 +5 +7 +9 +11 +13 +Think of a way to use this to find out the number of matchsticks in +Step 33 (without continuing to write the numbers). +As each time 2 matchsticks are being added, finding out how many +2s will be added in Step 33 will help. Look at the table below and try to +find out. +Step Number +1 +2 +3 +4 +5 +6 +No. of +Matchsticks +3 +5 +7 +9 +11 +13 +3 + 2 +3 + 2 + 2 +3 + 2 + 2 + 2 +3 + 2 + 2 + 2 + 2 +The number of matchsticks needed to make 33 triangles (Step 33) is +_________. Similarly, find the number of matchsticks needed for Step 84 +and Step 108. +1 +2 +3 +4 +100 + +Expressions Using Letter-Numbers +What could be an expression describing the rule/formula to find out +the number of matchsticks at any step? +The pattern is such that in Step 10, nine 2s and an added 3 (3 + 2 × +9) gives the number of matchsticks; in Step 11, ten 2s and an added +3 (3 + 2 × 10) gives the number of matchsticks. For step y, what is +the expression? +It is: one less than y (i.e. y – 1) 2s and a 3. +Therefore, the expression is +3 + 2 × (y – 1). +This expression gives the number of matchsticks in Step y. Now we +can find the number of matchsticks at any step quickly. +You might have already noticed that there is a 2 in the first step also, +3 = 1 + 2. Using this, the expression we get is +2y + 1. +Does the above expression also give the number of matchsticks at each +step correctly? Are these expressions the same? +We can check by simplifying the expression 3 + 2 × (y – 1). +3 + 2 × (y – 1) = 3 + 2y – 2 += 2y + 1. +Both expressions are the same. +There is a different way to count, or see the pattern. Let us take a +look at the picture again. +Matchsticks are placed in two orientations — (a) horizontal ones at +the top and bottom, and (b) the ones placed diagonally in the middle. +For example, in step 2 there are 2 matchsticks placed horizontally and +3 matchsticks placed diagonally. +What are these numbers in Step 3 and Step 4? +How does the number of matchsticks change in each orientation as the +steps increase? Write an expression for the number of matchsticks at +Step ‘y’ in each orientation. Do the two expressions add up to 2y + 1? +101 + +Ganita Prakash | Grade 7 +Figure it Out +For the problems asking you to find suitable expression(s), first try to +understand the relationship between the different quantities in the +situation described. If required, assume some values for the unknowns +and try to find the relationship. +1. One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. +If x plates of Jowar roti and y plates of pulao were ordered in a day, +which expression(s) describe the total amount in rupees earned +that day? +(a) 30x + 20y +(b) +(30 + 20) × (x + y) +(c) 20x + 30y +(d) +(30 + 20) × x + y +(e) 30x – 20y +2. Pushpita sells two types of flowers on Independence day: champak +and marigold. ‘p’ customers only bought champak, ‘q’ customers +only bought marigold, and ‘r’ customers bought both. On the same +day, she gave away a tiny national flag to every customer. How +many flags did she give away that day? +(a) p + q + r +(b) +p + q + 2r +(c) 2 × (p + q + r) +(d) +p + q + r + 2 +(e) p + q + r + 1 +(f) +2 × (p + q) +3. A snail is trying to climb along the wall of a deep well. During the +day it climbs up ‘u’ cm and during the night it slowly slips down +‘d’ cm. This happens for 10 days and 10 nights. +(a) Write an expression describing how far away the snail is +from its starting position. +(b) What can we say about the snail’s movement if d > u? +4. Radha is preparing for a cycling race and practices daily. The +first week she cycles 5 km every day. Every week she increases +the daily distance cycled by ‘z’ km. How many kilometers would +Radha have cycled after 3 weeks? +5. In the following figure, observe how the expression w + 2 becomes +4w + 20 along one path. Fill in the missing blanks on the remaining +paths. The ovals contain expressions and the boxes contain +operations. +Try +This +w + 2 +×3 +–5 ++3 +×4 +w + 5 +4w + 20 +–4 +–8 ++3 +×4 +3w – 6 +w - 3 +102 + +Expressions Using Letter-Numbers +6. A local train from Yahapur to Vahapur stops at three stations at +equal distances along the way. The time taken in minutes to travel +from one station to the next station is the same and is denoted by t. +The train stops for 2 minutes at each of the three stations. +(a) If t = 4, what is the time taken to travel from Yahapur to +Vahapur? +(b) What is the algebraic expression for the time taken to travel +from Yahapur to Vahapur? [Hint: Draw a rough diagram to +visualise the situation] +7. Simplify the following expressions: +(a) 3a + 9b – 6 + 8a – 4b – 7a + 16 +(b) 3 (3a – 3b) – 8a – 4b – 16 +(c) 2 (2x – 3) + 8x + 12 +(d) 8x – (2x – 3) + 12 +(e) 8h – (5 + 7h) + 9 +(f) 23 + 4(6m – 3n) – 8n – 3m – 18 +8. Add the expressions given below: +(a) 4d – 7c + 9 and 8c – 11 + 9d +(b) – 6f + 19 – 8s and – 23 + 13f + 12s +(c) 8d – 14c + 9 and 16c – (11 + 9d) +(d) 6f – 20 + 8s and 23 – 13f – 12s +(e) 13m – 12n and 12n – 13m +(f) – 26m + 24n and 26m – 24n +9. Subtract the expressions given below: +(a) 9a – 6b + 14 from 6a + 9b – 18 +(b) – 15x + 13 – 9y from 7y – 10 + 3x +(c) 17g + 9 – 7h from 11 – 10g + 3h +(d) 9a – 6b + 14 from 6a – (9b + 18) +(e) 10x + 2 + 10y from –3y +8 – 3x +(f) 8g + 4h – 10 from 7h – 8g + 20 +10. Describe situations corresponding to the following algebraic +expressions: +(a) 8x + 3y +(b) 15x – 2x +11. Imagine a straight rope. If it is cut once as shown in the picture, we +get 2 pieces. If the rope is folded once and then cut as shown, we +103 + +Ganita Prakash | Grade 7 +get 3 pieces. Observe the pattern and find the number of pieces if +the rope is folded 10 times and cut. What is the expression for the +number of pieces when the rope is folded r times and cut? +12. Look at the matchstick pattern below. Observe and identify the +pattern. How many matchsticks are required to make 10 such +squares. How many are required to make w squares? +13. Have you noticed how the colours change in a traffic signal? The +sequence of colour changes is shown below. +Find the colour at positions 90, 190, and 343. Write expressions to +describe the positions for each colour. +14. Observe the pattern below. How many squares will be there in Step +4, Step 10, Step 50? Write a general formula. How would the formula +change if we want to count the number of vertices of all the squares? +1 +2 +3 +4 +5 +................... +104 + +Expressions Using Letter-Numbers +1 +1 +5 +9 +13 +2 +2 +6 +10 +14 +3 +3 +7 +11 +15 +4 +4 +8 +12 +16 +15. Numbers are written in a particular +sequence in this endless 4-column grid. +(a) Give expressions to generate all the +numbers in a given column (1, 2, 3, +4). +(b) In which row and column will the +following numbers appear: +(i) +124 +(ii) +147 +(iii) 201 +(c) What number appears in row r and +column c? +(d) Observe the positions of multiples of 3. +Do you see any pattern in it? List other patterns that you see. +• Algebraic expressions are used in formulas to model patterns +and mathematical relationships between quantities, and to make +predictions. +�� Algebraic expressions use not only numbers but also letter-numbers. +The rules for manipulating arithmetic expressions also apply to +algebraic expressions. These rules can be used to reduce algebraic +expressions to their simplest forms. +• Algebraic expressions can be described in ordinary language, and +vice versa. Patterns or relationships that are easily written using +algebra can often be long and complex in ordinary language. This is +one of the advantages of algebra. +SUMMARY +Math +Talk +105 + +1 + +Page No. 83 + How much should she pay if she buys 8 coconuts and 9 kg jaggery? +Ans: Total cost: ₹ 820 +Page No. 84 + Figure it Out +1. Write formulas for the perimeter of: (a) triangle with all sides equal. +(b) a regular pentagon (as we have learnt last year, we use the word +‘regular’ to say that all sidelengths and angle measures are equal) +(c) a regular hexagon +Ans: +(a) Let length of each side of the triangle = a units +Perimeter of Triangle = 3a units +(b) Let length of each side of pentagon = a units +Perimeter of Regular Pentagon = 5a units +(c) Let length of each side of regular hexagon = a units +Perimeter of Regular Hexagon = 6a units + +2. Munirathna has a 20 m long pipe. However, he wants a longer +watering pipe for his garden. He joins another pipe of some length +to this one. Give the expression for the combined length of the pipe. +Use the letter–number ‘k’ to denote the length in meters of the other +pipe. +Ans: (20 + k) meters + + + + +Chapter – 4 +Expressions Using Letter–Numbers + +2 + +3. What is the total amount Krithika has, if she has the following +numbers of notes of ₹100, ₹20 and ₹5? Complete the following table: +Ans: +No. of ₹100 +Notes +No. of ₹20 +notes +No. of ₹5 +notes +Expression and total amount +(in ₹) +3 +5 +6 +3 × 100 + 5 × 20 + 6 × 5 += 430 +6 +4 +3 +6 × 100 + 4 × 20 + 3 × 5 += 695 +8 +4 +z +8 × 100 + 4 × 20 + z × 5 += 880 + 5z +x +y +z +x × 100 + y × 20 + z × 5 += 100x + 20y + 5z + +4. Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller +mill to start running. Once it is running, each kg of grain takes 8 +seconds to grind into powder. Which of the expressions below +describes the time taken to complete grind ‘y’ kg of grain, assuming +the machine is off initially? +(a) 10 + 8 + y +(b) (10 + 8) × y (c) 10 × 8 × y + +(d) 10 + 8 × y +(e) 10 × y + 8 +Ans: (d) 10 + 8 × y +5. Write algebraic expressions using letters of your choice. (a) 5 more +than a number (b) 4 less than a number (c) 2 less than 13 times a +number (d) 13 less than 2 times a number +Ans: +Let letter ‘d’ represent the number, then +(a) d + 5 +(b) d – 4 +(c) 13 × d – 2 +(d) 2 × d – 13 + +3 + +Page No. 85 +6. Describe situations corresponding to the following algebraic +expressions: +(a) 8 × x + 3 × y +Ans: +A shopkeeper sells a pen for ₹x and a notebook for ₹y. Abha buys 8 pens +and 3 notebooks. What will be the total cost of her purchase? + +(b) 15 × j – 2 × k +Ans: +A factory makes 15 chairs every day. If the factory works for j days, it will +make 15 × j chairs. But 2 chairs break daily for k days. How many good +chairs remain after j days and k days breakages? +Try for more situations +7. In a calendar month, if any 2 × 3 grid full of dates is chosen as shown +in the picture, write expressions for the dates in the blank cells if +the bottom middle cell has date ‘w’. +Ans: + +Page No. 89 + If c = ₹50, find the total amount earned by the sale of pencils. +Ans: 18c = ₹900. + +Page No. 92 + Fill the blanks below by replacing the letter–numbers by numbers; +an example is shown. +Then compare the values that 5u and 5 + u take. + +4 + +Ans: + After filling in the two diagrams, do you think the two expressions +are equal? +Ans: + +The expressions 10y – 3 and 10(y – 3) are not equal. +Page No. 93-94 + Figure it Out + Add the numbers in each picture below. Write their corresponding +expressions and simplify them. Try adding the numbers in each +picture in a couple different ways and see that you get the same thing. + +Ans: +Some of the ways are + +5 + +(i) Method 1: Group by Like Terms +5y + x + x + 5y + (– 6) + 2 += (5y + 5y) + (x + x) + (– 6 + 2) += 10y + 2x – 4 +Method 2: Group by Rows +Top row: 5y + (– 6) + x +Bottom row: x + 2 + 5y +Combine: (5y + 5y) + (x + x) + (– 6 + 2) + =10y + 2x – 4 +Final Number: 10y + 2x – 4 +Try some other ways! +(ii) Method 1: Group by Term Type +2p appears 4 times: 8p +3q appears 4 times: 12q +Final number: 3 + (– 2) + (– 2) + 3 = 2 +Total: 8p + 12q + 2 +Method 2: Group by Columns +Column 1: 2p + 3q +Column 2: 3q + 2p +Column 3: (– 2) + 3 + 2p + 3q = 1 + 2p + 3q +Column 4: 3 + (– 2) + 3q + 2p = 1 + 3q + 2p +Final Number: 8p + 12q + 2 +Try some other ways! +(iii) Method 1: Group by Term Type +–5g appears 4 times: –20g +5k appears 12 times: 60k +Total: –20g + 60k +Method 2: Group by Columns +First column: –5g + 5k + 5k + (–5g) = –10g + 10k +Middle columns 2 and 3 all 5k = 8 × 5k = 40k +Last column: –5g + 5k + 5k + (–5g) = –10g + 10k +Total of all: = –10g + 10k + 40k –10g + 10k += –20g + 60k +Try some other ways! + +6 + + Simplify each of the following expressions: +(a) p + p + p + p, p + p + p + q, p + q + p – q +(b) p − q + p − q, p + q – p + q +(c) p + q − (p + q), p – q – p – q +(d) 2d – d – d – d, 2d – d – d – c +(e) 2d – d − (d − c), 2d − (d − d) – c +(f) 2d – d – c – c +Ans: +(a) 4p, 3p + q, 2p + +(b) 2p – 2q, 2q + +(c) 0, –2q +(d) – d, – c + + +(e) c, 2d – c + +(f) d – 2c +Some simplifications of algebraic expressions are done below. The +expression on the right–hand side should be in its simplest form. + +● Observe each of them and see if there is a mistake. +● If you think there is a mistake, try to explain what might have gone +wrong. +● Then, simplify it correctly. +Ans: +Sr. No. +Expression +Simplest Form +Correct Simplest +form +1 +3a + 2b +5 +3a + 2b +2 +3b – 2b –b +0 +0 +3 +6 (p + 2) +6p + 8 +6p + 12 +4 +(4x + 3y) – (3x + 4y) +x +5 +x – y +5 +5 – (2 – 6z) +3 – 6z +3 + 6z +6 +2 + (x + 3) +2x – 6 +x + 5 +7 +2y + (3y – 6) +–y + 6 +5y – 6 +8 +7p – p + 5q – 2q +7p + 3q +6p + 3q +9 +5 (2w + 3x + 4w) +10w + 15x + 20w +30w + 15x +10 +3j + 6k + 9h + 12 +3 (j + 2k + 3h + 4) +3j + 6k + 9h + 12 +11 +4 (2r + 3s + 5) +– 20 – 8r – 12s +8r + 12s + 20 + +7 + +Page No. 95 + Take a look at all the corrected simplest forms (i.e. brackets are +removed, like terms are added, and terms with only numbers are +also added). Is there any relation between the number of terms and +the number of letter–numbers these expressions have? +Ans: +Sr. +No. +(A) +Expression +(B) +Correct Simplest +form +(C) +Number of +terms in (B) +(D) +Number of +letters in (B) +1 +3a + 2b +3a + 2b +2 +2 +2 +3b – 2b –b +0 +1 +0 +3 +6 (p + 2) +6p +12 +2 +1 +4 +(4x + 3y) – (3x + 4y) +x – y +2 +2 +5 +5 – (2 – 6z) +3 + 6z +2 +1 +6 +2 + (x + 3) +x + 5 +2 +1 +7 +2y + (3y – 6) +5y – 6 +2 +1 +8 +7p – p + 5q – 2q +6p + 3q +2 +2 +9 +5 (2w + 3x + 4w) +30w + 15x +2 +2 +10 +3j + 6k + 9h + 12 +3j + 6k + 9h + 12 +4 +3 +11 +4 (2r + 3s + 5) +8r + 12s + 20 +3 +2 +No. of letters ≤ no of terms + +Page No. 96 + Find the formulas of the number machines below and write the +expression for each set of inputs. +Ans: (i) Formula: Two is subtracted from the sum of the two numbers. +Algebraic expression = a + b – 2 + +8 + + +(ii) +Formula: One is added to the multiplication of two numbers. +Algebraic expression= (a × b) + 1 = ab + 1 +4th expression: 10 × 3 + 1 = 30 + 1 = 31 +5th expression: a × b + 1 + +Page No. 97 + Use this to find what design appears at positions 99, 122, and 148. +Ans: +1. Design C appears at position 99. +2. Design B appears at position 122. +3. Design A appears at position 148. + +Page No. 99 + Will this always happen? How do you show this? +Ans: Let’s build a similar 3×3 grid with a Centre number a. + +a – 7 + +a – 1 +a +a + 1 + +a + 7 + +Here, the sum is +a – 7 + a – 1 + a + a +1 + a + 7 = 5a (5times the number in the Centre.) + +9 + +Page No. 101. + What are these numbers in Step 3 and Step 4? +Ans: +Step 3: No. of matchsticks placed horizontally-3 +No. of matchsticks placed diagonally-4 +Step 4: No. of matchsticks placed horizontally-4 +No. of matchsticks placed diagonally-5 + How does the number of matchsticks change in each orientation as +the steps increase? Write an expression for the number of +matchsticks at Step ‘y’ in each orientation. Do the two expressions +add up to 2y + 1? +Ans: +For horizontal placement: 1, 2, 3, 4, … +At the yth step, the number of horizontal matchsticks = y. +For diagonal placement: 2, 3, 4, 5, … +At the yth step, the number of diagonal matchsticks = y + 1. +Yes. +Page No. 102. + Figure it Out +1. One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. If x +plates of Jowar roti and y plates of pulao were ordered in a day, which +expression(s) describe the total amount in rupees earned that day? +(a) 30x + 20y + +(b) (30 + 20) × (x + y) + +(c) 20x + 30y +(d) (30 + 20) × x + y +(e) 30x – 20y +Ans: (a) 30x + 20y. +2. Pushpita sells two types of flowers on Independence Day: champak and +marigold. ‘p’ customers only bought champak, ‘q’ customers only +bought marigold, and ‘r’ customers bought both. On the same day, she +gave away a tiny national flag to every customer. How many flags did +she give away that day? +(a) p + q + r + +(b) p + q + 2r + + +(c) 2 × (p + q + r) +(d) p + q + r + 2 +(e) p + q + r + 1 + + +(f) 2 × (p + q) +Ans: (a) p + q + r + +10 + +3. A snail is trying to climb along the wall of a deep well. During the day it +climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This +happens for 10 days and 10 nights. +(a) Write an expression describing how far away the snail is from its +starting position. +(b) What can we say about the snail’s movement if d > u? +Ans: +(a) 10 (u − d) cm +(b) If d > u, the snail slips down more than it climbs. + This means the snail will never reach the top. +4. Radha is preparing for a cycling race and practices daily. The first week +she cycles 5 km every day. Every week she increases the daily distance +cycled by ‘z’ km. How many kilometers would Radha have cycled after +3 weeks? +Ans: 105 + 21z km +5. In the following figure, observe how the expression w + 2 becomes 4w + +20 along one path. Fill in the missing blanks on the remaining paths. +The ovals contain expressions and the boxes contain operations. +Ans: + +Page No. 103 +6. A local train from Yahapur to Vahapur stops at three stations at equal +distances along the way. The time taken in minutes to travel from one +station to the next station is the same and is denoted by t. The train +stops for 2 minutes at each of the three stations. (a) If t = 4, what is the + +11 + +time taken to travel from Yahapur to Vahapur? (b) What is the +algebraic expression for the time taken to travel from Yahapur to +Vahapur? [Hint: Draw a rough diagram to visualise the situation] +Ans: +(a) 22 minutes +(b) The algebraic expression for the time taken to travel from Yahapur to +Vahapur = 4t + 6 minutes. +7. Simplify the following expressions: +(a) 3a + 9b – 6 + 8a – 4b – 7a + 16 +(b) 3 (3a – 3b) – 8a – 4b –16 +(c) 2 (2x – 3) + 8x + 12 +(d) 8x – (2x – 3) + 12 +(e) 8h – (5 + 7h) + 9 +(f) 23 + 4 (6m – 3n) – 8n – 3m –18 +Ans: +(a) 4a + 5b + 10 +(b) a – 13b – 16 +(c) 12x + 6 +(d) 6x + 15 +(e) h + 4 +(f) 5 + 21m – 20n +8. Add the expressions given below: +(a) 4d – 7c + 9 and 8c – 11 + 9d +(b) – 6f + 19 – 8s and – 23 + 13f + 12s +(c) 8d – 14c + 9 and 16c – (11 + 9d) +(d) 6f – 20 + 8s and 23 – 13f – 12s +(e) 13m – 12n and 12n –13m +(f) – 26m + 24n and 26m –24n + Ans: +(a) 13d + c – 2 +(b) 7f + 4s – 4 +(c) 2c – d – 2 +(d) –7f – 4s + 3 +(e) 0 +(f) 0 + +12 + +9. Subtract the expressions given below: +(a) 9a – 6b + 14 from 6a + 9b – 18 +(b) – 15x + 13 – 9y from 7y – 10 + 3x +(c) 17g + 9 – 7h from 11 – 10g + 3h +(d) 9a – 6b + 14 from 6a – (9b + 18) +(e) 10x + 2 + 10y from –3y +8 – 3x +(f) 8g + 4h – 10 from 7h – 8g + 20 +Ans: +(a) – 3a + 15b – 32 +(b) 16y + 18x – 23 +(c) 10h – 27g + 2 +(d) – (3a + 3b + 32) +(e) – 13y – 13x + 6 +(f) 3h – 16g + 30 +10. Describe +situations +corresponding +to +the +following +algebraic +expressions: +Ans: +(a) 8x + 3y +One of the examples is–Situation: A fruit seller sells mangoes at ₹8 each +and bananas at ₹3 each. If a customer buys x mangoes and y bananas. +The total cost would be 8x + 3y. +(b) 15x – 2x +Situation: A shopkeeper has 15 pencils in a packet. The cost of one pencil +is ₹x. Two pencils in the packet are not sold. Find the amount received for +this packet. +11. Imagine a straight rope. If it is cut once as shown in the picture, we get +2 pieces. If the rope is folded once and then cut as shown, we get 3 +pieces. Observe the pattern and find the number of pieces if the rope is +folded 10 times and cut. What is the expression for the number of pieces +when the rope is folded r times and cut? + + +13 + +Ans: If the rope is folded 10 times and cut, we get 10 + 2 = 12 pieces. +In the same way, when the rope is folded r times and cut, we get r + 2 pieces. +Page No. 104 +12. Look at the matchstick pattern below. Observe and identify the pattern. +How many matchsticks are required to make 10 such squares? How +many are required to make w squares? + +Ans: To make 10 squares, we need +4 + 3 × 9 = 31 matchsticks +To make w squares the number of matchsticks required is = 4 + 3 (w - 1) +13. Have you noticed how the colours change in a traffic signal? The +sequence of colour changes is shown below. Find the colour at positions +90, 190, and 343. Write expressions to describe the positions for each +colour. + +Ans: For Position 90: +𝟗𝟎 +𝟒 +The colour is Yellow. +For Position 190: +𝟏𝟗𝟎 +𝟒 +The colour is Yellow +For Position 343: +𝟑𝟒𝟑 +𝟒 +The colour is Green +Expressions: +In general, 4n – 3 positions for red light, 4n – 1 position for green light +and 2n position for yellow light. + +14 + +14. Observe the pattern below. How many squares will be there in Step 4, +Step 10, Step 50? Write a general formula. How would the formula +change if we want to count the number of vertices of all the squares? + +Ans: No. of Squares in Step 4 = 17 +No. of Squares in Step 10 = 41 +No. of Squares in Step 50 = 201 +So, general formula = 5 + (n – 1) × 4 = 4n + 1 +No. of vertices = 16n + 4. +Page No. 105 +15. Numbers are written in a particular sequence in this endless 4 – column +grid. + +Ans: +(a) Give expressions to generate all the numbers in a given column (1, 2, 3, 4). +Let r be the row number. +Column 1: +So, number in the rth row of column 1 = 4 × (r – 1) + 1 + +15 + +Column 2: 4 × (r – 1) + 2 +Column 3: 4 × (r – 1) + 3 +Column 4: 4 × (r – 1) + 4 +If c is the column number, +then the general formula to generate all numbers is= 4 × (r – 1) + c +(b) In which row and column will the following numbers appear: +(i) 124 + +(ii) 147 (iii) 201 +(i) We divide each number by 4 to find its row and column +124 ÷ 4 So, Quotient = 31 and remainder is 0 +∴ 124 = 4 × 31 + 0 or 4 × 30 + 4 +Comparing it with 4 × (r – 1) + c, we get +r – 1 = 30, c = 4 +So, r = 31 and c = 4 +(ii) 147 will appear at row 37 and column 3 +(iii) 201 will appear at row 51 and column 1. +(c) What number appears in row r and column c? +The number that appears in row r and column c is 4(r – 1) + c. +(d) Observe the positions of multiples of 3. +The pattern in the columns in which multiples of 3 appear, follow a +repeating sequence of column number: 3, 2, 1, 4, 3, 2, 1, ... as shown. + + +16 + +Do you see any pattern in it? List other patterns that you see. +1. All numbers in Column 4 are multiples of 4. +2. Even numbers always appear in column 2 and column 4. +3. Odd numbers always appear in column 1 and column 3. +Find some more!" +class_7,5,Parallel and Intersecting Lines,ncert_books/class_7/gegp1dd/gegp105.pdf,"PARALLEL AND +INTERSECTING +LINES +5 +5.1 Across the Line +Take a piece of square paper and fold it in different ways. Now, on +the creases formed by the folds, draw lines using a pencil and a scale. +You will notice different lines on the paper. Take any pair of lines and +observe their relationship with each other. Do they meet? If they do +not meet within the paper, do you think they would meet if they were +extended beyond the paper? +Fig. 5.1 +In this chapter, we will explore the relationship between lines on a +plane surface. The table top, your piece of paper, the blackboard, and +the bulletin board are all examples of plane surfaces. +Let us observe a pair of lines that meet each other. You will notice +that they meet at a point. When a pair of lines meet each other at a +point on a plane surface, we say that the lines intersect each other. +Let us observe what happens when two lines intersect. +How many angles do they form? +In Fig. 5.2, where line l intersects line m, we can see that four angles +are formed. + +Parallel and Intersecting Lines +m +a +b +l +c +d +Fig. 5.2 +Can two straight lines intersect at more than one point? +Activity 1 +Draw two lines on a plain sheet of paper so that they intersect. Measure +the four angles formed with a protractor. Draw four such pairs of +intersecting lines and measure the angles formed at the points of +intersection. +What patterns do you observe among these angles? +In Fig. 5.2, if ∠a is 120°, can you figure out the measurements of ∠b, ∠c +and ∠d, without drawing and measuring them? +We know that ∠a and ∠b together measure 180°, because when they +are combined, they form a straight angle which measures 180°. So, if +∠a is 120°, then ∠b must be 60°. +Similarly, ∠b and ∠c together measure 180°. So, if ∠b is 60°, then ∠c +must be 120°. And ∠c and ∠d together measure 180°. So, if ∠c is 120°, +then ∠d must be 60°. +Therefore, in Fig. 5.2, ∠a and ∠c measure 120°, and ∠b and ∠d +measure 60°. +When two lines intersect each other and form four angles, labelled +a, b, c and d, as in Fig. 5.2, then ∠a and ∠c are equal, and ∠b and ∠d +are equal! +Is this always true for any pair of intersecting lines? +Check this for different measures of ∠a. Using these measurements, +can you reason whether this property holds true for any measure of ∠a? +We can generalise our reasoning for Fig. 5.2, without assuming the +values of ∠a. +Since straight angles measure 180°, we must have ∠a + ∠b = ∠a + ∠d += 180°. Hence, ∠b and ∠d are always equal. Similarly, ∠b + ∠a = ∠b + +∠c = 180°, so ∠a and ∠c must be equal. +Adjacent angles, like ∠a and ∠b, formed by two lines intersecting +each other, are called linear pairs. Linear pairs always add up to 180°. +107 + +Ganita Prakash | Grade 7 +Opposite angles, like ∠b and ∠d, formed by two lines intersecting +each other, are called vertically opposite angles. Vertically opposite +angles are always equal to each other. +From the above reasoning, we conclude that whenever two lines +intersect, vertically opposite angles are equal. Such a justification is +called a proof in mathematics. +Figure it Out +List all the linear pairs and vertically opposite angles you +observe in Fig. 5.3: +Linear Pairs +∠a and ∠b, … +Pairs of Vertically +Opposite Angles +∠b and ∠d, … +Measurements and Geometry +You might have noticed that when you measure linear pairs, sometimes +they may not add up to 180°. Or, when you measure vertically opposite +angles they may be unequal sometimes. What are the reasons for this? +There could be different reasons: +• Measurement errors because of improper use of measuring +instruments — in this case, a protractor +• Variation in the thickness of the lines drawn. The “ideal” line in +geometry does not have any thickness! But it is not possible for us +to draw lines without any thickness +In geometry, we create ideal versions of “lines” and other shapes +we see around us, and analyse the relationships between them. For +example, we know that the angle formed by a straight line is 180°. +So, if another line divides this angle into two parts, both parts should +add up to 180°. We arrive at this simply through reasoning and not +by measurement. When we measure, it might not be exactly so, for +the reasons mentioned above. Still the measurements come out very +close to what we predict, because of which geometry finds widespread +application in different disciplines such as physics, art, engineering +and architecture. +5.2 Perpendicular Lines +Can you draw a pair of intersecting lines such that all four angles are +equal? Can you figure out what will be the measure of each angle? +c +b +a +d +l +m +Fig. 5.3 +108 + +Parallel and Intersecting Lines +m +l +Fig. 5.4 +If two lines intersect and all four angles are equal, then each angle +must be a right angle (90°). +Perpendicular lines are a pair of lines which intersect each other +at right angles (90°). In Fig. 5.4, we can say that lines l and m are +perpendicular to each other. +5.3 Between Lines +Observe Fig. 5.5 and describe the way the line segments meet or cross +each other in each case, with appropriate mathematical words (a point, +an endpoint, the midpoint, meet, intersect) and the degree measure of +each angle. +For example, line segments FG and FH meet at the endpoint F at an +angle 115.3°. +S +T +V +U +C +A +B +D +X +H +M +J +Y +I +L +R +P +O +Q +F +G +115.3° +Fig. 5.5 +Are line segments ST and UV likely to meet if they are extended? +Are line segments OP and QR likely to meet if they are extended? +Here are some examples of lines we notice around us. +109 + +Ganita Prakash | Grade 7 +What is common to the lines in the pictures above? They do not seem +likely to intersect each other. Such lines are called parallel lines. +Parallel lines are a pair of lines that lie on the same plane, and do +not meet however far we extend them at both ends. +Name some parallel lines you can spot in your classroom. +Parallel lines are often used in artwork and shading. +Which pairs of lines appear to be parallel in Fig. 5.6 below? +a +c +d +e +g +f +h +i +b +Fig. 5.6 +Note to the Teacher: It is important that the lines lie on the same plane. A line +drawn on a table and a line drawn on the board may never meet but that does +not make them parallel. + +110 + +Parallel and Intersecting Lines +5.4 Parallel and Perpendicular Lines in Paper +Folding +Activity 2 +Take a plain square sheet of paper (use a newspaper for this activity). +• How would you describe the opposite edges of the sheet? They +are _________________________ to each other. +• How would you describe the adjacent edges of the sheet? The +adjacent edges are _________________________ to each other. They +meet at a point. They form right angles. +• Fold the sheet horizontally in half. A new line is formed +(see Fig. 5.7). +• How many parallel lines do you see now? How does the new line +segment relate to the vertical sides? +Fig. 5.7 +• Make one more horizontal fold in the folded sheet. How many +parallel lines do you see now? +• What will happen if you do it once more? How many parallel +lines will you get? Is there a pattern? Check if the pattern extends +further, if you make another horizontal fold. +• Make a vertical fold in the square sheet. This new vertical line is +___________ to the previous horizontal lines. +• Fold the sheet along a diagonal. Can you find a fold that creates a +line parallel to the diagonal line? +111 + +Ganita Prakash | Grade 7 +Here is another activity for you to try. +• Take a square sheet of paper, fold it in the middle and unfold it. +• Fold the edges towards the centre line and unfold them. +• Fold the top right and bottom left corners onto the creased line to +create triangles. Refer to Fig. 5.8. +• The triangles should not cross the crease lines. +• Are a, b and c parallel to p, q and r respectively? Why or why not? +Fig. 5.8 +Notations +In mathematics, we use an arrow mark (>) to show that a set of lines is +parallel. If there is more than one set of parallel lines (as in Fig. 5. 9), the +second set is shown with two arrow marks and so on. Perpendicular +lines are marked with a square angle between them. + +90° +Fig. 5.9 +112 + +Parallel and Intersecting Lines +Figure it Out +1. Draw some lines perpendicular to the lines given on the dot paper +in Fig. 5.10. +Fig. 5.10 +2. In Fig. 5.11, mark the parallel lines using the notation given +above (single arrow, double arrow etc.). Mark the angle between +perpendicular lines with a square symbol. +(a) How did you spot the perpendicular lines? +(b) How did you spot the parallel lines? +Fig. 5.11 +3. In the dot paper following, draw different sets of parallel lines. +The line segments can be of different lengths but should have dots +as endpoints. +113 + +Ganita Prakash | Grade 7 +4. Using your sense of how parallel lines look, try to draw lines parallel +to the line segments on this dot paper. +a +b +c +e +f +g +h +d +Fig. 5.12 +(a) Did you find it challenging to draw some of them? +(b) Which ones? +(c) How did you do it? +5. In Fig. 5.13, which line is parallel to line a —– line b or line c? How +do you decide this? +Note to the Teacher: It is easier to draw vertical and horizontal lines and the +ones inclined at 45° (on rectangular dot sheets), but drawing a line parallel to +one which has a different orientation is slightly harder. Let students use their +intuition for this. +From previous exercises we observed that sometimes it is difficult +to be sure whether two lines are parallel. To determine this we use the +idea of transversals. +a +c +b +Fig. 5.13 +114 + +Parallel and Intersecting Lines +5.5 Transversals +We saw what happens when two lines intersect in different ways. Let +us explore what happens when one line intersects two different lines. +l +1 +4 +5 +8 +2 +3 +6 +7 +t +m +Fig. 5.14 +In Fig. 5.14, line t intersects lines l and m. t is called a transversal. +Notice that 8 angles are formed when a line crosses a pair of lines. +Is it possible for all the eight angles to have different measurements? +Why, why not? +What about five different angles — 6, 5, 4, 3 and 2? +In Fig. 5.14, since ∠1 and ∠3 are vertically opposite angles, they are +equal. Are there other pairs of vertically opposite angles? We can see +that there are a total of four pairs of vertically opposite angles and in +each pair, the angles are equal to each other. +Thus, when a transversal intersects two lines, it forms eight angles +with a maximum of four distinct angle measures. +5.6 Corresponding Angles +In Fig. 5.14, we notice that the transversal t forms two sets of angles — +one with line l and another with line m. There are angles in the first set +that correspond to angles in the second set based on their position. ∠1 +and ∠5 are called corresponding angles. Similarly, ∠2 and ∠6, ∠3 and +∠7, ∠4 and ∠8 are the corresponding angles formed when transversal +t intersects lines l and m. +Activity 3 +Draw a pair of lines and a transversal such that they form two +distinct angles. +115 + +Ganita Prakash | Grade 7 +Step 1: Draw a line l and a transversal t intersecting it at point X. +Fig. 5.15 +X +l +t + +Step 2: Measure ∠a formed by lines l and t (let us say it is 60°). + +a = 60° +l +X +t +Fig. 5.16 +How many distinct angles have formed now? +If one angle is 60°, the other angle of the linear pair should be 120°. +So, we already have two distinct angles. +So, when we draw another line intersecting the transversal t we +wish to form only two angles, 60° and 120°. +Step 3: Mark a point Y on line t. +a = 60° +l +X +t +Y +Fig. 5.17 +Step 4: Draw a line m through point Y that forms a 60° angle to line t. +This can be done either by copying ∠a with a tracing paper or you can +use a protractor to measure the angles. + +a = 60° +b = 60° +l +m +X +t +Y +Fig. 5.18 +What do you observe about lines l and m? Do they appear to be +parallel to each other? +Yes, they do appear to be parallel to each other. +Angles, ∠a and ∠b are corresponding angles formed by the transversal +t on lines l and m. These corresponding angles are equal to each other. +116 + +Parallel and Intersecting Lines +From this we can observe: +When the corresponding angles formed by a transversal on a pair +of lines are equal to each other, then the pair of lines are parallel +to each other. +Suppose, we have a transversal intersecting two parallel lines. What +can be said about the corresponding angles? +Activity 4 +Fig. 5.19 has a pair of parallel lines l and m (what is the notation used +in the figure to indicate they are parallel?) . Line t is the transversal +across these two lines. ∠a and ∠b are corresponding angles. Take a +tracing paper and trace ∠a on it. Now place this tracing paper over +∠b and see if the angles align exactly. You will observe that the angles +match. Check the other corresponding angles in the figure using a +protractor. Are all the corresponding angles equal to each other? +l +a +b +m +t +Fig. 5.19 +Corresponding angles formed by a transversal intersecting a pair +of parallel lines are always equal to each other. +Activity 5 +In Fig. 5.20, draw a transversal t to the lines l and m such that one pair +of corresponding angles is equal. You can measure the angles with a +protractor. +l +m +Fig. 5.20 +Are you finding it hard to draw a transversal such that the +corresponding angles are equal? +117 + +Ganita Prakash | Grade 7 +When a pair of lines are not parallel to each other, the corresponding +angles formed by a transversal can never be equal to each other. +5.7 Drawing Parallel Lines +Can you draw a pair of parallel lines using a ruler and a set square? +Fig. 5.21 shows how you can do it. +Draw a line l with a scale. By sliding your set square you can make +two lines perpendicular to line l. +Are these two lines parallel to each other? How are we sure that they +are parallel to each other? What angles are formed between these lines +and line l? +l +Fig. 5.21 +Since we used a set square, the angles measure 90°. The position +of the lines is different, but they make the same angle with l. If line l +is seen as a transversal to the two new lines, then the corresponding +angles measure 90°. +Fig. 5.22 +l +As we know these are corresponding angles and they are equal, we +can be sure that the lines are parallel. +Draw two more parallel lines using the long side of the set square as +shown in Fig. 5.22. +118 + +Parallel and Intersecting Lines +How do you know these two lines are parallel? Can you check if the +corresponding angles are equal? +Note to the Teacher: Students should be encouraged to check the equality of +corresponding angles both by using the tracing method and using protractors to +measure the angles. Pay attention to the language used to make the relationship +between corresponding angles and parallel lines. Equality of corresponding +angles is both necessary and sufficient for the pair of lines to be parallel to +each other. +Figure it Out +Can you draw a line parallel to l, that goes through point A? How will +you do it with the tools from your geometry box? Describe your method. +Fig. 5.23 +A +l +Making Parallel Lines through Paper Folding +Let us try to do the same with paper folding. For a line l (given as a +crease), how do we make a line parallel to l such that it passes through +point A? +We know how to fold a piece of paper to get a line perpendicular to l. +Now, try to fold a perpendicular to l such that it passes through point +A. Let us call this new crease t. +Now, fold a line perpendicular to t passing through A again. Let us +call this line m. The lines l and m are parallel to each other. +Fig. 5.24 +l +l +A +t +l +A +l +m +A +t +119 + +Ganita Prakash | Grade 7 +Why are lines l and m parallel to each other? +5.8 Alternate Angles +In Fig. 5.25, ∠d is called the alternate angle of ∠f, and ∠c is the alternate +angle of ∠e. +a +b +d +c +e +h +g +m +l +t +f +Fig. 5.25 +You can find the alternate angle of a given angle, say ∠f, by first +finding the corresponding angle of ∠f, which is ∠b and then finding the +vertically opposite angle of ∠b, which is ∠d. +Activity 6 +In Fig. 5.25, if ∠f is 120° what is the measure of its alternate angle ∠d? +We can find the measure of ∠d if we know ∠b because they are +vertically opposite angles. Remember, vertically opposite angles are +equal. +What is the measure of ∠b? It is 120° because it is the corresponding +angle of ∠f. +So, ∠d also measures 120°. +In fact, ∠f = ∠b irrespective of the measure of ∠f. Why? Because ∠b +is the corresponding angle of ∠f. +Similarly, ∠b = ∠d irrespective of the measure of ∠b. Why? Because +∠d is the vertically opposite angle of ∠b. So, it must always be the +case that +∠f = ∠d +Using our understanding of corresponding angles without any +measurements, we have justified that alternate angles are always equal. +Alternate angles formed by a transversal intersecting a pair of +parallel lines are always equal to each other. +Example 1: In Fig. 5.26, parallel lines l and m are intersected by the +transversal t. If ∠6 is 135°, what are the measures of the other angles? +120 + +Parallel and Intersecting Lines +l +t +m +2 +3 +4 +5 +6 +135° +7 +8 +1 +Fig. 5.26 +Solution: ∠6 is 135°, so ∠2 is also 135°, because it is the corresponding +angle of ∠6 and the lines l and m are parallel. +∠8 is 135°, because it is the vertically opposite angle of ∠6. ∠4 is 135° +because it is the corresponding angle of ∠8. +∠2 is 135° because is the vertically opposite angle of ∠4. So, ∠2, ∠4, +∠6, and ∠8 are all 135°. +∠5 and ∠6 are a linear pair, together they measure 180°. If ∠6 is +135°, then +∠5 = 180 – 135 = 45° +We can similarly find out that ∠1, ∠3, and ∠7 measure 45°. +Example 2: In Fig. 5.27, lines l and m are intersected by the transversal +t. If ∠a is 120° and ∠f is 70°, are lines l and m parallel to each other? +g +l +m +t +b +a +d +c +f +h +e +120° +70° +Fig. 5.27 +Solution: ∠a is 120°, so ∠b is 60° because ∠a and ∠b form a linear pair. +∠b is a corresponding angle of ∠f. If l and m are parallel, ∠b should be +equal to ∠f, however, they are not equal. +Therefore, lines l and m are not parallel to each other as the corresponding +angles formed by the transversal t are not equal to each other. +Example 3: In Fig. 5.28, parallel lines l and m are intersected by the +transversal t. If ∠3 is 50°, what is the measure of ∠6? +l +m +t +50° +2 +1 +4 +3 +5 +6 +7 +8 +Fig. 5.28 +121 + +Ganita Prakash | Grade 7 +Solution: ∠3 is 50°; therefore, ∠2 is 130°, because ∠2 and ∠3 form a +linear pair, and linear pairs always add up to 180°. +∠2 and ∠6 are corresponding angles, and they need to be equal since +lines l and m are parallel. +So, ∠6 is 130°. +Angles ∠3 and ∠6 are called interior angles. + +Is there a relation between ∠3 and ∠6? You could try to find the +relationship by taking different values for ∠3 and see what ∠6 is. +Once you find a relation, try to justify it or prove that this relation +holds always. You will find that the sum of the interior angles on +the same side of the transversal always add up to 180°. +Example 4: In Fig. 5.29, line segment AB is parallel to CD and AD is +parallel to BC. ∠DAC is 65° and ∠ADC is 60°. What are the measures of +angles ∠CAB, ∠ABC, and ∠BCD? +Solution: Let us observe the parallel lines AB and CD. AD is a +transversal of these two lines. +65° +60° +A +B +C +D +Fig. 5.29 +We know that the sum of the interior angles formed by a transversal +on a pair of parallel lines adds up to 180°. So +∠ADC + ∠DAB = 180° +60° + ∠DAB = 180°. +So ∠DAB = 120°. +Can we find ∠CAB from this? +∠DAB = ∠DAC + ∠CAB. +So 120° = 65° + ∠CAB. +So ∠CAB = 55°. +Let us observe the parallel line segments AD and BC. They are +intersected by a transversal CD. So, ∠ADC + ∠BCD = 180°, because they +are interior angles on the same side of the transversal. Since ∠ADC is +given as 60°, ∠BCD = 120° +Similarly, we find ∠ABC = 60°. +Therefore, in Fig. 5.29, ∠CAB = 55°, ∠ABC = 60°, and ∠BCD = 120°. +122 + +Parallel and Intersecting Lines +Figure it Out +1. Find the angles marked below. + +48° +a° +a= +81° +99° +d° +d= +58° +122° +58° +122° +g° +g= +52° +b° +b= +97° 83° +69° +e° +e= +75° +120° +h° +h= +j° +27° +97° +124° +j= +81° +99° +c° +c= +132° +f° +f= +54° +56° +i° +70° +i= +Fig. 5.30 +123 + +Ganita Prakash | Grade 7 +2. Find the angle represented by a. +a° +42° +100° +62° +a° +110° +35° +a° +67° +a° +Fig. 5.31 +3. In the figures below, what angles do x and y stand for? +65° +x° +y° +78° +53° +x° +Fig. 5.32 +4. In Fig. 5.33, ∠ABC = 45° and ∠IKJ = 78°. Find angles ∠GEH, ∠HEF, +∠FED +78° +45° +C +B +J +I +K +A +D +E +G +H +F +Fig. 5.33 +124 + +Parallel and Intersecting Lines +5. In Fig. 5.34, AB is parallel to CD and CD is parallel to EF. Also, EA is +perpendicular to AB. If ∠BEF = 55°, find the values of x and y. +55° +A +C +E +D +B +F +y° +x° +Fig. 5.34 +6. What is the measure of angle ∠NOP in Fig. 5.35? +40° +96° +L +N +M +O +Q +P +52° +a° +Fig. 5.35 +[Hint: Draw lines parallel to LM and PQ through points N and O.] +5.9 Parallel Illusions +There do not seem to be any parallel lines here. Or, are there? + + What causes these illusions? +125 + +Ganita Prakash | Grade 7 +• When two lines intersect, they form four angles. The vertically +opposite angles are equal and the linear pairs add up to 180°. +• When two lines intersect and the angles formed are 90° (i.e., all four +angles are equal), the lines are said to be perpendicular to each other. +• When two lines never intersect on a plane, they are called parallel +lines. +• When a line t intersects another pair of lines, it is called a transversal +and it forms 2 sets of 4 angles. Each of the 4 angles in the first set has +a corresponding angle in the second set. +• When a transversal intersects a pair of parallel lines, the +corresponding angles are equal. When a transversal intersects a pair +of lines and the corresponding angles are equal, then the pair of lines +is parallel. +• When a transversal intersects a pair of parallel lines, the alternate +angles are equal. +• The interior angles on the same side formed by a transversal +intersecting a pair of parallel lines always add up to 180°. +SUMMARY +126 + + +Page No. 107 + Can two straight lines intersect at more than one point? +Ans: No + What patterns do you observe among these angles? +Ans: +∠a = ∠c and ∠b = ∠d. +∠a + ∠b = 180° and ∠b + ∠c = 180°. +Find more such relations. + Is this always true for any pair of intersecting lines? +Ans: Yes +Page No. 108. + Figure it Out +List all the linear pairs and vertically opposite angles you observe in +Fig. 5.3: + + +Ans: +Linear Pairs +∠a and ∠b, ∠b and ∠c, ∠c and ∠d, ∠d +and ∠a. +Pairs of Vertically Opposite Angles +∠b and ∠d, ∠a and ∠c + + +Fig. 5.3 +Chapter – 5 +Parallel and Intersecting Lines + +Page No. 110 + Which pairs of lines appear to be parallel in Fig. 5.6 below? + + +Ans: a, i and h; c and g; d and f; e and b +Page No. 111 +Activity 2 +Take a plain square sheet of paper (use a newspaper for this +activity). +• +How would you describe the opposite edges of the sheet? They +are _______ to each other. +• +How would you describe the adjacent edges of the sheet? The +adjacent edges are ______ to each other. They meet at a point. +They form right angles. +• +Fold the sheet horizontally in half. A new line is formed (see +Fig. 5.7). +• +How many parallel lines do you see now? How does the new +line segment relate to the vertical sides? +Fig. 5.6 + + + +• +Make one more horizontal fold in the folded sheet. How many +parallel lines do you see now? +• +What will happen if you do it once more? How many parallel +lines will you get? Is there a pattern? Check if the pattern +extends further, if you make another horizontal fold. +• +Make a vertical fold in the square sheet. This new vertical line +is ___________ to the previous horizontal lines. +• +Fold the sheet along a diagonal. Can you find a fold that creates +a line parallel to the diagonal line? + +Ans: +Fill in the Blanks +● The opposite edges of the sheet are parallel to each other. +● The adjacent edges of the sheet are perpendicular to each other. +First Horizontal Fold +● We will see three parallel lines. The new line segment is perpendicular +to the vertical sides. +Subsequent Horizontal Folds +● After one more horizontal fold, we will see a total of five parallel lines. +The two original horizontal edges, the first fold line, and the second fold +line are all parallel to each other. +● If we fold it horizontally one more time, we will get nine parallel lines. +Yes, the pattern is 2n + 1, where n is the number of folds. + +Fig. 5.7 + +Vertical and Diagonal Folds +• This new vertical line is perpendicular to the previous horizontal lines. +• Yes, we can find a fold that creates a line parallel to the diagonal line. +Page No. 112 +Here is another activity for you to try. +• Take a square sheet of paper, fold it in the middle and unfold it. +• Fold the edges towards the centre line and unfold them. +• Fold the top right and bottom left corners onto the creased line to +create triangles. Refer to Fig. 5.8. +• The triangles should not cross the crease lines. +• Are a, b and c parallel to p, q and r respectively? Why or why not? + +Fig. 5.8 + +Ans: +• +Line segments a, b and c are parallel to Line segments p, q and r +respectively. +• +a and p lie on parallel lines. b, q are both respectively perpendicular +to these, so are parallel to each other. +• +c, r are both parallel because the triangular folds create lines +parallel to the same diagonal. + + + +Page No. 113-114 + Figure it Out +1. Draw some lines perpendicular to the lines given on the dot paper +in the Figure. + + +Fig. 5.10 + + +Ans: + + + + +2. In the given Fig.5.11, mark the parallel lines using the notation given +above (single arrow, double arrow, etc). Mark the angle between +perpendicular lines with a square symbol. +(a) How did you spot the perpendicular lines? +(b) How did you spot the parallel lines? + +Fig. 5.11 + + + + +Ans: +(a) The vertical and horizontal lines on the grid paper meet at a 90- +degree angle (a right angle) +(b) By identifying lines that always remain the same distance apart. +3. In the dot paper following, draw different sets of parallel lines. The +line segments can be of different lengths but should have dots as +endpoints. +Ans: Some of the lines are: + +4. Using your sense of how parallel lines look, try to draw lines parallel +to the line segments on this dot paper. + + + + +(a) Did you find it challenging to draw some of them? +(b) Which ones? +(c) How did you do it? +Ans: Some of the lines are +(a) Yes +(b) Line segments e, f, h, and g. +(c) Parallel lines are drawn by keeping them equidistant from the given +line. +5. In Fig 5.13, which line is parallel to line a — line b or line c? How do +you decide this? + + +Ans: In the given figure, line c is parallel to line a because these two lines are +always at the same distance apart. + + + +Fig. 5.13 + +Page No. 115 + + Is it possible for all the eight angles to have different measurements? +Why, why not? +Ans: No, because  1 =  3,  2 =  4,  5 =  7, and  6 =  8 (vertically opposite +angles are equal). +What about five different angles — 6, 5, 4, 3 and 2? +Ans: No, because  2 =  4. +Page No. 119 + Figure it Out +Can you draw a line parallel to l, that goes through point A? How will +you do it with the tools from your geometry box? Describe your +method. + +Ans: +Tools required: Ruler, Set-squares (right-angled triangle), Pencil, eraser +Steps: +1. Place the set square so that one side is along the line l. +2. Hold the ruler against the other side of the set square so that the +ruler won’t move. +3. Slide the set square along the ruler until one side reaches point A. +4. Draw a line along the edge of the set square through point A. + +5. This new line is parallel to line l and passes through point A. + +Page No. 120. + Why are lines l and m parallel to each other? + +Ans: +When we fold the paper, first we make line t perpendicular to l passing +through A. Then we make another line m perpendicular to t through point +A. Since lines l and m are perpendicular to t, so pair of corresponding +angles are equal i.e. 900. Thus, l and m are parallel. + + + + + + + +Page No. 123-125 + Figure it Out +1. Find the angles marked below. + +Ans: + a = 48° + + b = 52° + + c = 81° + + d = 99° + + e = 69° + + f = 48° + + g = 122° + + h = 75° + i = 54° + + j = 97° + +2. Find the angle represented by a. + +Ans: (i) 138° + +(ii) 118° + +(iii) 105° + +(iv) 23° + +3. In the figures below, what angles do x and y stand for? + +Ans: +(i) + x = 25°, + y = 155° +(ii) + x = 25° +4. In Figure, ABC = 45° and IKJ = 78°. Find angles GEH, HEF, FED. + +Ans: + GEH = 45°; + + HEF = 57°; + + FED = 78° + +5. In the Figure, AB is parallel to CD, and CD is parallel to EF. Also, EA is +perpendicular to AB. If  BEF = 55°, find the values of x and y. + +Ans:  x =  y = 125° [Corresponding Angles] +6. What is the measure of angle  NOP in the given figure? + +[Hint: Draw lines parallel to LM and PQ through points N and O] +Ans:  NOP = 108°" +class_7,6,Number Play,ncert_books/class_7/gegp1dd/gegp106.pdf,"6.1 Numbers Tell us Things +What do the numbers in the figure below tell us? +Remember the children from the Grade 6 textbook of mathematics? +Now, they call out numbers using a different rule. +What do you think these numbers mean? +The children rearrange themselves and each one says a number +based on the new arrangement. +Could you figure out what these numbers convey? Observe and try to +find out. +NUMBER PLAY +6 + +Ganita Prakash | Grade 7 +The rule is — each child calls out the number of children in front of +them who are taller than them. Check if the number each child says +matches this rule in both the arrangements. +Write down the number each child should say based on this rule for +the arrangement shown below. +Figure it Out +1. Arrange the stick figure cutouts given at the end of the book or +draw a height arrangement such that the sequence reads: +(a) 0, 1, 1, 2, 4, 1, 5 +(b) 0, 0, 0, 0, 0, 0, 0 +(c) 0, 1, 2, 3, 4, 5, 6 +(d) 0, 1, 0, 1, 0, 1, 0 +(e) 0, 1, 1, 1, 1, 1, 1 +(f) 0, 0, 0, 3, 3, 3, 3 +2. For each of the statements given below, think and identify if it +is Always True, Only Sometimes True, or Never True. Share your +reasoning. +(a) If a person says ‘0’, then they are the tallest in the group. +(b) If a person is the tallest, then their number is ‘0’. +(c) The first person’s number is ‘0’. +(d) If a person is not first or last in line (i.e., if they are standing +somewhere in between), then they cannot say ‘0’. +(e) The person who calls out the largest number is the shortest. +(f) What is the largest number possible in a group of 8 people? +128 + +Number Play +6.2 Picking Parity +Kishor has some number cards and is working on a puzzle: There +are 5 boxes, and each box should contain exactly 1 number card. The +numbers in the boxes should sum to 30. Can you help him find a way +to do it? + ++ ++ ++ ++ += 30 +Can you figure out which 5 cards add to 30? Is it possible? +There are many ways of choosing 5 cards from this collection. +Is there a way to find a solution without checking all possibilities? +Let us find out. +Add a few even numbers together. What kind of number do you get? +Does it matter how many numbers are added? +Any even number can be arranged in pairs without any leftovers. +Some even numbers are shown here, arranged in pairs. + +As we see in the figure, adding any number of even numbers +will result in a number which can still be arranged in pairs +without any leftovers. In other words, the sum will always be an +even number. +Now, add a few odd numbers together. What kind of number do you +get? Does it matter how many odd numbers are added? +Odd numbers can not be arranged in pairs. An odd number is one +more than a collection of pairs. Some odd numbers are shown below: + +129 + +Ganita Prakash | Grade 7 +Can we also think of an odd number as one less than a collection +of pairs? +This figure shows that the sum of two odd numbers must always +be even! This along with the other figures here are more examples +of a proof! +We can see that +two odd numbers added +together can always be +arranged in pairs. +What about adding 3 odd numbers? Can the resulting sum be arranged +in pairs? No. +Explore what happens to the sum of (a) 4 odd numbers, (b) 5 odd +numbers, and (c) 6 odd numbers. +Let us go back to the puzzle Kishor was trying to solve. There are +5 empty boxes. That means he has an odd number of boxes. All the +number cards contain odd numbers. +They should add to 30, which is an even number. Since, adding any +5 odd numbers will never result in an even number, Kishor cannot +arrange these cards in the boxes to add up to 30. +Two siblings, Martin and Maria, were born exactly one year apart. +Today they are celebrating their birthday. Maria exclaims that the sum +of their ages is 112. Is this possible? Why or why not? +As they were born one year apart, their ages will be (two) consecutive +numbers. Can their ages be 51 and 52? 51 + 52 = 103. Try some other +consecutive numbers and see if their sum is 112. +The counting numbers 1, 2, 3, 4, 5, ... alternate between even and odd +numbers. In any two consecutive numbers, one will always be even +and the other will always be odd! +What would be the resulting sum of an even number and an odd +number? We can see that their sum can’t be arranged in pairs and thus +will be an odd number. +130 + +Number Play +Since 112 is an even number, and Martin’s and Maria’s ages are +consecutive numbers, they cannot add up to 112. +We use the word parity to denote the property of being even or odd. +For instance, the parity of the sum of any two consecutive numbers is +odd. Similarly, the parity of the sum of any two odd numbers is even. +Figure it Out +1. Using your understanding of the pictorial representation of odd +and even numbers, find out the parity of the following sums: +(a) Sum of 2 even numbers and 2 odd numbers (e.g., even + even ++ odd + odd) +(b) Sum of 2 odd numbers and 3 even numbers +(c) Sum of 5 even numbers +(d) Sum of 8 odd numbers +2. Lakpa has an odd number of ₹1 coins, an odd number of ₹5 coins +and an even number of ₹10 coins in his piggy bank. He calculated +the total and got ₹205. Did he make a mistake? If he did, explain +why. If he didn’t, how many coins of each type could he have? +3. We know that: +(a) even + even = even +(b) odd + odd += even +(c) even + odd = odd +Similarly, find out the parity for the scenarios below: +(d) even – even = ___________________ +(e) odd – odd += ___________________ +(f) even – odd += ___________________ +(g) odd – even += ___________________ +Small Squares in Grids +In a 3 × 3 grid, there are 9 small squares, +which is an odd number. Meanwhile, in +a 3 × 4 grid, there are 12 small squares, +which is an even number. +Given the dimensions of a grid, can you +tell the parity of the number of small +squares without calculating the product? +131 + +Ganita Prakash | Grade 7 +Find the parity of the number of small squares in these grids: +(a) 27 × 13 +(b) 42 × 78 +(c) 135 × 654 +Parity of Expressions +Consider the algebraic expression: 3n + 4. For different values of n, the +expression has different parity: +n +Value of 3n + 4 +Parity of the Value +3 +13 +odd +8 +28 +even +10 +34 +even +Come up with an expression that always has even parity. +Some examples are: 100p and 48w – 2. Try to find more. +Come up with expressions that always have odd parity. +Come up with other expressions, like 3n + 4, which could have either +odd or even parity. +The expression 6k + 2 evaluates to 8, 14, 20,... (for k = 1, 2, 3,...) — many +even numbers are missing. +Are there expressions using which we can list all the even numbers? +Hint: All even numbers have a factor 2. +Are there expressions using which we can list all odd numbers? +We saw earlier how to express the nth term of the sequence of +multiples of 4, where n is the letter-number that denotes a position in +the sequence (e.g., first, twenty third, hundred and seventeenth, etc.). +What would be the nth term for multiples of 2? Or, what is the nth even +number? +Let us consider odd numbers. +What is the 100th odd number? +To answer this question, consider the following question: +132 + +Number Play +What is the 100th even number? +This is 2 × 100 = 200. +Does this help in finding the 100th odd number? Let us compare the +sequence of evens and odds term-by-term. +Even Numbers: +2, 4, 6, 8, 10, 12,... +Odd Numbers: +1, 3, 5, 7, 9, 11,... +We see that at any position, the value at the odd number sequence +is one less than that in the even number sequence. Thus, the 100th odd +number is 200 – 1 = 199. +Write a formula to find the nth odd number. +Let us first describe the method that we have learnt to find the odd +number at a given position: +(a) Find the even number at that position. This is 2 times the +position number. +(b) Then subtract 1 from the even number. +Writing this in expressions, we get +(a) 2n +(b) 2n – 1 +Thus, 2n is the formula that gives the nth even number, and 2n – 1 is +the formula that gives the nth odd number. +6.3 Some Explorations in Grids +Observe this 3 × 3 grid. It is filled following a simple +rule — use numbers from 1 – 9 without repeating +any of them. There are circled numbers outside the +grid. +Are you able to see what the circled numbers +represent? +The numbers in the yellow circles are the sums of the corresponding +rows and columns. +Fill the grids below based on the rule mentioned above: +9 +5 +13 +14 +18 +12 +9 +24 +4 +3 +24 +15 +6 +17 +16 +12 +4 +7 +5 +6 +1 +2 +3 +9 +8 +16 +9 +20 +15 +17 +13 +133 + +Ganita Prakash | Grade 7 +Make a couple of questions like this on your own and challenge your +peers. +Try solving the problem below. +You might have realised that it is not possible to +find a solution for this grid. Why is this the case? +The smallest sum possible is 6 = 1 + 2 + 3. The +largest sum possible is 24 = 9 + 8 + 7. Clearly, any +number in a circle cannot be less than 6 or +greater than 24. The grid has sums 5 and 26. +Therefore, this is impossible! +In the earlier grids which we solved, +Kishor noticed that the sum of all the +numbers in the circles was always 90. Also, +Vidya observed that the sum of the circled +numbers for all three rows, or for all three +columns, was always 45. Check if this is true +in the previous grids you have solved. +Why should the row sums and column sums +always add to 45? +From this grid, we can see that all the +row sums added together will be the same +as the sum of the numbers 1 – 9. This can be +seen for column sums as well. The sum of +the numbers 1 – 9 is +1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45. +A square grid of numbers is called a +magic square if each row, each column and +each diagonal, add up to the same number. +This number is called the magic sum. +Diagonals are shown in the picture. +Trying to create a magic square by randomly filling +the grid with numbers may be difficult! This is because +there are a large number of ways of filling a 3 × 3 grid +using the numbers 1 – 9 without repetition. In fact, it +can be found that there are exactly 3,62,880 such ways. +Surprisingly, the number of ways to fill in the grid can be +found without listing all of them. We will see in later years how to do this. +Instead, we should proceed systematically to make a magic square. +For this, let us ask ourselves some questions. +1. What can the magic sum be? Can it be any number? +6 +5 +21 +19 +26 +11 +9 +The 3 row sums +added together gives +45! So does adding +the column sums. +4 +7 +5 +6 +1 +2 +3 +9 +8 +4+6+3 +5+2+8 +3+9+8 +6+1+2 +4+7+5 +7+1+9 +134 + +Number Play +Let us focus, for the moment, only on the row sums. We have seen that +in a 3 × 3 grid with numbers 1 – 9, the total of row sums will always be +45. Since in a magic square the row sums are all equal, and they add up +to 45, they have to be 15 each. Thus, we have the following observation. +Observation 1: In a magic square made using the numbers 1 – 9, the +magic sum must be 15. +2. What are the possible numbers that could occur at the centre of a +magic square? +Let us consider the possibilities one by one. +Can the central number be 9? If yes, then 8 must +come in one of the other squares. For example, +In this, we must have 8 + 9 + other number = 15. +But this is not possible! The same issue will +occur no matter where we place 8. +So, 9 cannot be at the centre. Can the central +number be 1? +If yes, then 2 should come in one of the other +squares. +Here, we must have 2 + 1 + other number = 15. +But this is not possible because we are only +using the numbers 1 – 9. The same issue will +occur no matter where we place 1. +So, 1 cannot be at the centre, either. +Using such reasoning, find out which other +numbers 1 – 9 cannot occur at the centre. +This exploration will lead us to the following +interesting observation. +Observation 2: The number occurring at +the centre of a magic square, filled using 1 – 9, +must be 5. +Let us now see where the smallest number 1 +and the largest number 9 should come in a magic +square. Our second observation tells us that +they will have to come in one of the boundary +positions. Let us classify these positions into two categories: +8 +9 +1 +2 +5 +135 + +Ganita Prakash | Grade 7 +Can 1 occur in a corner position? For example, +can it be placed as follows? +If yes, then there should exist three ways of +adding 1 with two other numbers to give 15. +We have 1 + 5 + 9 = 1 + 6 + 8 = 15. Is any other +combination possible? +Similarly, can 9 can be placed in a corner position? +Observation 3: The numbers 1 and 9 cannot occur in any corner, so +they should occur in one of the middle positions. +Can you find the other possible positions for 1 and 9? +Now, we have one full row or column of the magic square! +Try completing it! +[Hint: First fill the row or columns containing 1 and 9] +Figure it Out +1. How many different magic squares can be made using the +numbers 1 – 9? +2. Create a magic square using the numbers 2 – 10. What strategy +would you use for this? Compare it with the magic squares +made using 1 – 9. +3. Take a magic square, and +(a) increase each number by 1 +(b) double each number +In each case, is the resulting grid also a magic square? How do the +magic sums change in each case? +4. What other operations can be performed on a magic square to +yield another magic square? +5. Discuss ways of creating a magic square using any set of 9 +consecutive numbers (like 2 – 10, 3 – 11, 9 – 17, etc.). +Generalising a 3 × 3 Magic Square +We can describe how the numbers within the magic square are related +to each other, i.e., the structure of the magic square. +1 +5 +15 +15 +15 +1 +5 +9 +1 +5 +9 +Math +Talk +Math +Talk +136 + +Number Play +Choose any magic square that you have made so far +using consecutive numbers. If m is the letter-number +of the number in the centre, express how other +numbers are related to m, how much more or less +than m. +[Hint: Remember, how we described a 2 × 2 grid of a +calendar month in the Algebraic Expressions chapter]. +Once the generalised form is obtained, share your observations +with the class. +Figure it Out +1. Using this generalised form, find a magic square if the centre +number is 25. +2. What is the expression obtained by adding the 3 terms of any row, +column or diagonal? +3. Write the result obtained by— +(a) adding 1 to every term in the generalised form. +(b) doubling every term in the generalised form +4. Create a magic square whose magic sum is 60. +5. Is it possible to get a magic square by filling nine + + + +non-consecutive numbers? +The First-ever 4 × 4 Magic Square +The first ever recorded 4 × 4 magic square is found in a 10th century +inscription at the Pārśhvanath Jain temple in Khajuraho, India, and is +known as the Chautīsā Yantra. +7 +12 +1 +14 +2 +13 +8 +11 +16 +3 +10 +5 +9 +6 +15 +4 +The first ever recorded 4 × 4 magic square, the Chautīsā Yantra, at Khajuraho, India +Chau̐ tīs means 34. Why do you think they called it the Chautīsā Yantra? +Every row, column and diagonal in this magic square adds up to 34. +Can you find other patterns of four numbers in the square that add up +to 34? +m +Math +Talk +Try +This +137 + +Ganita Prakash | Grade 7 +Magic Squares in History and Culture +The first magic square ever recorded, the Lo Shu Square, dates +back over 2000 years to ancient China. The legend tells of a +catastrophic flood on the Lo River, during which the gods sent +a turtle to save the people. The turtle carried a 3 × 3 +grid on its back, with the numbers 1 to 9 arranged in +a magical pattern. +Magic squares were studied in different parts of +the world at different points of time including India, Japan, +Central Asia, and Europe. +Indian mathematicians have worked extensively on magic +squares, describing general methods of constructing them. +The work of Indian mathematicians was not just limited to 3 × 3 and +4 × 4 grids, which we considered above, but also extended to 5 × 5 and +other larger square grids. We shall learn more about these in +later grades. +The occurrence of magic squares is not limited to scholarly +mathematical works. They are found in many places in India. +The picture to the right is of a 3 × 3 magic square found on +a pillar in a temple in Palani, Tamil Nadu. The temple dates +back to the 8th century CE. +3 × 3 magic squares can also be found in homes and shops in +India. The Navagraha Yantra is one such example shown below. + +2 +7 +6 +9 +5 +1 +4 +3 +8 +138 + +Number Play +Notice that a different magic sum is associated with each graha. A +picture of a Kubera Yantra is shown below: +6.4 Nature’s Favourite Sequence: The Virahāṅka– +Fibonacci Numbers! +The sequence 1, 2, 3, 5, 8, 13, 21, 34, . . . (Virahāṅka–Fibonacci Numbers) +is one of the most celebrated sequences in all of mathematics — it +occurs throughout the world of Art, Science, and Mathematics. Even +though these numbers are found very frequently in Science, it is +remarkable that these numbers were first discovered in the context of +Art (specifically, poetry)! +The Virahāṅka–Fibonacci Numbers thus provide a beautiful +illustration of the close links between Art, Science, and Mathematics. +Discovery of the Virahāṅka Numbers +The Virahāṅka numbers first came up thousands of years ago in the +works of Sanskrit and Prakrit linguists in their study of poetry! +In the poetry of many Indian languages, including, Prakrit, Sanskrit, +Marathi, Malayalam, Tamil, and Telugu, each syllable is classified as +either long or short. +A long syllable is pronounced for a longer duration than a short +syllable — in fact, for exactly twice as long. When singing such a poem, +a short syllable lasts one beat of time, and a long syllable lasts two +beats of time. +This leads to numerous mathematical questions, which the ancient +poets in these languages considered extensively. A number of important +mathematical discoveries were made in the process of asking and +answering these questions about poetry. +One of these particularly important questions was the following. +How many rhythms are there with 8 beats consisting of short syllables +(1 beat) and long syllables (2 beats)? That is, in how many ways can one +27 +20 +25 +22 +24 +26 +23 +28 +21 +139 + +Ganita Prakash | Grade 7 +fill 8 beats with short and long syllables, where a short syllable takes +one beat of time and a long syllable takes two beats of time. +Here are some possibilities: +long long long long +short short short short short short short short +short long long short long +long long short short long + + . + + . + + . +Can you find others? +Phrased more mathematically: In how many different ways can one +write a number, say 8, as a sum of 1’s and 2’s? +For example, we have: + + + + + +8 = 2 + 2 + 2 + 2, + + + + + +8 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1, + + + + + +8 = 1 + 2 + 2 + 1 + 2, + + + + + +8 = 2 + 2 + 1 + 1 + 2, + + + + + +etc. +Do you see other ways? +Here are all the ways of writing each of the numbers 1, 2, 3, and 4, +as the sum of 1’s and 2’s. +Different Ways +Number of Ways +n = 1 +1 +1 +n = 2 +1 + 1 +2 +2 +n = 3 +1 + 1 + 1 +1 + 2 +2 + 1 +3 +n = 4 +1 + 1 + 1 + 1 +1 + 1 + 2 +1 + 2 + 1 +2 + 1 + 1 +2 + 2 +5 +Try writing the number 5 as a sum of 1s and 2s in all possible ways +in your notebook! How many ways did you find? (You should find 8 +different ways!) Can you figure out the answer without listing down all +the possibilities? Can you try it for n = 8? +Here is a systematic way to write down all rhythms of short and long +syllables having 5 beats. Write a ‘1+’ in front of all rhythms having 4 +beats, and then a ‘2+’ in front of all rhythms having 3 beats. This gives +us all the rhythms having 5 beats: +140 + +Number Play +n = 5 +1 + 1 + 1 + 1 + 1 +1 + 1 + 1 + 2 +1 + 1 + 2 + 1 +1 + 2 + 1 + 1 +1 + 2 + 2 +2 + 1 + 1 + 1 +2 + 1 + 2 +2 + 2 + 1 +Thus, there are 8 rhythms having 5 beats! +The reason this method works is that every 5-beat rhythm must begin +with either a ‘1+’ or a ‘2+’. If it begins with a ‘1+’, then the remaining +numbers must give a 4-beat rhythm, and we can write all those down. +If it begins with a 2+, then the remaining number must give a 3-beat +rhythm, and we can write all those down. Therefore, the number of +5-beat rhythms is the number of 4-beat rhythms, plus the number of +3-beat rhythms. +How many 6-beat rhythms are there? By the same reasoning, it will +be the number of 5-beat rhythms plus the number of 4-beat rhythms, +i.e., 8 + 5 = 13. Thus, there are 13 rhythms having 6 beats. +Use the systematic method to write down all 6-beat rhythms, i.e., write +6 as the sum of 1’s and 2’s in all possible ways. Did you get 13 ways? +This beautiful method for counting all the rhythms of short syllables +and long syllables having any given number of beats was first given by +the great Prakrit scholar Virahāṅka around the year 700 CE. He gave +his method in the form of a Prakrit poem! For this reason, the sequence +1, 2, 3, 5, 8, 13, 21, 34, . . . is known as the Virahāṅka sequence, and +the numbers in the sequence are known as the Virahāṅka numbers. +Virahāṅka was the first known person in history to explicitly consider +these important numbers and write down the rule for their formation. +Other scholars in India also considered these numbers in the +same poetic context. Virahāṅka was inspired by earlier work of the +legendary Sanskrit scholar Piṅgala, who lived around 300 BCE. After +Virahāṅka, these numbers were also written about by Gopala (c. 1135 +CE) and then by Hemachandra (c. 1150 CE). +In the West, these numbers have been known as the Fibonacci +numbers, after the Italian mathematician who wrote about them in +the year 1202 CE — about 500 years after Virahāṅka. As we can see, +Fibonacci was not first, nor the second, not even the third person to +write about these numbers! Sometimes the term “Virahāṅka–Fibonacci +numbers” is used so that everyone understands what is being referred +to. +So, how many rhythms of short and long syllables are there having +8 beats? We simply take the 8th element of the Virahāṅka sequence: +1, 2, 3, 5, 8, 13, 21, 34, 55, ... +Thus, there are 34 rhythms having 8 beats. +141 + +Ganita Prakash | Grade 7 +Write the next number in the sequence, after 55. +We have seen that the next number in the sequence is given by +adding the two previous numbers. Check that this holds true for the +numbers given above. The next number is 34 + 55 = 89. +Write the next 3 numbers in the sequence: +1, 2, 3, 5, 8, 13, 21, 34 , 55, 89, ____, ____, ____, … +If you have to write one more number in the sequence above, can +you tell whether it will be an odd number or an even number (without +adding the two previous numbers)? +What is the parity of each number in the sequence? Do you notice any +pattern in the sequence of parities? +Today, the Virahāṅka–Fibonacci numbers form the basis of many +mathematical and artistic theories, from poetry to drumming, to +visual arts and architecture, to science. Perhaps the most stunning +occurrences of these numbers are in nature. For example, the number +of petals on a daisy is generally a Virahāṅka number. +How many petals do you see on each of these flowers? +A daisy with 13 +petals +A daisy with 21 +petals +A daisy with 34 +petals +There +are +many +other +remarkable +mathematical +properties of the Virahāṅka– +Fibonacci numbers that we +will see later, in mathematics +as well as in other subjects. +These +numbers +truly +exemplify the close connections +between +Art, +Science, +and +Mathematics. +6.5 Digits in Disguise +You have done arithmetic operations with numbers. How about doing +the same with letters? +142 + +Number Play +In the calculations below, digits are replaced by letters. Each letter +stands for a particular digit (0 – 9). You have to figure out which digit +each letter stands for. +T +T ++ T +UT +Here, we have a one-digit number that, when added to itself twice, +gives a 2-digit sum. The units digit of the sum is the same as the single +digit being added. +What could U and T be? Can T be 2? Can it be 3? +Once you explore, you will see that T = 5 and UT = 15. +Let us look at one more example shown on the right. +Here K2 means that the number is a 2-digit number +having the digit ‘2’ in the units place and ‘K’ in the tens place. K2 is +added to itself to give a 3-digit sum HMM. What digit should the letter +M correspond to? +Both the tens place and the units place of the sum have the same +digit. +What about H? Can it be 2? Can it be 3? +These types of questions can be interesting and fun to solve! Here +are some more questions like this for you to try out. Find out what each +letter stands for. +Share how you thought about each question with your classmates; +you may find some new approaches. +YY +B5 +KP +C1 ++ Z ++ 3D ++ KP ++ C +ZOO +ED5 +PRR +1FF +These types of questions are called ‘cryptarithms’ or ‘alphametics’. +Figure it Out +1. +A light bulb is ON. Dorjee toggles its switch 77 times. Will the +bulb be on or off? Why? +K2 ++ K2 +HMM +143 + +Ganita Prakash | Grade 7 +2. Liswini has a large old encyclopaedia. When she opened it, several +loose pages fell out of it. She counted 50 sheets in total, each printed +on both sides. Can the sum of the page numbers of the loose sheets +be 6000? Why or why not? +3. Here is a 2 × 3 grid. For each row and column, +the parity of the sum is written in the circle; ‘e’ +for even and ‘o’ for odd. Fill the 6 boxes with 3 +odd numbers (‘o’) and 3 even numbers (‘e’) to +satisfy the parity of the row and column sums. +4. Make a 3 × 3 magic square with 0 as the magic +sum. All numbers can not be zero. Use negative numbers, as needed. +5. Fill in the following blanks with ‘odd’ or ‘even’: +(a) Sum of an odd number of even numbers is ______ +(b) Sum of an even number of odd numbers is ______ +(c) Sum of an even number of even numbers is ______ +(d) Sum of an odd number of odd numbers is +______ +6. What is the parity of the sum of the numbers from 1 to 100? +7. Two consecutive numbers in the Virahāṅka sequence are 987 and +1597. What are the next 2 numbers in the sequence? What are the +previous 2 numbers in the sequence? +8. Angaan wants to climb an 8-step staircase. His playful rule is that +he can take either 1 step or 2 steps at a time. For example, one of +his paths is 1, 2, 2, 1, 2. In how many different ways can he reach +the top? +9. What is the parity of the 20th term of the Virahāṅka sequence? +10. Identify the statements that are true. +(a) The expression 4m – 1 always gives odd numbers. +(b) All even numbers can be expressed as 6j – 4. +(c) Both expressions 2p + 1 and 2q – 1 describe all odd numbers. +(d) The expression 2f + 3 gives both even and odd numbers. +11. Solve this cryptarithm: + +UT ++ TA +TAT +o +e +o +e +e +144 + +Number Play +In this chapter, we have explored the following: +• In the first activity, we saw how to represent information about how +a sequence of numbers (e.g., height measures) is arranged without +knowing the actual numbers. +• We learnt the notion of parity — numbers that can be arranged in +pairs (even numbers) and numbers that cannot be arranged in pairs +(odd numbers). +• We learnt how to determine the parity of sums and products. +• While exploring sums in grids, we could determine whether filling +a grid is impossible by looking at the row and column sums. We +extended this to construct magic squares. +• We saw how Virahāṅka numbers were first discovered in history +through the Arts. The Virahāṅka sequence is 1, 2, 3, 5, 8, 13, 21, 34, +55, ... +• We became math-detectives through cryptarithms, where digits are +replaced by letters. +SUMMARY +I just heard +that 14,70,369 +people got +married last +year. +Wait, shouldn’t +it be an even +number? +145 + +1 + + +Page No- 128 + Write down the number each child should say based on this rule for the +arrangement shown below. + +The rule is: +Each child calls out the number of children standing in front of them who are +taller than they are. Let’s assign the numbers from left to right for the 7 +children: +Child 1: From left → 0 +Child 2: From left → 0 +Child 3: From left → 1 +Child 4: From left → 0 +Child 5: From left → 3 +Child 6: From left → 0 +Child 7: From left → 3 + Figure it out +1. Arrange the stick figure cutouts given at the end of the book or draw a +height arrangement such that the sequence reads: +(a) 0, 1, 1, 2, 4, 1, 5 +(b) 0, 0, 0, 0, 0, 0, 0 +(c) 0, 1, 2, 3, 4, 5, 6 +(d) 0, 1, 0, 1, 0, 1, 0 +(e) 0, 1, 1, 1, 1, 1, 1 +(f) 0, 0, 0, 3, 3, 3, 3 +CHAPTER – 6 +Number Play + +2 + +Ans: +(a) From the cutouts at the back of the book we have → F, C, B, G, A, D, E +May also like to draw this way — + + + + + + + +Try for other subparts +2. For each of the statements given below, think and identify if it is Always +True, Only Sometimes True, or Never True. Share your reasoning. +(a) If a person says ‘0’, then they are the tallest in the group. +(b) If a person is the tallest, then their number is ‘0’. +(c) The first person’s number is ‘0’. +(d) If a person is not first or last in line (i.e., if they are standing +somewhere in between), then they cannot say ‘0’. +(e) The person who calls out the largest number is the shortest. +(f) +What is the largest number possible in a group of 8 people? +Ans: +(a) +Only sometimes true +(b) Always true +(c) +Always true +(d) Only sometimes true +(e) +Only sometimes true +(f) +The largest possible number in a group of 8 people is 7 + +Page No- 131 + Figure it out +1. Using your understanding of the pictorial representation of odd and even +numbers, find out the parity of the following sums: +(a) Sum of 2 even numbers and 2 odd numbers (e.g., even + even + odd + +odd) +(b) Sum of 2 odd numbers and 3 even numbers + +3 + +(c) Sum of 5 even numbers +(d) Sum of 8 odd numbers +Ans: +(a) +Even number +(b) Even number +(c) +Even number +(d) Even number +2. Lakpa has an odd number of ₹1 coins, an odd number of ₹5 coins and an +even number of ₹10 coins in his piggy bank. He calculated the total and got +₹205. Did he make a mistake? If he did, explain why. If he didn’t, how many +coins of each type could he have? +Ans: +Lakpa must have made a mistake! +The total can never be ₹205 with the given coin counts. + +3. We know that: +(a) even + even = even +(b) odd + odd = even +(c) even + odd = odd +Similarly, find out the parity for the scenarios below: +(d) even – even = _____ +(e) odd – odd = _____ +(f) even – odd = _____ +(g) odd – even = _____ +Ans: +(d) Even +(e) Even +(f) Odd +(g) Odd + + + +4 + + Given the dimensions of a grid, can you tell the parity of the number of +small squares without calculating the product? + +Ans: +The parity of the number of small squares in a grid (with m rows and n +columns) is: +(i) +Even, if at least one of the numbers (either m or n) is even. +(ii) Odd, if both numbers (m and n) are odd. +(iii) Even, if both numbers (m and n) are even. + +Page No- 132 + Find the parity of the number of small squares in these grids: +(a) 27 × 13 +(b) 42 × 78 +(c) 135 × 654 +Ans: +(a) Odd +(b) Even +(c) Even + Come up with an expression that always has odd parity. +Ans: +Some such expressions are 2m + 1, 8a + 3; 4n + 3, 6n + 5 etc. + Come up with other expressions, like 3n + 4, which could have either odd +or even parity. +Ans: + +5 + +(i) If n is odd, then 3n + 4 is odd then some expressions of odd parity would be— +5n + 2, 9n + 4 etc. +Some expressions of even parity — 2n + 4, 4n + 6 etc. +(ii) If n is even, then 3n + 4 is even then all expression will be even parity — +(5n – 2), n + 4 etc. + + Are there expressions using which we can list all odd numbers? +Ans: 2n-1 is the expression that can list all odd numbers. + What would be the nth term for multiples of 2? Or, what is the nth even +number? +Ans: 2n + Fill the grids below based on the rule mentioned above: + +Ans: + +Try to find out other possible ways. + + + +6 + +Page No- 136 + Figure it out +1. How many different magic squares can be made using the numbers 1 – 9? +Ans: There is exactly one unique magic square with numbers 1 – 9, if we ignore +rotations and reflections. +8 +1 +6 +3 +5 +7 +4 +9 +2 + +2. Create a magic square using numbers 2 – 10. What strategy would you use +for this? Compare it with the magic squares made using 1 – 9. +Ans: We will add 1 to every number in the magic square in Q.1 + +9 +2 +7 +4 +6 +8 +5 +10 +3 + +3. Take a magic square, and +(a) Increase each number by 1 +(b) Double each number +In each case, is the resulting grid also a magic square? How do the magic +sums change in each case? +Ans: Consider the magic square. +8 +1 +6 +3 +5 +7 +4 +9 +2 + +7 + +(a) After increasing each number by 1: + +9 +2 +7 +4 +6 +8 +5 +10 +3 + +This is still a magic square. +(b) After Doubling each number: + +16 +2 +12 +6 +10 +14 +8 +18 +4 + +This is still a magic square +In case (a), magic sum is increased by 3 times that constant. (i.e., 1) +In case (b), magic sum is multiplied by that constant. (i.e., 2) +4. What other operations can be performed on a magic square to yield +another magic square? +Ans: +Operations like Addition, subtraction, multiplication and division can be +performed on a magic square to yield another magic square. +5. Discuss ways of creating a magic square using any set of 9 consecutive +numbers (like 2 – 10, 3 – 11, 9 – 17, etc.). +Ans: +Magic square with numbers 1 – 9: + +8 +1 +6 +3 +5 +7 +4 +9 +2 + +8 + + +Magic square with numbers 2 – 10 can be created by adding 1 to each number +of the magic square 1 – 9. + +9 +2 +7 +4 +6 +8 +5 +10 +3 + +Magic square with numbers 3 – 11 can be created by adding 2 to each number +of the magic square 1 – 9. +Magic square with numbers 9 – 17 can be created by adding 8 to each number +of the magic square 1 – 9. + Choose any magic square that you have made so far using consecutive +numbers. If m is the letter-number of the number in the centre, express +how other numbers are related to m, how much more or less than m. [Hint: +Remember, how we described a 2 × 2 grid of a calendar month in the +Algebraic Expressions chapter]. +Ans: Taking ‘m’ as the letter-number of the number in the centre and expressing +other numbers in relation to ‘m’. + +m + 3 +m – 4 +m + 1 +m – 2 +m +m + 2 +m – 1 +m + 4 +m – 3 + +Page No- 137 + Figure it out +1. Using this generalised form, find a magic square if the centre number is +25. +Ans: +28 +21 +26 +23 +25 +27 +24 +29 +22 + +9 + + +2. What is the expression obtained by adding the 3 terms of any row, +column, or diagonal? +Ans: The expression obtained = 3 × m +where m is the letter - number representing the number in the centre. +3. Write the result obtained by +(a) Adding 1 to every term in the generalised form. +(b) Doubling every term in the generalised form. +Ans: +(a) +m + 4 m – 3 m + 2 +m – 1 m + 1 m + 3 +m +m + 5 m – 2 +(b) +2m + 6 +2m – 8 +2m + 2 +2m – 4 +2m +2m + 4 +2m – 2 +2m + 8 +2m – 6 + +4. Create a magic square whose magic sum is 60. +Ans: +23 +16 +21 +18 +20 +22 +19 +24 +17 + +5. Is it possible to get a magic square by filling nine non-consecutive +numbers? +Ans: +Yes, it is possible. +24 +3 +18 +9 +15 +21 +12 +27 +6 +Try some more. + +10 + + +Page No- 142 + Write the next 3 numbers in the sequence: +1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ____, ____, ____, … +If you have to write one more number in the sequence above, can you tell +whether it will be an odd number or an even number (without adding the +two previous numbers)? +Ans: 55 + 89 = 144, 89 + 144 = 233, 144 + 233 = 3779 +The next number will be an even number. + What is the parity of each number in the sequence? Do you notice any +pattern in the sequence of parities? +The parity patterns — +Sequence: 1(O), 2(E), 3(O), 5(O), 8(E), 13(O), 21(O), 34(E), … +The pattern is: Odd, Even, Odd, Odd, Even, Odd, Odd, Even…. +The group odd, even, odd repeats. + 3rd term = 1st term + 2nd term = O + E = O; + 4th term = 2nd term + 3rd term = E + O = O; + 5th term = 3rd term + 4th term = O + O = E… + +Page No- 143-44 + Figure it out +1. A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or +off? Why? +Ans. The bulb will be OFF. + +2. Liswini has a large old encyclopaedia. When she opened it, several loose +pages fell out of it. She counted 50 sheets in total, each printed on both +sides. Can the sum of the page numbers of the loose sheets be 6000? Why +or why not? +Ans. Each sheet has two pages, one with an even number and the other with an odd +number. +Since there are 50 sheets, so there will be fifty pages with an even number and +50 pages with an odd number +The sum of the even numbers is even. +The sum of the odd numbers taken even number of times is also even. +Thus, 6000 being even, it is possible that sum of all the page numbers of the +loose sheets could be 6000. + +11 + + +3. Here is a 2 × 3 grid. For each row and column, the parity of the sum is +written in the circle; ‘e’ for even and ‘o’ for odd. Fill the 6 boxes with 3 odd +numbers (‘o’) and 3 even numbers (‘e’) to satisfy the parity of the row and +column sums. + +Ans: One of the ways is – + +Try more combinations. +4. Make a 3 × 3 magic square with 0 as the magic sum. All numbers cannot +be zero. Use negative numbers, as needed. +Ans: One of the such magic squares is – +3 +– 4 +1 +– 2 +0 +2 +– 1 +4 +– 3 + +5. Fill in the following blanks with ‘odd’ or ‘even’: +(a) Sum of an odd number of even numbers is ______ +(b) Sum of an even number of odd numbers is ______ +(c) Sum of an even number of even numbers is ______ +(d) Sum of an odd number of odd numbers is ______ +Ans: +(a) Even. +(b) Even. + +12 + +(c) Even. +(d) Odd. +6. What is the parity of the sum of numbers from 1 to 100? +Ans: The parity is even. [Hint– Add numbers from 1 to 100.] +7. Two consecutive numbers in the Virahanka sequence are 987 and 1597. +What are the next 2 numbers in the sequence? What are the previous 2 +numbers in the sequence? +Ans: +The next two numbers are: +987 + 1597 = 2584 +1597 + 2584 = 4181 +The previous two numbers are: +1597 – 987 = 610 +987 – 610 = 377 +The sequence is …, 377, 610, 987, 1597, 2584, 4181, … +8. Angaan wants to climb an 8 - step staircase. His playful rule is that he can +take either 1 step or 2 steps at a time. For example, one of his paths is 1, 2, +2, 1, 2. In how many different ways can he reach the top? +Ans: +Different Ways +Number of ways +1, 1, 1, 1, 1, 1, 1, 1 +1 +1, 1, 1, 1, 1, 1, 2 +1, 1, 1, 1, 1, 2, 1 +. +. +. +7 +1, 1, 1, 1, 2, 2 +1, 1, 1, 2, 1, 2 +. +. +. +15 +1, 1, 2, 2, 2 +1, 2, 2, 2, 1 +. +. +. +10 +2, 2, 2, 2 +1 +So, the total ways in which Angaan reaches the top = 1 + 7 + 15 + 10 + 1 = 34 +ways. + +13 + + +9. What is the parity of the 20th term of the Virahanka sequence? +Ans: +Consider the Virahanka sequence given below: +1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, …. +The pattern of odd/even in Virahanka numbers is: +O, E, O, O, E, O … +So the 20th term of the Virahanka sequence is even. + +10. Identify the true statements. +(a) +The expression 4m – 1 always gives odd numbers. +(b) All even numbers can be expressed as 6j – 4. +(c) +Both expressions 2p + 1 and 2q – 1 describe all odd numbers. +(d) The expression 2f + 3 gives both even and odd numbers. +Ans: +(a) +True +(b) +False +(c) +False +(d) +False + +11. Solve this cryptarithm: + +Ans: +U = 9, T = 1 and A = 0." +class_7,7,A Tale of Three Intersecting Lines,ncert_books/class_7/gegp1dd/gegp107.pdf,"A TALE OF THREE +INTERSECTING +LINES +7 +A triangle is the most basic closed shape. As we know, it consists of: +• three corner points, that we call the vertex of the triangle, and +• three line segments or the sides of the triangle that join the pairs +of vertices. +Triangles come in various shapes. Some of them are shown below. +B +C +∆ ABC +A +X +Z +Y +∆ YZX +∆WUV +V +U +W +∆BAU +A +B +U +Observe the symbol used to denote a triangle and how the triangles +are named using their vertices. While naming a triangle, the vertices +can come in any order. +The three sides meeting at the corners give rise to three angles that +we call the angles of the triangle. For example, in ∆ABC, these angles +are ∠CAB, ∠ABC, ∠BCA, which we simply denote as ∠A, ∠B and ∠C, +respectively. +What happens when the three vertices lie on a straight line? +7.1 Equilateral Triangles +Among all the triangles, the equilateral triangles are the most symmetric +ones. These are triangles in which all the sides are of equal lengths. Let +us try constructing them. +Construct a triangle in which all the sides are of length 4 cm. + +A Tale of Three Intersecting Lines +How did you construct this triangle and what tools did you use? Can +this construction be done only using a marked ruler (and a pencil)? +Constructing this triangle using just a ruler is certainly possible. But +this might require several trials. Say we draw the base — let us call it +AB — of length 4 cm (see the figure below), and mark the third point C +using a ruler such that AC = 4 cm. This may not lead to BC also having +a length of 4 cm. If this happens, we will have to keep making attempts +to mark C till we get BC to be 4 cm long. +A +B +C +4 cm? +4 cm +4 cm +How do we make this construction more efficient? +Recall solving a similar problem in the previous year using a compass +(in the Chapter ‘Playing with Constructions’). We had to mark the top +point of a ‘house’ which is 5 cm from two other points. The method we +used to get that point can also be used here. +After constructing AB = 4 cm, we can do the following. +Step 1: Using a compass, construct a sufficiently long arc of radius 4 cm +from A, as shown in the figure. The point C is somewhere on this arc. +How do we mark it? +A +B +4 cm +147 + +Ganita Prakash | Grade 7 +Step 2: Construct another arc of radius 4 cm from B. +A +B +4 cm +Let C be the point of intersection of the arcs. +The construction ensures that both AC and BC are of length 4 cm. Can +you see why? +Step 3: Join AC and BC to get the required equilateral triangle. +A +B +4 cm +C +7.2 Constructing a Triangle When its Sides are +Given +How do we construct triangles that are not equilateral? +Construct a triangle of sidelength 4 cm, 5 cm and 6 cm. +As in the previous case, this triangle can also be constructed using +just a marked ruler. But it will involve several trials. +148 + +A Tale of Three Intersecting Lines +How do we construct this triangle more efficiently? +Choose one of the given lengths to be the base of the triangle: say 4 +cm. Draw the base. Let A and B be the base vertices, and call the third +vertex C. Let AC = 5 cm and BC = 6 cm. +Fig. 7.1 +Like we did in the case of equilateral triangles, let us first get all the +points that are at a 5 cm distance from A. These points lie on the circle +whose centre is A and has radius 5 cm. The point C must lie somewhere +on this circle. How do we find it? +4 cm +A +B +5 cm +Fig. 7.2 +We will make use of the fact that the point C is 6 cm away from B. +Construct an arc of radius 6 cm from B. +A +B +5 cm +6 cm +Fig. 7.3 +149 + +Ganita Prakash | Grade 7 +The required point C is one of the points of intersection of the two +circles. +The reason why the point of +intersection is the third vertex is the +same as for equilateral triangles. This +point lies on both the circles. Hence +its distance from A is the radius of +the circle centred at A (5 cm) and its +distance from B is the same as the +radius of the circle centred at B (6 cm). +Let us summarise the steps of +construction, noting that constructing +full circles is not necessary to get the +third vertex (see Fig. 7.2 and 7.3). +Step 1: Construct the base AB with one of the side lengths. Let us choose +AB = 4 cm (see Fig. 7.1). +Step 2: From A, construct a sufficiently long arc of radius 5 cm +(see Fig. 7.2). +Step 3: From B, construct an arc of radius 6 cm such that it intersects +the first arc (see Fig. 7.3). +Step 4: The point where both the arcs meet is the required third +vertex C. Join AC and BC to get ∆ABC. +Construct +Construct triangles having the following sidelengths (all the units are +in cm): +(a) 4, 4, 6 +(b) 3, 4, 5 +(c) 1, 5, 5 +(d) 4, 6, 8 +(e) 3.5, 3.5, 3.5 +We have seen that triangles having all three equal sides are called +equilateral triangles. Those having two equal sides are called isosceles +triangles. +Figure it Out +1. Use the points on the circle and/or the centre to +form isosceles triangles. +A +B +C +150 + +A Tale of Three Intersecting Lines +2. Use the points on the circles and/or their centres to form isosceles +and equilateral triangles. The circles are of the same size. +A +B +A and B are the centres of +circles of the same size +A +B +C +A, B, and C are the centres of +circles of the same size +Are Triangles Possible for any Lengths? +Can one construct triangles having any given sidelengths? Are there +lengths for which it is impossible to construct a triangle? Let us explore +this. +Construct a triangle with sidelengths 3 cm, 4 cm, and 8 cm. +What is happening? Are you able to construct the triangle? +Here is another set of lengths: 2 cm, 3 cm, and 6 cm. Check if a triangle +is possible for these sidelengths. +Try to find more sets of lengths for which a triangle construction +is impossible. See if you can find any pattern in them. +We see that a triangle is possible for some sets of lengths and +not possible for others. How do we check if a triangle exists for a given +set of lengths? One way is to actually try to construct the triangle and +check if it is possible. Is there a more efficient way to check this? +Triangle Inequality +Consider the lengths 10 cm, 15 cm and 30 cm. Does there exist a triangle +having these as sidelengths? +To tackle this question, let us study a property of triangles. Imagine +a small plot of plain land having a tent, a tree, and a pole. Imagine +you are at the entrance of the tent and want to go to the tree. Which is +the shorter path: (i) the straight-line path to the tree (the red path) or +(ii) the straight-line path from the tent to the pole, followed by the +straight-line path from the pole to the tree (the yellow path)? +Math +Talk +151 + +Ganita Prakash | Grade 7 +Clearly, the direct straight-line path from the tent to the tree is shorter +than the roundabout path via the pole. In fact, the direct straight-line +path is the shortest possible path to the tree from the tent. +Will the direct path between any two points be shorter than the +roundabout path via a third point? Clearly, the answer is yes. +Can this understanding be used to tell something about the existence +of a triangle having sidelengths 10 cm, 15 cm and 30 cm? +Let us suppose that there is a triangle for this set of lengths. Remember +that at this point we are not sure about the existence of the triangle but +we are only supposing that it exists. Let us draw a rough diagram. +Fig. 7.4 +152 + +A Tale of Three Intersecting Lines +Does everything look right with this triangle? +If this triangle were possible, then the direct path between any two +vertices should be shorter than the roundabout path via the third +vertex. Is this true for our rough diagram? +Let us consider the paths between B and C. +Direct path length = BC = 10 cm +What is the length of the roundabout path via the vertex A? It is the +sum of the lengths of line segments BA and AC. +Roundabout path length = BA + AC = 15 cm + 30 cm = 45 cm +Is the direct path length shorter than the roundabout path length? Yes. +Let us now consider the paths between A and B. +Direct path length = AB = 15 cm +Finding the length of the roundabout path via the vertex C, we get +Roundabout path length = AC + CB = 30 cm + 10 cm = 40 cm +Is the direct path length shorter than the roundabout path length? Yes. +Finally, consider paths between C and A. +Direct path length = CA = 30 cm +Roundabout path length = CB + BA = 10 cm + 15 cm = 25 cm +Is the direct path length shorter than the roundabout path length? In +this case, the direct path is longer, which is absurd. Can such a triangle +exist? No. +Therefore, a triangle having sidelengths 10 cm, 15 cm and 30 cm +cannot exist. +We are thus able to see without construction why a triangle for the +set of lengths 10 cm, 15 cm and 30 cm cannot exist. We have been able +to figure this out through spatial intuition and reasoning. +Recall how we used similar intuition and reasoning to discover +properties of intersecting and parallel lines. We will continue to do +this as we explore geometry. +Can we say anything about the existence of a triangle having sidelengths +3 cm, 3 cm and 7 cm? Verify your answer by construction. +“In the rough diagram in Fig. 7.4, is it possible to assign +lengths in a different order such that the direct paths are +always coming out to be shorter than the roundabout +paths? If this is possible, then a triangle might exist.” +Is such rearrangement of lengths possible in the triangle? +Math +Talk +153 + +Ganita Prakash | Grade 7 +Figure it Out +1. We checked by construction that there are no triangles having +sidelengths 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if +you could have found this without trying to construct the triangle. +2. Can we say anything about the existence of a triangle for each of +the following sets of lengths? +(a) 10 km, 10 km and 25 km +(b) 5 mm, 10 mm and 20 mm +(c) 12 cm, 20 cm and 40 cm +You would have realised that using a rough figure and comparing +the direct path lengths with their corresponding roundabout path +lengths is the same as comparing each length with the sum of the other +two lengths. There are three such comparisons to be made. +3. For each set of lengths seen so far, you might have noticed that + +in at least two of the comparisons, the direct length was less +than the sum of the other two (if not, check again!). For example, +for the set of lengths 10 cm, 15 cm and 30 cm, there are two +comparisons where this happens: +10 < 15 + 30 +15 < 10 + 30 +But this doesn’t happen for the third length: 30 > 10 + 15. +Will this always happen? That is, for any set of lengths, will there be at +least two comparisons where the direct length is less than the sum of +the other two? Explore for different sets of lengths. +Further, for a given set of lengths, is it possible to identify which +lengths will immediately be less than the sum of the other two, without +calculations? +[Hint: Consider the direct lengths in the increasing order.] +Given three sidelengths, what do we need to compare to check for the +existence of a triangle? +When each length is smaller than the sum of the other two, we say +that the lengths satisfy the triangle inequality. For example, the set 3, +4, 5 satisfies the triangle inequality whereas, the set 10, 15, 30 does not +satisfy the triangle inequality. +We have seen that lengths such as 10, 15, 30 that do not satisfy the +triangle inequality cannot be the sidelengths of a triangle. +Try +This +154 + +A Tale of Three Intersecting Lines +Does a triangle exist with sidelengths 4 cm, 5 cm and 8 cm? +This satisfies the triangle inequality: +8 < 4 + 5 = 9 +Why do we not need to check the other two sides? +This means that all the direct path lengths are less than the +roundabout path lengths. Does this confirm the existence of a triangle? +If one of the direct path lengths had been longer, we could have +concluded that a triangle would surely not exist. But in this case, we +can only say that a triangle may or may not exist. +For the triangle to exist, the arcs that we construct to get the third +vertex must intersect. Is it possible to determine that this will happen +without actually carrying out the construction? +Visualising the construction of circles +Let us imagine that we start the construction by constructing the +longest side as the base. Let AB be the base of length 8 cm. The next +step is the construction of sufficiently long arcs corresponding to the +other two lengths: 4 cm and 5 cm. +Instead of just constructing the arcs, let us complete the full circles. +Suppose, we construct a circle of radius 4 cm with A as the centre. +Now, suppose that a circle of radius 5 cm is constructed, centred at B. +Can you draw a rough diagram of the resulting figure? +Note that in the figure below, AX = 4 cm and AB = 8 cm. So, what is +BX? Does this length help in visualising the resulting figure? +155 + +Ganita Prakash | Grade 7 +Since BX = 4 cm, and the radius of the circle centred at B is 5 cm, it is +clear that the circles will intersect each other at two points. +Fig. 7.5: Circles intersecting each other at two points +What does this tell us about the existence of a triangle? The points +A and B along with either of the points of intersection of the circles +will give us the required triangle. Thus, there exists a triangle having +sidelengths 4 cm, 5 cm and 8 cm. +Figure it Out +1. Which of the following lengths can be the sidelengths of a triangle? +Explain your answers. Note that for each set, the three lengths have +the same unit of measure. +(a) 2, 2, 5 +(b) 3, 4, 6 +(c) 2, 4, 8 +(d) 5, 5, 8 +(e) 10, 20, 25 +(f) +10, 20, 35 +(g) 24, 26, 28 +We observe from the previous problems that whenever there is +a set of lengths satisfying the triangle inequality (each length < +sum of the other two lengths), there is a triangle with those three +lengths as sidelengths. +Will triangles always exist when a set of lengths satisfies the triangle +inequality? How can we be sure? +156 + +A Tale of Three Intersecting Lines +We can be sure of the existence of a triangle only if we can show that +the circles intersect internally (as in Fig. 7.5) whenever the triangle +inequality is satisfied. But are there other possibilities when the two +circles are constructed? Let us visualise and study them. +The following different cases can be conceived: +A +B +Case 1: Circles touch each other +A +B +Case 2: Circles do not intersect +A +B +Case 3: Circles intersect each other internally +Note that while constructing the circles, we take +(a) +the length of the base AB = longest of the given length +(b) the radii of the circles to be the smaller two lengths. +Which of the above-mentioned cases will lead to the formation of a +triangle? Clearly, triangles are formed only when the circles intersect +each other internally (Case 3). +Let us study each of these cases by finding the relation between the +radii (the smaller two lengths) and AB (longest length). +Case 1: Circles touch each other at a point +For this case to happen, +sum of the two radii = AB +or +sum of the two smaller lengths = longest length +157 + +Ganita Prakash | Grade 7 +A +B +X +Case 2: Circles do not intersect internally +For this case to happen, what should be the relation between the radii +and AB? +It can be seen from the figure that, +sum of the two radii < AB +or +sum of the the two smaller lengths < longest length +A +B +Case 3: Circles intersect each other +A +radius 1 +radius 2 +B +158 + +A Tale of Three Intersecting Lines +AB is composed of one radius and a part of the other. So, +sum of the two radii > AB, +or +sum of the two smaller lengths > longest length +Can we use this analysis to tell if a triangle exists when the lengths +satisfy the triangle inequality? +If the given lengths satisfy the triangle inequality, then the sum of +the two smaller lengths is greater than the longest length. This means +that this will lead to Case 3 where the circles intersect internally, and +so a triangle exists. +How will the two circles turn out for a set of lengths that do not satisfy +the triangle inequality? Find 3 examples of sets of lengths for which +the circles: +(a) +touch each other at a point, +(b) do not intersect. +Frame a complete procedure that can be used to check the existence of +a triangle. +Conclusion +If a given set of three lengths satisfies the triangle inequality, then a +triangle exists having those as sidelengths. If the set does not satisfy the +triangle inequality, then a triangle with those sidelengths does not exist. +Figure it Out +1. Check if a triangle exists for each of the following set of lengths: +(a) 1, 100, 100 +(b) +3, 6, 9 +(c) 1, 1, 5 +(d) +5, 10, 12 +2. Does there exist an equilateral triangle with sides 50, 50, 50? In +general, does there exist an equilateral triangle of any sidelength? +Justify your answer. +3. For each of the following, give at least 5 possible values for the +third length so there exists a triangle having these as sidelengths +(decimal values could also be chosen): +(a) 1, 100 +(b) 5, 5 +(c) 3, 7 +159 + +Ganita Prakash | Grade 7 +See if you can describe all possible lengths of the third +side in each case, so that a triangle exists with those sidelengths. +For example, in case (a), all numbers strictly between 99 and +101 would be possible. +7.3 Construction of Triangles When Some Sides +and Angles are Given +We have seen how to construct triangles when their sidelengths are +given. Now, we will take up constructions where in place of some +sidelengths, angle measures are given. +Two Sides and the Included Angle +How do we construct a triangle if two sides and the angle included +between them are given? Here are some examples of measurements +showing the included angle. +4 cm +60° +3 cm +5 cm +30° +6 cm +40° +3 cm +2 cm +Construct a triangle ABC with AB = 5 cm, AC = 4 cm and ∠A= 45°. +Let us take one of the given sides, AB, as the base of the triangle. +45° +A +B +5 cm +Step 1: Construct a side AB of length 5 cm. +Step 2: Construct ∠A = 45° by drawing the +other arm of the angle. +Step 3: Mark the point C on the other arm +such that AC = 4 cm. +Step 4: Join BC to get the required triangle. +Try +This +45° +A +B +C +5 cm +4 cm +160 + +A Tale of Three Intersecting Lines +Figure it Out +1. Construct triangles for the following measurements where the +angle is included between the sides: +(a) 3 cm, 75°, 7 cm +(b) 6 cm, 25°, 3 cm +(c) 3 cm, 120°, 8 cm +We have seen that triangles do not exist for all sets of sidelengths. Is +there a combination of measurements in the case of two sides +and the included angle where a triangle is not possible? Justify +your answer using what you observe during construction. +Two Angles and the Included Side +In this case, we are given two angles and the side that is a part of both +angles, which we call the included side. Here are some examples of +such measurements: +40° +50° +5 cm +4 cm +30° +45° +20° +50° +6 cm +Construct a triangle ABC where AB = 5 cm, ∠A = 45° and ∠B = 80°. +Let us take the given side as the base. +45° +80° +A +B +C +5 cm +Step 1: Draw the base AB of length 5 cm. +Step 2: Draw ∠A and ∠B of measures 45°, and +80° respectively. +Step 3: The point of intersection of the two new +line segments is the third vertex C. +Math +Talk +45° +A +B +C +5 cm +80° +161 + +Ganita Prakash | Grade 7 +Figure it Out +1. Construct triangles for the following measurements: +(a) 75°, 5 cm, 75° +(b) 25°, 3 cm, 60° +(c) 120°, 6 cm, 30° +Do triangles always exist? +Do triangles exist for every combination of two angles and their +included side? Explore. +As in the case when we are given all three sides, it turns out that +there is not always a triangle for every combination of two angles and +the included side. +Find examples of measurements of two angles with the included side +where a triangle is not possible. +Let us try to visualise such a situation. Once the base is drawn, try to +imagine how the other sides should be so that they do not meet. +Here are some obvious examples. +If the two angles are greater than or equal to a right angle (90°), then +it is clear that a triangle is not possible. +Now we make one of the base angles an acute angle, say 40°. What +are the possible values that the other angle should take so that the lines +don’t meet? +40° +C +A +B +l +It is clear that if the line from B is “inclined” sufficiently to the right, +then it will not meet the line l. +(a) Try to find a possible ∠B (marked in the figure) for this +to happen. +(b) What could be smallest value of ∠B for the lines to not meet? +Math +Talk +162 + +A Tale of Three Intersecting Lines +40° +C +A +B +l +The blue line is the line with the least rightward bend that doesn’t +meet the line ‘l’ +Visually, it is clear that the line that creates the smallest ∠B has to be +the one parallel to l. Let us call this parallel line m. +Can you tell the actual value of ∠B be in this case? +[Hint: Note that AB is the transversal.] +We have seen that when two lines are parallel, the internal angles +on the same side of the transversal add up to 180°. So ∠B = 140°. +So, for what values of ∠B, does a triangle not exist? Does the length +AB play any part here? +From the discussion above, it can be seen that the length AB does not +play any part in deciding the existence of a triangle. We can say that a +triangle does not exist when ∠B is greater than or equal to 140°. +Figure it Out +1. For each of the following angles, find another angle for which a +triangle is (a) possible, (b) not possible. Find at least two different +angles for each category: +(a) 30° +(b) 70° +(c) 54° + +(d) 144° +2. Determine which of the following pairs can be the angles of a +triangle and which cannot: +(a) 35°, 150° +(b) 70°, 30° +(c) 90°, 85° +(d) 50°, 150° +Like the triangle inequality, can you form a rule that describes the two +angles for which a triangle is possible? +Can the sum of the two angles be used for framing this rule? +163 + +Ganita Prakash | Grade 7 +When the sum of two given angles is less than 180°, a triangle exists +with these angles. If the sum is greater than or equal to 180°, there is +no triangle with these angles. +Let us take two angles, say 60° and 70°, whose sum is less than 180°. +Let the included side be 5 cm. +What could the measure of the third angle be? Does this measure +change if the base length is changed to some other value, say 7 cm? +Construct and find out. +In general, once the two angles are fixed, does the third angle depend +on the included sidelength? Try with different pairs of angles and +lengths. +The measurements might show that the sidelength has no effect (or +a very small effect) on the third angle. With this observation, let us see +if we can find the third angle without carrying out the construction +and measurement. +Try experimenting with different triangles to see if there is a +relation between any two angles and the third one. To find this +relation, what data will you keep track of and how will you +organise the data you collect? +Consider a triangle ABC with ∠B= 50° and ∠C= 70°. Let us see how we +can find ∠A without construction. +50° +70° +A +C +B +We saw that the notion of parallel lines was useful to determine that +the sum of any two angles of a triangle is +less than 180°. Parallel lines can be used to +find the third angle, ∠BAC as well. +Let us suppose we construct a line XY +parallel to BC through vertex A. +We can see new angles being formed here: +∠XAB, and ∠YAC. What are their values? +Since the line XY is parallel to BC, +∠XAB = ∠B and ∠YAC = ∠C, because they are alternate angles of the +transversals AB and AC. +Math +Talk +50° +70° +A +C +B +Fig. 7.6 +X +Y +164 + +A Tale of Three Intersecting Lines +Therefore, ∠XAB = 50°, and ∠YAC = 70°. Can we find ∠BAC from this? +We know that ∠XAB, ∠YAC and ∠BAC together form 180°. So +∠XAB + ∠YAC + ∠BAC = 180° +50° + ∠BAC + 70° = 180° +120° + ∠BAC = 180° +Thus, ∠BAC = 60° +Now construct a triangle (taking BC to be of any suitable length) and +verify if this is indeed the case. +Figure it Out +1. Find the third angle of a triangle (using a parallel line) when two of +the angles are: +(a) 36°, 72° +(b) 150°, 15° +(c) 90°, 30° +(d) 75°, 45° +2. Can you construct a triangle all of whose angles are equal to 70°? If +two of the angles are 70° what would the third angle be? If all the +angles in a triangle have to be equal, then what must its measure +be? Explore and find out. +3. Here is a triangle in which we know +∠B = ∠C and ∠A = 50°. Can you find ∠B and ∠C? +50° +A +B +C +Angle Sum Property +What can we say about the sum of the angles of any triangle? +Consider a triangle ABC. To find the sum of its angles, we can use +the same method of drawing a line parallel to the base: construct a line +through A that is parallel to BC. +Try +This +165 + +Ganita Prakash | Grade 7 +We need to find ∠A + ∠B + ∠C. +We know that ∠B = ∠XAB, ∠C = ∠YAC. +So, ∠A + ∠B + ∠C = ∠A + ∠XAB + ∠YAC + + + + += 180° as together they form a straight angle. +Thus we have proved that the sum of the three angles in any triangle +is 180°! This rather surprising result is called the angle sum property +of triangles. +X +A +B +C +Y +Fig. 7.7 +Take some time to reflect upon how we figured out the angle sum +property. In the beginning, the relationship between the third angle +and the other two angles was not at all clear. However, a simple idea of +drawing a line parallel to the base through the top vertex (as in Fig.  7.7) +suddenly made the relationship obvious. This ingenious idea can be +found in a very influential book in the history of mathematics called +‘The Elements’. This book is attributed to the Greek mathematician +Euclid, who lived around 300 BCE. +This solution is yet another example of how creative thinking plays +a key role in mathematics! +There is a convenient way of verifying the angle sum property by +folding a triangular cut-out of a paper. Do you see how this shows that +the sum of the angles in this triangle is 180°? +166 + +A Tale of Three Intersecting Lines +Exterior Angles +The angle formed between the extension of a side of a triangle and +the other side is called an exterior angle of the triangle. In this figure, +∠ACD is an exterior angle. +A +B +C +D +Exterior Angle +Find ∠ACD, if ∠A = 50°, and ∠B = 60°. +From the angle sum property, we know that +50° + 60° + ∠ACB = 180° +110° + ∠ACB = 180° +So, ∠ACB = 70° +So, ∠ACD = 180 ° – 70° = 110°, +since, ∠ACB and ∠ACD together form a straight angle. +Find the exterior angle for different measures of ∠A and ∠B. Do you +see any relation between the exterior angle and these two angles? +[Hint: From angle sum property, we have ∠A + ∠B + ∠ACB = 180°.] +We also have ∠ACD + ∠ACB = 180°, since they form a straight angle. +What does this show? +7.4 Constructions Related to Altitudes of Triangles +There is another set of useful measurements with respect to a triangle —  +the height of each of its vertices with respect to the opposite sides. +In the world around us, we talk of the +heights of various objects: the height of a +person, the height of a tree, the height of +a building, etc. What do we mean by the +word ‘height’? +Consider a triangle ABC. What is the +height of the vertex A from its opposite +side BC, and how can it be measured? +Let AD be the line segment from A drawn perpendicular to BC. The +length of AD is the height of the vertex A from BC. The line segment AD +is said to be one of the ‘altitudes’ of the triangle. The other altitudes +A +C +D +B +Fig. 7.8 +167 + +Ganita Prakash | Grade 7 +are BE and CF in the figure below: the perpendiculars drawn from the +other vertices to their respective opposite sides. +A +B +C +E +F +Whenever we use the word height of the triangle, we generally +refer to the length of the altitude to whatever side we take as base (this +altitude is AD in the case of Fig. 7.8). +What would the altitude from A to BC be in this triangle? +A +B +C +A +B +D +C +We extend BC and then drop the perpendicular from A to this line. +Altitudes Using Paper Folding +Cut out a paper triangle. Fix one of the sides as the base. Fold it in +such a way that the resulting crease is an altitude from the top vertex +to the base. Justify why the crease formed should be perpendicular to +the base. +Construction of the Altitudes of a Triangle +Construct an arbitrary triangle. Label the vertices A, B, C taking BC to +be the base. +Construct the altitude from A to BC, +Constructing the altitude using just a ruler is not accurate. To get a +more precise angle of 90°, we use a set square along with a ruler. +Can you see how to do this? +168 + +A Tale of Three Intersecting Lines +Step 1: Keep the ruler aligned to the base. Place the set square on the +ruler as shown, such that one of the edges of the right angle touches +the ruler. +A +B +C +Step 2: Slide the set square along the ruler till the vertical edge of the +set square touches the vertex A. +B +C +A +Step 3: Draw the altitude to BC through A using the vertical edge of the +set square. +Does there exist a triangle in which a side is also an altitude? +Visualise such a triangle and draw a rough diagram. +We see that this happens in triangles where one of the angles is a +right angle. +Triangles having one right angle are called right-angled triangles +or simply right triangles. +Altitude from A to BC +A +B +C +169 + +Ganita Prakash | Grade 7 +7.5 Types of Triangles +In our study of triangles, we have encountered the following types of +triangles; equilateral, isosceles, scalene and right-angled triangles. +Did you spot any other type of triangle? +The classification of triangles as equilateral and isosceles was based +on equality of sides. +Equilateral triangles have sides of equal length. Isosceles triangles +have two sides of equal length. Scalene triangles have sides of three +different lengths. +Can a similar classification be done based on equality of angles? Is +there any relation between these two classifications? We will answer +these questions in a later chapter. +We used angle measures when classifying a triangle as a right- +angled triangle. +What are the other types of triangles based on angle measures? +A classification of triangles based on their angle measures is acute- +angled, right-angled and obtuse-angled triangles. We have already +seen what a right-angled triangle is. It is a triangle with one right angle. +Similarly, an obtuse-angled triangle has one obtuse angle. +What could an acute-angled triangle be? Can we define it as a +triangle with one acute angle? Why not? +In an acute-angled triangle, all three angles are acute angles. +Figure it Out +1. Construct a triangle ABC with BC = 5 cm, AB = 6 cm, CA = 5 cm. +Construct an altitude from A to BC. +2. Construct a triangle TRY with RY = 4 cm, TR = 7 cm, ∠R = 140°. +Construct an altitude from T to RY. +Equilateral Triangle +Isosceles Triangle +Scalene Triangle +Math +Talk +170 + +A Tale of Three Intersecting Lines +3. Construct a right-angled triangle ∆ABC with ∠B = 90°, +AC = 5 cm. How many different triangles exist with these +measurements? + +[Hint: Note that the other measurements can take any values. Take +AC as the base. What values can ∠A and ∠C take so that the other +angle is 90°?] +4. Through construction, explore if it is possible to construct an +equilateral triangle that is (i) right-angled (ii) obtuse-angled. +Also construct an isosceles triangle that is (i) right-angled +(ii) obtuse-angled. +• Use of a compass simplifies the construction of triangles when the +sidelengths are given. +• A set of three lengths where the length of each is smaller than the +sum of the other two is said to satisfy the triangle inequality. +• Sidelengths of a triangle satisfy triangle inequality, and, if a given set +of lengths satisfy the triangle inequality, a triangle can be constructed +with those sidelengths. +• Triangles can be constructed when the following measurements are +given: +(a) +two of the sides and their included angle +(b) two angles and the included side +• The sum of the angles of a triangle is always 180°. +• An altitude of a triangle is a perpendicular line segment from a +vertex to its opposite side. +• Equilateral triangles have sides of equal length. Isosceles triangles +have two sides of equal length. Scalene triangles have sides of three +different lengths. +• Triangles are classified based on their angle measures as +acute-angled, right-angled and obtuse-angled triangles. +SUMMARY +Try +This +171 + +Ganita Prakash | Grade 7 +Shortest Path in a Box! +There is a spider in a corner of a box. It wants to reach the farthest +opposite corner (marked in the figure). Since it +cannot fly, it can reach the opposite point only by +walking on the surfaces of the box. What is the +shortest path it can take? +Take a cardboard box and mark the path that +you think is the shortest from one corner to its +opposite corner. Compare the length of this path with that of the +paths made by your friends. +Hint: +172 + +1 + + +Page No. 150 + Figure it out +1. Use the points on the circle and/or the centre to form isosceles +triangles. + +Ans: One Such diagram is +Try more + +Page No. 151 +2. Use the points on the circles and/or their centres to form isosceles +and equilateral triangles. The circles are of the same size. + + + +Ans: +Take points C, C1, C2, C3 … on the circle with centre B. The triangles ABC, +AC1B, AC2B, … are isosceles triangle. Similarly for points D1, D2, D3 … ��ADB, +∆AD1B, ∆AD2B, … are isosceles triangles. +∆ABC and ∆ADB are equilateral triangle. +Similarly it can be done for the other figure. + + +Chapter – 7 +A Tale of Three Intersecting Lines + +2 + +Page No. 154 + Figure it out +1. We checked by construction that there are no triangles having +sidelengths 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if you +could have found this without trying to construct the triangle. +Ans: Here sum of the two sides (i.e 3 cm and 4 cm) is less than the third side +(i.e 8 cm). Similarly for sides of lengths 2cm, 3cm and 6cm, 2 + 3 < 6. So, in +both cases triangles are not possible. +2. Can we say anything about the existence of a triangle for each of the +following sets of lengths? +Ans: +(a) 10 km, 10 km and 25 km +Ans: A triangle does not exist + +(b) 5 mm, 10 mm and 20 mm +Ans: A triangle does not exist + +(c) 12 cm, 20 cm and 40 cm +Ans: A triangle does not exist + +3. Will this always happen? That is, for any set of lengths, will there be +at least two comparisons where the direct length is less than the sum +of the other two? Explore for different sets of lengths. +Ans: Some sets are 7, 10, 15; 12, 14, 18 + +Page No. 156 + Figure it out +1. Which of the following lengths can be the sidelengths of a triangle? +Explain your answers. Note that for each set, the three lengths have +the same unit of measure. +Ans: +(a) 2, 2, 5 +Ans: Since 2 + 2 < 5. +So, these are not the side lengths of a triangle. + +3 + +(b) 3, 4, 6 +Ans: Since 3 < 4 + 6, 4 < 6 + 3 and 6 < 3 + 4. +These are the side lengths of a triangle. +(c) 2, 4, 8 +Ans: 8 > 2 + 4. +So, these are not the side lengths of a triangle. +(d) 5, 5, 8 +Ans: Since 5 < 5 + 8, 8 < 5 + 5. +These are the side lengths of a triangle. +(e) 10, 20, 25 +Ans: Since 10 < 20 + 25, 20 < 25 + 10 and 25 < 10 + 20. +These are the side lengths of a triangle. +(f) 10, 20, 35 +Ans: 35 > 10 + 20. +So, these are not the side lengths of a triangle. +(g) 24, 26, 28 +Ans: Since 24 < 26 + 28, 26 < 28 + 24 and 28 < 24 + 26. +These are the side lengths of a triangle. + +Page No. 159 + Figure it out +1. Check if a triangle exists for each of the following set of lengths: +(a) 1, 100, 100 +Ans: A triangle exists. +(b) 3, 6, 9 +Ans: A triangle cannot exists. +(c) 1, 1, 5 +Ans: A triangle cannot be constructed. +(d) 5, 10, 12 +Ans: A triangle exists. + +4 + +2. Does there exist an equilateral triangle with sides 50, 50, 50? In +general, does there exist an equilateral triangle of any sidelength? +Justify your answer. +Ans: There exists an equilateral triangle with sides 50, 50, 50 because each +length < sum of the other two lengths. There exists an equilateral +triangle of any side length. +3. For each of the following, give at least 5 possible values for the third +length so there exists a triangle having these as sidelengths (decimal +values could also be chosen): +(a) 1, 100 +Ans: 99.5, 100, 100.7, 100.5, 99.4. +(b) 5, 5 +Ans: 5, 4, 3, 4.9, 1 +(c) 3, 7 +Ans: 5, 8, 7, 6.4, 6 +Try some more lengths. +Page No. 161 + We have seen that triangles do not exist for all sets of sidelengths. Is +there a combination of measurements in the case of two sides and the +included angle where a triangle is not possible? Justify your answer +using what you observe during construction. +Ans: If the included angle is greater than or equal to 180°, then a triangle is +not possible. +For example– +(a) 4cm, 210°, 6cm +(b) 3cm, 180°, 7cm + + + + +5 + +Page No. 163 + Figure it out +1. For each of the following angles, find another angle for which a +triangle is (a) possible, (b) not possible. Find at least two different +angles for each category: + +Triangle is possible +Triangle is not possible +(a) +30° +Another angle should be less than +150°. It could be 50° or 90°. +Find the third angle in each case. +Another angle should be +greater than or equal to 150°. +It could be 150° or 170°. +(b) +70° +Another angle should be less than +110°. It could be 60° or 80°. +Find the third angle in each case. +Another angle should be +greater than or equal to 110°. +It could be 140° or 120°. +(c) +54° +Another angle should be less than +126°. It could be 90° or 64°. +Find the third angle in each case. +Another angle should be +greater than or equal to 126°. +It could be 134° or 154°. +(d) +144° +Another angle should be less than +36°. It could be 20° or 35°. +Find the third angle in each case. +Another angle should be +greater than or equal to 36°. +It could be 36° or 90°. +2. Determine which of the following pairs can be the angles of a +triangle and which cannot: +(a) 35°, 150° +Ans: This pair cannot form angles of a triangle. +(b) 70°, 30° +Ans: This pair of angles can be the angles of a triangle. + (c) 90°, 85° +Ans: This pair of angles can be the angles of a triangle. +(d) 50°, 150° +Ans: This pair cannot form angles of a triangle. + + +6 + + Like the triangle inequality, can you form a rule that describes the +two angles for which a triangle is possible? +Ans: Rules for two angles A and B to form a triangle is 0° < A + B < 180°. + +Page No. 165 + Figure it out +1. Find the third angle of a triangle (using a parallel line) when two of +the angles are: +Process for (a) is given. Try for rest. +(a) 36°, 72° +Ans: Consider ∆ABC, Draw line XY parallel to line BC so AB and AC become +transversals. + Therefore XAB = B = 36° and YAC = C = 72° (Alternate Angles) +On line XY — +XAB + BAC + YAC = 180° +So, BAC = 72° + +(b) 150°, 15° +Ans: 15° +(c) 90°, 30° +Ans: 60° +(d) 75°, 45° +Ans: 60° + + +7 + + +2. Can you construct a triangle all of whose angles are equal to 70°? If +two of the angles are 70° what would the third angle be? If all the +angles in a triangle have to be equal, then what must its measure +be? Explore and find out. + +Ans: +• No, such triangle is not possible. +• 40° +• 60° +3. Here is a triangle in which we know, +B = C and A = 50°. Can you find B and C? + + +Ans: B = C = 65°" +class_7,8,Working with Fractions,ncert_books/class_7/gegp1dd/gegp108.pdf,"WORKING WITH +FRACTIONS +8 +8.1 Multiplication of Fractions +Aaron walks 3 kilometres in 1 hour. +How far can he walk in 5 hours? +This is a simple question. We know +that to find the distance, we need to +find the product of 5 and 3, i.e., we +multiply 5 and 3. +Distance covered in 1 hour = 3 km. +Therefore, +Distance covered in 5 hours += 5 × 3 km += 3 + 3 + 3 + 3 + 3 km += 15 km. +Aaron’s pet tortoise walks at a much slower pace. It can walk only 1 +4 +kilometre in 1 hour. How far can it walk in 3 hours? +Here, the distance covered in an hour is a fraction. This does not +matter. The total distance covered is calculated in the same way, as +multiplication. +Distance covered in 1 hour = 1 +4 km. +Distance in +1 hour +Distance in 3 hour +1 +4 km +0 + +Ganita Prakash | Grade 7 +Therefore, distance covered in 3 hours = 3 × 1 +4 km += 1 +4 + 1 +4 + 1 +4 km += 3 +4 km. +The tortoise can walk 3 +4 km in 3 hours. +Let us consider a case where the time spent walking is a fraction of +an hour. +We saw that Aaron can walk 3 kilometres in 1 hour. How far can he +walk in 1 +5 hours? +We continue to calculate the total distance covered through +multiplication. +Distance covered in 1 +5 hours = 1 +5 × 3 km. +Finding the product: +Distance covered in 1 hour = 3 km. +In 1 +5 hours, distance covered is equal to the length we get by dividing +3 km into 5 equal parts, which is 3 +5 km. +This tells us that 1 +5 × 3 = 3 +5 . +How far can Aaron walk in 2 +5 hours? +Distance in 1 hour +Distance in 1 +5 +hour +0 +3 km +174 + +Working with Fractions +Once again, we have — +Distance covered = 2 +5 × 3 km. +Finding the product: +1. We can first find the distance covered in 1 +5 hours. +2. Since, the duration 2 +5 is twice 1 +5, we multiply this distance by 2 to +get the total distance covered. +Here is the calculation. +Distance covered in 1 hour = 3 km. +1. Distance covered in 1 +5 hour + += The length we get by dividing 3 km in 5 equal parts + + + += 3 +5 km. +2. Multiplying this distance by 2, we get + +2 × 3 +5 = 6 +5 km. +From this we can see that +2 +5 × 3 = 6 +5. +Discussion +We did this multiplication as follows: +• First, we divided the +multiplicand, 3 , by the +denominator of the multiplier, 5, to get 3 +5. +Distance in 1 hour +Distance in +1 +5 hour +Distance in 2 +5 hour +Multiplier +Multiplicand +2 +5 + × 3 +175 + +Ganita Prakash | Grade 7 +• We then multiplied the result by the numerator of the multiplier, +that is 2, to get 6 +5. +Thus, +whenever +we +need +to +multiply a fraction and a whole +number, we follow the steps above. +Example +1: +A +farmer +had +5 +grandchildren. She distributed 2 +3 acre +of land to each of her grandchildren. +How much land in all did she give to her grandchildren? +5 × 2 +3 = 2 +3 + 2 +3 + 2 +3 + 2 +3 + 2 +3 = 10 +3 . +Example 2: 1 hour of internet time costs ₹8. How much will 1 1 +4 hours +of internet time cost? +1 1 +4 hours is 5 +4 hours (converting from a mixed fraction). +Cost of 5 +4 hour of internet time = 5 +4 × 8 += 5 × 8 +4 += 5 × 2 += 10. +It costs ₹10 for 1 1 +4 hours of internet time. +Figure it Out +1. Tenzin drinks 1 +2 glass of milk every day. How many glasses of milk +does he drink in a week? How many glasses of milk did he drink in +the month of January? +2. A team of workers can make 1 km of a water canal in 8 days. So, in +one day, the team can make ___ km of the water canal. If they work +5 days a week, they can make ___ km of the water canal in a week. +3. Manju and two of her neighbours buy 5 litres of oil every week +and share it equally among the 3 families. How much oil does each +family get in a week? How much oil will one family get in 4 weeks? +4. Safia saw the Moon setting on Monday at 10 pm. Her mother, who is +a scientist, told her that every day the Moon sets 5 +6 hour later than +Multiplier +Multiplicand +2 +5 × 3 +3 ÷ 5 = 3 +5 +2 × 3 +5 = 6 +5 +2 +5 +2 +5 +176 + +Working with Fractions +the previous day. How many hours after 10 pm will the moon set +on Thursday? +5. Multiply and then convert it into a mixed fraction: +(a) 7 × 3 +5 +(b) 4 × 1 +3 +(c) 9 +7 × 6 +(d) 13 +11 × 6 +So far, we have learnt multiplication of a whole number with a +fraction, and a fraction with a whole number. What happens when +both numbers in the multiplication are fractions? +Multiplying Two Fractions +We know, that Aaron’s pet tortoise can walk only 1 +4 km in 1 hour. How +far can it walk in half an hour? +Following +our +approach +of +using +multiplication to solve such problems, we +have, +Distance covered in 1 +2 hour = 1 +2 × 1 +4 km. +Finding the product: +Distance covered in 1 hour = 1 +4 km. +Therefore, the distance covered in 1 +2 an hour is the length we get by +dividing 1 +4 into 2 equal parts. +To find this, it is useful to represent fractions using the unit square +to stand for a “whole”. +Hour +Distance +1 +1 +4 +1 +2 +? +1 +4 +Unit square as a “whole” +1 +4 of the whole +177 + +Ganita Prakash | Grade 7 +Now we divide this 1 +4 into 2 equal parts. What do we get? +What fraction of the whole is shaded? +Since the whole is divided into 8 equal parts +and one of the parts is shaded, we can say that 1 +8 +of the whole is shaded. So, the distance covered +by the tortoise in half an hour is 1 +8 km. +This tells us that 1 +2 × 1 +4 = 1 +8. +If the tortoise walks faster and it can cover 2 +5 km in 1 hour, how far will +it walk in 3 +4 of an hour? +Distance covered = 3 +4 × 2 +5. +Finding the product: +(i) +First find the distance covered in 1 +4 of an hour. +(ii) +Multiply the result by 3, to get the distance covered in 3 +4 of an +hour. +(i) +Distance in km covered in 1 +4 of an hour + + = The quantity we get by dividing 2 +5 into +4 equal parts. +Taking the unit square as the whole, the +shaded part (in Fig.   8.1) is a region we get +when we divide 2 +5 into 4 equal parts. +How much of the whole is it? +The whole is divided into +5 rows and 4 columns, +creating 5 × 4 = 20 equal parts. +Number of these parts shaded = 2. +So, the distance covered in 1 +4 of an hour = 2 +20 . +Fig. 8.1 +2 +5 +whole +? +1 +4 divided into 2 equal parts +178 + +Working with Fractions +(ii) +Now, we need to multiply 2 +20 by 3. +Distance covered in 3 +4 of an hour = 3 × 2 +20 += 6 +20 . +So, 3 +4 × 2 +5 = 6 +20 = 3 +10. +Discussion +In the case of a fraction multiplied by +another fraction, we follow a method +similar to the one we used, when +we multiplied a fraction by a whole +number. We multiplied as follows: +2 +5 ÷ 4 = 2 +20 +Divide the multiplicand by 4. +3 +4 +3 × 2 +20 = 6 +20 =  Divide the multiplicand by 3. +3 +4 +Using this understanding, multiply 5 +4 × 3 +2. +First, let us represent 3 +2, taking the unit square as +the whole. Since, the fraction 3 +2 is one whole and a +half, it can be seen as follows: +Following the steps of multiplication, we need to +first divide this fraction 3 +2 into 4 equal parts. It can +be done as shown in the Fig. 8.2 with the yellow +shaded region representing the fraction obtained +by dividing 3 +2 into 4 equal parts. What is its value? +We see that the whole is divided into — +2 rows and 4 columns, +creating 2 × 4 = 8 equal parts. +Number of parts shaded = 3. +So the yellow shaded part = 3 +8. +Multiplier +Multiplicand +3 +4 × 2 +5 +Math +Talk +Fig. 8.2 +whole +3 +2 +179 + +Ganita Prakash | Grade 7 +Now, the next step is multiplying this result by 5. This gives the +product of 5 +4 and 3 +2: +5 +4 × 3 +2 = 5 × 3 +8 = 15 +8 . +Connection between the Area of a Rectangle and Fraction +Multiplication +In the Fig. 8.3, what is the length and breadth of the shaded rectangle? +Since we started with a unit square (of side 1 unit), the length and +breadth are 1 +2 unit and 1 +4 unit. +What is the area of this rectangle? We see +that 8 such rectangles give the square of area +1 square unit. So, the area of each rectangle +is 1 +8 square units. +Do you see any relation between the area and +the product of length and breadth? +The area of a rectangle of fractional sides equals the product +of its sides. +In general, if we want to find the product of two fractions, we can +find the area of the rectangle formed with the two fractions as its sides. +Figure it Out +1. Find the following products. Use a unit square as a whole for +representing the fractions: +(a) 1 +3 × 1 +5 +(b) 1 +4 × 1 +3 +(c) 1 +5 × 1 +2 +(d) 1 +6 × 1 +5 + +Now, find 1 +12 × 1 +18. +Fig. 8.3 +1 +8 +180 + +Working with Fractions +Doing this by representing the fractions using a unit square is +cumbersome. Let us find the product by observing what we did in the +above cases. +In each case, the whole is divided into rows and columns. +The number of rows is the denominator of the multiplicand, which +is 18 in this case. +The number of columns is the denominator +of the multiplier, which is 12 in this case. +Thus, the whole is divided into 18 × 12 equal +parts. +So, 1 +18 × 1 +12 = +1 +(18×12) = 1 +216. +Thus, +when +two +fractional +units +are +multiplied, their product is +1 +(product of denominators). +We express this as: +1 +b × 1 +d = +1 +b × d  . +2. Find the following products. Use a unit square as a whole for +representing the fractions and carrying out the operations. +(a) 2 +3 × 4 +5 +(b) 1 +4 × 2 +3 +(c) 3 +5 × 1 +2 +(d) +4 +6 × 3 +5 +Multiplying Numerators and Denominators +Now, find 5 +12 × 7 +18. +Like the previous case, let us find the product +by performing the multiplication, step by step. +First, the whole is divided into 18 rows and 12 +columns creating 12 × 18 equal parts. +The value we get by dividing 7 +18 into 12 equal +parts is +7 +(12 × 18) . +1 +18 +12 parts +18 parts +7 +18 +18 parts +12 parts +181 + +Ganita Prakash | Grade 7 +Then, we multiply this result by 5 to get +the product. This is (5 × 7) +(12 × 18). +So, 5 +12 × 7 +8 = (5 × 7) +(12 × 18) = 35 +216. +From this we can see that, in general, +a +b × c +d = a × c +b × d . +This formula was first stated in this general form by Brahmagupta in +his Brāhmasphuṭasiddhānta in 628 CE. +The formula above works even when the multiplier or multiplicand +is a whole number. We can simply rewrite the whole number as a +fraction with denominator 1. For example, +3 × 3 +4 can be written 3 +1 × 3 +4 += 3×3 +1×4 = 9 +4. +And, +3 +5 × 4 can be written 3 +5 × 4 +1 += 3 × 4 +5 × 1 = 12 +5 . +Multiplication of Fractions — Simplifying to Lowest Form +Multiply the following fractions and express the product in its lowest +form: +12 +7 × 5 +24 +Instead of multiplying the numerators (12 and 5) and denominators +(7 and 24) first and then simplifying, we could do the following: +12 +7 × 5 +24 = 12 × 5 +7 × 24 +We see that both the circled numbers have a common factor of 12. +We know that a fraction remains the same when the numerator and +denominator are divided by the common factor. In this case, we can +divide them by 12. +7 +18 +18 parts +12 parts +5 +182 + +Working with Fractions +12 × 5 +7 × 24 = 1 × 5 +7 × 2 = 5 +14 +1 +2 + +. +Let us use the same technique to do one more multiplication. +14 +15 × 25 +42 +14 × 25 +15 × 42 = 1 × 5 +3 × 3 = 5 +9 . +1 +3 +3 +5 +When multiplying fractions, we can first divide the numerator +and denominator by their common factors before multiplying the +numerators and denominators. This is called cancelling the common +factors. +A Pinch of History +In India, the process of reducing a fraction to its lowest terms — known +as apavartana — is so well known that it finds mention even in a +non-mathematical work. A Jaina scholar Umasvati (c. 150 CE) used it as +a simile in a philosophical work. +Figure it Out +1. A water tank is filled from a tap. If the tap is open for 1 hour, 7 +10 of +the tank gets filled. How much of the tank is filled if the tap is open +for +(a) 1 +3 hour ____________ +(b) 2 +3 hour ____________ +(c) 3 +4 hour ____________ +(d) 7 +10 hour ____________ +(e) For the tank to be full, how long should the tap be running? +2. The government has taken 1 +6 of Somu’s land to build a road. +What part of the land remains with Somu now? She gives half of +Try +This +183 + +Ganita Prakash | Grade 7 +the remaining part of the land to her daughter Krishna and 1 +3 of +it to her son Bora. After giving them their shares, she keeps the +remaining land for herself. +(a) What part of the original land did Krishna get? +(b) What part of the original land did Bora get? +(c) What part of the original land did Somu keep for herself? +3. Find the area of a rectangle of sides 3 3 +4 ft and 9 3 +5 ft. +4. Tsewang plants four saplings in a row in his garden. The distance +between two saplings is 3 +4 m. Find the distance between the first +and last sapling. [Hint: Draw a rough diagram with four saplings +with distance between two saplings as 3 +4 m] +5. Which is heavier: 12 +15 of 500 grams or 3 +20 of 4 kg? +Is the Product Always Greater than the Numbers +Multiplied? +Since, we know that when a number is multiplied by 1, the product +remains unchanged, we will look at multiplying pairs of numbers +where neither of them is 1. +When we multiply two counting numbers greater than 1, say 3 and +5, the product is greater than both the numbers being multiplied. +3 × 5 = 15 +The product, 15, is more than both 3 and 5. +But what happens when we multiply 1 +4 and 8? +1 +4 × 8 = 2 +In the above multiplication the product, 2, is greater than 1 +4, but less +than 8. +What happens when we multiply 3 +4 and 2 +5 ? +3 +4 × 2 +5 = 6 +20 +Let us compare this product 6 +20 with the numbers 3 +4 and 2 +5. For this, +184 + +Working with Fractions +let us express 3 +4 as 15 +20 and 2 +5 as 8 +20. +From this we can see that the product is less than both the numbers. +When do you think the product is greater than both the numbers +multiplied, when is it in between the two numbers, and when is it +smaller than both? +[Hint: The relationship between the product and the numbers +multiplied depends on whether they are between 0 and 1 or they +are greater than 1. Take different pairs of numbers and observe their +product. For each multiplication, consider the following questions.] +Situation +Multiplication +Relationship +Situation 1 +Both numbers are +greater than 1 +(e.g., 4 +3 × 4) +The product (16 +3 ) is +greater than both the +numbers +Situation 2 +Both numbers are +between 0 and 1 +(e.g., 3 +4 × 2 +5) +The product ( 3 +10) is +less than both the +numbers +Situation 3 +One number is +between 0 and 1, +and one number is +greater than 1 +(e.g., 3 +4 × 5) +The product (15 +4 ) is +less than the number +greater than 1 and +greater than the +number between 0 +and 1 +Create more such examples for each situation and observe the +relationship between the product and the numbers being multiplied. +What can you conclude about the relationship between the numbers +multiplied and the product? Fill in the blanks: +• When one of the numbers being multiplied is between 0 and 1, +the product is ____________ (greater/less) than the other number. +• When one of the numbers being multiplied is greater than 1, the +product is _____________ (greater/less) than the other number. +185 + +Ganita Prakash | Grade 7 +Order of Multiplication +We know that 1 +2 × 1 +4 = 1 +8. +Now, what is 1 +4 × 1 +2 ? +That is 1 +8 too. +In general, note that the area of a rectangle remains the same even if +the length and breadth are interchanged. +The order of multiplication does not matter. Thus, +a +b × c +d = c +d × a +b. +This can also be seen from Brahmagupta’s formula for multiplying +fractions. +8.2 Division of Fractions +What is 12 ÷ 4? You know this already. +But can this problem be restated as a +multiplication problem? +What should be multiplied by 4 to get +12? That is, +4 × ? = 12 +1 +2 +1 +4 +1 +8 +Dividend +Divisor +Quotient +12 ÷ 4 = 3 +1 +4 +1 +2 +1 +8 +186 + +Working with Fractions +We can use this technique of converting division into multiplication +problems to divide fractions. +What is 1 ÷ 2 +3 ? +Let us rewrite this as a multiplication problem +2 +3 × ? = 1 +What should be multiplied by 2 +3 to get the product 1? +If we somehow cancel out the 2 and the 3, we are left with 1. +So, +1 ÷ 2 +3 = 3 +2. +Let us try another problem: +3 ÷ 2 +3. +This is the same as +2 +3 × ? = 3. +Can you find the answer? +We know what to multiply 2 +3 by to get 1. We just need to multiply that +by 3 to get 3. So, +So, +3 ÷ 2 +3 = 3 +2 × 3 = 9 +2. +What is 1 +5 ÷ 1 +2 ? +Rewriting it as a multiplication problem, we have +1 +2 × ? = 1 +5. +2 +3 × 3 +2 = 1 +Answer +2 +3 × 3 +2 × 3 = 3 +Answer +187 + +Ganita Prakash | Grade 7 +How do we solve this? +So, +1 +5 ÷ 1 +2 = 2 × 1 +5 = 2 +5. +What is 2 +3 ÷ 3 +5 ? +Rewriting this as multiplication, we have +3 +5 × ? = 2 +3. +How will we solve this? +So, +2 +3 ÷ 3 +5 = 5 +3 × 2 +3 = 10 +9 . +Discussion +In each of the division problems above, observe how we found the +answer. Can we frame a rule that tells us how to divide two fractions? +Let us consider the previous problem. +In every division problem we have +a dividend, divisor and quotient. The +technique we have been using to get +the quotient is: +1. First, find the number which gives +1 when multiplied by the divisor. +We see that the resulting number is +a fraction whose numerator is the +divisor’s denominator and denominator is the divisor’s numerator. + +For the divisor 3 +5 this fraction is 5 +3. We call 5 +3 the reciprocal of 3 +5. +When we multiply a fraction by its reciprocal, we get 1. So, the first +step in our technique is to find the divisor’s reciprocal. +1 +2 × 2 × 1 +5 = 1 +5 +Answer +3 +5 × 5 +3 × 2 +3 = 2 +3 +Answer +2 +3 ÷ 3 +5 +Divisor +Quotient +Dividend += 5 +3 × 2 +3 = 10 +9 +188 + +Working with Fractions +2. We then multiply the dividend with this reciprocal to get the +quotient. + +Summarising, to divide two fractions: +• Find the reciprocal of the divisor +• Multiply this by the dividend to get the quotient. + So, + a +b ÷ c +d = d +c × a +b = d × a +c × b. +This can be rewritten as: +a +b ÷ c +d = a +b × d +c = a × d +b × c. +As with methods and formulas for addition, subtraction, and +multiplication of fractions that you learnt earlier, this method and +formula for division of fractions, in this general form, was first explicitly +stated by Brahmagupta in his Brāhmasphuṭasiddhānta (628 CE). +So, to evaluate, for example, 2 +3 ÷ 3 +5 using Brahmagupta’s formula +above, we write: +2 +3 ÷ 3 +5 = 2 +3 × 5 +3 = 2 × 5 +3 × 3 = 10 +9 . +Dividend, Divisor and the Quotient +When we divide two whole numbers, say 6 ÷ 3, we get the quotient 2. +Here the quotient is less than the dividend. +6 ÷ 3 = 2, 2 < 6 +But what happens when we divide 6 by 1 +4 ? +6 ÷ 1 +4 = 24. +Here the quotient is greater than the dividend! +What happens when we divide 1 +8 by 1 +4 ? +1 +8 ÷ 1 +4 = 1 +2. +Here too the quotient is greater than the dividend. +When do you think the quotient is less than the dividend and when +is it greater than the dividend? +Is there a similar relationship between the divisor and the quotient? +189 + +Ganita Prakash | Grade 7 +Use your understanding of such relationships in multiplication to +answer the questions above. +8.3 Some Problems Involving Fractions +Example 3: Leena made 5 cups of tea. She used 1 +4 litre of milk for this. +How much milk is there in each cup of tea? +Leena used 1 +4 litres of milk in 5 cups of tea. So, in 1 cup of tea the +volume of milk should be: +1 +4 ÷ 5. +Writing this as multiplication, we have: +5 × (milk per cup) = 1 +4. +We perform the division as follows as per Brahmagupta’s method: +The reciprocal of 5 (the divisor) is 1 +5. +Multiplying this reciprocal by the dividend ( 1 +4 ), we get +1 +5 × 1 +4 = 1 +20. +So, each cup of tea has 1 +20 litre of milk. +Example 4: Some of the oldest examples of working with non-unit +fractions occur in humanity’s oldest geometry texts, the Śhulbasūtra. +Here is an example from Baudhāyana’s Śhulbasūtra (c. 800 BCE). +Cover an area of 7 1 +2 square units with square bricks each of whose +sides is 1 +5 units. +1 +4 litres of 5 cups +Milk in +1 cup +190 + +Working with Fractions +How many such square bricks are needed? +Each square brick has an area of 1 +5 × 1 +5 = 1 +25 square units. +The total area to be covered is 7 1 +2 sq. units = 15 +2 sq. units. +As (Number of bricks) × (Area of a brick) = Total Area, +Number of bricks = 15 +2 ÷ 1 +25. +The reciprocal of the divisor is 25. +Multiplying the reciprocal by the dividend, we get +25 × 15 +2 = 25 × 15 +2 + = 375 +2 . +Example 5: This problem was posed by Chaturveda Pṛithūdakasvāmī +(c. +860 +CE) +in +his +commentary +on +Brahmagupta’s +book +Brāhmasphuṭasiddhānta. +Four fountains fill a cistern. The first fountain can fill the cistern in a +day. The second can fill it in half a day. The third can fill it in a quarter of +a day. The fourth can fill the cistern in one fifth of a day. If they all flow +together, in how much time will they fill the cistern? +Let us solve this problem step by step. +In a day, the number of times — +• the first fountain will fill the cistern is 1÷ 1 = 1 +• the second fountain will fill the cistern is 1 ÷ 1 +2 = _____ +• the third fountain will fill the cistern is 1 ÷ 1 +4= _____ +• the fourth fountain will fill the cistern is 1 ÷ 1 +5 = _____ +The number of times the four fountains together will fill the cistern +in a day is ___ + ____ + ___ + ____ = 12. +Thus, the total time needed by the four fountains to fill the cistern +together is 1 +12 days. +191 + +Ganita Prakash | Grade 7 +Fractional Relations +Here is a square with some lines drawn inside. +Fig. 8.4 +What fraction of area of the whole square does the shaded region +occupy? +There are different ways to solve this problem. Here is one of them: +Let the area of the whole square be 1 square unit. +We can see that the top right square (in Fig. 8.5), occupies 1 +4 of the +area of the whole square. +Fig. 8.5 +Area of red square = 1 +4 square units. +Math +Talk +Fig, 8.6 +192 + +Working with Fractions +Let us look at this red square. The area of the triangle inside it +(coloured yellow) is half the area of the red square. So, +the area of the yellow triangle = 1 +2 × 1 +4 = 1 +8 square units. +What fraction of this yellow triangle is shaded? +The shaded region occupies 3 +4 of the area of the +yellow triangle. Are you able to see why? +The area of shaded part = 3 +4 × 1 +8 = 3 +32 square units. +Thus, the shaded region occupies 3 +32 of the area of the whole square. +In each of the figures given below, find the fraction of the big square +that the shaded region occupies. +We will solve more interesting problems of this kind in a later +chapter. +A Dramma-tic Donation +The +following +problem +is +translated +from +Bhāskarāchārya’s +(Bhāskara II’s) book, Līlāvatī, written in 1150 CE. +“O wise one! A miser gave to a beggar 1 +5 of 1 +16 of 1 +4 of 1 +2 of 2 +3 of 3 +4 of +a dramma. If you know the mathematics of fractions well, tell me O +child, how many cowrie shells were given by the miser to the beggar.” +Dramma refers to a silver coin used in those times. The tale says that +1 dramma was equivalent to 1280 cowrie shells. Let’s see what fraction +of a dramma the person gave: +( +1 +2 × 2 +3 × 3 +4 × 1 +5 × 1 +16 × 1 +4) +th + part of a dramma. +193 + +Ganita Prakash | Grade 7 +Evaluating it gives +6 +7680 . +Upon simplifying to its lowest form, we get +6 +7680 = +1 +1280. +So, one cowrie shell was given to the beggar. +You can see in the answer Bhāskarāchārya’s humour! The miser had +given the beggar only one coin of the least value (cowrie). +Around the 12th century, several types of coins were in use in +different kingdoms of the Indian subcontinent. Most commonly used +were gold coins (called dinars/gadyanas and hunas), silver coins (called +drammas/tankas), copper coins (called kasus/panas and mashakas), and +cowrie shells. The exact conversion rates between these coins varied +depending on the region, time period, economic conditions, weights of +coins and their purity. +Gold coins had high-value and were used in large transactions and +to store wealth. Silver coins were more commonly used in everyday +transactions. Copper coins had low-value and were used in smaller +transactions. Cowrie shells were the lowest denomination and were +used in very small transactions and as change. +If we assume 1 gold dinar = 12 silver drammas, 1 silver dramma = 4 +copper panas, 1 copper pana = 6 mashakas, and 1 pana = 30 cowrie shells, +1 copper pana = 1 +48 gold dinar ( +1 +12 × 1 +4) +1 cowrie shell =____ copper panas +1 cowrie shell =____ gold dinar. +A Pinch of History  +As you have seen, fractions are an important type of number, playing +a critical role in a variety of everyday problems that involve sharing +and dividing quantities equally. The general notion of non-unit fractions +as we use them today — equipped with the arithmetic operations of +addition, subtraction, multiplication, and division — developed largely in +India. The ancient Indian geometry texts called the Śhulbasūtra — which +go back as far as 800 BCE, and were concerned with the construction +of fire altars for rituals — used general non-unit fractions extensively, +including performing division of such fractions as we saw in Example 3. +Fractions even became commonplace in the popular culture of +India as far back as 150 BCE, as evidenced by an offhand reference to +the reduction of fractions to lowest terms in the philosophical work of +the revered Jain scholar Umasvati. +194 + +Working with Fractions +General rules for performing arithmetic operations on fractions — +in essentially the modern form in which we carry them out today — +were first codified by Brahmagupta in his Brāhmasphuṭasiddhānta in +628 CE. We have already seen his methods for adding and subtracting +general fractions. For multiplying general fractions, Brahmagupta +wrote: +“Multiplication of two or more fractions is obtained by taking the +product of the numerators divided by the product of the denominators.” +(Brāhmasphuṭasiddhānta, Verse 12.1.3) +That is, + + + +a +b × c +d = a × c +b × d . +For division of general fractions, Brahmagupta wrote: +“The division of fractions is performed by interchanging the numerator +and denominator of the divisor; the numerator of the dividend is then +multiplied by the (new) numerator, and the denominator by the (new) +denominator.” +Bhāskara II in his book Līlāvatī in 1150 CE clarifies Brahmagupta‘s +statement further in terms of the notion of reciprocal: +“Division of one fraction by another is equivalent to multiplication of +the first fraction by the reciprocal of the second.” (Līlāvatī, Verse 2.3.40) +Both of these verses are equivalent to the formula: +a +b ÷ c +d = a +b ×d +c = a × d +b × c . +Bhāskara I, in his 629 CE commentary Āryabhaṭīyabhāṣhya on +Aryabhata’s 499 CE work, described the +geometric interpretation of multiplication +of fractions (that we saw earlier) in terms +of the division of a square into rectangles +via equal divisions along the length and +breadth. +Many +other +Indian +mathematicians, +such as Śhrīdharāchārya (c. 750 CE), +Mahāvīrāchārya (c. 850 CE), Caturveda +Pṛithūdakasvāmī (c. 860 CE), and Bhāskara II +(c. 1150 CE) developed the usage of arithmetic +of fractions significantly further. +The Indian theory of fractions and +arithmetic operations on them was +transmitted to, and its usage developed further, by Arab and African +mathematicians such as al-Hassâr (c. 1192 CE) of Morocco. The theory +was then transmitted to Europe via the Arabs over the next few +Bhāskara I’s visual explanation that +1 +5 x 1 +4 = 1 +20 + +1 +20 +1 +5 +1 +4 +195 + +Ganita Prakash | Grade 7 +centuries, and came into general use in Europe in only around the +17th century, after which it spread worldwide. The theory is indeed +indispensable today in modern mathematics. +Figure it Out +1. Evaluate the following: +2. For each of the questions below, choose the expression that +describes the solution. Then simplify it. +(a) Maria bought 8 m of lace to decorate the bags she made for +school. She used 1 +4 m for each bag and finished the lace. How +many bags did she decorate? +(i) +8 × 1 +4 +(ii) +1 +8 × 1 +4 +(iii) 8 ÷ 1 +4 +(iv) +1 +4 ÷ 8 +(b) 1 +2 meter of ribbon is used to make 8 badges. What is the +length of the ribbon used for each badge? +(i) +8 × 1 +2 +(ii) +1 +2 ÷ 1 +8 +(iii) 8 ÷ 1 +2 +(iv) +1 +2 ÷ 8 +(c) A baker needs 1 +6 kg of flour to make one loaf of bread. He has +5 kg of flour. How many loaves of bread can he make? +(i) +5 × 1 +6 +(ii) +1 +6 ÷ 5 +(iii) 5 ÷ 1 +6 +(iv) +5 × 6 +3 ÷ 7 +9 +14 +4 ÷ 2 +2 +3÷ 2 +3 +14 +6 ÷ 7 +3 +4 +3 ÷ 3 +4 +7 +4 ÷ 1 +7 +8 +2 ÷ 4 +15 +1 +5 ÷ 1 +9 +1 +6 ÷ 11 +12 +3 2 +3 ÷ 1 3 +8 +196 + +Working with Fractions +3. If 1 +4 kg of flour is used to make 12 rotis, how much flour is used to +make 6 rotis? +4. Pāṭīgaṇita, a book written by Sridharacharya in the 9th century +CE, mentions this problem: “Friend, after thinking, what sum will +be obtained by adding together 1 ÷ 1 +6 , 1 ÷ 1 +10 , 1 ÷ 1 +13 , 1 ÷ 1 +9 , and +1 ÷ 1 +2 +”. What should the friend say? +5. Mira is reading a novel that has 400 pages. She read 1 +5 of the pages +yesterday and 3 +10 of the pages today. How many more pages does +she need to read to finish the novel? +6. A car runs 16 km using 1 litre of petrol. How far will it go using 2 3 +4 +litres of petrol? +7. Amritpal decides on a destination for his vacation. If he takes a +train, it will take him 5 1 +6 hours to get there. If he takes a plane, it +will take him 1 +2 hour. How many hours does the plane save? +8. Mariam’s grandmother baked a cake. Mariam and her cousins +finished 4 +5 of the cake. The remaining cake was shared equally by +Mariam’s three friends. How much of the cake did each friend get? +9. Choose the option(s) describing the product of (  +565 +465 × 707 +676 ): +(a) > 565 +465 +(b) +< 565 +465 +(c) > 707 +676 +(d) +< 707 +676 +(e) > 1 +(f) +< 1 +10. What fraction of the whole square is shaded? +197 + +Ganita Prakash | Grade 7 +11. A colony of ants set out in search of food. + +As they search, they keep splitting equally +at each point (as shown in the Fig. 8.7) +and reach two food sources, one near a +mango tree and another near a sugarcane +field. What fraction of the original group +reached each food source? +12. What is 1 – 1 +2 ? + +(1 – 1 +2) × (1 – 1 +3) ? + +(1 – 1 +2) × (1 – 1 +3) × (1 – 1 +4) × (1 – 1 +5) ? + +(1 – 1 +2) × (1 – 1 +3) × (1 – 1 +4) × (1 – 1 +5) × (1 – 1 +6) × (1 – 1 +7) × (1 – 1 +8) × (1 – 1 +9) × (1 – 1 +10) ? + +Make a general statement and explain. +• Brahmagupta’s formula for multiplication of fractions: + +a +b × c +d = a × c +b × d . +• When multiplying fractions, if the numerators and denominators +have some common factors, we can cancel them first before +multiplying the numerators and denominators. +• In multiplication — when one of the numbers being multiplied is +between 0 and 1, the product is less than the other number. If one of +the numbers being multiplied is greater than 1, then the product is +greater than the other number. +• The reciprocal of a fraction a +b is b +a. When we multiply a fraction by its +reciprocal, the product is 1. +• Brahmagupta’s formula for division of fractions: +a +b ÷ c +d = a +b × d +c = a × d +b × c. +• In division — when the divisor is between 0 and 1, the quotient is +greater than the dividend. When the divisor is greater than 1, the +quotient is less than the dividend. +SUMMARY +Fig. 8.7 +198 + +Working with Fractions +Chess Puzzles —  +Non-attacking Queens +Chess is a popular 2-player strategy game. This game has its origins +in India. It is played on an 8 × 8 chequered grid. There are 2 sets of +pieces — black and white — one set for each player. Find out how +each piece should move and the rules of the game. +Here is a famous chess-based puzzle. From its current position, +a Queen piece can move along the horizontal, vertical or diagonal. +Place 4 Queens such that no 2 queens attack each other. For example, +the arrangement below is not valid as the queens are in the line of +attack of each other. +Now, place 8 queens on this 8 × 8 grid so that no 2 queens attack +each other! +199 + +Learning Material Sheets + + +Note + + +Page No. 176 + Figure it out +1. Tenzin drinks +𝟏 +𝟐 glass of milk every day. How many glasses of milk +does he drink in a week? How many glasses of milk did he drink in +the month of January? +Ans: Quantity of milk that Tenzin drinks in a week = +7 +2 glasses. +Quantity of milk Tenzin drinks in the month of January which has 31 +days = +31 +2 glasses. +2. A team of workers can make 1 km of a water canal in 8 days. So, in one +day, the team can make ___ km of the water canal. If they work 5 days +a week, they can make ___ km of the water canal in a week. +Ans: +Length of water canal made in 1 day = +1 +8 km. +Length of water canal made in 5 days = +5 +8 km. +3. Manju and two of her neighbours buy 5 litres of oil every week and +share it equally among the 3 families. How much oil does each family +get in a week? How much oil will one family get in 4 weeks? +Ans: +Oil each family gets in 1 week = +5 +3 litres. +Oil each family gets in 4 weeks = +20 +3 litres. +4. Safia saw the Moon setting on Monday at 10 pm. Her mother, who is a +scientist, told her that every day the Moon sets +𝟓 +𝟔 hour later than the +previous day. How many hours after 10 pm will the moon set on +Thursday? +Ans: +Time at which the moon sets on Thursday (i.e. 3 days after Monday) += 10 + 2 + +1 +2 = 12 +1 +2 am Or 12:30 am + + +CHAPTER – 8 +Working with Fractions + +Page No. 177 +5. Multiply and then convert it into a mixed fraction: +(a) 7 × +𝟑 +𝟓 +Ans: 7 × +3 +5 = +21 +5 = 4 +1 +5 + +(b) 4 × +𝟏 +𝟑 +Ans: 4 × +1 +3 = +4 +3 = 1 +1 +3 + +(c) +𝟗 +𝟕 × 6 +Ans: +9 +7 × 6 = +54 +7 = 7 +5 +7 + +(d) +𝟏𝟑 +𝟏𝟏 × 6 +Ans: +13 +11 × 6 = +78 +11 = 7 +1 +11 +Page No. 180 + Figure it out +1. Find the following products. Use a unit square as a whole for +representing the fractions: +(a) +𝟏 +𝟑 × +𝟏 +𝟓 + +Ans: +1 +15 + + +(b) +𝟏 +𝟒 × +𝟏 +𝟑 + +Ans: +1 +12 + + +(c) +𝟏 +𝟓 × +𝟏 +𝟐 + +Ans: +1 +10 + +(d) +𝟏 +𝟔 × +𝟏 +𝟓 + +Ans: +1 +30 + +Page No. 181 +2. Find the following products. Use a unit square as a whole for +representing the fractions and carrying out the operations. +(a) +𝟐 +𝟑 × +𝟒 +𝟓 +Ans: +8 +15 +(Draw the figure in the same manner as shown in question 1) +(b) +𝟏 +𝟒 × +𝟐 +𝟑 +Ans: +2 +12 = +1 +6 +(c) +𝟑 +𝟓 × +𝟏 +𝟐 +Ans: +3 +10 +(d) +𝟒 +𝟔 × +𝟑 +𝟓 +Ans: +12 +30 = +2 +5 +Page No. 183 + Figure it out +1. A water tank is filled from a tap. If the tap is open for 1 hour, +𝟕 +𝟏𝟎 of the +tank gets filled. How much of the tank is filled if the tap is open for +(a) +𝟏 +𝟑 hour ____________ +Ans: +7 +30 part + +(b) +𝟐 +𝟑 hour ____________ +Ans: +14 +30 part + +(c) +𝟑 +𝟒 hour ____________ +Ans: +21 +40 part + + +(d) +𝟕 +𝟏𝟎 hour ____________ +Ans: +49 +100 part + +(e) For the tank to be full, how long should the tap be running? +Ans: +10 +7 hours + +2. The government has taken +𝟏 +𝟔 of Somu’s land to build a road. What part +of the land remains with Somu now? She gives half of the remaining +part of the land to her daughter Krishna and +𝟏 +𝟑 of it to her son Bora. +After giving them their shares, she keeps the remaining land for +herself. +(a) What part of the original land did Krishna get? +(b) What part of the original land did Bora get? +(c) What part of the original land did Somu keep for herself? +Ans: Part of land that remains with somu = +5 +6 +(a) +5 +12 +(b) +5 +18 +(c) +5 +36 + +Page No. 184 +3. Find the area of a rectangle of sides 3 +𝟑 +𝟒 ft and 9 +𝟑 +𝟓 ft. +Ans: 36 sq. ft. +4. Tsewang plants four saplings in a row in his garden. The distance +between two saplings is +𝟑 +𝟒 m. Find the distance between the first and +last sapling. +Ans: 2 +1 +4 m +5. Which is heavier: +𝟏𝟐 +𝟏𝟓 of 500 grams or +𝟑 +𝟐𝟎 of 4 kg? +Ans: +3 +20 of 4 kg is heavier than +12 +15 of 500 gm. + + +Page No. 191 +In a day, the number of times — + +(a) the first fountain will fill the cistern is 1÷ 1 = 1 + + + +(b) the second fountain will fill the cistern is 1 ÷ +𝟏 +𝟐 = _____ + +Ans: 2 times + + + +(c) the third fountain will fill the cistern is 1 ÷ +𝟏 +𝟒 = _____ + +Ans: 4 times + + + +(d) the fourth fountain will fill the cistern is 1 ÷ +𝟏 +𝟓 = _____ + +Ans: 5 times + +The number of times the four fountains together will fill the cistern in +a day is ___ + ____ + ___ + ____ = 12. +Ans: The number of times the four fountains together will fill the cistern +in a day is + + = 1 + 2 + 4 + 5 = 12 + +Page No. 193 + In each of the figures given below, find the fraction of the big square +that the shaded region occupies. + + + +(a) +(b) +Ans: The shaded region of Fig. (a) occupies +3 +8 of the area of whole square. +The red shaded part of Fig. (b) occupies +1 +16 of the area of whole square. + + + + +Page No. 196 + Figure it out +1. Evaluate the following: +3 ÷ +7 +9 = +27 +7 +14 +4 ÷ 2 = +7 +4 +2 +3 ÷ +2 +3 = 1 +14 +6 ÷ +7 +3 = 1 +4 +3 ÷ +3 +4 = +16 +9 +7 +4 ÷ +1 +7 = +49 +4 +8 +2 ÷ +4 +15 = 15 + +1 +5 ÷ +1 +9 = +9 +5 +1 +6 ÷ +11 +12 = +2 +11 +3 +2 +3 ÷ 1 +3 +8 = 2 +2 +3 + + +2. For each of the questions below, choose the expression that describes +the solution. Then simplify it. +(a) Maria bought 8 m of lace to decorate the bags she made for school. +She used +𝟏 +𝟒 m for each bag and finished the lace. How many bags +did she decorate? +Ans: +(iii) 8 ÷ +1 +4 +(b) +𝟏 +𝟐 meter of ribbon is used to make 8 badges. What is the length of +the ribbon used for each badge? +Ans: +(iv) +1 +2 ÷ 8 +(c) +A baker needs +𝟏 +𝟔 kg of flour to make one loaf of bread. He has 5 kg +of flour. How many loaves of bread can he make? +Ans: +(iii) 5 ÷ +1 +6 + +Page No. 197 +3. If +𝟏 +𝟒 kg of flour is used to make 12 rotis, how much flour is used to make +6 rotis? +Ans: +1 +8 kg + + +4. Patiganita, a book written by Sridharacharya in the 9th century C.E., +mentions this problem: “Friend, after thinking, what sum will be +obtained by adding together +1 ÷ +𝟏 +𝟔, 1 ÷ +𝟏 +𝟏𝟎, 1 ÷ +𝟏 +𝟏𝟑, 1 ÷ +𝟏 +𝟗, and 1 ÷ +𝟏 +𝟐 ”. What should the friend say? +Ans: 40 + +5. Mira is reading a novel that has 400 pages. She read +𝟏 +𝟓 of the pages +yesterday and +𝟑 +𝟏𝟎 of the pages today. How many more pages does she +need to read to finish the novel? +Ans: 200 pages. + +6. A car runs 16 km using 1 litre of petrol. How far will it go using 2 +𝟑 +𝟒 +litres of petrol? +Ans: 44 km. + +7. Amritpal decides on a destination for his vacation. If he takes a train, +it will take him 5 +𝟏 +𝟔 hours to get there. If he takes a plane, it will take +him +𝟏 +𝟐 hour. How many hours does the plane save? +Ans: 4 +2 +3 hour + +8. Mariam’s grandmother made a cake. Mariam and her cousins finished +𝟒 +𝟓 of the cake. The remaining cake was shared equally by Mariam’s +three friends. How much of the cake did each friend get? +Ans: +1 +15 of the whole cake. + +9. Choose the option(s) describing the product of ( +𝟓𝟔𝟓 +𝟒𝟔𝟓 × +𝟕𝟎𝟕 +𝟔𝟕𝟔 ): +Ans: (a), (c) and (e) are correct. + +10. What fraction of the whole square is shaded? + +Ans: +3 +32 of the whole square. + + +11. A colony of ants set out in search of food. As they search, they keep splitting +equally at each point (as shown in the figure 8.7) and reach two food +sources, one near a mango tree and another near a sugarcane field. What +fraction of the original group reached each food source? +Ans: Mango trees = +29 +32 and Sugarcane trees = +3 +32 + +12. What is 1 – +𝟏 +𝟐 ? + (1 − +1 +2) +Ans: +1 +2 + + (1 − +1 +2) × (1 − +1 +3)? +Ans: +1 +3 + +(1 − +1 +2) × (1 − +1 +3) × (1 − +1 +4) × (1 − +1 +5)? +Ans: +1 +5 + +(1 − +1 +2) × (1 − +1 +3) × (1 − +1 +4) × (1 − +1 +5) × (1 − +1 +6) × (1 − +1 +7) × (1 − +1 +8) × (1 − +1 +9) × (1 − +1 +10)? +Ans: +1 +10 + +Make a general statement and explain. +Ans: (1 − +1 +2) × (1 − +1 +3) × (1 − +1 +4) × (1 − +1 +5) . . . . . . . . . . . (1 − +1 +𝑛) = +1 +𝑛" +class_7,9,Geometric Twins,ncert_books/class_7/gegp1dd2/gegp201.pdf,"GEOMETRIC +TWINS +1 +1.1 Geometric Twins +The symbol on this signboard needs to be recreated on another board. +How do we do it? +One way is to trace the outline of this symbol on tracing paper to +reconstruct the figure. But this is difficult for big symbols. What else can +we do? +Can we take some measurements that would allow us to exactly recreate +this figure? If yes, what measurements should we take? +Let us name the corner points of this symbol as shown. +A +C +B +Are the arm lengths AB and BC sufficient to exactly recreate this figure? + +Ganita Prakash | Grade 7 | Part-II +Suppose these lengths are AB = 4 cm, BC = 8 cm. We observe that +several such symbols can be constructed with the same lengths. +A +B +C +A +B +C +A +B +C +A +B +C +To get the exact replica, would it help to take any other measurement? +The measure of ∠ABC, along with the two arm lengths AB and BC, fix +the shape and size of this figure. +Can you draw the symbol if it is known that AB = 4 cm, BC = 8 cm, and +∠ABC = 80°? +These three measurements can help us create an exact replica of the +symbol on the signboard. Figures that are exact copies of each other +or have the same shape and size are said to be congruent. Congruent +figures can be superimposed exactly, one over the other. +Two congruent figures are shown below. You could use a tracing +paper to trace the first figure and superimpose it on the second one. You +will find that they fit exactly, one over the other. +Note that while checking for congruence, a figure can be rotated or +flipped before superimposing it on the other figure. So, the following +pairs of figures are also congruent to each other. +2 + +Geometric Twins +Let us get back to the symbol we saw on the signboard. Suppose there +are two such symbols that look identical and we need to confirm that they +are indeed congruent. Can we use their measurements to verify this? +If it is known that both symbols have the same arm lengths, can it be +concluded that the two symbols are congruent? +We have seen that there can exist several such non-congruent figures +with different angles between the given arm lengths. Fixing the angle +determines the shape and size of the figure. +Thus, if both symbols have the same arm lengths and angle, we can be +sure that the figures are congruent. +Figure it Out +1. Check if the two figures are congruent. +2. Circle the pairs that appear congruent. +3. What measurements would you take to create a figure congruent to +a given: +(a) Circle +(b) Rectangle +3 + +Ganita Prakash | Grade 7 | Part-II +Using this, state how would you check if two — +(a) Circles are congruent? +(b) Rectangles are congruent? +4. How would we check if two figures like the one below are congruent? +Use this to identify whether each of the following pairs are congruent. +1.2 Congruence of Triangles +Meera and Rabia have been asked to make a cardboard cutout identical +to a triangular frame they have in school. They see that the frame is too +big to be traced on a paper and replicated. +4 + +Geometric Twins +What do you think they can do? +Measuring the Sidelengths +Can certain measurements of the triangle be used for this? Using a +measuring tape, the girls measure the sides of the triangle to be 40 cm, +60 cm, and 80 cm. +Then, Rabia takes out her protractor to measure the angles. She is +stopped by Meera. +Meera: The angles of the triangle are not required! With the side +lengths we have measured, we can create a triangle congruent to this +one. +Do you agree with Meera? +Instead of the lengths being 40 cm, 60 cm, and 80 cm, suppose the +sidelengths had been 4 cm, 6 cm, 8 cm (this triangle can fit on our page). +Is this information sufficient to replicate the triangle with the same size +and shape? If yes, can you do so? +Rabia: If I were to construct this triangle, I would first draw a line +segment having one of the given lengths, say 6 cm, and then draw circles +from each of its end points with radii 4 cm and 8 cm. But the circles +would intersect at two points, forming two triangles: +∆ABE and ∆ABF +B +A +AB = 6 cm +E +F +Rabia: Do these two triangles have the same shape and size? If not, +then we will not be sure which of these would actually be congruent to +the original triangle we are trying to replicate. +5 + +Ganita Prakash | Grade 7 | Part-II +Examine whether ∆ABE and ∆ABF are congruent. +For this, you could use one or more of the following methods — tracing +and comparing, taking a cutout and superimposing, or observing that AB +acts as a line of symmetry due to the ‘sameness’ of the act of construction +above and below this line. +We see that ∆ABE and ∆ABF are congruent. From this general +construction, we can see that all triangles with the same sidelengths are +congruent. Hence, Meera was right when she said that the sidelengths +are sufficient to construct a congruent triangle. +Thus, we have the following result: +If two triangles have the same sidelengths, then they are congruent. +We call this the SSS (Side Side Side) condition for congruence. +Conventions to Express Congruence +The two triangles given below are congruent. How can these two triangles +be superimposed? Which vertices of ∆XYZ and ∆ABC should we overlap? +This has to be done so that the equal sides overlap. Figure out how. +C +A +B +Y +X +Z +Overlapping Vertex A over Vertex X, Vertex B over Vertex Y and Vertex +C over Vertex Z will ensure that equal sides overlap, making the triangles +fit exactly over each other. +Are there other ways of overlapping the vertices so that the triangles fit +exactly over each other? +The fact that these triangles are congruent shows that their respective +angles are equal: +∠A = ∠X, ∠B = ∠Y and ∠C = ∠Z +Thus, when two triangles are congruent, there are corresponding +vertices, sides and angles which fit exactly over each other when the +triangles are made to overlap. In this case, they are +(a) Corresponding Vertices: A and X, B and Y, C and Z +6 + +Geometric Twins +(b) Corresponding Sides: AB and XY, BC and YZ, AC and XZ +(c) Corresponding Angles: ∠A and ∠X, ∠B and ∠Y, ∠C and ∠Z +To capture this relation that exists when two triangles are congruent, +their congruence is written as follows: +∆ABC ≅ ΔXYZ +∆ABC ≅ ΔXYZ +c +o +r +r +e +s +p +o +n +d +s + +t +o +c +o +r +r +e +s +p +o +n +d +s + +t +o +c +o +r +r +es +p +o +n +d +s + +t +o +By writing this, we mean that: +• the first vertex in the name of ΔABC corresponds to the first vertex +in the name of ΔXYZ, +• the second vertex in the name of ΔABC corresponds to the second +vertex in the name of ΔXYZ, and +• similarly with the third vertices in the names of ΔABC and ΔXYZ. +By this convention, it is incorrect to write for these two triangles that +ΔACB ≅ ΔXYZ. +However, another correct way of saying it is +ΔACB ≅ ΔXZY. +Can you identify a pair of congruent triangles below? Why are they +congruent? +A +B +D +C +Fig.1.1 +Consider ΔABD and ΔCDB. Since ABCD is a rectangle, we have +AB = CD +AD = CB +If the remaining sides of ΔABD and ΔCDB have the same length then +the SSS condition is satisfied, confirming the congruence of the two +triangles. Is this the case? +7 + +Ganita Prakash | Grade 7 | Part-II +The remaining side is a common side BD, so the SSS condition holds. +Hence, the triangles are congruent. +We know the corresponding sides of the two triangles. We have to +identify the corresponding vertices. Can they be the following? +ΔABD ΔCDB +A C +B B +D D +Verify this by superimposing paper cutouts of the triangles obtained +from the rectangle ABCD (Fig. 1.1). +We see that this correspondence lays the side AB of ΔABD over the +side CB of ΔCDB. But these sides need not be equal, and hence, this +superimposition will not establish congruence. +Identify the correct correspondence of vertices and express the +congruence between the two triangles. +Figure it Out +1. Suppose ΔHEN is congruent to ΔBIG. List all the other correct ways +of expressing this congruence. +2. Determine whether the triangles are congruent. If yes, express the +congruence. +R +E +D +3.5 cm +5 cm +6 cm +J +M +A +3.5 cm +5 cm +6 cm +8 + +Geometric Twins +3. In the figure below, AB = AD, CB = CD. +Can you identify any pair of congruent triangles? If yes, explain why +they are congruent. +Does AC divide ∠BAD and ∠BCD into two equal parts? Give reasons. +A +B +C +D +4. In the figure below, are ΔDFE and ΔGED congruent to each other? It +is given that DF = DG and FE = GE. +F +G +D +E +Measuring the Angles +Instead of measuring the three sidelengths of the triangular frame, if +Meera and Rabia measure the three angles, can they recreate the triangle +exactly? +Suppose the angles are 30°, 70°, and 80°. Can we create an exact copy of +the frame with this? +As we see, we can draw many triangles with these measurements that +are not congruent. +9 + +Ganita Prakash | Grade 7 | Part-II +80° +30° +B +C +A +70° +80° +30° +A +C +B +70° +30° +80° +A +C +B +70° +These triangles are seen to have the same shape, but not the same +size. Hence, two triangles that have the same set of angles need not +be congruent. +Measuring Two Sides and the Included Angle +ΔABC and ΔXYZ are two triangles such that +AB = XY = 6 cm, AC = XZ = 5 cm, and ∠A = ∠X = 30° +Are they congruent? +To check this, we need to see if there can exist non-congruent triangles +with the given measurements. +These measurements correspond to the case of two sides and the +included angle. We have seen how to construct a triangle given these +measurements. +Construct a triangle having the above measurements. +Compare it with the triangles constructed by your classmates. Are +the triangles all congruent? Explain why all such triangles with these +measurements are congruent. +Thus, when two sides and the included angle of two triangles are +equal, the two triangles are congruent. +This is referred to as the SAS (Side Angle Side) condition for +congruence. +Measuring Two Sides and a Non-included Angle +What if two sides and a non-included angle are equal? +ΔABC and ΔXYZ are two triangles such that +AB = XY = 6 cm, AC = XZ = 4 cm, and ∠B = ∠Y = 30° +Are they congruent? +Can there exist non-congruent triangles having these measurements? +Construct and find out. +10 + +Geometric Twins +Looking at a rough diagram helps in planning the construction. +R +P +Q +30° +6 cm +4 cm +How does one construct a triangle having these measurements? +Step 1: Draw the base PQ of length 6 cm. +Step 2: Draw a line l from P that makes an angle of 30° with PQ. +P +Q +l +6 cm +30° +Step 3: Draw a sufficiently long arc from Q of radius 4 cm cutting the +line l. +P +Q +6 cm +30° +R +S +How do we find the required triangle from this figure? +A point of intersection of the arc and the line l gives the third point of +the required triangle. But we see that the arc intersects the line l at two +different points R and S. +Both ΔPQR and ΔPQS satisfy the given measurements. +l +11 + +Ganita Prakash | Grade 7 | Part-II +R +P +Q +30° +6 cm +4 cm +s +P +Q +30° +6 cm +4 cm +Hence, we can draw two non-congruent triangles with the given +measurements. +This is called the SSA (Side Side Angle) condition. We have seen that +SSA condition does not guarantee congruence. +We have examined cases using two sides and an angle for determining +congruence. Can we use two angles and a side? +Let us first take the case of two angles and the included side. +Two Angles and the Included Side +ΔABC and ΔXYZ are two triangles with, +BC = YZ = 5 cm, ∠B = ∠Y = 50° and ∠C = ∠Z = 30°. +Are they congruent? +Can there exist non-congruent triangles having these measurements? +Construct and find out. +We have seen how to construct a triangle when we are given two +angles and the included side. +This construction should make it clear that all the triangles +having these measurements must be congruent to each other. Hence, +ΔABC ≅ ΔXYZ. +This condition is referred to as the ASA (Angle Side Angle) condition +for congruence. +In the figure, Point O is the midpoint of AD and BC. What can one say +about the lengths AB and CD? +A +C +D +B +O +12 + +Geometric Twins +We have, +AO = OD (as O is the midpoint of AD) +BO = OC (as O is the midpoint of BC). +Are there any other equal sides or angles? +We also have, +∠AOB = ∠DOC, as they are vertically opposite angles. +We see that the SAS condition (two sides and the included angle) is +satisfied, and so we can conclude that the triangles are congruent. +What are the Corresponding Vertices? +As we need AO and OD to overlap, and BO and CO to overlap for the +triangles to exactly fit over each other, the corresponding vertices +in ΔAOB and ΔDOC are A and D, O and O (vertex common to both the +triangles), and B and C. Thus, +ΔAOB ≅ ΔDOC. +AB and DC are corresponding sides as they overlap when the triangles +are superimposed. Thus, their lengths are equal. +Figure it Out +1. Identify whether the triangles below are congruent. What conditions +did you use to establish their congruence? Express the congruence. +A +7 cm +B +5 cm +C +47° +X +Y +Z +7 cm +5 cm +47° +2. Given that CD and AB are parallel, and AB = CD, what are the other +equal parts in this figure? (Hint: When the lines are parallel, the +alternate angles are equal. Are the two resulting triangles congruent? +If so, express the congruence.) +13 + +Ganita Prakash | Grade 7 | Part-II +D +A +B +C +O +3. Given that ∠ABC = ∠DBC and ∠ACB = ∠DCB, show that +∠BAC = ∠BDC. Are the two triangles congruent? +A +D +C +B +4. Identify the equal parts in the following figure, given that ∠ABD = +∠DCA and ∠ACB = ∠DBC. +A +B +C +D +Measuring Two Angles and a Non-Included Side +The following triangles ΔABC and ΔXYZ are such that ∠A = ∠X = 35°, +∠C = ∠Z = 75°, and BC = YZ = 4 cm. Are the triangles congruent? Give +a reason. +14 + +Geometric Twins +A +B +C +4 cm +75° +35° +X +Y +Z +4 cm +75° +35° +Fig.1.2 +How do we proceed with this problem? Here is a method. +What are the measures of ∠B and ∠Y? +We know that the sum of the angles of a triangle is 180°. +So ∠B + 35° + 75° = 180°, +or ∠B + 110° = 180° +Thus, ∠B = 70°. +Similarly, ∠Y is also 70°. +Thus, we have ∠B = ∠Y. +Does this help in showing that ΔABC and ΔXYZ are congruent? +A +B +C +4 cm +75° +70° +35° +70° +X +Y +Z +4 cm +75° +35° +These two triangles now satisfy the ASA condition with +∠B = ∠Y +BC = YZ +∠C = ∠Z +So, ΔABC ≅ ΔXYZ. +15 + +Ganita Prakash | Grade 7 | Part-II +In Fig. 1.2, the equalities are between two angles and the +non-included side of the two triangles. This condition is referred to as +the AAS (Angle Angle Side) condition. +As we have seen, the AAS condition guarantees congruence. +We have seen that the SSA condition doesn’t always guarantee +congruence. However, there are some special cases when SSA does +guarantee congruence. Here is one such important case. +Measuring Two Sides in a Right Triangle +ΔABC and ΔXYZ are right-angled triangles such that BC = YZ = 4 cm, +∠B = ∠Y = 90° and AC = XZ=5cm. Are they congruent? +Can there exist non-congruent triangles having these measurements? +Construct and find out. +Looking at the rough diagram helps in planning the construction. +P +Q +R +4 cm +5 cm +Step 1: Draw the base QR of length 4 cm. +Step 2: Draw a line l perpendicular to QR from Q. +Step 3: From R, cut an arc on line l of radius 5 cm. +Step 4: Let P be the point at which the arc intersects the line l. Join PR. +P +Q +R +4 cm +ΔPQR is the required triangle. +16 + +Geometric Twins +Consider the downward extension of line l below QR. Would the arc +from R meet this line downwards as well (as in the case of triangle +construction when the sidelengths are given)? If so, would this lead to a +triangle whose size and shape are different from ΔPQR, and yet has the +given measurements? +It can be seen that the other triangle we get below is also congruent to +ΔPQR. Why? Therefore, all triangles having these measurements will be +congruent to each other. +Thus, we conclude that ΔABC ≅ ΔXYZ. +In the case that we have considered, the parts that are equal to their +corresponding parts in another triangle are +(a) the right angle +(b) two other sides, one of which is opposite to the right angle. This +side is called the hypotenuse. +This is called the RHS (Right Hypotenuse Side) condition, and is one +more condition for congruence. +Conditions that are sufficient to guarantee congruence +From the discussions so far, we can see that two triangles are congruent +if any of the following conditions are satisfied: +(a) SSS condition +(b) SAS condition +(c) ASA condition +(d) AAS condition +(e) RHS condition +1.3 Angles of Isosceles and Equilateral Triangles +Congruence is a very powerful tool for studying properties of geometric +figures. Let us use it to discover an important property of isosceles +triangles. +ΔABC is isosceles with AB = AC, and ∠A = 80. What can we say about ∠B +and ∠C? +80° +A +B +C +Construct the altitude from A to BC. +17 + +Ganita Prakash | Grade 7 | Part-II +We have, + + + + + + AB = AC (given) +∠ADB = ∠ADC = 90° (from construction) +AD is a common side of the two triangles ΔADB and ΔADC. +A +D +B +C +Thus, the triangles satisfy the RHS condition. Hence, ΔADB ≅ ΔADC. +This shows that ∠B = ∠C, as they are corresponding parts of congruent +triangles. +Thus, the angles opposite to equal sides are equal. +Can you use this fact to find ∠B and ∠C? +Angles in an Equilateral Triangle +Equilateral triangles are those in which all the sides have equal lengths. +What can we say about their angles? +We can use the recently discovered fact that angles opposite to equal +sides are equal. +A +B +C +The sides AB and AC are equal. So ∠B = ∠C. +A +B +C +18 + +Geometric Twins +Similarly, the sides AB and BC are equal. So ∠A = ∠C. +So, all the three angles of an equilateral triangle are equal, just like +their sides. +What could be their measures? +As the three angles should add up to 180°, we have +3 × angle in an equilateral triangle = 180°. +So each angle is 60°. +Verify this by construction. +Thus, just using the notion of congruence, we have deduced that the +angles of an equilateral triangle are all 60°. +Congruent Triangles in Real Life: Congruent triangles can be seen +in various constructions and designs from ancient to modern times. +Here are a few examples. +World-famous Louvre Museum +in Paris +World-famous Egyptian Pyramid of Giza +Rangoli design +Dome design +19 + +Ganita Prakash | Grade 7 | Part-II +Rabindra Setu or Howrah Bridge +Describe the congruent triangles you see in each picture. +Figure it Out +1. ΔAIR ≅ ΔFLY. Identify the corresponding vertices, sides and angles. +2. Each of the following cases contains certain measurements taken +from two triangles. Identify the pairs in which the triangles are +congruent to each other, with reason. Express the congruence +whenever they are congruent. +(a) +AB = DE +(b) AB = EF + +BC = EF + +∠A = ∠E + +CA = DF + +AC = ED +(c) +AB = DF +(d) ∠A = ∠D + +∠B = ∠D = 90° + +∠B = ∠E + +AC = FE + +AC = DF +(e) +AB = DF + +∠B = ∠F + +AC = DE +3. It is given that OB = OC, and OA = OD. Show that AB is parallel to CD. +[Hint: AD is a transversal for these two lines. Are there any equal +alternate angles?] +D +B +A +C +O +20 + +Geometric Twins +4. ABCD is a square. Show that ΔABC ≅ ΔADC. Is ΔABC also congruent +to ΔCDA? +A +D +B +C +Give more examples of two triangles where one triangle is +congruent to the other in two different ways, as in the case above. +Can you give an example of two triangles where one is congruent to +the other in six different ways? +5. Find ∠B and ∠C, if A is the centre of the circle. +120° +A +C +B +6. Find the missing angles. As per the convention that we have been +following, all line segments marked with a single ‘|’ are equal to each +other and those marked with a double ‘|’ are equal to each other, etc. +34° +68° +98° +44° +46° +90° +56° +30° +44° +34° +34° +56° +K +R +A +C +U +L +V +B +F +D +21 + +Ganita Prakash | Grade 7 | Part-II +• Figures that have the same shape and size are said to be congruent. +These figures can be superimposed so that one fits exactly over the other. +• While verifying congruence, a figure can be rotated or flipped to make +it fit exactly over the other figure via superimposition. +• When two triangles have the same sidelengths, we say that the SSS +(Side Side Side) condition is satisfied. The SSS condition guarantees +congruence. +• When two sides and the included angle of one triangle are equal to the +two sides and the included angle of another triangle, we say that the +SAS (Side Angle Side) condition is satisfied. The SAS condition also +guarantees congruence. +• When two angles and the included side of one triangle are equal to the +two angles and the included angle of another triangle, we say that the ASA +(Angle Side Angle) condition is satisfied. The ASA condition guarantees +congruence. Congruence holds even if the side is not included between +the angles AAS (Angle Angle Side) condition. +• In a right-angled triangle, the side opposite to the right angle is called +the hypotenuse. +• When a side and a hypotenuse of a right-angled triangle are equal to a +side and the hypotenuse of another right-angled triangle, we say that the +RHS (Right Hypotenuse Side) condition is satisfied. The RHS condition +also guarantees congruence. +• Two triangles need not be congruent if two sides and a non-included +angle are equal. +• In a triangle, angles opposite to equal sides are equal. +• The angles in an equilateral triangle are all 60°. +SUMMARY +22 + +Expression Engineer! +Draw lines and split the region consisting of white squares into 6 +smaller congruent regions." +class_7,10,operations with integers,ncert_books/class_7/gegp1dd2/gegp202.pdf,"OPERATIONS +WITH INTEGERS +2 +2.1 A Quick Recap of Integers +Rakesh’s Puzzle: A Number Game +Rakesh gives you a challenge. +“I have thought of two numbers”, he says. +“Their sum is 25, and their difference is 11.” +Can you tell me the two numbers? +You don’t need to use any formulas. Just try different pairs of numbers +and then check: +1. Do the two numbers add up to 25? +2. Is the difference between them 11? +(Remember: the difference means first number – second number.) +Write your guesses like this: +First Number +Second Number +Sum +Difference +10 +15 +25 +–5 +20 +5 +25 +15 +19 +6 +25 +13 +18 +7 +25 +11 +Did you find the right pair? +Now that you’ve found the correct pair, Rakesh gives you a second +challenge: +“Think of two numbers whose sum is 25, but their difference is –11.” +Use the same method. Try different pairs of numbers and fill in the +table again. You will notice that if you swap the numbers from the first + +Operations with Integers +puzzle, you get the answer to Rakesh’s second puzzle. That is, the first +number is 7 and the second is 18! +Figure it Out +Let us try to find a few more pairs of numbers from their sums and +differences: +(a) Sum = 27, Difference = 9 +(b) Sum = 4, Difference = 12 +(c) Sum = 0, Difference = 10 +(d) Sum = 0, Difference = – 10 +(e) Sum = – 7, Difference = – 1 +(f) Sum = – 7, Difference = – 13 +Choose a partner and take turns to play this game. In each turn, one +of you can think of two integers, and give their sum and difference; the +other person must then figure out the integers. After some practice, you +can perform this magic trick for your family members and surprise +them! +Carrom Coin Integers +A carrom coin is struck to move it to the right. Each strike moves the +coin a certain number of units of distance rightwards based on the force +of the strike. +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +To begin with, the coin is at point 0. If the coin is struck twice, with the +first strike moving it by 4 units and the second strike moving it by 3 +units, what will be the final position of the coin? +It is clear that the coin will be 4 + 3 = 7 units from 0. +If the coin is struck twice, and if the two movements are known, can you +give a formula for the final position of the coin? +If the first strike moves the coin ‘a’ +units to the right and if the second +strike moves the coin ‘b’ units to +the right, then the final position is +P = a + b, where P is the distance of the +coin from the starting point 0. +P = a + b +0 +P +b +a +25 + +Ganita Prakash | Grade 7 | Part-II +Now, suppose the coin can be struck to move it in either direction — +left or right. +The coin is at 0. If it is struck twice (the direction of the two strikes may +be the same or different) can you give a formula for the final position of +the coin? +One way to do this is to consider different cases of the directions of +the strikes +• both are rightward, +• both are leftward, +• the first one is rightward and the second one is leftward, and +• the first one is leftward and the second one is rightward. +An efficient way to model this situation is to use positive and negative +integers. First, let us model the line on which the coin moves as a +number line. +–5 +–4 +–3 +–2 +–1 +0 +1 +2 +3 +4 +5 +6 +. . . +. . . ++ve +–ve +Let us consider the rightward movement positive and the leftward +movement negative. +Suppose the first strike moves the coin rightward by 5 units from 0, +and the second strike leftward by 7 units, then we take the +First Movement = 5 units. +Second Movement = – 7 units. +What is the final position of the coin? +This can be found by simply adding the two movements: 5 + (– 7) = – 2. +So the coin is at –2, or it is 2 units to the left of 0. +In general, if the first strike moves the coin ‘a’ units (which is positive +if the strike is to the right and negative if the strike is to the left), and the +second strike ‘b’ units (which is positive if the strike is to the right and +negative if the strike is to the left), then the final position ‘P’, after the +two strikes, is again P = a + b. +Based on this new model, answer the following questions: +1. If the first movement is – 4 and the final position is 5, what is the +second movement? +0 +–ve ++ve +26 + +Operations with Integers +2. If there are multiple strikes causing movements in the order 1, – 2, 3, +– 4, …, – 10, what is the final position of the coin? +By modeling the movements as numbers, both positive and negative, +we are able to capture two pieces of information — the distance +(magnitude) and the direction (rightwards or leftwards). For example, +when we say the movement is – 4, the magnitude is 4 and the direction +is leftward. +From the figures below, what can you conclude about the magnitudes +of a and b compared to each other, and what are their directions? +Remember to start from 0. +1. +2. +3. +Try +This +b +a +P +0 ++ve +–ve +b +a +P +0 ++ve +–ve +b +a +P +0 ++ve +–ve +27 + +Ganita Prakash | Grade 7 | Part-II +In addition to the number line, we used the token model to understand +integers. We used this model to perform addition and subtraction in +Grade 6. Let us do a quick recall. We used green ( +) token to represent +positive 1 and red ( +) token to represent negative 1, that is –1. Together +they make zero, as they cancel each other out. +Find (+7) – (+18). +To subtract 18 from 7, i.e., (+7) – (+18), we need to remove 18 positives +from 7 positives. +But there are not enough tokens to remove 18 positives! +We put in enough zero pairs so that we can remove 18 positives. +How many? +We have 7 positives and we need 11 more. So we need to put in 11 +zero pairs: +We can now remove 18 positives. +What is left? There are 11 negatives, meaning –11. +So, 7 – 18 = –11. +We had also seen that subtracting a number is the same as adding its +additive inverse. +Using tokens, argue out the following statements. +(a) 7 – 18 = 7 + (– 18) (additive inverse of 18 is – 18) +(b) 4 – (– 12) = 4 + 12 (additive inverse of – 12 is 12) +Math +Talk +Note to the Teacher: Tokens of different shapes may be used for positive and +negative numbers for visually challanged students. +28 + +Operations with Integers +Additive inverse of an integer a is represented as – a. So the additive +inverse of 18 is represented as – (18) = – 18, and the additive inverse of +– 18 is represented as – (– 18) = 18. +2.2 Multiplication of Integers +We used the token model to represent addition and subtraction of +integers. We now explore how to model multiplication of integers using +tokens. +Suppose we put some positive tokens into an empty bag as shown in +the figure. +How many positives are in the bag now? +There are 8 positives in the bag. We can +see this as adding 2 positives to the bag +4 times. Thus, the operation is, 4 × 2 = 8. +We have seen this kind of multiplication +of positive integers before. Can we use +tokens to give meaning to multiplications +like 4 × (– 2)? +Let us see how. +For every new operation, we start +with an empty bag. +4 × (–2) can be interpreted as placing +2 negatives into an empty bag 4 times. +We use red tokens for negatives, so we +place 2 negatives into an empty bag 4 +times. +There are now 8 red tokens or +8 negatives in the bag, meaning – 8. +4 × (– 2) = (– 8). +Similarly find the values of 4 × (– 6) and 9 × (– 7)? How can we interpret +(– 4) × 2? +When the multiplier is positive, we place tokens into the bag. When +the multiplier is negative, we remove tokens from the bag. +So, for (– 4) × 2, we need to remove two positives or two green tokens +from the bag 4 times. +Two green tokens 4 times = 8 +4 × 2 = 8 +Two red tokens 4 times = – 8 +4 × (– 2) = – 8 +Multiplier +Multiplicand +Product +29 + +Ganita Prakash | Grade 7 | Part-II +Why are we trying to remove green tokens and not red tokens? +But there are no tokens in the bag, because we start with an +empty bag. Just as in the case of subtraction, to remove 2 positives from +an empty bag, we need to first place 2 zero pairs inside and then remove +the 2 positives. We need to do this 4 times. +Remove 2 green tokens from the zero pairs, 4 times +(–  4) × 2 = –  8 +After removing the positives, 8 negatives are left in the bag. This is – 8. +This shows that (– 4) × 2 = – 8. +What happens when both the integers in the multiplication are negative? +How do we model (–4) × (–2) with tokens? +For (– 4) × (– 2), we need to remove 2 negatives from the bag 4 times. +Since there are no red tokens in the bag, we need to place 2 zero pairs +and remove 2 negatives, and we need to do this 4 times. +Remove 2 red tokens from the zero pairs, 4 times +(–  4) × (– 2) = 8 +8 positives are left in the bag. This is + 8. +So, +– 4 × – 2 = + 8. +So far, we have established the following results by using tokens: +4 × 2 = 8, +4 × (– 2) = – 8, +(– 4) × 2 = – 8, and +(– 4) × (– 2) = 8. +Math +Talk +30 + +Operations with Integers +Figure it Out +1. Using the token interpretation, find the values of: +(a) 3 × (– 2) +(b) (– 5) × (– 2) +(c) (– 4) × (– 1) +(d) (– 7) × 3 +2. If 123 × 456 = 56088, without calculating, find the value of: +(a) (– 123) × 456 +(b) (– 123) × (– 456) +(c) (123) × (– 456) +3. Try to frame a simple rule to multiply two integers. + +Consider the numbers represented by the following tokens: +(a) +(b) +(c) +We can see that all of them represent the number (– 2). Now, take 4 +times each of these token sets. That is, place each set into the empty bag +4 times. +What integer do we get as the final answer in each case? Do we +get different answers because the sets look different, or the same +answer because they all represent – 2? +Check this for 5 × 4, by taking different token sets corresponding to 4. +We have seen that – 4 × 2 is the number obtained by removing 2 +positive tokens from the empty bag 4 times. +We know that removing or subtracting a number is the same as adding +its inverse. +Using this, can – 4 × 2 be defined through a process of addition of +tokens instead of removal of tokens? +Math +Talk +Math +Talk +Math +Talk +31 + +Ganita Prakash | Grade 7 | Part-II +Since removing 2 positive tokens is the same as +adding 2 negative tokens, – 4 × 2 can also be obtained +by adding 2 negative tokens to the empty bag 4 times. +(–  4) × 2 +Patterns in Integer Multiplication +We have explored the multiplication of integers in cases where the +multiplier is positive, when it is negative, when the multiplicand +is positive and when it is negative. Using this understanding, let us +construct a sequence of multiplications and observe the patterns. +What do you notice in this pattern? Can you +describe it? +We can see that, when the multiplicand is +positive, for every unit decrease in the multiplier +the product decreases by the multiplicand. +4 × 3 = 12 +3 × 3 = 9 +2 × 3 = 6 +1 × 3 = 3 +0 × 3 = 0 +–3 +–3 +–3 +–3 +Will this pattern continue when the multiplier goes +below zero and becomes a negative number? +Yes indeed! The same pattern continues when +the multiplier becomes a negative number. +What is the pattern when the multiplicand is a +negative integer? +4 × (–3) = –12 +3 × (–3) = –9 +2 × (–3) = –6 +1 × (–3) = –3 +0 × (–3) = 0 ++3 ++3 ++3 ++3 +4 × 3 = 12 +3 × 3 = 9 +2 × 3 = 6 +1 × 3 = 3 +0 × 3 = 0 +–1 × 3 = –3 +–2 × 3 = –6 +–3 × 3 = –9 +–3 +–3 +–3 +–3 +–3 +–3 +–3 +This is the inverse of the previous pattern. When the multiplicand is +negative, for every unit decrease of the multiplier, the product increases +by the multiplicand. +Will this pattern continue when the multiplier goes below zero and +becomes a negative integer? +Yes! +32 + +Operations with Integers +Even when the multiplier is negative, +the same pattern is observed. When the +multiplicand is negative, for every unit +decrease in the multiplier, the product +increases by the multiplicand. +We can see from these patterns that, what +is true for multiplication when the integers +are positive, is also true when the integers +are negative. +4 × (–3) = –12 +3 × (–3) = –9 +2 × (–3) = –6 +1 × (–3) = –3 +0 × (–3) = 0 +(–1) × (–3) = 3 +(–2) × (–3) = 6 +(–3) × (–3) = 9 ++3 ++3 ++3 ++3 ++3 ++3 ++3 +With this understanding of multiplication of integers, let us look at +the times 3 tables when the multipliers and multiplicands are positive, +and when they are negative. +1 × 3 = 3 +2 × 3 = 6 +3 × 3 = 9 +4 × 3 = 12 +5 × 3 = 15 +6 × 3 = 18 +7 × 3 = 21 +8 × 3 = 24 +9 × 3 = 27 +10 × 3 = 30 +–1 × 3 = –3 +– 2 × 3 = –6 +–3 × 3 = –9 +–4 × 3 = –12 +–5 × 3 = –15 +–6 × 3 = –18 +–7 × 3 = –21 +–8 × 3 = –24 +–9 × 3 = –27 +–10 × 3 = –30 +1 × –3 = –3 +2 × –3 = –6 +3 × –3 = –9 +4 × –3 = –12 +5 × –3 = –15 +6 × –3 = –18 +7 × –3 = –21 +8 × –3 = –24 +9 × –3 = –27 +10 × –3 = –30 +–1 × –3 = 3 +–2 × –3 = 6 +–3 × –3 = 9 +–4 × –3 = 12 +–5 × –3 = 15 +–6 × –3 = 18 +–7 × –3 = 21 +–8 × –3 = 24 +–9 × –3 = 27 +–10 × –3 = 30 +We observe the following: +• The magnitude of the product does not change with the change in +the signs of the multiplier and the multiplicand. +• When both the multiplier and the multiplicand are positive, the +product is positive. +• When both the multiplier and the multiplicand are negative, the +product is positive. +• When one of the multiplier or the multiplicand is positive and the +other is negative, their product is negative. +Figure it Out +Find the following products. +(a) 4 × (– 3) +(b) (– 6) × (– 3) +33 + +Ganita Prakash | Grade 7 | Part-II +(c) (– 5) × (– 1) +(d) (– 8) × 4 +(e) (– 9) × 10 +(f) 10 × (– 17) +Consider the expression 1 × a. We know that the value of this expression +is ‘a’ for all positive integers. +Is this true for all negative integers too? +Using the token model, we put ‘a’ negatives into the bag just once. After +this, the bag contains ‘a’ negatives. For example, if ‘a’ is – 5 (5 negatives), +then the bag contains 5 negatives, i.e., –5. So, +1 × a = a (for all integers a, both positive and negative). +What is the value of the expression – 1 × a? +When ‘a’ is positive, then from our observations on integer +multiplication, the product has the same magnitude as ‘a’ but is negative. +When ‘a’ is negative, then the product has the same magnitude as ‘a’ +but is positive. +In each case, we notice that the product is the additive inverse of the +multiplicand ‘a’. Thus, +– 1 × a = – a (for all integers a). +In the case of integers, is the product the same when we swap the +multiplier and the multiplicand? Try this for some numbers. +Observe the following pairs of multiplications (fill in the blanks where +needed): + 3 × – 4 = –12 +– 4 × 3 = –12 +– 30 × 12 = _______ + 12 × – 30 = _______ +–15 × – 8 = 120 +– 8 × –15 = 120 + 14 × – 5 = –70 +– 5 × _____ = – 70 +What do you notice in these pairs of multiplication statements? +The product is the same when we ‘swap’ the multiplier and +multiplicand. Earlier, we have seen a similar property with addition. +Will this always happen? +The magnitude of the product does not change when the multiplier +and the multiplicand are swapped. This is because the magnitude of +the product depends only on the magnitudes of the multiplier and the +multiplicand, and we know that the product of two positive integers +does not change when the numbers being multiplied are swapped. +34 + +Operations with Integers +Does the sign of the product change if we swap the multiplier and +multiplicand? +If both are positive or both are negative, the product is positive before +and after swapping. So the sign does not change in this case. +If one is positive and the other negative, the product is negative before +and after swapping. So the sign does not change in this case either. +Hence, the product does not change when the multiplier and +multiplicand are swapped, whatever their signs may be. +Thus, multiplication is commutative for integers. +In general, for any two integers, a and b, we can say that +a × b = b × a. +Brahmagupta’s Rules for Multiplication and Division of Positive and +Negative Numbers +Just like for addition and subtraction of integers, Brahmagupta in his +Brāhmasphuṭasiddhānta (628 CE) also articulated explicit rules for +integer multiplication and division. He used the notions of fortune +(dhana) for positive values and debt (ṛṇa) for negative values. In his +Brāhmasphuṭasiddhānta (18.30-32), Brahmagupta wrote: +“The product or quotient of two fortunes is a fortune. +The product or quotient of two debts is a fortune. +The product or quotient of a debt and a fortune is a debt. +The product or quotient of a fortune and a debt is a debt.” +This represented the first time that rules for multiplication and +division of positive and negative numbers were articulated, and was an +important step in the development of arithmetic and algebra! +Example 1: An exam has 50 multiple choice questions. 5 marks are given +for every correct answer and 2 negative marks for every wrong answer. +What are Mala’s total marks if she had 30 correct answers and 20 wrong +answers? +Solution: We use positive and negative integers. The mark for each +correct answer is a positive integer 5 and for each wrong answer is a +negative integer – 2. +Marks for 30 correct answers = 30 × 5. +Marks for 20 wrong answers = 20 × (– 2). +Thus the arithmetic expression for 30 correct answers and 20 wrong +answers is: +30 × 5 + 20 × (– 2) += 150 + (– 40) += 110. +Mala got 110 marks in the exam. +35 + +Ganita Prakash | Grade 7 | Part-II +What are the maximum possible marks in the exam? What are the +minimum possible marks? +Example 2: There is an elevator in a mining shaft that moves above +and below the ground. The elevator’s positions above the ground are +represented as positive integers and positions below the ground are +represented as negative integers. +(a) The elevator moves 3 metres +per minute. If it descends +into the shaft from the +ground +level +(0), +what +will be its position after +one hour? +(b) If it begins to descend from +15 m above the ground, +what will be its position +after 45 minutes? +Solution: +Solution to part (a) of the +question: +Method 1: +We +can +model +this +using +subtraction. +The elevator moves at 3 metres per minute. So in one hour it moves +180 metres (60 × 3). If it started at ground level (0 metres) and descended, +we should subtract 180 from 0. +0 – 180 = (–180). +So, it will reach the (–180) metre position, which is 180 metres below +the ground. +Method 2: +Let us say that the speed and direction of the elevator are represented +by an integer (metres per minute). It is +3 when it is moving up and it is +(–3) when moving down. +Since the elevator is moving down, the speed is (–3) metres per minute. +It moves for 60 minutes. So it goes +60 × (–3) = (–180). +36 + +Operations with Integers +The position of the elevator after 60 minutes is 180 metres below the +ground level. +Find the solution to part (b) using Method 1 described above. +Solution to part (b) using Method 2: +Starting Position = 15. +Distance Travelled = The elevator moves down at the speed of 3 metres +per minute for 45 minutes, that is, (45 × (–3)). So, +Ending Position = 15 + (45 × (–3)) += 15 + (–(45×3)) += 15 + (–135) += (–120). +The elevator will be 120 metres below the ground. +A Magic Grid of Integers +A grid containing some numbers is given below. Follow the steps as +shown until no number is left. +8 +–4 +12 +–6 +–28 +14 +–42 +21 +12 +–6 +18 +–9 +20 +–10 +30 +–15 +Circle any number +Circle any unstruck number +Strike out the row and the coloumn +containing that number +When there are no more unstruck numbers, stop. Multiply the +circled numbers. +An example is shown below. +8 +–4 +12 +–6 +–28 14 –42 21 +12 +–6 +18 +–9 +20 –10 30 –15 +Round 1 +8 +–4 +12 +–6 +–28 14 –42 21 +12 +–6 +18 +–9 +20 –10 30 –15 +Round 2 +8 +–4 +12 +–6 +–28 14 –42 21 +12 +–6 +18 +–9 +20 –10 30 –15 +Round 3 +8 +–4 +12 +–6 +–28 14 –42 21 +12 +–6 +18 +–9 +20 –10 30 –15 +Round 4 +37 + +Ganita Prakash | Grade 7 | Part-II +Try again , and choose different numbers this time. What product did +you get? Was it different from the first time? Try a few more times with +different numbers! +Play the same game with the grid below. What answer do you get? +8 +–4 +12 +–6 +–28 +14 +–42 +21 +12 +–6 +18 +–9 +20 +–10 +30 +–15 +What is so special about these grids? +Is the magic in the numbers or the way +they are arranged or both? Can you +make more such grids? +Division of Integers +We have earlier seen how division can be converted into multiplication. +For example, (– 100) ÷ 25 can be reframed as, ‘what should be multiplied +to 25 to get (– 100)?’. That is, +25 × ? = (– 100). +We know that +25 × (– 4) = (– 100). +Therefore, +(– 100) ÷ 25 = (– 4). +Similarly, (– 100) ÷ (– 4) can be reframed as, ‘What should be multiplied +to (– 4) to get (– 100)?’ +(– 4) × ? = (– 100). +We know that +(– 4) × 25 = (– 100). +Therefore, +(– 100) ÷ (– 4) = 25. +Similarly, we know that +(– 25) × (– 2) = 50. +Therefore, +50 ÷ (– 25) = (– 2). +Try +This +38 + +Operations with Integers +Can you summarise the rules for integer division looking at the +above pattern? +In general, for any two positive integers a and b, where b ≠ 0, we can +say that +a ÷ – b = – (a ÷ b), +– a ÷ b = – (a ÷ b), and +– a ÷ – b = a ÷ b. +Figure it Out +1. Find the values of: +(a) 14 × (– 15) +(b) – 16 × (– 5) +(c) 36 ÷ (– 18) +(d) (– 46) ÷ (– 23) +2. A freezing process requires that the room temperature be lowered +from 32°C at the rate of 5°C every hour. What will be the room +temperature 10 hours after the process begins? +3. A cement company earns a profit of ₹8 per bag of white cement sold +and a loss of ₹5 per bag of grey cement sold. [Represent the profit/ +loss as integers.] +(a) The company sells 3,000 bags of white cement and 5,000 bags of +grey cement in a month. What is its profit or loss? +(b) If the number of bags of grey cement sold is 6,400 bags, what is +the number of bags of white cement the company must sell to +have neither profit nor loss. +4. Replace the blank with an integer to make a true statement. +(a) (– 3) × _____ = 27 +(b) 5 × _____ = (– 35) +(c) _____ × (– 8) = (– 56) +(d) _____ × (– 12) = 132 +(e) _____ ÷ (– 8) = 7 +(f) _____ ÷ 12 = – 11 +Expressions Using Integers +What is the value of the expression 5 × – 3 × 4? Does it matter whether we +multiply 5 × – 3 and then multiply the product with 4, or if we multiply +– 3 × 4 first and then multiply the product with 5? +(5 × – 3) × 4 += – 15 × 4 += – 60. + 5 × (– 3 × 4) += 5 × – 12 += – 60. +39 + +Ganita Prakash | Grade 7 | Part-II +Take a few more examples of multiplication of 3 integers and check this +property. What do you observe? +We can see that the product is the same when we ‘group’ the +multiplications in these two ways. So, integer multiplication is associative, +just like integer addition. +In general, for any three integers a, b, and c, +a × (b × c) = (a × b) × c. +In the expression 5 × – 3 × 4, try to multiply 5 and 4 first and then +multiply the product with – 3: (5 × 4) × – 3. +5 × 4 = 20, and 20 × – 3 = – 60. +This also gives the same product. +Are there orders in which 5 × – 3 × 4 can be evaluated? Will the product +be the same in all these cases? +Multiply the expression 25 × – 6 × 12 in all the different orders and check +if the product is the same in all cases. +The product remains the same when 3 or more numbers are multiplied +in any order. +Look at the following series of multiplications: +– 1 × – 1 = 1 +– 1 × – 1 × – 1 = – 1 +– 1 × – 1 × – 1 × – 1 = 1 +– 1 × – 1 × – 1 × – 1 × – 1 = 1 +When –1 is multiplied 2 or 4 times the product is positive. +When it is multiplied 3 or 5 times the product is negative. +Can you generalise these statements further? +Using this understanding of multiplication of many integers, can you +give a simple rule to find the sign of the product of many integers? +Now, consider the expression 5 × (4 + (– 2)). As in the case of positive +integers, is this expression equal to 5 × 4 + 5 × (– 2)? +We see that it does. Recall that we call this property the distributive +property. +Check if the distributive property holds for (– 2) × (4 + (– 3)) (that is, if +this expression equals (– 2) × 4 + (– 2) × (– 3)), and for a few other such +expressions of your choice. +What do you observe? We see that the distributive property seems to +hold for integers, as well. Will this always happen? +Math +Talk +40 + +Operations with Integers +In the case of positive integers, we +used a rectangular arrangement of +objects to visually understand why +the distributive property holds. We +can use the same setup even in the +case of integers by using green tokens +for positive numbers and red tokens +for negative numbers. For example, +consider the following rectangular +arrangement of tokens — +We see that the overall arrangement +represents 4 × (2 + (– 3)), and it is clear +that this also equals the sum of 4 × 2 +and 4 × (–3). +4 �� 2 +4 × (– 3) +4 × (2 + (– 3)) +Can you visually show the distributive property for an expression +like –4 × (2 + (–3))? [Hint: Use the fact that multiplying a number by +–4 is adding the inverse of the number 4 times.] +Thus, for any integers a, b, and c, we have +Try +This +a × (b + c) = (a × b) + (a × c). +Pick the Pattern +Two pattern machines are given below. Each machine takes 3 numbers, +does some operations and gives out the result. +Find the operations being done by Machine 1. +41 + +Ganita Prakash | Grade 7 | Part-II +The operation done by Machine 1 is +(first number) + (second number) – (third number). +Written as an expression, this will be a + b – c, where a is the first +number, b is the second number, and c is the third number. +For example, 5 + 8 – 3 = 10, and (– 4) + (– 1) – (– 6) = 1. +So, the result of the last group will be, (– 10) + (– 12) – (– 9) = _______. +Find the operations being done by Machine 2 and fill in the blank. +Make your own machine and challenge your peers in finding its +operations. +Figure it Out +1. Find the values of the following expressions: +(a) (– 5) × (18 + (– 3)) +(b) (– 7) × 4 × (– 1) +(c) (– 2) × (– 1) × (– 5) × (– 3) +2. Find the values of the following expressions: +(a) (– 27) ÷ 9 +(b) 84 ÷ (– 4) +(c) (– 56) ÷ (– 2) +3. Find the integer whose product with (– 1) is: +(a) 27 +(b) – 31 +(c) – 1 +(d) 1 +(e) 0 +4. If 47 – 56 + 14 – 8 + 2 – 8 + 5 = – 4, then find the value of – 47 + 56 – 14 ++ 8 – 2 + 8 – 5 without calculating the full expression. +5. Do you remember the Collatz Conjecture from last year? Try a +modified version with integers. The rule is — start with any number; +if the number is even, take half of it; if the number is odd, multiply it +by – 3 and add 1; repeat. An example sequence is shown below. +–7 + 22 + 11 + 32 + –16 + –8 + –4 +–2 + –1 + 4 + 2 + 1 +Math +Talk +42 + +Operations with Integers + Try this with different starting numbers: (– 21), (– 6), and so on. + Describe the patterns you observe. +6. In a test, (+ 4) marks are given for every correct answer and (– 2) +marks are given for every incorrect answer. +(a) Anita answered all the questions in the test. She scored 40 +marks even though 15 of her answers were correct. How many +of her answers were incorrect? How many questions are in the +test? +(b) Anil scored (– 10) marks even though he had 5 correct answers. +How many of his answers were incorrect? Did he leave any +questions unanswered? +7. Pick the pattern — find the operations done by the machine shown +below. +8. Imagine you’re in a place where the temperature drops by 5°C each +hour. If the temperature is currently at 8°C, write an expression +which denotes the temperature after 4 hours. +9. Find 3 consecutive numbers with a product of (a) – 6, (b) 120. +10. An alien society uses a peculiar currency called ‘pibs’ with just two +denominations of coins — a+13 pibs coin and a – 9 pibs coin. You +have several of these coins. Is it possible to purchase an item that +costs + 85 pibs? +Math +Talk +43 + +Ganita Prakash | Grade 7 | Part-II + Yes, we can use 10 coins of +13 pibs and 5 coins of – 9 pibs to make +a total of + 85. Using the two denominations, try to get the following +totals: +(a) + 20 +(b) + 40 +(c) – 50 +(d) + 8 +(e) + 10 +(f) – 2 +(g) + 1 +[Hint: Writing down a few multiples of 13 and 9 can help.] +(h) Is it possible to purchase an item that costs 1568 pibs? +11. Find the values of: +(a) (32 × (– 18)) ÷ ((– 36)) +(b) (32 ) ÷ ((– 36) × (– 18)) +(c) (25 × (– 12)) ÷ ((45) × (– 27)) +(d) (280 × (– 7)) ÷ ((– 8) × (– 35)) +12. Arrange the expressions given below in increasing order. +(a) (– 348) + (– 1064) +(b) (– 348) – (– 1064) +(c) 348 – (– 1064) +(d) (– 348) × (– 1064) +(e) 348 × (– 1064) +(f) 348 × 964 +13. Given that (– 548) × 972 = – 532656, write the values of: +(a) (– 547) × 972 +(b) (– 548) × 971 +(c) (– 547) × 971 +14. Given that 207 × (– 33 + 7) = – 5382, write the value of – 207 × (33 – 7) += _________. +15. Use the numbers 3, – 2, 5, – 6 exactly once and the operations ‘+’, ‘–’, +and ‘×’ exactly once and brackets as necessary to write an expression +such that — +(a) the result is the maximum possible +(b) the result is the minimum possible +16. Fill in the blanks in at least 5 different ways with integers: +(a) + + + × + = –36 +(b) ( + – + ) × + = 12 +(c) ( + – ( + – + )) = –1 +Try +This +44 + +Operations with Integers +• When two integers are multiplied, the product is positive when both +the +multiplier +and +multiplicand +are +positive, or when both +are +negative. +The +product is negative if +one of them is positive +and +the +other +is +negative. +• When two integers are +divided, the quotient is +positive when both the +dividend and divisor +are positive, or both are +negative. The quotient +is negative when one of them is positive and the other is negative. +• Integer multiplication is commutative, i.e., for any two integers a and b, +a × b = b × a. +• Integer multiplication is associative, i.e., for any three integers, a, b, +and c, +a × (b × c) = (a × b) × c. +• Integer multiplication is distributive over addition, i.e., for any three +integers, a, b, and c, +a × (b + c) = (a × b) + (a × c). +– 9 +– 6 +– 3 +3 +3 +6 +9 +– 6 +– 4 +– 2 +2 +2 +4 +6 +– 3 +– 2 +– 1 +1 +1 +2 +3 +– 3 +– 2 +– 1 +× +1 +2 +3 +3 +2 +1 +1 +– 1 +– 2 +– 3 +6 +4 +2 +2 +– 2 +– 4 +– 6 +9 +6 +3 +3 +– 3 +– 6 +– 9 +SUMMARY +Terhüchü +Terhüchü is a game played in Assam and Nagaland. The board has 16 +squares and diagonals are marked as shown in the following figure. +This is usually roughly scratched on a large piece of stone or just +drawn on mud. There are 2 players, and each player has a set of 9 +coins placed as shown. The coins in one set look different from those +in the other set. +45 + +Objective +The goal is to capture all the opponent’s coins. The first player to do +so is the winner. A player may also win by blocking any legal move by +their opponent. If a draw seems unavoidable, the player with more +coins wins. +Gameplay +• The starting position of the game is as shown above. +• Players take turns. In each turn, they can move a single coin along +a line, in any direction, to a neighbouring vacant intersection. +Or, if an opponent’s coin is at a neighbouring intersection, and +there is a vacant intersection just beyond it, they can jump over +the opponent’s coin and land in the vacant intersection. +• If a player jumps over an opponent’s coin, it is considered captured +and is removed from the board. Multiple captures in one move +are allowed, and the direction can change after each jump. +• Inside the triangular corners, which are outside the main +square, a coin may skip an intersection and move straight to the +next one. That is, it can jump over an empty intersection and go +to the one beyond it." +class_7,11,finding common ground,ncert_books/class_7/gegp1dd2/gegp203.pdf,"FINDING COMMON +GROUND +3 +3.1 The Greatest of All +Sameeksha is building her new house. The main room of the house is 12 +ft by 16 ft. She feels that the room would look nice if the floor is covered +with square tiles of the same size. She also wants to use as few tiles as +possible, and for the length of the tile to be a whole number of feet. What +size tiles should she buy? +Let us explore how to find the largest square tile that can be used. The +breadth of the room is 12 ft and the length is 16 ft. +For the tiles to fit the breadth of the room exactly, the side of the tile +should be a factor of 12. Similarly, for the tiles to fit the length of the +room exactly, the side should be a factor of 16. So the side of the tile +should be a factor of both 12 and 16. What are the common factors of 12 +and 16? +The factors of 12 are 1, 2, 3, 4, 6, 12. The factors of 16 are 1, 2, 4, 8, 16. +The common factors are 1, 2, and 4. +So, the square tiles can have sides 1 ft, 2 ft, and 4 ft. Among these, she +should use the largest sized square tile. Can you explain why? +Therefore, she needs tiles of size 4 ft. + +Ganita Prakash | Grade 7 | Part-II +How many tiles of this size should she purchase? +What if Sameeksha did not insist on the length of the tile to be a +whole number of feet and the length could be a fractional number +of feet? Would the answer change? +The Highest Common Factor (HCF) of two or more numbers, is the +highest (or greatest) of their common factors. It is also known as the +Greatest Common Divisor (GCD). +In the previous problem, 4 is the HCF of 12 and 16. +We can draw rough diagrams like the one shown on the previous page to +visualise the given scenario. It may help in understanding and solving. +Lekhana purchases rice from two farms and sells it in the market. She +bought 84 kg of rice from one farm and 108 kg from the other farm. She +wants the rice to be packed in bags, so each bag has rice from only one +farm and all bags have the same weight that is a whole number of kg. If +she wants to use as few bags as possible, what should the weight of each +bag be? +To divide 84 kg of rice into bags of equal weight, we need the factors +of 84. Similarly, for 108, we need the factors of 108. +Factors of 84 — 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84. +Factors of 108 — 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108. +Since, Lekhana wants to use bags of the same weight for both farms, +the weight of the bag should be a common factor. The common factors +of 84 and 108 are +1, 2, 3, 4, 6, and 12. +She can use any of these weights to pack rice from both farms in bags +of equal weight. But, she wants to minimise the number of bags. +Which weight should she choose to minimise the number of bags? +Do you remember the ‘Jump Jackpot’ game from Grade 6 (see the +chapter ‘Prime Time’)? Grumpy places a treasure on a number and +Jumpy chooses a jump size and tries to collect the treasure. In each case +below, the two numbers upon which treasures are kept are given. Find +the longest jump size (starting from 0) using which Jumpy can land on +both the numbers having the treasure. +(a) 14 and 30 +(b) 7 and 11 +(c) 30 and 50 +(d) 28 and 42 +Is the longest jump size for the numbers the same as their HCF? Explain +why it is so. +Try +This +48 + +Finding Common Ground +So far, we have been listing all the factors to find the HCF. This can +become cumbersome for numbers with many factors, as you would +have observed for the numbers 30 and 50, and 28 and 42. +Sometimes, we may also miss some factors which can lead to errors. +Can this process be simplified? Can it be made more reliable? +It turns out that using prime factorisation can simplify the process. +We will start by revisiting primes and prime factorisation. +Primes +Recall that a prime is a number greater than 1 that has only 1 and the +number itself as its factors. Last year, we tried to find patterns amongst +primes between 1 to 100. We also came across the Sieve of Eratosthenes — +a method to list all primes. +Prime Factorisation +Any number can be written as a product of primes — keep rewriting +composite factors until only primes are left. +Recall that we call this the prime factorisation of a given number. For +example, we can find the prime factorisation of 90 as follows: +90 = 3 × 3 × 2 × 5. +The number 90 could also have been factorised as 3 × 30 or 2 × 45 or in +a few other different ways. Will these all lead to the same prime factors? +Remarkably, the resulting prime factors are always the same with +perhaps only a change in their order. For example, if we consider +factorising 3 × 30 further, we get +90 = 3 × 30 = 3 × 2 × 15 = 3 × 2 × 3 × 5, +and we have arrived at the same prime factors, just in a different order. +Note that the prime factorisation of a prime number is the prime +number itself. +Procedure for Prime Factorisation +Can you see what is happening below? +3—— 105 +5—— 35 +7 +2—— 30 +3—— 15 +5 +49 + +Ganita Prakash | Grade 7 | Part-II +Each circled number is the product of the numbers that are to its left +and below. For example, 30 = 2 × 15, 15 = 3 × 5. Each time, a composite +number is factorised so that, at least one factor is prime. We stop when +we reach a prime number. +Can you write the prime factorisation of 105 and 30 using these two +figures? +We collect the prime factors along the left and the one +at the bottom. We then construct the prime factorisation +as shown in the figure. +While carrying out factorisation, the circles are usually +left out and the steps are carried out in this format: +3 +105 + 5 +35 + +  7 +2 +30 + 3 +15 + +  5 +Let us call this method the division method. +Try finding the prime factorisation of 1200 using the method above. +If we had used the earlier method, our calculation would have been as +follows: +1200 = 40 × 30 = 5 × 8 × 5 × 6 = … +Which calculation is easier to carry out? +Factors of a Number Using Prime Factorisation +From the prime factorisation of a number, we can construct all its factors. +This can be used to simplify the procedure for finding the HCF of two +numbers. +Consider the number 840 and its prime factorisation 2 × 2 × 2 × 3 × 5 × 7. +Is 2 × 2 × 7 = 28 a factor of 840? +If yes, what should it be multiplied by to get 840? +To answer these questions, we can reorder the prime factors of 840 +as follows (recall that reordering factors does not change the product): +840 = (2 × 2 × 7) × (2 × 3 × 5). +So, +840 = 28 × 30. +Thus, (2 × 2 × 7) = 28 is a factor of 840, and it should be multiplied by +30 (2 × 3 × 5) to get 840. +3—— 105 +5—— 35 +7 +105 = 3 × 5 × 7 +50 + +Finding Common Ground +Similarly, is 2 × 7 = 14 a factor of 840? Why or why not? +Is 2 × 2 × 2 a factor of 840? Why or why not? +Is 3 × 3 × 3 a factor of 840? Why or why not? +Can we use this idea to list down all the possible factors of a number +using just its prime factors? +Find the factors of 225 using prime factorisation. +Factorising 225 to its primes, we get +225 = 5 × 5 × 3 × 3. +We have seen that any ‘subpart’ of this factorisation +gives us a factor. Let us systematically form these +subparts. +Prime factors: 3, 5. +Combination of two prime factors: 3 × 3 = 9, 5 × 5 = 25, 3 × 5 = 15. +Combination of three prime factors: 3 × 3 × 5 = 45, 3 × 5 × 5 = 75. +Combination of four prime factors: 3 × 3 × 5 × 5 = 225. Adding 1 to this +list of factors, we see that the factors of 225 are 1, 3, 5, 9, 15, 25, 45, 75, 225. +Check that all the factors of 225 occur in this list. +Figure it Out +List all the factors of the following numbers: +(a) 90 +(b) 105 +(c) 132 +(d) 360 (this number has 24 factors) +(e) 840 (this number + +has 32 factors) +After observing a few prime factorisations, Anshu claims “The larger +a number is, the longer its prime factorisation will be”. +What do you think of Anshu’s claim? +We can see that it is not true. For example, look at the prime +factorisations of 96 and 121. +96 = 2 × 2 × 2 × 2 × 2 × 3 +121 = 11 × 11. +5 225 + 5 45 + 3 9 + 3 3 + + 1 +51 + +Ganita Prakash | Grade 7 | Part-II +In mathematics, statements or claims made without proof or verification +are called ‘conjectures’. Anshu’s claim is a conjecture. We disproved +this conjecture by finding a counterexample, i.e., an example where the +conjecture is false. +Finding the HCF of Numbers Using Prime Factorisation +We now see how to make use of the observations made so far to find +common factors and the Highest Common Factor (HCF). +Example 1: Find the common factors, and the HCF of 45 and 75. +Calculate the prime factorisation for both numbers: +45 = 3 × 3 × 5 +75 = 3 × 5 × 5. +We have seen: +Factors of 45 are subparts of factors occurring in 3 × 3 × 5 and factors +of 75 are subparts of factors occurring in 3 × 5 × 5. +So the common factors should be subparts of both the factorisations. +Can you write them down? +While exploring or solving problems you might also get some conjectures. +You can try to reason and verify these and also share with the class. +3, 5, 3 × 5 = 15 are subparts of both, and hence, they are the common +factors along with 1. The highest among them is 15. So, their HCF = 15. +Example 2: Find the common factors, and the HCF of 112 and 84. +Calculating prime factorisations, we get, +112 = 2 × 2 × 2 × 2 × 7 and +84 = 2 × 2 × 3 × 7. +3 × 3 × 5 + +3 +3 × 5 × 5 +3 × 3 × 5 + +5 +3 × 5 × 5 +3 × 3 × 5 +3 × 5 +3 × 5 × 5 +52 + +Finding Common Ground +Finding the common subparts, we get +The highest among the common factors, HCF, is 2 × 2 × 7 = 28. +Example 3: Find the common factors and the HCF of 96 and 275. +We have +96 = 2 × 2 × 2 × 2 × 2 × 3 +275 = 5 × 5 × 11. +There is no subpart that is common amongst these two factorisations. +So, 1 is the only common factor. It is also their HCF. +Figure it Out +Find the common factors and the HCF of the following numbers: +(a) 50, 60 +(b) +140, 275 +(c) 77, 725 +(d) +370, 592 +(e) 81, 243 +How do we directly find the HCF without listing all the factors? +Example 4: Find the HCF of 30 and 72. +30 = 2 × 3 × 5 +72 = 2 × 2 × 2 × 3 × 3 +We need the largest common subpart to find the HCF. Clearly, it will +contain only those primes that occur in both the factorisations: 2 and 3 +in this case. +How many 2s will it contain? +2 × 2 × 2 × 2 × 7 +2 +2 × 2 × 3 × 7 +2 × 2 × 2 × 2 × 7 +7 +2 × 2 × 3 × 7 +2 × 2 × 2 × 2 × 7 +2 × 2 +2 × 2 × 3 × 7 +2 × 2 × 2 × 2 × 7 + +2 × 7 +2 × 2 × 3 × 7 +2 × 2 × 2 × 2 × 7 + +2 × 2 × 7 +2 × 2 × 3 × 7 +53 + +Ganita Prakash | Grade 7 | Part-II +The prime factorisation of 30 has only one 2. So, the largest common +subpart should contain only one 2. +How many 3s will it contain? +The prime factorisation of 30 has only one 3. So, the largest common +subpart should contain only one 3. +This has been carried out below. +30 = 2 × 3 × 5 +72 = 2 × 2 × 2 × 3 × 3 +HCF = 2 × 3 = 6 +Thus, to find the HCF, we identify the common primes and find the +minimum number of times each of them appear in the factorisations of +the given numbers. +Example 5: Find the HCF of 225 and 750. +225 = 3 × 3 × 5 × 5 +750 = 2 × 3 × 5 × 5 × 5 +Common primes: 3 and 5. Let us find the largest common subpart. +How many 3s will it contain? +750 contains only one 3, which is the minimum number of 3 across +both numbers. So the largest common subpart should contain only one 3. +How many 5s will it contain? +225 contains the minimum number of 5s +which occurs two times. So, the largest common +subpart should contain two 5s, i.e., 5 × 5. +To find the HCF of more than 2 numbers, a similar method of finding +the largest common subpart of the common primes can be followed. +Figure it Out +1. Find the HCF of the following numbers: +(a) 24, 180 +(b) +42, 75, 24 +(c) 240, 378 +(d) +400, 2500 +(e) 300, 800 +2. Consider the numbers 72 and 144. Suppose they are factorised into +composite numbers as: 72 = 6 × 12 and 144 = 8 × 18. Seeing this, can +one say that these two numbers have no common factor other than +1? Why not? +225 = 3 × 3 × 5 × 5 +750 = 2 × 3 × 5 × 5 × 5 +HCF = 3 × 5 × 5 = 75 +54 + +Finding Common Ground +3.2 Least, but not Last! +Anshu and Guna make torans out of strips of cloth. Multiple strips are +placed one next to another to make a toran. Anshu uses strips of length +6 cm and Guna uses strips of 8 cm length. If both have to make torans of +the same length, what is the smallest possible length, the torans could be? +What is the length of the shortest toran that they can both make? +Anshu uses cloth strips of 6 cm each. Any toran he makes will be a +multiple of 6. So, the length of the toran +could be 6, 12, 18, 24, 30, 36, 42, 48, 54 cm, +and so on. +Similarly, any toran Guna makes should +be a multiple of 8. So, the length of the toran +he makes could be 8, 16, 24, 32, 40, 48, 56, +64, 72 cm, and so on. +From this, we can see that if both have to +make torans of the same length, the length +of the toran should be a common multiple of 6 and 8. +From the two lists, we can see that 24 and 48 are two of the common +multiples of 6 and 8. So, 24 cm and 48 cm are lengths of toran that Anshu +and Guna can both stitch. +24 is the smallest among them. So, 24 cm is the length of the shortest +toran that both can stitch. 24 is the lowest number among all the common +multiples of 6 and 8. +What about the largest common multiple? Does such a number exist? +A sweet shop gives out free gajak to school +children on Mondays. Today is a Monday +and Kabamai enjoyed eating the gajak. But +she visits the sweet shop once every 10 days. +When is the next time she would be able to +get free gajak from Sweet shop? (Answer in +number of days.) +As we saw before, imagining or visualising the scenario helped us to see +that it can be solved using multiples of the strips’ lengths. +Since the shop gives free sweets every Monday, it will give free sweets +again after +7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, ... days. +These are multiples of 7. +Gajak is a sweet made from +sesame seeds, jaggery and ghee +55 + +Ganita Prakash | Grade 7 | Part-II +Kabamai will arrive at the sweet shop again after +10, 20, 30, 40, 50, 60, 70, ... days. +These are multiples of 10. +When will Kabamai eat free sweets again? It will happen on days +common to the sequences of multiples above. It can be seen that this +will first happen after 70 days. +Notice that, here too, 70 is the lowest among all the common multiples +of 7 and 10. +For both these problems the solution was the lowest common multiple. +The Lowest Common Multiple (LCM) of two or more given numbers +is the lowest (or smallest or least) of their common multiples. +Do you remember the ‘Idli-Vada’ game from Grade 6 (see chapter +‘Prime Time’)? Two numbers are chosen and whenever players come +to their multiples, ‘idli’ or ‘vada’ should be called out depending on +whose multiple the number is. If the number happens to be a common +multiple, then ‘idli-vada’ should be called out. In each problem below, +the two numbers corresponding to ‘idli’ and ‘vada’ are given. Find the +first number for which ‘idli-vada’ will be called out: +(a) 4 and 6 +(b) 7 and 11 +(c) 14 and 30 +(d) 15 and 55 +Is the answer always the LCM of the two numbers? Explain. +As in the case of the HCF, the process of finding the LCM by listing down +the multiples may get tedious for larger numbers, as you would have +seen for questions (c) and (d) above. +Prime factorisation can simplify the process of finding the LCM as well. +How do we find the LCM of two numbers using their prime factors? +Finding LCM through Prime Factorisation +We have seen that every factor of a number is formed by taking a subpart +of its prime factorisation. We used this fact to come up with a method to +find the HCF of two numbers. In a similar manner, we can come up with +a method to find the LCM. +We begin by comparing the prime factorisations of a number and a +multiple of that number. For example, let us take 36 and its multiple +648 (=36 × 18). +56 + +Finding Common Ground +We get, +36 = 2 × 2 × 3 × 3, +648 = 36 × 18 = (2 × 2 × 3 × 3) × (2 × 3 × 3). +What do you observe? We can see that the prime factors of the multiple +contain the prime factors of the number along with some more prime +factors. Will this happen with every multiple? +Can this be used to find the LCM? +Example 6: Find the LCM of 14 and 35. +We get, +14 = 2 × 7 +35 = 5 × 7. +Common multiples should contain each prime factor as a subpart: +2 × 7 as a subpart and 5 × 7 as a subpart. +For example, +2 × 7 × 5 × 7 × 3 is a common multiple of 14 and 35. +2 × 2 × 5 × 7 × 7 × 11 is another common multiple. +Is 2 × 3 × 5 × 7 also a common multiple? +What is the lowest among all the common multiples of 14 and 35? +It is 2 × 5 × 7 = 70 because 2 × 5 × 7 contains 14 = 2 × 7 as well as +35 = 5 × 7, and removing any number from 2 × 5 × 7 will give a number +that is not a common multiple of 14 and 35. +Example 7: Find the LCM of 96 and 360. +We have, +96 = 2 × 2 × 2 × 2 × 2 × 3 +360 = 2 × 2 × 2 × 3 × 3 × 5. +Every common multiple should contain the prime factors of both 96 +and 360. For the LCM, we will need the smallest such number. Let us +build the LCM looking at each prime factor. +The prime factors appearing in both the numbers are 2, 3 and 5. +Now, we shall find out how many occurrences there are for each +prime factor. +How many 2s should the LCM contain? +The factorisation of 96 contains 2 × 2 × 2 × 2 × 2 (five occurrences of 2s) +and the factorisation of 360 contains 2 × 2 × 2 (three occurrences of 2s). +Choosing five occurrences of 2s as part of LCM will contain both these +subparts. +57 + +Ganita Prakash | Grade 7 | Part-II +Choosing more than five occurrences of 2s will give a common +multiple; but it will not be the lowest. Are you able to see why? +How many 3s should the LCM contain? +The factorisation of 96 contains 3 (one occurrence of 3) and the +factorisation of 360 contains 3 × 3 (two occurrences of 3s). Choosing two +occurrences of 3s as part of LCM will contain both these subparts. +How many 5s should the LCM contain? +The factorisation of 96 doesn’t have any 5s and the factorisation of +360 has one occurrence of 5. So, we choose one occurrence of 5 to be a +part of the LCM. +Thus, the LCM of 96 and 360 will be 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440. +To build the LCM of two numbers, we can identify all the prime factors +and find the maximum number of times each of them occur in either of +the factorisations. This process can be extended to find the LCM of two +or more numbers. +Figure it Out +Find the LCM of the following numbers: +(a) 30, 72 +(b) 36, 54 +(c) 105, 195, 65 +(d) 222, 370 +3.3 Patterns, Properties, and a Pretty Procedure! +The HCF of 6 and 18 is 6, which is one of the two given numbers. +Find more such number pairs where the HCF is one of the two numbers. +How can we describe such pairs of numbers? +We can see that it happens when one number is a factor of the other. This +also means the other number will +be a multiple of the first number! +Such a statement describing a +pattern or a property that holds in +all possible cases is called a general +statement. This process is called +generalisation. +Such a generalisation can also +be described using algebra. Let us +see how. +Be my factor, I’ll +be your multiple! +58 + +Finding Common Ground +If n is a number, then any multiple of n can be written as a positive +integer multiplied by n. For example, if we take n and 5n (short for 5 × +n), then 5n is a multiple of n, and n is a factor of 5n. +The HCF of n and 5n = n. +For number pairs satisfying this property (i.e., one of the numbers is the +HCF), +(a) if m is a number, what could be the other number? +(b) if 7k is a number, what could be the other number? +Figure it Out +1. Make a general statement about the HCF for the following pairs +of numbers. You could consider examples before coming up +with general statements. Look for possible explanations of why +they hold. +(a) Two consecutive even numbers +(b) Two consecutive odd numbers +(c) Two even numbers +(d) Two consecutive numbers +(e) Two co-prime numbers +Share your observations with the class. +2. The LCM of 3 and 24 is 24 (it is one of the two given numbers). +(a) Find more such number pairs where the LCM is one of the two +numbers. +(b) Make a general statement about such numbers. Describe such +number pairs using algebra. +3. Make a general statement about the LCM for the following pairs of +numbers. You could consider examples before coming up with these +general statements. Look for possible explanations of why they hold. +(a) Two multiples of 3 +(b) Two consecutive even numbers +(c) Two consecutive numbers +(d) Two co-prime numbers +What happens to the HCF of two numbers if both numbers are doubled? +Take some pairs of numbers and explore. Are you able to see why the +HCF will also double? +Math +Talk +59 + +Ganita Prakash | Grade 7 | Part-II +If both numbers are doubled, then both numbers get an extra factor of +2 in their prime factors. This 2 will be included as a factor in the largest +common subpart, and so the HCF will double. For example, consider the +numbers 270 and 50. +270 = 2 × 3 × 3 × 3 × 5 +50 = 2 × 5 × 5 +HCF = 2 × 5 = 10 +Let us double these numbers to get 540 and 100. +540 = 2 × 2 × 3 × 3 × 3 × 5 +100 = 2 × 2 × 5 × 5 +HCF = 2 × 2 × 5 =20. +Consider the following two multiples of 14 — 14 × 6, 14 × 9. What is +their HCF? +Clearly, 14 is a common factor. Is it also the highest common factor? +To see it, let us calculate the prime factorisations. +14 × 6 = 2 × 7 × 2 × 3 + 14 +14 × 9 = 2 × 7 × 3 × 3 +14 +HCF = 14 × 3 = 42. +Here are some more numbers where both numbers are multiples of the +same number. Find their HCF: +(a) 18 × 10, 18 × 15 +(b) 10 × 38, 10 × 21 +(c) 5 × 13, 5 × 20 +(d) 12 × 16, 12 × 20 +In which of these cases is the HCF the same as the common multiplier, +like problem (b) where the HCF is 10? Explore a few more examples of +this type to understand when this happens. +Efficient Procedures for HCF and LCM +See the procedure on the right. Can you explain how it +has been carried out? +2 +84, 180 + 2 42, 90 + 3 21, 45 + 7, 15 +60 + +Finding Common Ground +This is similar to the procedure for prime factorisation. At each step, +the two numbers are divided by a common prime factor, and the two +quotients are written down in the next row. This continues till we get +two numbers that do not have any common prime factors. +How do we use this to find the HCF of 84 and 180? Explore. +[Hint: Observe that 84 = 2 × 2 × 3 × 7, and 180 = 2 × 2 × 3 × 15 similar to +prime factorisation] +Find the HCF in the following cases. +HCF = 2 × 5 × 5 × 3 +HCF = 2 × 5 × 7 +2 +300, 150 + 5 150, 75 + 5 30, 15 + 3 6, 3 + 2, 1 +2 +630, 770 + 5 315, 385 + 7 63, 77 + 9, 11 + +This procedure not only gives the HCF but can also be used to find the +LCM! Can you see how? +2 +300, 150 + 5 150, 75 + 5 30, 15 + 3 6, 3 + 2, 1 +LCM = 2 × 5 × 5 × 3 × 2 × 1 +2 +630, 770 + 5 315, 385 + 7 63, 77 + 3 9, 11 + 3, 11 +LCM = 2 × 5 × 7 × 3 × 3 × 11 +Why are these the LCMs? +[Hint: Will the product of the factors marked as the LCM of 300 and 150 +contain the prime factorisations of both 300 and 150? Is this the smallest +such number?] +Guna says “I found a better way to factorise to find +HCF/LCM. This is faster than what was taught in class!” +“For the numbers 300 and 150, I can first directly divide +both numbers by 50. +The HCF will be 50 × 3. +50 300, 150 + 3  6,  3 + + 2,  1 +61 + +Ganita Prakash | Grade 7 | Part-II +The LCM will be 50 × 3 × 2 × 1”. +Anshu also tried to remove the bigger common factors. +“For 630 and 770, I will divide both numbers by 10 first. +Now, I can divide them by 7. +The HCF will be 10 × 7 = 70 +The LCM will be 10 × 7 × 9 × 11 = 6930”. +Can you see why this works? +We need not restrict ourselves to dividing only one prime factor at a +time. Both numbers can be divided by whatever common factor we are +able to identify. +You can try this method for these pairs of numbers. +(a) 90 and 150 +(b) 84 and 132 +Property Involving both the HCF and the LCM +Which is greater — the LCM of two numbers or their product? +You could analyse the above statement using examples. Then try to reason +or prove, why the LCM is never greater than the product of the numbers. +[Hint: Is the product also a common multiple of the two numbers?] +There is an interesting relation between the product of two numbers +and their HCF and LCM. +Consider the numbers 105 and 95. Find their LCM. +Factorising them into their primes: +105 = 3 × 5 × 7 +95 = 5 × 19 +LCM = 3 × 5 × 7 × 19. +Let us consider the product in the factorised form: +105 × 95 = 3 × 5 × 5 × 7 × 19 +Is the LCM a factor of the product? If yes, what should it be multiplied +with to get the product? It can be seen that +105 × 95 = LCM × 5. +Explore whether the LCM is a factor of the product in the following cases. +If yes, identify the number that the LCM should be multiplied by to get +the product. Do you see any pattern? Use these numbers: +(a) 45, 105 +(b) 275, 352 +(c) +222, 370 +10 630, 770 + 7 63, 77 + + 9, 11 +Try +This +62 + +Finding Common Ground +Do you see that, in each case, the number by which the LCM is multiplied +to get the product is actually the HCF? +Thus, our observations seem to suggest the following: +HCF × LCM = Product of the two numbers. +Why does this happen? Can you give an explanation or proof? +[Hint: Consider the prime factorisation of the given numbers. +Among their prime factors, some are common to both factorisations, +and the rest occur in only one of them. Between the HCF and the LCM, +see how the common and non-common prime factors get distributed. +In the product, observe how these two kinds of prime factors occur. +Compare them.] +Explore whether this property holds when 3 numbers are considered. +Figure it Out +1. In the two rows below, colours repeat as shown. When will the blue +stars meet next? +2. (a) Is 5 × 7 × 11 × 11 a multiple of 5 × 7 × 7 × 11 × 2? + +(b) +Is 5 × 7 × 11 × 11 a factor of 5 × 7 × 7 × 11 × 2? +3. Find the HCF and LCM of the following (state your answers in the +form of prime factorisations): +(a) 3 × 3 × 5 × 7 × 7 and 12 × 7 × 11 +(b) 45 and 36 +4. Find two numbers whose HCF is 1 and LCM is 66. +5. A cowherd took all his cows to graze in the fields. The cows came to +a crossing with 3 gates. An equal number of cows passed through +each gate. Later at another crossing with 5 gates again an equal +number of cows passed through each gate. The same happened at +the third crossing with 7 gates. If the cowherd had less than 200 cows, +how many cows did he have? (Based on the folklore mathematics +from Karnataka.) +Try +This +Try +This +63 + +Ganita Prakash | Grade 7 | Part-II +6. The length, width, and height of a box are 12 cm, 18 cm, and 36 cm +respectively. Which of the following sized cubes can be packed in +this box without leaving gaps? +(a) 9 cm +(b) +6 cm +(c) 4 cm +(d) +3 cm +(e) 2 cm +7. Among the numbers below, which is the largest number that perfectly +divides both 306 and 36? +(a) 36 +(b) +612 +(c) 18 +(d) +3 +(e) 2 +(f) +360 +8. Find the smallest number that is divisible by 3, 4, 5 and 7, but +leaves a remainder of 10 when divided by 11. +9. Children are playing ‘Fire in the Mountain’. When the number 6 was +called out, no one got out. When the number 9 was called out, no one +got out. But when the number 10 was called out, some people got +out. How many children could have been playing initially? +(a) 72 +(b) +90 +(c) 45 +(d) +3 +(e) 36 +(f) +None of these +10. Tick the correct statement(s). The LCM of two different prime +numbers (m, n) can be: +(a) Less than both numbers +(b) In between the two numbers +(c) Greater than both numbers +(d) Less than m × n +(e) Greater than m × n +11. A dog is chasing a rabbit that has a head start of 150 feet. It jumps +9 feet every time the rabbit jumps 7 feet. In how many leaps does +the dog catch up with the rabbit? +12. What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, +10? Do you remember the answer from Grade 6, Chapter 5? +13. Here is a problem posed by the ancient Indian Mathematician +Mahaviracharya (850 C.E.). Add together 8 +15, 1 +20, 7 +36, 11 +63 and 1 +21. +What do you get? How can we find this sum efficiently? +Try +This +Try +This +Math +Talk +64 + +Finding Common Ground +• Last year, we looked at common multiples, and common factors, and +were also introduced to the amazing world of primes! +• In this chapter, we learnt a method to find the prime factorisation of +a number. +• Finding all the factors of a number from its prime factorisation is easy +but quite tedious — we have to list every possible subpart! +• The Highest Common Factor (HCF) is the highest among all the +common factors of a group of numbers. + ƒ +Every common factor is contained in the prime factorisation of +the number. + ƒ +To find the HCF, we include the minimum number of occurances +of each prime across the prime factorisation of all the numbers. +• The Lowest Common Multiple (LCM) is the lowest among all the +common multiples of a group of numbers. + ƒ +Every common multiple contains the prime factorisation of the +numbers. + ƒ +To find the LCM, we include the highest number of occurrences +of each prime across the prime factorisations of all the numbers. +• We explored more about HCF and LCM; we discovered related +properties and patterns when numbers are consecutive, even, +co-prime, etc. +• We learnt a procedure to get both the HCF and the LCM at the same +time! We also saw how to make this even quicker! +• We learned some terms that are used when discussing mathematics, +such as ‘conjecture’ and ‘generalisation’. +If I start writing +this number, +how long could +it take me? +The largest prime found so far has +4,10,24,320 digits! It was discovered on +October 12, 2024. +SUMMARY +65 + +Mystery Colours! +You might have noticed and wondered about these different circle +designs around the page numbers on each page! +The picture below shows all the designs for the numbers from +1 to 100. +Try to decode the colour scheme for each number. +There are several interesting patterns here. +Share your observations with your classmates. +Extending this scheme, colour the page numbers from 101 – 110." +class_7,12,another peek beyond the point,ncert_books/class_7/gegp1dd2/gegp204.pdf,"ANOTHER PEEK +BEYOND THE POINT +4 +4.1 A Quick Recap of Decimals +Recall that decimals are the natural extension of the Indian place value +system to represent decimal fractions ( 1 +10, 1 +100, +1 +1000, and so on) and +their sums. +For example, 27.53 refers to a quantity that has: +• +2 Tens +• +7 Units (Ones) +• +5 Tenths +• +3 Hundredths +We have already learned how to multiply and divide fractions. In this +chapter, we will learn how to perform these operations with decimals. +You will see that the procedures for multiplying and dividing decimals +are natural extensions of the procedures for multiplying and dividing +counting numbers. +Jonali and Pallabi play a game. Jonali +says a fraction and Pallabi gives the +equivalent decimal. Write Pallabi’s +answer in the blank spaces. +Jonali goes to the market to +buy spices. She purchases 50 g of +Cinnamon, 100 g of Cumin seeds, 25 g +of Cardamom and 250 g of Pepper. +Express each of the quantities in +kilograms by writing them in terms of +fractions as well as decimals. +The fractions Jonali gave Pallabi +have denominators 10, 100, 1000, and +so on. + +Ganita Prakash | Grade 7 | Part-II +Write the following fractions as a sum of fractions and also as decimals: +Fraction +Expanding the +Numerator +Sum of one-tenths, +one-hundredths, +one-thousands,... +Decimals +254 +1000 +200 +1000 + 50 +1000 + +4 +1000 += 2 +10 + 5 +100 + +4 +1000 +0.2 + 0.05 + 0.004 +0.254 +847 +10000 +173 +100 +23 +1000 +Can you give a simple rule to divide any number by a number of the +form 1 followed by zeroes — 10, 100, 1000, etc.? For example, 123 +10 +, +24 +100 + or 678 +1000 + ? Look for a pattern in the previous problems. +Here is one such rule. Let us consider the example 123 ÷ 10. +Step 1: Write the dividend as it is and place a decimal point at the end. +123. +Step 2: Count the number of zeroes in the divisor. +10 + 1 zero +Step 3: Move the decimal point from Step 1 left by the same number of +places as the count from Step 2. Add zeroes in front if needed. +Math +Talk +12.3 +68 + +Another Peek Beyond the Point +Examples: +24 ÷ 100 = 0.24 +678 ÷ 1000 = 0.678 +12÷1000 = 0.012 +12345÷1000 = 12.345 +4.2 Decimal Multiplication +Example 1: Arshad goes to a stationery shop and purchases 5 pens. If +one pen costs ₹9.5 (9 rupees and 50 paisa), how much should he pay the +shopkeeper? +What operation must we use here? +We have to multiply 9.5 by 5, which is the same as adding 9.5, 5 times. +That is 9.5 × 5 = 9.5 + 9.5 + 9.5 + 9.5 + 9.5 = 47.5. +We can also directly multiply the numbers by converting them into +fractions. +9.5 is 95 +10 and 5 is 5 +1 as a fraction. +The cost of 5 pens = 5 +1 × 95 +10 . +Recall that, to find the product of two fractions, we multiply the +numerators and multiply the denominators. +5 +1 × 95 +10 = 5 × 95 +1 × 10 += 475 +10 + = 47.5. +The cost of 5 pens is ₹47.5. +Example 2: A car travels 12.5 km per litre of petrol. What is the distance +covered with 7.5 litres of petrol? +We have to multiply 12.5 by 7.5. +The distance covered = 12.5 × 7.5 += 125 +10 × 75 +10 = 125×75 +10×10 = 9375 +100 = 93.75 +The distance covered is 93.75 km. +69 + +Ganita Prakash | Grade 7 | Part-II +Can the product of two decimals be a natural number? +Can the product of a decimal and a natural number be a natural +number? +Example 3: The distance between Ajay's school and his home is 827 m. +He walks to school in the morning and then walks back home in the +evening, 6 days a week. How much does he walk in a week? Answer in +kilometres. +Each way between school and home, Ajay walks 827 metres, +i.e., 0.827 km. +So, in a day he walks, +0.827 × 2 = 827 +1000 × 2 = 827 × 2 +1000 = 1654 +1000 = 1.654 +He goes to school 6 days a week. So, in a week, he walks +1.654 × 6 = 1654 +1000 × 6 = 1654 × 6 +1000 + = 9924 +1000 = 9.924 +Ajay walks 9.924 km a week. +Example 4: Find the area of the given +rectangle. +Area of the rectangle = +5.7 × 13.3 = 57 +10 × 133 +10 = 7581 +100 = 75.81 +The area is 75.81 sq cm. +Observe the number of digits after the decimal point in the multiplier, +the multiplicand and the product. Also note the number of zeroes in the +denominator. +Examples +No. of digits after the decimal point in: +Multiplier +Multiplicand +Product +9.5 × 5 += 95 × 5 +10 + = 475 +10 = 47.5 +1 +0 +1 +12.5 × 7.5 += 125 +10 × 75 +10 = 9375 +100 = 93.75 +1 +1 +2 +Math +Talk +5.7 cm +13.3 cm +70 + +Another Peek Beyond the Point +1.64 × 6 += 164 +100 × 6 = 984 +100 = 9.84 +2 +0 +2 +5.7 × 13.35 += 57 +10 × 1335 +100 += 57 × 1335 +10 × 100 = 76095 +1000 +=76.095 +1 +2 +3 +Suppose we know that 596 × 248 = 147808, can you immediately +write down the product of 5.96 × 24.8? +By looking at the above examples, can you frame a rule to multiply two +decimals? +Multiplication of decimals is the same as the multiplication of their +corresponding fractions. When multiplying fractions, we multiply the +numerators and denominators, respectively. +The product of the numerators = Product of the numbers with the +decimal points removed. +Since both the denominators are of the form 1, 10, 100, 1000, … , the +product of the denominators is also of the form 1 followed by zeroes. +The number of zeroes in the product is the sum of the number of zeroes +in each denominator. +In the product, the decimal point is placed based on the total number +of zeroes in the denominator. +So, to multiply two decimals, we can multiply the two numbers +obtained by removing the decimal point, and then place the decimal +point appropriately as shown below. +To evaluate 5.96 × 24.8: +Math +Talk +596 × 248 = 147808 +2 decimal +places +1 decimal +place +3 decimal +places +2 + 1 = 3 decimal places +5.96 × 24.8 +147.808 += +} +} +} +} +71 + +Ganita Prakash | Grade 7 | Part-II +Example 5: Let us use the above rule to find the product of 5.8 and 1.24. +Let us first multiply 58 and 124. +The product is 7192. +The sum of the number of digits after +the decimal point in the multiplier and +multiplicand is 3. So, the product is 7.192. +Verify this by converting the multiplier +and multiplicand into fractions. +Is the Product Always Greater than the Numbers +Multiplied? +Recall multiplication of two fractions. Unlike counting numbers, when +two fractions are multiplied the product is not always greater than or +equal to both numbers. Let us examine the case of the product of two +decimals. +2.25 × 8 = 18. +In the above multiplication, the product (18) is greater than 2.25 +and 8. +But in, 0.25 × 8 = 2, the product (2) is greater than 0.25 but less than 8. +In the case of 0.25 × 0.8 = 0.2, the product (0.2) is less than both 0.25 +and 0.8. +When is the product of two decimals greater than both the numbers? +When is it less than both the numbers? +Since decimals are just representations of fractions, the relationship +between the numbers multiplied and the product are similar to fractions. +Situation +Multiplication +Relationship +Situation 1 +Both numbers are greater +than 1 (e.g., 3.4 × 6.5) +The product (22.1) is greater +than both the numbers. +Situation 2 +Both numbers are between 0 +and 1 (e.g., 0.75 × 0.4) +The product (0.3) is less +than both the numbers. +Situation 3 +One number is between 0 +and 1 and one number is +greater than 1 (e.g., 0.75 × 5) +The product (3.75) is less +than the number greater +than 1 and greater than the +number between 0 and 1. +1 digit after +decimal point +2 digits after +decimal point +5.8 × 1.24 +72 + +Another Peek Beyond the Point +Figure it Out +1. Recall that a tenth is 0.1, a hundredth is 0.01, and so on. Find the +following products in tenths, hundredths and so on: +(a) 6 × 4 tenths = 24 tenths +(b) 7 × 0.3 +(c) 9 × 5 hundredths +2. Find the products: +(a) 27.34 × 6 +(b) 4.23 × 3.7 +(c) 0.432 × 0.23 +3. Thejus needs 1.65 m of cloth for a shirt. How many metres of cloth +are needed for 3 shirts? +4. Meenu bought 4 notebooks and 3 erasers. The cost of each book was +₹15.50 and each eraser was ₹2.75. How much did she spend in all? +5. The thickness of a rupee coin is 1.45 mm. What is the total height of +the cylinder formed by placing 36 rupee coins one over the other? +Write the answer in centimeters. +6. The price of 1 kg of oranges is ₹56.50. What is the price of +2.250 kg of oranges? Can we write 56.50 as 56.5 and 2.250 as 2.25 +and multiply? Will we get the same product? Why? +7. Dwarakanath purchases notebooks at a wholesale price of ₹23.6 +per piece and sells each notebook at ₹30/-. How much profit does he +make if he sells 50 books in a week? +8. Given that 18 × 12 = 216, find the products: +(a) 18 × 1.2 +(b) +18 × 0.12 +(c) 1.8 × 1.2 +(d) +0.18 × 0.12 +(e) 0.018 × 0.012 +(f) +1.8 × 12 +In which of the cases above is the product less than 1? +9. In which of the following multiplications is the product less than 1? +Can you find the answer without actually doing the multiplications? +(a) 7 × 0.6 +(b) +0.7 × 0.6 +(c) 0.7 × 6 +(d) +0.07 × 0.06 +10. Multiplying the following numbers by 10, 100 and 1000 to complete +the table. +Math +Talk +73 + +Ganita Prakash | Grade 7 | Part-II +× 10 +× 100 +× 1000 +5.7 +23.02 +0.92 +0.306 +24.67 +4.3 Decimal Division +Example 6: Anuja has a 3.9 m length +of ribbon and she wants to cut it into +10 equal pieces. What is the length of +each piece in decimal? +Since there are 10 pieces, we +can find the length of each piece by +dividing 3.9 by 10. +So, what is 3.9 ÷ 10? +Let’s convert 3.9 into fraction 3.9 = 39 +10. +Hence, 3.9 ÷ 10 is the same as +dividing 39 +10 by 10. +Recall that, dividing a fraction by +a divisor is the same as multiplying +the fraction by the reciprocal of the +divisor. The reciprocal of 10 is 1 +10 +. +So, 39 +10 ÷ 10 = 39 +10 × 1 +10 = 39 +100 = 0.39. +Thus, the length of each piece of +ribbon is 0.39 m. +What is the length of each piece if the ribbon is cut into 100 equal pieces? +3.9 ÷ 100 = 39 +10 × 1 +100 = 39 +1000 = 0.039. +When the ribbon is cut into 100 equal pieces, the length of each piece +is 0.039 m. +What is the length of +each piece in decimal? +74 + +Another Peek Beyond the Point +What is 0.039 m in centimetres and millimetres? +By looking at these divisions, we can frame a simple rule for dividing +decimals by 1, 10, 100, 1000, and so on. +When we divide a decimal by 1, 10, 100, 1000, and so on, we can just +move the decimal point to the left by as many places as there are zeroes +in the divisor! +Decimal +÷ 10 +÷ 100 +÷ 1000 +÷ 10000 +18.7 +1.87 +0.187 +0.0187 +0.00187 +21.1 +0.13 +2.146 +0.0058 +Example 7: Neenu has 29 metres of red ribbon and wants to share it +equally with Anu. What is the length of +ribbon that each of them will get? +Since the ribbon needs to be divided into +two equal parts, each girl will get a piece of +29 ÷ 2 metres. +If each of them gets 14 m, then 1 m +remains. If we divide 1 m among the two, +each will get another 1 +2 m. +How do we convert 1 +2 into a decimal? +It is easy to express a fraction as a decimal +if the denominator is 1, 10, 100, 1000, etc. +So, can we find a fraction equivalent to 1 +2 +with such a denominator? +1 +2 = 1 × 5 +2 × 5 = 5 +10 (multiplying the numerator and denominator by 5). +75 + +Ganita Prakash | Grade 7 | Part-II +We know that the fraction 5 +10 can be represented as a decimal 0.5. +So, each girl will get 14 m and an additional 0.5 m of ribbon. +Hence, the length of ribbon each will get is 14 + 0.5 = 14.5 m. +Now, what if the ribbon was shared between four friends instead +of 2? +So, each will get 29 ÷ 4 m, that is 29 +4 m. +Now, the denominator of the fraction is 4. To convert a fraction to a +decimal, it helps if the denominator is of the form 1, 10, 100, 1000, and +so on. Can we find a fraction equivalent to 29 +4 with such a denominator? +Is 4 a factor of 10? No. Is it a factor of 100? Yes. 4 × 25 = 100. So we +can get an equivalent fraction of 29 +4 by multiplying the numerator and +denominator by 25. +29 × 25 +4 × 25 = 725 +100 = 7.25 +So each of the 4 friends will get 7.25 m of ribbon. +Division Using Place Value +We have seen how to divide two counting numbers to get a decimal +quotient. We first represented the division as a fraction. Then we found +an equivalent fraction, with the denominator being of the form 1, 10, 100, +1000, and so on. It was then easy to represent this equivalent fraction as +a decimal. +Now, let us look at the division using place value procedure to calculate +the decimal quotient. +Suppose we want to write the quotient 10 +3 as a decimal. Can we convert +this fraction to an equivalent fraction with a denominator such as 1, 10, +100, 1000, etc.? +It is not possible. So, we need a more general method to divide any +two counting numbers. Let us see how we can use division using place +value for this. +Let us start with a quick recap of division using place value. +76 + +Another Peek Beyond the Point +Example 8: Find the value of 1324 ÷ 4. +1324 ÷ 4 + Divide 1324 into 4 equal parts. +1 Thousand + 3 Hundreds + 2 Tens + 4 Ones +1 Thousand ÷ 4 + Not possible without regrouping. +Regroup 1 Thousand into 10 Hundreds. +10 Hundreds + 3 Hundreds = 13 Hundreds. +13 Hundreds + 2 Tens + 4 Ones +13 Hundreds ÷ 4 + Each part gets 3 Hundreds, and 1 Hundred remains. +13 Hundreds + 2 Tens + 4 Ones +3 Hundreds +3 Hundreds +3 Hundreds +3 Hundreds +1 Hundred +4) 1324 (0 + 3 + ___ + ___ + –12 + 1 +Th +H +T +O +Regroup 1 Hundred into 10 Tens. +10 Tens + 2 Tens = 12 Tens. +77 + +Ganita Prakash | Grade 7 | Part-II +12 Tens ÷ 4 + Each part gets 3 Tens. +4 Ones ÷ 4 + Each part gets 1 Ones. +So, 1324 ÷ 4 = 0 Thousands + 3 Hundreds + 3 Tens + 1 Ones = 331. +We also call this division using place values as ‘long division’. +Division with a Decimal Quotient +Now, let us use this understanding of long division to find the value of +1325 ÷ 4. +1325 ÷ 4 + Divide 1325 into 4 equal parts. +We can follow the same steps as in the previous problem. +12 Tens + 4 Ones +3 Hundreds +3 Tens +3 Hundreds +3 Tens +3 Hundreds +3 Tens +3 Hundreds +3 Tens +Th H +T +O +4) 1324 (0 + 3 + 3 + ___ + –12 + 12 + –12 + 0 +4 Ones +3 Hundreds +3 Tens +1 Ones +3 Hundreds +3 Tens +1 Ones +3 Hundreds +3 Tens +1 Ones +3 Hundreds +3 Tens +1 Ones +4) 1324 (0 + 3 + 3 + 1 + –12 + 12 + –12 + 04 + – 4 + 0 +Th H +T +O +78 + +Another Peek Beyond the Point +4) 1325 (0 + 3 + 3 + 1 + –12 + 12 + –12 + 05 + – 4 + 1 +Th H +T +O +5 Ones +3 Hundreds +3 Tens +1 Ones +3 Hundreds +3 Tens +1 Ones +3 Hundreds +3 Tens +1 Ones +3 Hundreds +3 Tens +1 Ones +1 Ones +We are left with 1 Ones. +It is not clear how to divide 1 Ones into 4 equal parts. But we can +regroup this as 10 tenths. +10 Tenths +3 Hundreds +3 Tens +1 Ones +3 Hundreds +3 Tens +1 Ones +3 Hundreds +3 Tens +1 Ones +3 Hundreds +3 Tens +1 Ones +10 Tenths ÷ 4 + Each part gets 2 Tenths and 2 Tenths remain. +10 Tenths +3 Hundreds +3 Tens +1 Ones +2 Tenths +3 Hundreds +3 Tens +1 Ones +2 Tenths +3 Hundreds +3 Tens +1 Ones +2 Tenths +3 Hundreds +3 Tens +1 Ones +2 Tenths +2 Tenths +4) 1325 (0 + 3 + 3 + 1 + 2 + –12 + 12 + –12 + 05 + – 4 + 10 + – 8 + 2 +Th H +T +O +Tenths +Tens +Ones +Tenths +79 + +Ganita Prakash | Grade 7 | Part-II +We are left with 2 tenths. +To divide 2 Tenths into 4 equal parts, we have to regroup them as 20 +Hundredths. +20 Hundredths +3 Hundreds +3 Tens +1 Ones +2 Tenths +3 Hundreds +3 Tens +1 Ones +2 Tenths +3 Hundreds +3 Tens +1 Ones +2 Tenths +3 Hundreds +3 Tens +1 Ones +2 Tenths +20 Hundredths ÷ 4 + Each part gets 5 Hundredths. +20 Hundredths +3 Hundreds +3 Tens +1 Ones +2 Tenths +5 Hundredths +3 Hundreds +3 Tens +1 Ones +2 Tenths +5 Hundredths +3 Hundreds +3 Tens +1 Ones +2 Tenths +5 Hundredths +3 Hundreds +3 Tens +1 Ones +2 Tenths +5 Hundredths +2 Tenths +4) 1325 (0 + 3 + 3 + 1 + 2 + 5 + –12 + 12 + –12 + 05 + – 4 + 10 + – 8 + 20 + – 20 + 0 +Th H +T +O +Tenths +Hundredths +Tens +Ones +Tenths +Hundredths +So, 1325 ÷ 4 = 0 Thousands + 3 Hundreds + 3 Tens + 1 Ones + 2 Tenths ++ 5 Hundredths. This we know is 331.25, thus, +1325 ÷ 4 = 331.25. +Can we verify this by finding an equivalent fraction for 1325 +4 +? +To get an equivalent fraction such that the denominator is of the +form 1, 10, 100, 1000, and so on, we can multiply the numerator and +denominator by 25. +1325 × 25 +4 × 25 + = 33125 +100 = 331.25. +80 + +Another Peek Beyond the Point +Thus, the procedure of division using place value can be extended to +find quotients with decimal values. Ones can be regrouped as tenths, +tenths can be regrouped as hundredths and so on. +Example 9 : Find the value of 237 ÷ 8. +237 ÷ 8 + Divide 2 Hundreds + 3 Tens + 7 Ones into 8 equal parts. +To divide 2 Hundred into 8 equal parts we need to regrouped them +as 20 Tens. +20 Tens + 3 Tens = 23 Tens. +23 Tens ÷ 8 + 2 Tens, and 7 Tens remain. +8) 237 (0 2 + –16 + 7 +H T O +Tens +7 Tens can be regrouped as 70 Ones. +70 Ones + 7 Ones = 77 Ones. +77 Ones ÷ 8 + 9 Ones, and 5 Ones remain. +8) 237 (0 2 9 + –16 + 77 + – 72 + 5 +H T O +Tens +Ones +To divide 5 Ones into 8 equal parts we need to regroup them as 50 Tenths. +When we regroup Ones into Tenths, we place a decimal point in +the quotient. +50 Tenths ÷ 8 + 6 Tenths, and 2 Tenths remain. +Remember, when we +regroup Ones into Tenths we +need to place a decimal point +in the quotient. +8) 237 (0 2 9 . 6 + 16 + 77 + 72 + 50 + 48 + 2 +H T O +Tenths +Tens +Ones +Tenths +2 Tenths cannot be divided into 8 +equal parts. So we need to regroup +them as 20 Hundredths. +20 +Hundredths +÷ +8 + +2 +Hundredths, and 4 Hundredths +remain. +8) 237 (0 2 9 . 6 2 + 16 + 77 + 72 + 50 + 48 + 20 + 16 + 4 +H T O +Tenths +Hundredths +Tens +Ones +Tenths +Hundredths +81 + +Ganita Prakash | Grade 7 | Part-II +To divide 4 Hundredths into 8 equal parts we need to regroup them +as 40 Thousandths. +40 Thousandths ÷ 8 + 5 Thousandths. +8) 237 (0 2 9 . 6 2 5 + 16 + 77 + 72 + 50 + 48 + 20 + 16 + 40 + 40 + 0 +H T O +Tenths +Hundredths +Tens +Ones +Tenths +Hundredths +Thousandths +Thousandths +Thus, 237 ÷ 8 = 29.625. +Division with a Decimal Dividend +Example 10: A shopkeeper has 9.5 kg of sugar and he wants to pack it +equally in 4 bags. What is the weight of each bag of sugar? +To find the weight of each bag we need to divide 9.5 by 4. +4) 9.5 (2 . 3 7 5 + 8 + 15 + 12 + 30 + 28 + 20 + 20 + 0 +O +Tenths +Hundredths +Ones +Tenths +Hundredths +Thousandths +Thousandths +Again, we place the decimal point in the quotient before we divide the +tenths.Each bag of sugar weighs 2.375 kg. +82 + +Another Peek Beyond the Point +Example 11: What is the value of 0.06 ÷ 5? +0.06 + 0 Ones + 0 Tenths + 6 Hundredths +0 Ones ÷ 5 + 0 Ones. +When we move from Ones to Tenths we need +to place the decimal point in the quotient. +0 Tenths ÷ 5 + 0 Tenths. +6 Hundredths ÷ 5 + 1 Hundredths, and +1 Hundredth remains. We need to regroup 1 +Hundredth into 10 Thousandths. +10 Thousandths ÷ 5 + 2 Thousandths. +So, the quotient is 0.012. +Figure it Out +1. Find the quotient by converting the denominator into 1, 10, 100 or +1000 and verify the solution by the long division method (division +by place value). +(a) 18 +5 +(b) 415 +4 +(c) 1217 +2 + +(d) 4827 +8 +2. Choose the correct answer: +(a) 1526 +4 + = + (i) 38.15 +(ii) 380.15 + (iii) 381.5 +(iv) 381.05 +(b) 3567 +8 + = + (i) 4458.75 +(ii) 44.5875 + (iii) 445.875 +(iv) 4458.75 +3. What is the quotient? +(a) 132 ÷ 4 = +(b) 13.2 ÷ 4 = +(c) 1.32 ÷ 4 = +(d) 0.132 ÷ 4 = +4. What is the quotient? +(a) 126 ÷ 8 = +(b) 12.6 ÷ 8 = +(c) 1.26 ÷ 8 = +(d) 0.126 ÷ 8 = +(e) 0.0126 ÷ 8 = +5) 0.06 (0 . 0 1 2 + 0 + 00 + 00 + 06 + 5 + 10 + 10 + 0 +O +Tenths +Hundredths +Ones +Tenths +Hundredths +Thousandths +Thousandths +Remember, +when we +regroup Ones +into Tenths we +need to place +a decimal +point in the +quotient. +83 + +Ganita Prakash | Grade 7 | Part-II +Division with a Decimal Divisor +Example 12: Ravi went from Pune to Matheran by scooter in 2.5 hours. +The distance was 126 km. What was his average speed? +We can get the average speed by dividing the distance by the time +taken. +126 ÷ 2.5. +When the divisor is a decimal, we convert the divisor into a fraction. +126 ÷ 25 +10 = 126 × 10 +25 = 1260 +25 . +With the long division procedure we find the quotient is 50.4. +So, the average speed at which Ravi travelled was 50.4 km per hour. +Example 13: Find 4.68 ÷ 1.3. +Again converting the divisor into a fraction we get +4.68 ÷ 13 +10 = 4.68 × 10 +13 = 46.8 +13 . +Now, what about 4.68 ÷ 0.13? +4.68 ÷ 0.13 = 4.68 ÷ 13 +100 = 4.68 × 100 +13 = 468 +13 . +What do you notice in these cases? +When the divisor is a decimal, we can convert it into a counting +number by suitably multiplying it by 10, 100, 1000, and so on. We must +also multiply the dividend by the same number. Thus, +4.68 +0.13 = 4.68 × 100 +0.13 × 100 = 468 +13 . +Once we convert the divisor into a counting number, we can then +follow the division using place value procedure to find the quotient. +Does This Ever End? +Can you calculate 10 ÷ 3? Try dividing using long division. +10 + 1 Tens + 0 Ones. +Step 1: Regroup 1 Tens into 10 Ones. 10 Ones ÷ 3 + 3 Ones, and 1 Ones +remain. +Step 2: Regroup 1 Ones as 10 Tenths. 10 Tenths ÷ 3 + 3 Tenths, and 1 +Tenths remain. +84 + +Another Peek Beyond the Point +Step 3: Regroup 1 Tenths as 10 Hundredths. 10 Hundredths ÷ 3 + 3 +Hundredths , and 1 Hundredths remain. +Step 4: Regroup 1 Hundredths as 10 Thousandths. +10 Thousandths ÷ 3 + 3 Thousandths, and +1 Thousandths remain. Regroup 1 Thousandths as +10 TenThousandths. +This never seems to end! Each time we divide by 3, there +is a remainder of 1. +Will this process end? +In long division, we see that at each step we get a +remainder of 1. So, the process will never end! +So, 10 ÷ 3 cannot be expressed using a finite number of +digits in the decimal form. +10 ÷ 3 = 3.333 ... +There are decimal divisions where the quotient never ends! We will +explore such numbers in greater detail in a later class. +Can you find the quotients of 10 ÷ 9, and 100 ÷ 11? +Now divide 1 by 7 (1 ÷ 7). +Will this end? +Note all the remainders we get. It starts with 1, then +3, then 2, then 6, and so on. Let us represent this as a +chain. +1 + 3 + 2 + 6 + 4 + 5 +1 + 3 ... +5 + 4 + 6 + 2 + 3 + 1 +What do you observe? Can you explain why this +division never ends? +Not only do the remainders repeat in a cycle, the +digits of the quotient also repeat in a cycle! +0.142857 142857 14… +A Magic Number: 142857 +Let us consider the number 142857 that arose when +dividing 1 by 7. Multiply 142857 by numbers from 1 to 6. +3) 10 ( 3.333 ... + 9 + 10 + 9 + 10 + 9 + 10 + 9 + ... +7) 1 ( 0.142857142857... + 0 + 10 + 7 + 30 + 28 + 20 + 14 + 60 + 56 + 40 + 35 + 50 + 49 + 10 + 7 + 30 + 28 + 20 + 14 + 60 + 56 + 40 + 35 + 50 + 49 + 1 +85 + +Ganita Prakash | Grade 7 | Part-II +What are the products? What do you notice? +You get the same number back, but with the digits cycled around! +Multiply 142857 by 7. What do you observe? +Are there other such numbers? Yes! +To find one such number, you can find 1 ÷ 17 in decimal, and use the +repeating block of digits. +Are there infinitely many such “cyclic” numbers? That is, can we +keep finding more cyclic numbers, or do they eventually stop? In 1927, +the Austrian mathematician Emil Artin conjectured (guessed) that there +must be infinitely many such numbers. However, even today, nearly a +century later, this conjecture remains unsolved—despite a lot of research +on the question by many mathematicians! +Dividend, Divisor, and Quotient +When we divide two counting numbers, the quotient is always less +than the dividend. For example, 128 ÷ 4 = 32, and 32 (quotient) < 128 +(dividend). +But what happens when we divide 128 by 0.4? +128 ÷ 0.4 = 320. +The quotient is greater than the dividend. +Will the quotient be always greater than the dividend when the +divisor is a decimal? Try it out with different values of the divisor. +Describe the relationship between the dividend, divisor, and the +quotient. Create a table for capturing this relationship in different +situations, like we did for multiplication. +Figure it Out +1. Express the following fractions in decimal form: +(a) 2 +5 +(b) 13 +4 +(c) +4 +50 +(d) 5 +8 +2. Find the quotients: +(a) 24.86 ÷ 1.2 +(b) 5.728 ÷ 1.52 +Try +This +Math +Talk +86 + +Another Peek Beyond the Point +3. Evaluate the following using the information 156 × 12 = 1872. +(a) 15.6 × 1.2 = __________ +(b) 187.2 ÷ 1.2 = __________ +(c) 18.72 ÷ 15.6 = __________ +(d) 0.156 × 0.12 = __________ +4. Evaluate the following: +(a) 25 ÷ ______ = 0.025 + +(b) 25 ÷ ______ = 250 +(c) 25 ÷ ______ = 2.5 + +(d) 25 ÷ 10 = 25 × _____ +(e) 25 ÷ 0.10 = 25 × ______ +(f) +25 ÷ 0.01 = 25 × ______ +5. Find the quotient: +(a) 2.46 ÷ 1.5 = + +(b) 2.46 ÷ 0.15 = +(c) 2.46 ÷ 0.015 = + +Is the quotient obtained in 24.6 ÷ 1.5 the same as the quotient +obtained in 2.46 ÷ 0.15? +6. A 4 m long wooden block has to be cut into 5 pieces of equal length. +What is the length of each piece? +7. If the perimeter of a regular polygon with 12 sides is 208.8 cm, what +is the length of its side? +8. 3 litres of watermelon juice is shared among 8 friends equally. How +much watermelon juice will each get? Express the quantity of juice +in millilitres. +9. A car covers 234.45 km using 12.6 litres of petrol. What is the distance +travelled per litre? +10. 13.5 kg of flour (aata) was distributed equally among 15 students. +How much flour did each student receive? +1 +2 = 0.5 +1 +2 × 2 = 0.25 +1 +2 × 2 × 2 = 0.125 +1 +2 × 2 × 2 × 2 = 0.0625 +1 +2 × 2 × 2 × 2 × 2 = ? +1 +5 = 0.2 +1 +5 × 5 = 0.04 +1 +5 × 5 × 5 = 0.008 +1 +5 × 5 × 5 × 5 = 0.0016 +1 +5 × 5 × 5 × 5 × 5 = ? +What pattern do you observe? Why are 2 and 5 related in this way? +Math +Talk +87 + +Ganita Prakash | Grade 7 | Part-II +4.4 Look Before You Leap! +Did you know that it takes the Earth +365.2422 days to go around the Sun +and not 365 days? For our convenience, +we consider 365 days as a year in +a calendar. We are talking about +Gregorian calendar. +This means that, after one calendar +year or 365 days, the Earth still needs +0.2422 more days to complete one full +revolution around the Sun. This doesn’t +seem like much. But what happens +after 100 such calendar years? +Using our understanding of decimal multiplication, +0.2422 × 100 = 24.22 days. +After 100 calendar years, the Earth will need 24.22 more days to +complete its 100th revolution around the Sun. +In your Science classes, you have learnt that we experience seasons +because of the Earth’s tilt in axis and its revolution around the Sun. If our +calendar does not accurately indicate the position of the Earth around +the Sun, our seasons and our annual calendar will not match! +To correct this problem, the idea of a leap year was introduced. Every +fourth year, one additional day is added to the calendar year. +Making an Adjustment +Add an extra day +every 4 years. +Hey, +I have not +yet finished! +88 + +Another Peek Beyond the Point +Is the year divisible by 4? +The year has 366 days +The year has 365 days +Yes +No +Do you know which month has this extra day? +Let us see how this solution works. +Year 1 +Year 2 +Year 3 +Year 4 +Year 5 +Year 6 +Year 7 +Year 8 +… +365 +365 +365 +366 +365 +365 +365 +366 +… +Looking at the above sequence, the number of days after 4 calendar +years is +4 × 365 + 1 = 1461 days. +What is the number of days that the Earth needs to make 4 full +revolutions around the Sun? +4 × 365.2422 = 1460.9688 days. +With this new scheme of adding one extra day every 4th year, what +is the number of days in 100 calendar years? Can you write an +expression to calculate that number? +Here is one way to form the expression. Each calendar year has 365 +days. In 100 calendar years, the number of days is 100 × 365. But years +that are divisible by 4 have one extra day. +How many years are divisible by 4 in 100 years? +So, the number of days in 100 calendar years is, +(100 × 365 + 100 +4 × 1) = 36,525 days. +Can you form different expressions for the same question? +The actual number of days that the Earth takes to go around the +Sun 100 times is, +100 × 365.2422 = 36,524.22 days. +Thus, by adding a day every fourth year, after 100 years, the calendar +days are more than the actual number of days taken by the Earth to go +around the Sun. We have overcompensated. +Math +Talk +Math +Talk +89 + +Ganita Prakash | Grade 7 | Part-II +So, the good people who designed calendars decided that they will not +add 1 extra day in every hundredth year! +Making Another Adjustment +No extra day in the +100th year. +Is the year divisible by 100? +The year has 365 days +Is the year divisible by 4? +Yes +No +The year has 366 days +The year has 365 days +Yes +No +Can you write an expression for the number of days in 100 calendar +years with this new adjustment? +We saw that there are 25 years divisible by 4 in 100 years. 100 +is also divisible by 4, but we have to exclude it. So only 24 years have +366 days, and the rest (76 years) have 365 days. So the expression can be +written as, +( 100 +4 – 100 +100 ) × 366 + (100 – ( 100 +4 – 100 +100 ))× 365 += (24 × 366) + (76 × 365) = 36,524 days. +This is close to 36524.22 days but is it close enough? What happens +after 1000 years with this adjustment? +Math +Talk +90 + +Another Peek Beyond the Point +If we follow this new scheme of adding 1 day every four years, but not +in the 100th year, the number of calendar days in 1000 years is +36524 × 10 = 3,65,240 days. +The number of days the Earth takes to go around the Sun 1000 times is +1000 × 365.2422 = 3,65,242.2 days. +So, there is a difference of 2.2 days. +To bridge this gap, it was decided that every 400th year would be a +leap year! +Add an extra day +every 400th year. +Is the year divisible by 400? +The year has 366 days +Is the year divisible by 100? +Yes +No +The year has 365 days +The year has 366 days +Is the year divisible by 4? +The year has 365 days +Yes +Yes +No +No +With this scheme, let us calculate the number of days in 1000 +calendar years. +91 + +Ganita Prakash | Grade 7 | Part-II +In 1000 calendar years, how many years are divisible by 400? 2. +In 1000 calendar years, how many years are divisible by 100 but not +divisible by 400? 10 – 2 = 8. +In 1000 calendar years, how many years are divisible by 4 but not +divisible by 100 and 400? 250 – 10 = 240. +The rest of the years are 1000 – (2 + 8 + 240) = 750. +So, the total number of days in 1000 calendar years is +(750 × 365) + (240 × 366) + (8 × 365) + (2 × 366) = 3,65,242 days. +The Earth needs 3,65,242.2 days to go around the Sun 1000 times. +Hence, in 1000 years, the calendar year is slightly shorter (by 0.2 +days) than the actual number of days the Earth takes to go around +the Sun. +The calendar makers did not want to bother about a small difference +that would happen in 1000 years! So, they left the scheme of leap years +as is … +Making Yet Another Adjustment +With this final scheme of leap years can you +calculate the number of calendar days in 10,000 +years and the number of actual days the Earth +will take to make 10,000 revolutions around the +Sun? What is the difference? If there is a big +difference, can you suggest a way to fix this +problem? +Try +This +92 + +Another Peek Beyond the Point +Do you wonder how people figured out that the Earth completes one +revolution around the Sun in exactly 364.2422 days? +Investigate how traditional calendars in India managed to +consistently align the days in the calendar with astronomical events +like the Earth going around the Sun or even the positions of the stars +in the sky accurately. +Figure it Out +1. A 210 gram packet of peanut chikki costs ₹70.5, while a 110 gram +packet of potato chips costs ₹33.25. Which is cheaper? +2. Write the decimal number at the arrow mark: +3.1 +3.2 +2.15 +2.17 +3. Shyamala bought 3 kg bananas +at ₹30/- per kg. She counted 35 +bananas in all. She sells each +banana for ₹5/-. How much +profit does she make selling all +the bananas? +4. A teacher placed textbooks that are 2.5 cm thick on a bookshelf. The +teacher wanted to place 80 textbooks on the shelf. The bookshelf is +160 cm long. How many books could be placed on the shelf? Was +there any space left? If yes, how much? +Try +This +93 + +Ganita Prakash | Grade 7 | Part-II +5. Fill in the following blanks appropriately: +1 cm = 10 mm +1 m = 100 cm +1 km = 1000 m +1 kg = 1000 g +1 g = 1000 mg +1 l = 1000 ml +5.5 km = _________ m +35 cm = ________ m +14.5 cm = _______ mm +68 g = ________ kg +9.02 m = ________ mm 125.5 ml = _______ l +6. The following problem was set by Sridharacharya in his book, +Patiganita. “6 1 +4 is divided by 2 1 +2, and 60 1 +4 is divided by 3 1 +2. Tell the +quotients separately.” Can you try to solve it by converting the +fractions into decimals? +7. Fill the boxes in at least 2 different ways: +(a) + × + = 2.4 +(b) + × + = 14.5 +8. Find the following quotients given that 756 ÷ 36 = 21: +(a) 75.6 ÷ 3.6 +(b) 7.56 ÷ 0.36 +(c) 756 ÷ 0.36 +(d) 75.6 ÷ 360 +(e) 7560 ÷ 3.6 +(f) 7.56 ÷ 0.36 +9. Find the missing cells if each cell represents a÷b: + b a +1517 +151.7 +15.17 +1.517 +15170 +37 +41 + + + + +3.7 + + +4.1 + + +0.37 + + + + + +0.037 + +4100 + + + +370 + + + + + +10. Using the digits 2, 4, 5, 8, and 0 fill the boxes + + . + × + . + +to get the: +(a) maximum product +(b) minimum product +Math +Talk +94 + +Another Peek Beyond the Point +(c) product greater than 150 +(d) product nearest to 100 +(e) product nearest to 5 +11. Sort the following expressions in increasing order: +(a) 245.05 × 0.942368 +(b) 245.05 × 7.9682 +(c) 245.05 ÷ 7.9682 +(d) 245.05 ÷ 0.942368 +(e) 245.05 +(f) 7.9682 +• In this chapter, we learnt procedures to perform decimal multiplication +and division. +• For decimal multiplication, we first multiply the multiplier and +multiplicand as counting numbers. The number of decimal digits in +the product is the total number of decimal digits in the multiplier and +multiplicand. +• Division of decimals uses the same procedure, i.e., division using +place value (long division), as with counting numbers. The regrouping +continues after the Ones place to Tenths, Hundredths, Thousandths, +and so on. When the Ones are regrouped to Tenths, a decimal point is +placed in the quotient. +• There are decimal divisions where the quotient never ends. After each +regrouping and dividing there is always a remainder! +SUMMARY +  +33 +35 +24 +22 +21 +26 +13 +11 +27 +9 +18 +7 +5 +40 +1 +Puzzle +32 +33 +35 +36 +37 +31 +34 +24 +22 +38 +30 +25 +23 +21 +12 +39 +39 +26 +20 +13 +11 +27 +28 +14 +19 +9 +10 +15 +16 +18 +8 +2 +17 +7 +16 +3 +5 +4 +40 +1 +Solution +Hidato +95 + +In Hidato, a grid of cells is given. It is usually square-shaped, like +Sudoku or Kakuro, but it can also include hexagons or any shape +that forms a tessellation. It can have inner holes (like a disc), but it is +made of only one piece. +Usually, in every Hidato puzzle the lowest and the highest numbers +are given on the grid. +Your task is to fill the grid such that there is a continuous path +of consecutive numbers from the lowest to the highest number. The +next number must be in any one of the adjacent cells, including +diagonally adjacent cells. +The grid comes pre-filled with some numbers (with values +between the smallest and the highest) to ensure that these puzzles +have a single solution. +Try solving the following Hidato puzzles. +20 +4 +14 +8 +15 +9 +25 +1 +36 +2 +4 +6 +13 +34 +12 +30 +31 +17 +26 +23 +21 +39 +1 +9 +7 +10 +5 +17 +15 +44 +49 +28 +54 +42 +50 +29 +26 +35 +34 +25 +23 +39 +1 +56" +class_7,13,connecting the dots...,ncert_books/class_7/gegp1dd2/gegp205.pdf,"CONNECTING +THE DOTS… +5 +5.1 Of Questions and Statements +Your teacher tells you that they are meeting two of their childhood +friends this evening. One is 5 feet tall and the other is 6 feet tall. What is +your guess as to each friend’s gender based on this information? +You might have guessed that the 5-foot-tall person is a woman and +the 6-foot-tall person is a man. There is a chance that you are wrong. But +experience tells us that 5-foot-tall men and 6-foot-tall women are rare. +We have seen that, more often, men are taller than women. +The above is a simple example of statistical thinking. We regularly +come across statements like — +Jemimah’s batting has +been very consistent over +the past year. We can +expect a century from her +in tomorrow’s match. +I take about +15 minutes +to cycle from +school to home. +I think my pen +might last for 2 +more weeks; it is +time to get a new +one soon. +The population of +their village has +reduced by about +100 in the last +decade. +Since I started +to eat fruits and +vegetables more +frequently, I am +able to run 2 km +more each day. +David spends +about 7 hours +daily in the school. + +Ganita Prakash | Grade 7 | Part-II +We call these statistical statements. Simply put, a statistical statement +is a claim or summary about some phenomenon, expressed in terms of +numerical values, proportions, probabilities, or predictions. +A statistical question is a question that can be answered by collecting +data. For example, “How tall are Grade 7 students in our school?” is a +statistical question. We expect that not all Grade 7 students have the same +height, but we can collect data, analyse it, and make conclusions about +the heights that do occur. The question “Typically, are onions costlier +in Yahapur or Wahapur?” is also a statistical question. Prices can vary +over time. Therefore, answering this question requires us to look at data, +analyse it, and come to conclusions making suitable statistical statements. +Which of the following are statistical questions? +(a) What is the price of a tennis ball in India? +(b) How old are the dogs that live on this street? +(c) What fraction of the students in your class like walking up a hill? +(d) Do you like reading? +(e) Approximately how many bricks are in this wall? +(f) Who was the best bowler in the match yesterday? +(g) What was the rainfall pattern in Barmer last year? +The term statistics refers to the study of collecting, organising, +analysing, interpreting, and presenting data. In this chapter, we shall +encounter some statistical questions and learn how analysing data and +graphs can help answer them. +5.2 Representative Values +The runs scored by Shubman and Yashasvi in a cricket series are given +in the table below. Who do you think performed better? +Match 1 +Match 2 +Match 3 +Match 4 +Shubman +0 +17 +21 +90 +Yashasvi +67 +55 +18 +35 +Shreyas says, “Both their performances are similar +since Yashasvi scored more in the first and second +matches, whereas Shubman scored more in the third +and fourth”. +Math +Talk +98 + +Connecting the Dots… +Vaishnavi says, “I think Shubman performed better because he scored +the highest number of run in a match — 90!”. +Shreyas says, “No! Yashasvi batted better since the +total number of runs he made is 175, while Shubman +made only 128”. +Vaishnavi says, “Oh! Also, Yashasvi’s batting is more +consistent — the difference between his maximum +score and minimum score is lower”. +The table below shows the runs scored by these two +players in another series. Who do you think performed +better in this series? +Match 1 +Match 2 +Match 3 +Match 4 +Match 5 +Shubman +23 +07 +10 +52 +18 +Yashasvi +26 +53 +02 +- +15 +Vaishnavi says, “Here, Shubman performed better since his total is +110 runs, while Yashasvi’s total is 96 runs”. +What do you think of Vaishnavi’s statement? +Shreyas says, “But Yashasvi made 96 runs in 4 matches and Shubman +made 110 runs in 5 matches”. +So, how do we say who performed better? It is often not simple to compare +two groups of numbers and clearly say that one is better than the other. +Can a single number act as a representative of a group of numbers? +For example, can we represent Shubman’s or Yashasvi’s batting in +this series with one number? Discuss. +We saw one way already — the total of the values in the group! But, +if the group sizes are different, then the total may not be an appropriate +measure to compare. +In some matches, a player could have scored more and in other matches +less. A representative number for the group can be found by balancing +out these highs and lows. For example, we can add up the runs scored in +all the matches and divide the total by the number of matches played. We +call this value the ‘average’ or ‘arithmetic mean’ of the given data. +Here, the average number of runs scored by a player in a match += [Total runs scored by the player in all the matches] ÷ [Number of +matches played]. +Average number of runs scored by Shubman in a match = 110 ÷ 5 = 21 runs. +Average number of runs scored by Yashasvi in a match = 96 ÷ 4 = 24 runs. +In this series, Yashasvi’s average number of runs is higher than Shubman’s. +Math +Talk +99 + +Ganita Prakash | Grade 7 | Part-II +The Average or Arithmetic Mean (A.M.), or simply Mean, is calculated +as follows: +Mean = Sum of all the values in the data +Number of values in the data . +Average as Fair-Share +The average can also be understood as fair-share or equal-share. +Shreyas and 4 of his friends have collected the following numbers of +guavas: 3, 8, 10, 5, and 4. Parag and 5 of his friends have collected the +following numbers of guavas: 5, 4, 6, 3, 4, and 8. Each group will share +their guavas equally amongst themselves. In which group will each +member get a bigger share of guavas? +To find this out, we first find out how many guavas each group has +collected. Then we divide this total by the number of people in the group +to get each member’s share. +Shreyas’s group has collected 3 + 8 + 10 + 5 + 4 = 30 guavas. Each +member of Shreyas’s group gets 30 ÷ 5 = 6 guavas. +Parag’s group has collected 5 + 4 + 6 + 3 + 4 + 8 = 30 guavas. Each +member of Parag’s group gets 30 ÷ 6 = 5 guavas. +So, the members of Shreyas’s group get 1 more guava each than the +members of Parag’s group. +3 + 8 + 10 + 5 + 4 = 30 +6 × 5 = 30 +5 + 4 + 6 + 3 + 4 + 8 = 30 +6 × 5 = 30 +3 +6 +5 +5 +5 +5 +5 +5 +5 +8 +6 +4 +10 +6 +6 +5 +6 +3 +4 +6 +4 +8 +Vaishnavi tracks the number of Hibiscus flowers blooming in her garden +each day. The data for the last few days’ is 2, 7, 9, 4, 3. What +is the average number of Hibiscus flowers blooming per +day in Vaishnavi’s garden? +The average = (the total number of Hibiscus flowers +bloomed) ÷ (number of days) += (2 + 7 + 9 + 4 + 3) ÷ 5 += 5. +100 + +Connecting the Dots… +On an average, 5 Hibiscus flowers bloom daily. +In this case the average tells us the number of flowers blooming each +day, if an equal number of flowers bloomed daily. +One of the terms used for the Arithmetic Mean in ancient Indian +mathematics is samamiti (mean measure): ‘sama’ means equal. Some +terms used for the Arithmetic Mean in Indian texts include — samarajju +(mean measure of a line segment) by Brahmagupta (628 CE), samīkaraṇa +(levelling, equalising) by Mahāvīrācārya (850 CE) sāmya (equality, +impartiality, equability towards) by Śrīpati (1039 CE) and samamiti +(mean measure) by Bhāskarācārya (1150 CE) and Gaṇeṣa (1545 CE). The +terminology shows that ancient Indian scholars perceived the Arithmetic +Mean as the ‘common’ or ‘equalising’ value that is a representative +measure of a collection of values. +Figure it Out +1. Shreyas is playing with a bat and a ball — +but not cricket. He counts the number of +times he can bounce the ball on the bat +before it falls to the ground. The data for +8 attempts is 6, 2, 9, 5, 4, 6, 3, 5. Calculate +the average number of bounces of the ball +that Shreyas is able to make with his bat. +2. Try the activity above on your own. +Collect data for 7 or more attempts and find the average. +3. Identify a flowering plant in your neighbourhood. Track the +number of flowers that bloom every day over a week during +its flowering season. What is the average number of flowers that +bloomed per day? +4. Two friends are training to run a 100 m race. Their running times +over the past week are given in seconds — Nikhil: 17, 18, 17, 16, 19, +17, 18; Sunil: 20, 18, 18, 17, 16, 16, 17. Who on average ran quicker? +5. The enrolment in a school during six consecutive years was as +follows: 1555, 1670, 1750, 2013, 2040, 2126. Find the mean enrolment +in the school during this period. +Know Your Onions! +The table shows the monthly price of onions, in rupees per kilogram +(kg), at two towns. Where are onions costlier, according to you? +Try +This +Math +Talk +101 + +Ganita Prakash | Grade 7 | Part-II +Month +Yahapur +Month +Wahapur +January +25 +January +19 +February +24 +February +17 +March +26 +March +23 +April +28 +April +30 +May +30 +May +38 +June +35 +June +35 +July +39 +July +42 +August +43 +August +39 +September +49 +September +53 +October +56 +October +60 +November +59 +November +52 +December +44 +December +42 +Khushboo: ‘I think Wahapur is costlier because it has the highest price +of ₹60.’ +Nafisa: ‘I added the prices of all months in each location - Yahapur’s +total is 458, whereas Wahapur’s total is 450.’ +Vishal: ‘Wahapur is costlier since it has 3 numbers in the 50s’. +Sampat: ‘I compared the prices in each month in both locations. Prices +in Yahapur are higher for 6 months, prices in Wahapur are higher for +5 months, and the prices are the same for 1 month. So, I feel Yahapur is +costlier.’ +Jithin: ‘I noticed that the difference between the highest and lowest +prices in Yahapur is 59 – 24 = 35, and in Wahapur it is 60 – 17 = 43.’ +Data can be described and compared by referring to its minimum +value, maximum value, the average value, the sum total of all its values, +and the difference between the maximum and minimum values. +Can you think of any other ways to compare the data? +To study data, we can visualise it in multiple ways. One way is shown +below — it is called a dot plot. Dot plots show data points as dots on a +line, helping us visualise variability and patterns in data. In the following +figure, each dot represents the monthly price of onions. +102 +17 +102 + +Connecting the Dots… +Yahapur +10 +20 +30 +40 +50 +60 +10 +20 +30 +40 +50 +60 +One of the values is +19. Here it refers to +one of the monthly +prices of Wahapur +Two of the values are +42. Both belonging to +Wahapur +Two of the values are 39. +One belonging to each of +Yahapur and Wahapur +Wahapur +The prices in Yahapur are in green and those in Wahapur are in +purple. The horizontal line shows the prices from 10 to 60 (instead of +starting from 0 as there are no values from 0 – 10 or above 60). The dots +on the vertical line give the number of occurrences of a data value. +Notice the equal spacing between the units along the horizontal as well +as the vertical lines. +Does this visualisation capture all the data presented in the tables +earlier? +Looking at it, can we tell the price of onions in Yahapur in the month of +January? +This method of presentation orders or sorts the data, but it loses the +original (month-wise) sequence of the values. However, it allows us to +group the data however we wish, just as Vishal did. For instance, there +are 2 data values between 11 – 20 for Wahapur, while Yahapur has none. +This representation makes it easier to observe the variation in the data — +where and how the data is clustered or spread out. We can easily see +that the prices in Wahapur are more spread out than those in Yahapur. +It is also easy to spot the highest and lowest values. +We can also use the average as one of the ways to compare the prices +at these two places. +103 + +Ganita Prakash | Grade 7 | Part-II +Find the average price of onions at Yahapur and Wahapur. +A statement such as, “The price of onions is ₹35 per kilo”, may not +trigger any further questions. But looking at variations in data, like the +prices of onions over a year in Yahapur and Wahapur, can spark one’s +curiosity. For example, one might be curious to know more about the +two locations. +You might wonder — +Do the seasons +affect the price of +onions? +How much do +onion prices +vary across +shops in the +same area? +What other +commodities +might have +similar +patterns? +How do the price +fluctuations impact +farmers, consumers, +and the industry? +Where are these +two locations? Are +they close to each +other or far apart? +What are the +factors that +determine the +price of onions? +What else do you wonder about? +You can discuss questions that you are curious about with your +peers, teachers, or family members to find answers. +Observing and trying to make sense of data can reveal interesting things. +It can also trigger our curiosity in different directions. +Averages Around Us +The Arithmetic Mean is frequently used in statistics, mathematics, +experimental sciences, economics, sociology, sports, biology and diverse +Math +Talk +13 104 + +Connecting the Dots… +disciplines as a representative of data. It is popular partly because the +definition of the arithmetic mean is simple and easy to understand. Some +statements involving averages in different scenarios are shown below: +Wheat yield averages +4.7 tonnes per hectare in +Punjab vs. 2.9 tonnes per +hectare in Bihar. +3126 is the average +number of Indian long +films released annually +between 2017 – 2024. +My scooty's average +mileage this year is about +45 kilometers per liter. +An average Indian citizen +generates 0.45 kg of waste +per day. +The average rainfall per +day in Jharkhand in the +month of July is 37.2mm. +Smartphone users check +their phone 58 times a day +on average. +Outliers and Medians +Does the average always give a reasonable summary of the values in a +collection? If not, what is an alternative? Let us find out. +Height of a Family +The heights of the family members of Yaangba and Poovizhi are as +follows: +Yaangba’s family: 169 cm, 173 cm, 155 cm, 165 cm, 160 cm, 164 cm. +Poovizhi’s family: 170 cm, 173 cm, 165 cm, 118 cm, 175 cm. +Find the average height of each family. Can we say that Yaangba’s +family is taller than Poovizhi’s family? +Math +Talk +105 + +Ganita Prakash | Grade 7 | Part-II +The average height of Poovizhi’s family (160.2 cm) is less than that +of Yaangba’s family (164.3 cm). Although most members in Poovizhi’s +family are taller, their family’s average height is less because one child +is much younger and not as tall as the rest of the family. Their average +height, 160.2 cm, is less than the heights of 4 out of 5 members. Here, the +average doesn’t seem to represent the data very well. +125 +135 +145 +155 +165 +175 +115 +Mean = 160.2 –– +125 +135 +145 +155 +165 +175 +115 +Mean = 164.3 –– +Can you think of any other number that can represent the data better? +One way is to sort the data and pick the number in the middle. This +number is called the Median. +To find the median height of Poovizhi’s family, we first sort the +heights — +118, 165, 170, 173, 175. +The middle number in this sorted data is 170. Therefore, the median +height is 170 cm. +Let us find the median height of Yaangba’s family. Sorting the heights, +we get +155, 160, 164, 165, 169, 173. +Since the median is the number in the middle, it will have an equal +number of values less than it and greater than it. This data does not have +a single middle number because it has an even number of values (6). In +such cases, we take the average of the two middle numbers in the sorted +data. Therefore, the median height of Yaangba’s family is (164 + 165) ÷ 2 += 164.5 cm. +125 +135 +145 +155 +165 +175 +115 +Mean = 164.3 –– +Median = 164.5 - - - +125 +135 +145 +155 +165 +175 +115 +Mean = 160.2 –– +Median = 170 - - - +In this case, does the median represent the heights of the families better +than the average? +53 +106 + +Connecting the Dots… +In Poovizhi’s family, +the height of the youngest +child is quite different +from the heights of the +rest of the family. We call +such a value an outlier. +Outliers are values which +significantly deviate from +the rest of the values in +the data. +Notice how the mean and the median are close to each other in +Yaangba’s data, in the absence of any outlier. +In Poovizhi’s data, because of the outlier, the mean is much lower +than the median. +Find the mean and median in Poovizhi’s data without the outlier value +118. What change do you notice? +Are you a bookworm? +After the summer vacation, a class teacher asked his class how many +short stories they had read. Each student answered the number of +stories read on a piece of paper, as shown below. Find the mean and +median number of short stories read. Before calculating them, can you +guess whether the mean will be less than or greater than the median? +Mark the data, the mean, and the median on the dot plot below. +0 +5 +10 +15 +20 +25 +30 +35 +40 +The median value 6 means that half of the class members have read +6 or more stories. +107 + +Ganita Prakash | Grade 7 | Part-II +Which of the values would you consider an outlier? +Find the mean and median in the absence of the outlier. What change +do you notice? +The average may not always be an appropriate representative of +data that has outliers. A very high or a very low outlier can significantly +impact the sum, thus affecting the average. For example, the 118 cm +height in Poovizhi’s family is an outlier at the lower end of the data. And +the count of 40 short stories read is an outlier at the higher end of the +data. In these cases, we saw that the median was not affected much by +the outliers. +Are We on the Same Page? +Do you read newspapers? Have you noticed how +many pages a newspaper has on different days of +the week — is it the same or different? +The list below shows the number of pages for a +particular newspaper from Monday to Sunday: 16, +18, 20, 22, 26, 16, 10. +Mark the data, the mean, and the median on the +dot plot below. +0 +5 +10 +15 +20 +25 +In the three examples we considered — the heights, short-stories, +and newspaper pages — observe the variability in data when: +(a) the mean and median are close to each other +(b) the mean and median are comparatively far apart, with + +mean < median +(c) the mean and median are comparatively far apart, with + +mean > median +When the data is more balanced or uniformly spread out the mean, +and the median appear to be close to each other. When the outlier is +on the lower end, the mean appears to shift in that direction, i.e., +mean < median. When the outlier is on the higher end, the mean appears +to shift in that direction, i.e., mean > median. +Math +Talk +108 + +Connecting the Dots… +Discuss the effect on the mean and median when outliers are +present on both sides. You may take some example data to +examine and explain this. +Math +Talk + Mean and Median are called measures of central tendency, i.e., the +tendency of the values to pile up around a particular value. In other +words, they represent the ‘centre’ of the data. +Of Ends and the Essence +As we have just seen, the mean and the median can give different +perspectives on the data. As part of analysing data, it can also be valuable +to look at the variability in the given data, i.e., its extremes (minimum +and maximum values). +How Tall is Your Class? +Suppose you are asked the question, “How tall is your class?” What +would you say? +The table below shows the heights of students in a Grade 5 class in +centimeters. +Boys +147, 135, 130, 154, 128, 135, 134, 158, 155, 146, 146, 142, 140, 141, 144, 145, 150 +Girls +143, 136, 150, 144, 154, 140, 145, 148, 156, 150, 150 +We can visualise the data using a dot plot, identify the ends and +patterns, and look at the variability. We can also find the measures of +central tendency. The dot plot for the whole class, followed by the dot +plots for boys and girls, respectively, are shown. The mean and the +median are also shown for each collection. +125 +Whole +class +Mean = 144.4 +Median = 145 +Mean = 142.94 +Median = 144 +Mean = 146.9 +Median = 148 +Boys +Girls +125 +125 +130 +130 +130 +135 +135 +135 +140 +140 +140 +145 +145 +145 +150 +150 +150 +155 +155 +155 +109 + +Ganita Prakash | Grade 7 | Part-II +What can we infer from the dot plots and the central tendency measures? +The following points can help answer the question of how tall the +class is. +• +The boys’ heights are more spread out and are between 128 and +158. The girls’ heights lie between 136 and 156. Both the tallest and +shortest in the class are boys. +• +Yet, the boys’ average height is less than the whole class average, and +also less than the girls’ average height. We can say girls are taller +than boys in this class. Of course, this doesn’t mean every girl is +taller than every boy! +• +For boys’ heights, mean < median (142.94 < 144) indicating a +small influence of values on the lower side. For girls’ heights too, +mean < median (146.9 < 148) indicating a small influence of values +on the lower side. +How many students are taller than the class’ average height? +How many boys are taller than the class’ average height? +How long is a minute? +Two groups of children were asked to estimate +the length of 1 minute. They start by closing their +eyes and then open when they think 1 minute +has passed. Of course, they are not supposed to +count while their eyes are closed. The dot plots +below show after how many seconds the children opened their eyes. +Discuss how well both the groups fared at this activity. Describe and +compare the variability in data and their central tendency. +Group B +Mean = 59.28 +Median = 59.5 +Group A +Mean = 58.21 +Median = 60 +40 +40 +45 +45 +50 +50 +55 +55 +60 +60 +65 +65 +70 +70 +Math +Talk +11 +110 + +Connecting the Dots… +In yesterday’s +match, the median +runs scored by +England’s players was +0, and yet the team +scored 407 for the loss +of 10 wickets. +Zero Median Runs Scored! +In a cricket match, can a team’s median runs scored by a player be +0 but the team’s total score be 407/10? +Accounting for the extras, the average runs scored by a player in +this innings is (407 – 19) ÷ 11 = 35.27. +Zero vs. No value +Suppose a player scores 57, 13, 0, 84, — , 51, 27 in a series. Notice that the +player played Match 3 and scored 0 runs whereas the player did not play +Match 5. So, we consider the total number of matches to be 6 and not 7. +We calculate their average runs scored per match as +(57 + 13 + 0 + 84 + 51 + 27) ÷ 6. +Sita has a mango tree in her backyard. The number of mangoes the +tree gave every month over the last year, from January to December, is +0, 0, 8, 24, 41, 16, 5, 0, 0, 0, 0, 0 respectively. If we want to find the mean +or median number of mangoes per month, it would be appropriate +to consider only the (summer) months when mangoes are expected +to grow. +111 + +Ganita Prakash | Grade 7 | Part-II +A Mean Foot +In the early 1500s in Europe, the basic unit of +land measurement was the rod, defined as 16 +feet long. At that time, a foot meant the length +of a human foot! But foot sizes vary, so whose +foot could they measure? To solve this, 16 adult +males were asked to stand in a line, toe to heel, +and the length of that line was considered the +16-foot rod. After the rod was determined, it was +split into 16 equal sections, each representing +the measure of a single foot. In essence, this was +the arithmetic mean of the 16 individual feet, +even though the term ‘mean’ was not mentioned +anywhere. +Jacob Kobel’s depiction of the +determination of 1 foot +Figure it Out +1. Find the median of onion prices in Yahapur and Wahapur. +2. Sanskruti asked her class how many domestic animals and pets each +had at home. Some of the students were absent. The data values are +0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, —, 10, 25, 2, — , 2, 4. Find the mean +and median. How would you describe this data? +3. Rintu takes care of a date-palm tree farm in Habra. The heights of +the trees (in feet) in his farm are given as: 50, 45, 43, 52, 61, 63, 46, +55, 60, 55, 59, 56, 56, 49, 54, 65, 66, 51, 44, 58, 60, 54, 52, 57, 61, 62, +60, 60, 67. Fill the dot plot, and mark the mean and median. How +would you describe the heights of these palm trees? Can you think +of quicker ways to find the mean? How many trees are shorter than +the average height? +0 +5 +10 +15 +20 +25 +30 +35 +40 +45 +50 +55 +60 +65 +70 +75 +80 +4. The daily water usage from a tap was measured. The usage in liters +for the first few days are: 5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, 7.4. +(a) Can the mean or median daily usage lie between 25 and 30? +Justify your claim using the meaning of mean and median. +(b) Can the mean or median be lesser than the minimum value or +greater than the maximum value in a data? +5. The weights of a few newborn babies are given in kgs. Fill the dot +plot provided below. Analyse and compare this data. +112 + +Connecting the Dots… +Boys +3.5 +4.1 +2.6 +3.2 +3.4 +3.8 +Girls +4.0 +3.1 +3.4 +3.7 +2.5 +3.4 +0 +0.5 +1 +1.5 +2 +2.5 +3 +3.5 +4 +4.5 +6. The dot plots of heights of another section of Grade 5 students of the +same school are shown below. Can you share your observations? +What can we infer from the dot plots and the central tendency +measures? +Mean = 141.21 +Median = 142.5 +Mean = 142.05 +Median = 143 +Mean = 140.14 +Median = 140 +Boys +Whole +class +Girls +125 +125 +125 +130 +130 +130 +135 +135 +135 +140 +140 +140 +145 +145 +145 +150 +150 +150 +155 +155 +155 +Compare the heights of the two sections. Share your observations. +7. The +weights +of +some +sumo +wrestlers and ballet dancers are: +Sumo wrestlers: 295.2 kg, 250.7 +kg, 234.1 kg, 221.0 kg, 200.9 kg. +Ballet dancers: 40.3 kg, 37.6 kg, +38.8 kg, 45.5 kg, 44.1 kg, 48.2 kg. +Approximately how many times +heavier +is +a +sumo +wrestler +compared to a ballet dancer? +Math +Talk +113 + +Ganita Prakash | Grade 7 | Part-II +5.3 Visualising Data +We can often understand data more clearly if it is presented as a picture. +This is called data visualisation. Last year, we saw how to visualise data +using graphs. Let us explore visualisation further. +Clubbing the Columns +Earlier, we looked at the monthly onion prices in Yahapur and Wahapur. +Jan +Feb +Mar +Apr +May +Jun +Jul +Aug +Sep +Oct +Nov +Dec +Yahapur +25 +24 +26 +28 +30 +35 +39 +43 +49 +56 +59 +44 +Wahapur +19 +17 +23 +30 +38 +35 +42 +39 +53 +60 +52 +42 +Two column graphs for this data are given below. +60 +Monthly onion prices in Yahapur +50 +40 +30 +20 +10 +0 +Jan +Feb +Mar +Apr +May +Jun +Jul +Aug +Sep +Oct +Nov +Dec +Jan +Feb +Mar +Apr +May +Jun +Jul +Aug +Sep +Oct +Nov +Dec +60 +50 +40 +30 +20 +10 +0 +Monthly onion prices in Wahapur +114 + +Connecting the Dots… +The two graphs can also be combined into a single graph. We just draw +the bars side by side! Verify if the data in the table matches the graph below. +60 +50 +40 +30 +20 +10 +0 +Monthly onion prices in Yahapur and Wahapur +Yahapur +Wahapur +Price (in ₹) +Jan +Feb +Mar +Apr +May +Jun +Jul +Aug +Sep +Oct +Nov +Dec +We use different colours to clearly separate the data from the two +places. This is called a clustered column graph. Since it has two columns +in each cluster, we also call it a double column graph. +What is the scale used in this graph? +The relative heights of the bars tell +us where onions are costlier in each +month. We can also visually estimate +the difference by referring to the +markings along the vertical line. +Is it now easier to compare month-wise +prices in both places? +10…9…8…7…6…5…4…3…2…1…Take Off! +You might have heard about scientific probes (like Chandrayaan-3 +launched in 2023 by ISRO or the Voyager-1 launched in 1977 by NASA), +observational satellites (like Aryabhata launched in 1975 by ISRO or +Sputnik-1 launched in 1957 by the Soviet Space program), or about human +spaceflights to the International Space Station. All space missions are +The dots and slanted lines +within the bars help people who +find difficulty in distinguishing +colours. It is also useful when +things are printed in greyscale +(black-and-white). +115 + +Ganita Prakash | Grade 7 | Part-II +launched using rockets. Look at the graph below showing the number of +worldwide rocket launches by different organisations. +Share your observations (you may take the teacher's help to identify the +countries these organisations belong to). +Source: https://www.statista.com/chart/29410/number-of-worldwide-rocket-launches-by- +companies-and-space-agencies/ +Often there is a lot of information in graphs and it may be difficult to +understand. We can follow a 2-step process +to simplify making sense of the data in +graphs. +Step 1: Identify what is given +Notice how the graph is organised, what +scale is used, and what patterns the data +shows. +I wish +they had shown +guidelines for 5, 10, and +15 along the horizontal +line. It would help in +estimating the smaller +numbers. +116 + +Connecting the Dots… +• +For each organisation, the numbers of rocket launches for the years +2021, 2022, and 2023 are shown as three adjacent bars. The scale +used is 1 unit length = 20 rockets. Notice the numbers at the bottom. +• +The ‘Others’ category indicates multiple organisations worldwide +that are clubbed together to keep the graph short. +• +Note that in the double bar graph of onion prices, the months are +shown in order, i.e., January to December, to observe the change over +time, whereas in this case, a change in the order of organisations +does not affect the meaning. +Increasing year on year, +bar length of 2022 is +almost double that of +2021 +Decreasing year on +year, bar length of 2023 +is almost half of 2022 +Increase and +then decrease +Perhaps 61 +SpaceX +CASC +Roscosmos +Arianespace +Rocket Lab +ISRO +Galactic Energy +Expace +0 +20 +40 +60 +80 +100 +United Launch +Alliance +The 2023 count +of objects in this +category is around 25 +Step 2: Infer from what is given +Analyse and interpret each of your observations. +• +We can say that the USA, China, and Russia are the leading rocket +launching countries in the given time period. +• +SpaceX launched about twice the number of rockets in 2022 +compared to 2021. And it launched about 35 more rockets in 2023 +compared to 2022. +• +The number of rockets launched by Arianespace decreased every year. +• +United Launch Alliance launched more rockets in 2022 than in +2021. They launched fewer rockets in 2023 than in both the years +2022 and 2021. +• +Other organisations launched about 25 rockets in 2023. +117 + +Ganita Prakash | Grade 7 | Part-II +Identify which of the following statements can be justified using this +data. +(a) All organisations launched more rockets than the previous years. +(b) Only an organisation from the USA launched more than 50 rockets +in a single year. +(c) The total number of rockets launched by France in all 3 years is less +than 40. +(d) The average number of rockets launched by CASC in these 3 years +is around 40. +(e) ISRO launched more rockets than Galactic Energy in these 3 years. +(f) Russia launched more than 60 rockets in these 3 years. +List the organisations that have consistently launched more rockets +every year. +Estimate the total number of rockets launched worldwide in 2023. + (a) less than 200 +(b) 200 to 400 + (c) 400 to 600 +(d) more than 600 +We may have many questions after looking at the graph. We might +wonder why USA launches so many more rockets than other countries. +Or we might be eager to look at the data from previous and later years. +What are you curious to know after looking at this graph? +Summer and Winter at the Same Time +The tables below show data related to weather in two cities in different +countries. The numbers given are in hours. Can you guess what the data +might be related to? +City 1 +Jan +Feb +Mar +Apr +May +Jun +Jul +Aug +Sep +Oct +Nov +Dec +210 +257 +372 +441 +536 +564 +555 +465 +394 +310 +222 +186 +City 2 +Jan +Feb +Mar +Apr +May +Jun +Jul +Aug +Sep +Oct +Nov +Dec +459 +384 +381 +327 +304 +276 +295 +318 +369 +409 +435 +468 +118 + +Connecting the Dots… +The data shows the monthly hours of daylight (i.e., the Sun is at least +partly above the horizon) in these two cities over the year. Based on this +data, a clustered bar graph showing the average daylight hours per day +in each month is given below. This average is obtained by dividing the +monthly daylight hours by the number of days in the month. +Jan +Feb +Mar +Apr +May Jun +Jul +Aug +Sep +Oct +Nov +Dec +0 +5 +10 +15 +20 +Average daily sunshine hours in two cities +Number of Hours per day +City 1 +City 2 +Let us follow the 2-step process to identify and interpret the +information presented. +Step 1: Identify what is given +Notice how the graph is organised, what scale is used, and what patterns +the data shows. +• +The horizontal line shows the months of the year. The vertical +line shows the average daylight hours per day, using the scale +1 unit = 5 hours. The month of June has the maximum value for City +1 and the minimum value for City 2. +Step 2: Infer from what is given +Analyse and interpret each of your observations. Share appropriate +summary and conclusion statements. +• +The average number of daylight hours per day in City 1 increases +from January, reaching a maximum of about 17 – 18 hours in June. It +then decreases, reaching a minimum of about 6 hours in December. +119 + +Ganita Prakash | Grade 7 | Part-II +• +The average number of daylight hours per day in City 2 decreases +from January, reaching a minimum of about 9 hours in June. It then +increases, reaching a maximum of about 15 hours in December. +• +The maximum and minimum values in City 1 are more extreme than +those of City 2. That is, the maximum number of daylight hours per +day of City 1 is more than that of City 2, and the minimum number +of daylight hours per day of City 1 is less than that of City 2. +• +In June, City 1 experiences daylight for about 3 +4 th of the full day (24 +hours), whereas during December – January, it only experiences +daylight for about 1 +4 th of the full day. +Does this give some idea of where these two cities are located? +City 1 and City 2 are located away from the Equator in the Northern +and Southern hemispheres, respectively. City 1 is Helsinki, Finland, and +City 2 is Wellington, New Zealand. These are also shown on the map. In +June, the Northern Hemisphere is tilted towards the Sun, resulting in +longer daylight hours; it is summertime here. Meanwhile, the Southern +Hemisphere is tilted away from the Sun, leading to shorter days; it is +winter time here. The inverted seasonal daylight pattern is due to the +cities’ location in opposite hemispheres. The large variation in the data +is because they are away from the Equator. +120 + +Connecting the Dots… +Is there anything more that you wish to explore? +I got to know that +during summer +near the poles one +can see the Sun +even at midnight!! +Very interesting! +If we are there at +that time, we can +go sightseeing even +at midnight! +©timeanddate.com/Brendan Goodenough +The Midnight Sun at Andøya, Norway. +All it Takes is a Minute +Have you ever missed watching a cricket match? You can catch up in +a minute by looking at a graph. You might have seen graphs like the +following one. +Math +Talk +121 + +Ganita Prakash | Grade 7 | Part-II +1 +2 +3 +4 +5 +5 +0 +6 +7 +8 +9 +10 +Overs +Runs per over +10 +11 +12 +13 +14 +15 +15 +16 +17 +18 +19 +20 +The horizontal line lists the overs starting from 1, and the vertical line +indicates the runs scored in each over. The graph shows the number of +runs scored per over as a double bar graph — each bar corresponding +to a team. Let us call them the blue team (denoted by blue) and the red +team (denoted by red). The scale used for the runs per over is 1 unit = 5 +runs. The circles shown on top of the bars indicate that a wicket fell in +that over. +Answer the following questions based on the graph: +1. Can we tell who batted first? Who won the match? +2. How many runs did the blue team score in over 12? +3. In which over did the red team score the least number of runs? +4. Is it easy to tell the target set by the team batting first? +Figure it Out +1. The following infographic shows the speeds of a few animals in +air, on land, and in water. Can we call this graph a bar graph? +(a) What is the scale used in this graph? +(b) What did you find interesting in this infographic? What do +you want to explore further? +(c) Identify a pair of creatures where one’s speed is about +twice that of the other. +(d) Can we say that a sailfish is about 4 times faster than a +humpback whale? Can we say that a sailfish is the fastest +aquatic animal in the world? +Math +Talk +122 + +The peregrine falcon is the +world's fastest animal. It +hunts by diving at high +speed and striking a pigeon +or other bird in midair. +The Australian tiger beetle +moves at a blistering pace– +about 120 body lengths +per second. It's eyes can't +process images fast enough +to keep up, so at top speed +this beetle is running blind. + +Ganita Prakash | Grade 7 | Part-II +2. Preyashi asked her students ‘If you were to get a super power to +become aquatic (water-borne), aerial (air-borne), or spaceborne +which one would you choose?’. The responses are shown below. +Some chose none. Draw a double-bar graph comparing how both +grades chose each option. Choose an appropriate scale. +Grade 5 +w, a, a, a, w, n, s, a, n, w, a, a, a, a, a, w, w, s, a, a, n, w, a, a, n +Grade 9 +n, w, s, a, s, w, s, s, a, a, w, s, s, a, s, a, n, w, s, s, a, w, a, w, a +3. The temperature variation over two days in different months in +Jodhpur, Rajasthan, is given below. Draw a double-bar graph. Use +the scale 1 unit = 4°C. Can you guess which two months these days +might belong to? +12 am +3 am +6 am +9 am + 12 pm +3 pm +6 pm +9 pm +Day 1 +20°C +18°C +16°C +20°C +26°C +34°C +30°C +24°C +Day 2 +37°C +34°C +30°C +33°C +37°C +43°C +42°C +39°C +4. The following clustered-bar graph shows the number of electric +vehicles registered in some states every year from 2022 to 2024. +100000 +75000 +2022 +Number of Electric Vehicles Registered Per Year +2023 +2024 +Uttarakhand +West Bengal +Andhra +Pradesh +Odisha +Assam +Gujarat +Delhi +50000 +25000 +0 +124 + +Connecting the Dots… +(a) The data (rounded-off to thousands) for the states of Gujarat +and Delhi are given in the table below. Mark the corresponding +bars on the bar graph. (It is enough if you place the top of the +bars between the two appropriate vertical guidelines.) +2022 +2023 +2024 +Gujarat +69000 +89000 +78000 +Delhi +62000 +74000 +81000 +(b) Notice how the graph is organised, what scale is used, and what +patterns the data shows. +(c) How would you describe the change for various states between +2022 and 2024? +(d) Approximately how many more registrations did Assam get in +2023 compared to 2022? +(e) How many times more did the registrations in West Bengal +increase from 2022 to 2024? +(f) +Is this statement correct — ‘There were very few new +registrations in Uttarakhand in 2023 and 2024, as the increase +in the bar lengths is minimal’? +5.4 Data Detective +We put well-formed sentences one after the other to make a beautiful +story. In the same way, well-organised and well-presented data can tell +interesting stories, and can also expose new mysteries or help solve +mysteries! +Telling Tall Tales +Earlier, we saw data of two Grade 5 classrooms with heights of boys and +girls in each class. There, the average height of girls was more than boys +in one class and vice versa in the other class. +Following are the dot plots of heights of boys (in blue) and girls (in +orange) of Grades 6, 7 and 8 (in that order) of two different schools. +What do you notice? Share your observations. +Math +Talk +125 + +Ganita Prakash | Grade 7 | Part-II +120 +130 +140 +150 +160 +170 +120 +130 +140 +150 +160 +170 +120 +130 +140 +150 +160 +170 +120 +130 +140 +150 +160 +170 +120 +130 +140 +150 +160 +170 +120 +130 +140 +150 +160 +170 +120 +130 +140 +150 +160 +170 +120 +130 +140 +150 +160 +170 +120 +130 +140 +150 +160 +170 +120 +130 +140 +150 +160 +170 +120 +130 +140 +150 +160 +170 +120 +130 +140 +150 +160 +170 +Mean — += 134.8 +Mean — += 149.84 +Mean — += 137.78 +Mean — += 150.2 +Mean — += 141.8 +Mean — += 156.14 +Mean — += 141.83 +Mean — += 155.41 +Mean — += 149.35 +Mean — += 156.14 +Mean — += 147.81 +Mean — += 156.83 +School A +School B +Looking at this data, you might wonder: +“Why is there a considerable difference in heights in the same grades +across these two schools?” +“Where are these schools located?” +“How tall are students in Grades 6 to 8 in my school?” +“What is the average height of all Grade 6 boys and girls?” +We see that men are taller than women in general. But what about the +heights of boys and girls? Are boys taller than girls? Well, just by looking +at the data of one or two schools, we cannot generalise for all children +in our country, or around the world. +Let us look at some data (based on a survey) of the heights of boys +and girls of different ages in India over time. The following table shows +the average heights of boys and girls (in centimeters) across ages 5 to 19 +in the years 1989, 1999, 2009, and 2019. In each year, the first column +shows boys’ heights and the second column shows girls’ heights. +126 + +Connecting the Dots… +Age +1989 +1999 +2009 +2019 +5 +101.3 +100 +102.4 +101.7 +105.1 +104 +107.1 +107.2 +6 +107.5 +106 +108.7 +107.5 +111 +109.7 +113.1 +112.9 +7 +113 +111.4 +114.2 +112.6 +116.2 +114.8 +118.6 +118 +8 +118.1 +116.5 +119.2 +117.5 +120.9 +119.6 +123.5 +122.7 +9 +122.9 +121.7 +123.9 +122.4 +125.2 +124.5 +128.1 +127.6 +10 +127.5 +127.3 +128.3 +127.8 +129.4 +129.9 +132.6 +132.8 +11 +132.2 +133.4 +132.8 +133.6 +133.7 +135.7 +137 +138.6 +12 +137.7 +139 +138 +139.1 +138.9 +141.1 +142.2 +143.8 +13 +144.2 +143.2 +144.3 +143.1 +145.2 +145.1 +148.4 +147.7 +14 +150.6 +146.2 +150.5 +146.1 +151.5 +148 +154.4 +150.4 +15 +155.4 +148.5 +155.2 +148.4 +156.3 +150.1 +159 +152.4 +16 +158.9 +150.1 +158.7 +150.1 +159.9 +151.6 +162.3 +153.8 +17 +161.3 +151.2 +161.4 +151.3 +162.6 +152.6 +164.6 +154.7 +18 +162.9 +151.8 +163.2 +152.1 +164.3 +153 +166 +155.2 +19 +163.5 +151.9 +164.2 +152.4 +165.1 +153 +166.5 +155.2 +Spend sufficient time observing the data presented in this table. +Share your findings with the class. +These are some prompts for you to probe — +• +Changes in the heights of boys or girls of a certain age from 1989 to +2019. +• +The heights of boys vs. girls at different ages in a particular year. +• +Changes in height between successive ages in boys and girls in 2019. +Which of the following statements can be justified using the data? +1. The average heights of both boys and girls at every age increased +from 1989 to 2019. +2. The average height of 13-year-old girls in 1989 is more than the +average height of 14-year-old girls in 2009. +3. The average height of 15-year-old boys in 2019 is more than the +average height of 16-year-old boys in 1989. +4. All girls aged 13 are taller than all girls aged 11. +5. Throughout the age period 5 to 19, the average boy's height is more +than the average girl's height. +6. Boys keep growing even beyond age 19. +Math +Talk +127 + +Ganita Prakash | Grade 7 | Part-II +In 2019, between which two successive ages from 5 to 19 did boys +grow the most? Between which two successive ages from 5 to 19 +did girls grow most? +Suppose the average height of a newborn is 50 cm. Estimate the +average height of young children of ages 1 to 4. +Based on the trend observed in the table, write your estimates of +the heights of boys and girls for ages 5 to 19 in the year 2029. +Whenever you see data or some graph, look closely to know the story it +has to say and the mysteries it may hold. +You may want to look at the data of weights. Or you might be curious +to see if such patterns are present in other countries as well. You might +also wonder if humans were much shorter a few centuries ago! Also, are +people in some countries taller than others? The following visualisation +shows the change in average heights of 19-year-old boys and girls of +different countries from 1989 to 2019. +How is the graph organised? What information is presented? +What do you find interesting? +Notice that the vertical line starts from 145 cm — this helps give +a closer (zoomed in) look at the heights. +Height (in cm) +Math +Talk +Math +Talk +128 + +Connecting the Dots… +A Mean Decision! +Perhaps using the +average height of +the family to make +the door was not a +good idea! +Look at this picture +of the current +world’s tallest and +shortest humans +together. +Figure it Out +1. The dot plots below show the distribution of the number of pockets +on clothing for a group of boys and for a group of girls. +0 +1 +2 +3 +4 +5 +6 +7 +0 +1 +2 +3 +4 +5 +6 +7 +Boys +Girls +Number of Pockets +Number of Pockets + +Based on the dot plots, which of the following statements are true? +(a) The data varies more for the boys than for the girls. +(b) The median number of pockets for the boys is more than that +for the girls. +(c) The mean number of pockets for the girls is more than that for +the boys. +(d) The maximum number of pockets for boys is greater than that +for the girls. +2. The following table shows the points scored by each player in +four games: +Player +Game 1 +Game 2 +Game 3 +Game 4 +A +14 +16 +10 +10 +B +0 +8 +6 +4 +C +8 +11 +Did not play +13 +129 + +Ganita Prakash | Grade 7 | Part-II + +Now answer the following questions: +(a) Find the average number of points scored per game by A. +(b) To find the mean number of points scored per game by C, would +you divide the total points by 3 or by 4? Why? What about B? +(c) Who is the best performer? +3. The marks (out of 100) obtained by a group of students in a General +Knowledge quiz are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Another +group’s scores in the same quiz are 68, 59, 73, 86, 47, 79, 90, 93 and +86. Compare and describe both the groups performance using, mean +and median. +4. Consider this data collected from a survey of a colony. +Favourite +Sport +Cricket +Basket +Ball +Swimming +Hockey +Athletics +Watching +1240 +470 +510 +430 +250 +Participating +620 +320 +320 +250 +105 +Choose an appropriate scale and draw a double-bar graph. Write +down your observations. +5. Consider a group of 17 students with the following heights (in cm): +106, 110, 123, 125, 117, 120, 112, 115, 110, 120, 115, 102, 115, 115, +109, 115, 101. The sports teacher wants to divide the class into +two groups so that each group has an equal number of students: +one group has students with height less than a particular height +and the other group has students with heights greater than the +particular height. Suggest a way to do this. Can you guess the age +of these students based on the tabular data in the ‘Telling Tall +Tales’ section? +6. Describe the mean and median of heights of your class. You can +visualise the heights on a dot plot. +7. There are two 7th grade sections at a school. Each section has 15 boys +and 15 girls. In one section, the mean height of students is 154.2 cm. +From this information, what must be true about the mean height of +students in the other section? +(a) The mean height of students in the other section is 154.2 cm. +(b) The mean height of students in the other section is less than +154.2 cm. +130 + +Connecting the Dots… +(c) The mean height of students in the other section is more than +154.2 cm. +(d) The mean height of students in the other section cannot be +determined. +8. Standing tall in the storm. +(a) Write estimated values for the number of skyscrapers in +New York, Tokyo, and London. +(b) Are the following statements valid? +(i) +Only 12 cities have more skyscrapers than Mumbai. +(ii) Only 7 cities have fewer skyscrapers than Mumbai. +(iii) The tallest building in the world is in Hong Kong. +9. Estimate and then measure the objects listed in the following table. +Draw a double bar graph based on the data. How accurate were +your estimates? Find the average difference between the estimated +and measured values. +131 + +Ganita Prakash | Grade 7 | Part-II +Object +Estimate +(in cm) +Measure +(in cm) +Positive +Difference +Length of a pen +Length of an eraser +Length of your plam +Length of your geometry +box +Length of your math +notebook +10. Aditi likes solving puzzles. She recently started attempting the ‘Easy’ +level Sudoku puzzles. The time she took (in seconds) to solve these +puzzles are — 410, 400, 370, 340, 360, 400, 320, 330, 310, 320, 290, 380, +280, 270, 230, 220, 240. The first nine values +correspond to Week 1 and the rest to Week 2. +(a) Construct a dot plot below showing the data +for both weeks. +(b) Describe +the +mean, +median, +and +any +observations you may have about the data. +11. Individual Project: Pick at least one of the following: +(a) How Long is a Sentence? Pick any two textbooks from different +subjects. Choose any page with a lot of text from each book. +(i) +Use a dot plot to describe how many words the sentences +have on each page. +(ii) Compare the data of both the pages using mean and median. +(b) What is in a Name? Write down the names of all of your +classmates. The following are some interesting things you can +do with this data! +(i) +Find the mean and median name length (number of letters +in a name). +(ii) Visualise the data and describe its variability and central +tendency. +200 +Week 1 +Week 2 +220 +240 +260 +280 +300 +320 +340 +360 +380 +400 +132 + +Connecting the Dots… +(iii) Which starting letters are more popular? Which are less +popular? +(iv) What is the median starting letter? What does this say +about the number of names starting with the letters A – M +and N – Z? +(v) Plot a double-bar graph showing the number of boys’ names +and girls’ names that: +• start and end with vowels, +• start with vowels and end with consonants, +• start with consonants and end with vowels, +• start and end with consonants. +12. Individual project (long term): This requires collecting data over +2 weeks or more. +In and Out: Track how many times you step out of your house in a +day. Do this for a month. +(i) +Describe the variability and central tendency of this data. +Make a dot plot. +(ii) Do you find anything interesting about this data? Share +your observations. +(iii) You can ask any of your family members or friends to +do this as well. +13. Small-group project: Pick at least one of the following. Make groups +of 8 to 10. Collect data individually as needed. Put together everyone’s +data and do the appropriate analysis and visualisation. +(a) Our heights vs. our family’s heights: Collect the heights of your +family members. +(i) +Make a dot plot showing heights of just your family +members. Describe its variability and central tendency. +(ii) Make a double-bar graph showing each student’s height +next to their family’s mean height. +(iii) Look at everyone’s data and share your observations. +(b) Estimating time: Check the time and close your eyes. Open them +when you think 1 minute has passed (no counting). Note down +after how m-any seconds you opened your eyes. Collect this data +for yourself and for your family members. Repeat this activity +to estimate 3 minutes. +(i) +Make two dot plots (for 1 minute and 3 minutes) showing +estimates of just your family members. +Try +This +133 + +Ganita Prakash | Grade 7 | Part-II +(ii) Mark these on the respective dot plots. Describe its +variability and central tendency. +(iii) Make a double bar graph showing each family’s mean 1 +minute estimate and mean 3 minute estimate. +(iv) Look at everyone’s data and share your observations. +• Dot plots help us get a quick glimpse of the variability of the data — +minimum, maximum, range, and how data is clustered or spread out. +• The Arithmetic Mean = Sum of all the values in the data +Number of values in the data +. +• The Median is the number in the middle of any sorted data. If there +are an even number of values, then the median is the average of the +two middle numbers. +• We can describe and compare data in several ways including by +referring to the minimum, maximum, total, range, arithmetic mean, +and median. +• We learnt how to read and make clustered bar graphs. These graphs +can be used to compare and visualise values across categories and +across time. +• Examining data can lead to new questions and directions to probe +further. +SUMMARY +134 + +  +Connect the Dots… +A number lock has a 3-digit code. Find the code using the hints below. +One digit is correct and +well placed. +One digit is correct but +wrongly placed. +Nothing is correct. +One digit is correct but +wrongly placed. +Two digits are correct but +wrongly placed." +class_7,14,constructions and tilings,ncert_books/class_7/gegp1dd2/gegp206.pdf,"CONSTRUCTIONS +AND TILINGS +6 +6.1 Geometric Constructions +Eyes +Do you recall the ‘Eyes’ construction we did in Grade 6? +Of course, eyes can be drawn freehand, but we wanted to construct +them so that the lower arc and upper arc of each eye look symmetrical. +We relied on our spatial estimation to determine the two centres, A +and B (see the figure), from which we drew +the lower arc and upper arc respectively. +The arcs define a line XY that ‘supports’ +the drawing though it is not part of the final +figure. We can start with this supporting line +and systematically find the centres A and B. +For the eye to be symmetrical, or for the +supporting line to be the line of symmetry, the +upper and lower arcs should have the same +radius. In other words, we must have AX = BX. +Since AX = AY and BX = BY, this means +AX = AY = BX = BY +A +B +X +Y + +Constructions and Tilings +How do we find such A and B? +From X and Y, draw arcs above and below +XY, with the same radii. +The two points at which the arcs meet, +above and below XY, give us A and B, +respectively. +Use this to construct an eye. +In Fig. 6.1, join A and B with a line. Where +does AB intersect XY, and what is the angle +formed between them? +We observe that AB passes through the +midpoint of XY, and is also perpendicular to it. +A division of a line, or any geometrical +object, into two identical parts is called +bisection. A line that bisects a given line +and is perpendicular to it, is called the +perpendicular bisector. +Will the line joining the two points at which +the arcs meet, above and below XY, always be the perpendicular bisector +of XY, i.e., when XY is of any length, and the arcs are drawn using a +radius of any length? +This can be answered through congruence. Let us consider a line +segment XY. Find points A and B such that AX = AY = BX = BY. Draw the +lines AB, AX, AY, BX and BY. Let O be the point of intersection between +AB and XY. +Which two triangles should be congruent for AB to be the perpendicular +bisector of XY (that is, O is the midpoint of XY and AB is perpendicular +to XY)? +If we show that ∆AOX ≅ ΔAOY, then +OX = OY, and ∠AOX = ∠AOY because they are +corresponding parts of congruent triangles. +Since ∠AOX and ∠AOY together form a +straight angle, we have ∠AOX + ∠AOY = 180°. +Thus, ∠AOX = ∠AOY = 90°. This establishes +that O is the midpoint of XY and AB is +perpendicular to XY. +A +B +X +Y +Fig. 6.1 +A +B +X +Y +O +Common side +A +B +X +Y +O +137 + +Ganita Prakash | Grade 7 | Part-II +In ∆AOX and ΔAOY, we already know that AX = AY, and AO is common +to both triangles. +If we can show that ∠XAO = ∠YAO then, by +the SAS congruence condition, we can conclude +that the triangles are congruent. +To show this, we observe that ∆ABX ≅ ΔABY. +This is so because AX = AY, BX = BY, and AB is +common to both the triangles. +Thus, +we +have +∠XAB = ∠YAB, +or +∠XAO = ∠YAO because they are corresponding +parts of congruent triangles. +Hence, AB is the perpendicular bisector of XY. +We can have eyes of different shapes. +A +X +Y +O +Common side +B +138 + +Constructions and Tilings +How do we get these different shapes? Try! +One way is to choose two other points C and D such +that CX = CY = DX = DY. An eye of a different shape can be +drawn using these points. +Will C and D lie on the perpendicular bisector AB? +The points C and D are at the same distance from both +X and Y. We have just seen that joining any two such +points gives the perpendicular bisector of XY. Since XY +has only one perpendicular bisector, which is the line AB, +the points C and D must lie on the line AB. +Justify the following statement using the facts that we have established. + +Any point that has the same distance from X and Y lies on the +perpendicular bisector of XY. +Thus, eyes of different shapes can be drawn by suitably choosing +different pairs of points on the perpendicular bisector as centres to +construct the upper and lower arcs of the eyes. +Construction of Perpendicular Bisector +Given a line segment XY, how do we draw +its perpendicular bisector using only an +unmarked ruler and a compass? +We have seen that joining any two +points — one above XY and one below — that +are at equal distances from X and Y, gives +the perpendicular bisector of XY. This gives +a method to construct the perpendicular +bisector. +1. Taking some fixed radius, from X and +then Y, construct two sufficiently long +arcs above XY. Name the point where the +arcs meet as A. +A +X +Y +A +B +X +Y +C +D +139 + +Ganita Prakash | Grade 7 | Part-II +2. Using the same radius, from +X and then Y, construct two +sufficiently long arcs below XY. +Name the point where the arcs +meet as B. +3. AB is the required perpendicular +bisector. +A +B +X +Y +Thus, the perpendicular bisector can be constructed using the simplest +geometric tools — an unmarked ruler and a compass. We will use only +these two tools for all the other geometric constructions in this chapter, +unless there is a need for drawing lines of specific lengths in standard +units. +Figure it Out +1. When constructing the perpendicular bisector, is it necessary to +have the same radius for the arcs above and below XY? Explore this +through construction, and then justify your answer. + +[Hint 1: Any point that is of the same distance from X and Y lies on +the perpendicular bisector. + +Hint 2: We can draw the whole line if any two of its points are +known.] +2. Is it necessary to construct the pairs of arcs above and below XY? +Instead, can we construct both the pairs of arcs on the same side of +XY? Explore this through construction, and then justify your answer. +3. While constructing one pair of intersecting arcs, is it necessary that +we use the same radii for both of them ? Explore this +through construction, and then justify your answer. +4. Recreate this design using only a ruler and compass — +After completing the above design, you can use a colour pencil with a ruler +or compass to trace its boundary. This will make the design stand out from +the supporting lines and arcs. +140 + +Constructions and Tilings +This method of constructing the perpendicular bisector is not only +geometrically exact but also a practical way to construct it accurately. +This method to find the midpoint of a line segment is more accurate +than measuring the length using a marked scale. +Construction of a 90° Angle at a Given Point +Can +we +extend +the +method +of +constructing the perpendicular bisector +to construct a 90° angle at any point on +a line? Draw a line and mark a point O +on it. Construct a 90° angle at point O. +Find a segment of this line for which O +is the midpoint. +Extend the line on either side of O. +Using a compass, mark two points X +and Y at equal distance from O, so that +O is the midpoint of XY. +The perpendicular bisector of XY will +pass through O and is perpendicular +to the given line. +In this case, do we need to draw two +pairs of intersecting arcs to get the +perpendicular bisector of XY? +No, we don’t. We already have one +point, O, lying on the perpendicular +bisector. +Figures 6.2 and 6.3 describe the +steps to construct a 90° angle at a given +point on a line. +Construction Methods in Śulba-Sūtras +Ancient mathematicians from different civilizations, including India, +knew exact procedures to construct perpendiculars and perpendicular +bisectors. +In India, the earliest known texts containing these methods are the +Śulba-Sūtras. These are geometric texts of Vedic period dealing with the +construction of fire altars for rituals. The Śulba-Sūtras are part of one +of the six Vedāṅgas (a term that literally means ‘limbs of the Vedas’). +The Śulbas contain the methods that we developed earlier to construct +a perpendicular and the perpendicular bisector. All the construction +O +Fig. 6.2 +O +X +Y +( +) +O +X +Y +( +) +A +Fig. 6.3 +141 + +Ganita Prakash | Grade 7 | Part-II +methods in the Śulba-Sūtras make use of a different kind of compass +from what you would have used — a rope. A rope can be used to draw +circles or arcs. It can also be stretched to form a straight line. +In addition, the Śulba-Sūtras also contain other methods to construct +perpendicular lines. Here is an interesting construction of the +perpendicular bisector using a rope (Kātyāyana-Śulbasūtra 1.2). +Let XY be the given line segment, drawn on the ground, for which we +need to construct a perpendicular bisector. Fix a small pole or peg +vertically into the ground at each point X and Y. +Take a sufficiently long rope. Make two +loops at its ends. Without taking into account +the parts of the rope that has gone into the +loops, fold the rope into half to find and mark +its midpoint. +Fasten the two loops at the ends of the rope +to the poles at X and Y. +Pull the midpoint of the rope above XY, as +shown in the figure, such that the two parts +of the rope on either side are fully stretched. +Mark this position of the midpoint as A. +Now similarly pull the midpoint of the +rope below XY, as shown in the figure, such +that the two parts of the rope on either side +are fully stretched. Mark this position of the +midpoint as B. +AB is the required perpendicular bisector. +Figure it Out +1. Justify why AB in Fig. 6.4 is the perpendicular bisector. +2. Can you think of different methods to construct a 90° angle at a +given point on a line using a rope? +X +Y +A +Fig. 6.4 +A +X +Y +B +Math +Talk +142 + +Constructions and Tilings +Angle Bisection for a Design +How do we construct this figure? +Fig. 6.5 +The supporting lines for this +figure will look like this. +What is the angle between two adjacent lines? +We need the angle between every pair of adjacent lines to be equal. +Since 360° is equally divided into 8 parts, every angle is 45°. +How do we construct a 45° angle using only a ruler and a compass? +We know how to construct a 90° angle. If we can divide it into two +equal parts, or bisect it, then we get a 45° angle. +We will now develop a general method to +bisect any angle. Consider an angle ∠XOY. +X +Y +O +We can bisect it if we can draw two congruent +triangles ΔOBC and ΔOAC as shown in the figure. +Then ∠BOC = ∠AOC. +X +Y +O +A +B +C +143 + +Ganita Prakash | Grade 7 | Part-II +How do we construct these congruent triangles, given the angle? +If A and B are marked such that OA = OB, and if C is chosen such that +BC = AC, then by the SSS congruence condition, ΔOBC ≅ ΔOAC. So we can +bisect an angle as follows. +Steps for Angle Bisection +1. Mark points A and B such that OA = OB. +X +Y +O +B +A +2. Choosing any sufficiently long radius, cut +arcs from A and B, keeping the radius +same. Mark the point of intersection as C. +3. OC bisects ∠AOB. +X +Y +O +B +A +C +So, a 45° angle can be constructed by first constructing a 90° angle and +then bisecting it. +Figure it Out +1. Construct at least 4 different angles. Draw their bisectors. +2. Construct the 8-petalled figure shown in Fig. 6.5. +3. In Step 2 of angle bisection, if arcs of +equal radius are drawn on the other +side, as shown in the figure, will the +line OC still be an angle bisector? +Explore this through construction, +and then justify your answer. +X +Y +O +B +A +C +4. What are the other angles that can be constructed using angle +bisection? Can you construct 65.5° angle? +5. Come up with a method to construct the angle bisector using +a rope. +Math +Talk +144 + +Constructions and Tilings +6. Construct the following figure. + +How do we construct the petals so that they are of the maximum +possible size within a given square? +Repeating Units and Repeating Angles +Construct the following figure. +Fig. 6.6 +In this figure, there is a single unit repeating itself. To construct +this figure, we need to make exact copies of this unit in two different +orientations. +In order to make exact copies, all the units must have the same arm +lengths and the same angle between the arms. We can ensure equal arm +lengths using a compass, but how do we ensure equal angles? +Let us develop a method to create an exact copy of a given angle. +Draw an angle. Create a copy of this angle using only a ruler and +compass. +A +Math +Talk +145 + +Ganita Prakash | Grade 7 | Part-II +You might have developed your own method. Here is one simple +approach. +Steps of Construction to Copy an Angle +1. +A +X +2. +A +C +B +X +Draw an arc from A. +This gives us three points that form the isosceles triangle ∆ABC. +3. +A +C +B +X +Z +Draw an arc of the same radius from X. +4. +A +C +B +X +Z +Y +Measure BC using a compass. Transfer this length on the arc +from Z to get YZ = BC. +146 + +Constructions and Tilings +5. +A +C +B +X +Z +Y +By the SSS congruence condition, ∆ABC ≅ XYZ. So, ∠A = ∠X. +Figure it Out +1. Construct at least 4 different angles in different orientations without +taking any measurement. Make a copy of all these angles. +2. Construct the Fig. 6.6. +This procedure to copy an angle finds an important application in +constructing parallel lines using only a ruler and compass. +Construction of a Line Parallel to the Given Line +Recall that in the construction using a ruler and a set square, we +constructed equal corresponding angles to get parallel lines. +How do we implement this idea using a ruler +and a compass? +Suppose there is a line m to which we need +to draw a parallel line. We construct a line l +that intersects m. Line l will serve as a +transversal to line m and to the line parallel to +m that we are going to construct. +Let us choose a point B on l through which we +are going to draw the parallel line. This parallel +line must make the same corresponding angle, +as shown in the figure. This can be done by +copying the angle between m and l. +l +a +m +A +Fig. 6.7 +m +l +a +A +a +B +147 + +Ganita Prakash | Grade 7 | Part-II +Here is a step-by-step procedure for carrying this out. +m +l +A +B +C +D +Construct arcs of equal +radius from A and B +Fig. 6.8 +m +l +A +B +C +D +E +Transfer the length CD to +the arc from F +Fig. 6.9 +m +l +A +B +C +D +E +n +m || n +Fig. 6.10 +Figures 6.7–6.10 describe a method to construct a line parallel to the +given line. +Figure it Out +1. Construct 4 pairs of parallel lines in different orientations. +2. Construct the following figure. +A +B +C +D +F +E +G +H +S +T +U +V +W +X +Z +Y +F +F +148 + +Constructions and Tilings +Arch Designs +Trefoil Arch +Have you seen this kind of beautiful arch? +https://commons.wikimedia.org/wiki/ +File:Central_Park_A.jpg +Central park, New York City +https://commons.wikimedia.org/w/index. +php?curid=28374748 +Diwan-i-Aam, Red fort +How did they make these arches? +The first step is to be able to draw them on a plane surface such as +paper or stone. +Construct this arch shape on a piece of paper. +Let us think about the support lines this +figure will need. +For symmetry, we should have AB = CD, and ∠BAD = ∠CDA. +How would you construct these support lines? +A +D +C +B +A +D +Construct equal angles at A and D. +Mark B and C such that AB = CD. +Use these support lines to construct an arch. If required, adjust the radii +of the arcs to make the arch look more aesthetically pleasing. +A +D +C +B +149 + +Ganita Prakash | Grade 7 | Part-II +A Pointed Arch +Some arches look like this. +https://en.m.wikipedia.org/wiki/File:Diwan-i-Aam,_Red_Fort,_Delhi_-_2.jpg +Diwan-i-Aam, Red Fort +How do we construct this shape? +What supporting lines will you use to +draw this arch? +Remember ‘Wavy Wave’ from the +Grade 6 Textbook? +The supporting lines are just two line +segments of equal length. +If their midpoints are marked, will you be +able to construct a pointed arch? +Fig. 6.11 +150 + +Constructions and Tilings +Figure it Out +1. Use support lines in Fig. 6.11 to construct a pointed arch. Make +different arches, by changing the radius of the arcs. +2. Make your own arch designs. +Regular Hexagons +Recall that a regular polygon has equal sides and equal angles. A regular +polygon with 3 sides is an equilateral triangle, and a regular polygon +with 4 sides is a square. We have constructed these figures earlier. +How do we construct a regular pentagon (5-sided figure) and a +regular hexagon (6-sided figure)? To begin with, try to construct a +pentagon and hexagon with equal sidelengths. +To construct a regular pentagon, we first need to have a better +understanding of triangles and pentagons. We will discuss this in later +years. However, constructing a regular hexagon is within our reach! +Can we break a regular hexagon into smaller pieces that can be +constructed? +Regular Hexagon and Equilateral Triangles +What happens when we join the ‘opposite’ points of a +regular hexagon? Since a regular hexagon has equal +sides and angles, can we expect a figure like this? +Will all the triangles in the figure be equilateral +triangles? +To answer these questions, we will reverse our +approach. +Can six congruent equilateral triangles be placed +together as in Fig. 6.12? If yes, will it result in a regular hexagon? +Try +This +A +B +C +D +E +F +O +Fig. 6.12 +151 + +Ganita Prakash | Grade 7 | Part-II +If six congruent equilateral triangles can indeed be placed as shown +in Fig. 6.12, then the sides of the resulting hexagon are equal, and their +angles are 60 + 60 = 120° (how?). So what we really need to examine +is whether 6 congruent equilateral triangles can fit this way without +overlapping and without leaving any gaps around the centre. +We have defined a degree by taking the complete angle around a point +to be 360°. So all the angles around the centre should add up to 360°. +Consider this figure. Will the 70° angle fit into the gap? What is the gap +angle ∠AOI? +We have, +40° + 60° + 50° + 30° + 40° + 90° + gap angle = 360°. +Use this to determine whether the 70°angle +fits the gap. +Thus, if there are angles that add up to 360°, +their vertices can be joined together at a single +point such that +(a) the angles do not overlap, and +(b) they completely cover the region around +the point. +Since each angle in an equilateral triangle is +60°, six such angles add up to 360°. Therefore, +six congruent triangles can be arranged as +shown in Fig. 6.12. +In Fig. 6.12 can you explain why AOD, BOE and COF are straight lines? +Construct a regular hexagon with a sidelength 4 cm using a ruler and a +compass. +We can construct a regular hexagon more directly if we can construct +a 120° angle using a ruler and a compass. +How do we do it? +This can be done if we can construct a 60° angle. +60° +70° +40° +50°30° +40° +90° +? +A +B +C +D +E +F +I +O +G +H +If we +construct this, +we also +get this. +60° +120° +152 + +Constructions and Tilings +Construction of a 60° angle +How do we construct a 60° angle? +We get a 60° angle if we construct an equilateral triangle! We can use +the following steps for this. +Suppose we need a 60° angle at point A on a line segment AX. +Step 1 +B +X +A +Construct an arc with centre A and any radius. +Step 2 +C +B +X +A +With the same radius, cut another arc from B that meets the +first arc. Let C be the point at which the arcs meet. +We have ∠CAX = 60°. +Why is ∠CAX = 60°? Is there an equilateral triangle here? +We can use these ideas to construct a regular hexagon — +Construct a regular hexagon of sidelength 5 cm. +153 + +Ganita Prakash | Grade 7 | Part-II +Related Constructions +Construction of 30° and 15° angles +How will you construct 30° and 15° angles? +6-Pointed Star +Construct the following 6-pointed star. Note that it has a rotational +symmetry. +Hint: + +D +C +B +H +A +G +F +L +I +K +J +E + +Do you see a hexagon here? +Are the six triangles forming the 6 points of the star — ∆AGH, ∆BHI, ∆CIJ, +∆DJK, ∆ELK, ∆FLG — equilateral? Why? +[Hint: Find the angles.] +Figure it Out +1. Construct the following figures: +Math +Talk +(a) +An Inflexed Arc +(b) +The fun part about this figure is +that it can also be constructed using +only a compass! Can you do it? +154 + +Constructions and Tilings +(c) +(d) +(e) +2. Optical Illusion: Do you notice +anything interesting about the +following figure? How does this +happen? Recreate this in your +notebook. +3. Construct this figure. + +[Hint: Find the angles in this figure.] +4. Draw a line l and mark a point P anywhere outside the line. +Construct a perpendicular to the given line l through P. + +[Hint: Find a line segment on l whose perpendicular bisector +passes through P.] +6.2 Tiling +Tangrams are puzzles that originated in China. +They make use of 7 pieces obtained by dividing a +square as shown. +For the problems ahead, we need these 7 +tangram pieces. These are provided at the end of +the book. Or, by looking at the figure, you could +make cardboard cutouts of the pieces. +Try +This +A +G +C +B +D +E +F +155 + +Ganita Prakash | Grade 7 | Part-II +We can form interesting pieces by +rearranging the tangram pieces. Here is +an arrow. +Figure it Out +How can the tangram pieces be rearranged to form each of the +following figures? +156 + +Constructions and Tilings +Covering a region using a set of shapes, without gaps or overlaps, is +called tiling. +Consider a rectangular grid made of unit squares — +We call this a 4 × 6 grid, since it has 4 rows and 6 columns. +Can a 4 × 6 grid be tiled using multiple copies of 2 × 1 tiles? +We are allowed to rotate a 2 × 1 tile and use it. +Vertical tile +Horizontal tile +Here is one way. +Obviously, this is not the only tiling possible. +Can a 4 × 7 grid be tiled using 2 × 1 tiles? +What about a 5 × 7 grid? +To see that there is no way to tile a 5 × 7 grid using 2 × 1 tiles, observe +that this grid has 35 unit squares. Each tile covers exactly 2 unit squares. +Math +Talk +157 + +Ganita Prakash | Grade 7 | Part-II +Complete the justification. +Is an m × n grid tileable with 2 × 1 tiles, if both m and n are even? If yes, +come up with a general strategy to tile it. +One general strategy for this case is to cover each column with vertical +tiles. This is possible because the number of rows is even. +Even no. +of tiles +Is an m × n grid tileable with 2 × 1 tiles, if one of m and n is even and the +other is odd? If yes, come up with a general strategy to tile it. +Is an m × n grid tileable with 2 × 1 tiles, if both m and n are odd? Give +reasons. +Here is a 5 × 3 grid, with a unit square removed. +Now, it has an even number of unit squares. Is it +tileable with 2 × 1 tiles? +Is the following region tileable with 2 × 1 tiles? +Math +Talk +Math +Talk +158 + +Constructions and Tilings +What about this one? +Fig. 6.13 +Were you able to tile this? How can we be sure that this is not tileable? +Can you find another unit square that, when removed from a 5 × 3 grid, +makes it non-tileable? +There is an interesting way to look at these questions. For any tiling +problem of this kind, we can create a similar problem with the unit +squares coloured black and white so that black squares have only white +neighbours and white squares have only black neighbours. For the tiling +problem in Fig. 6.13, we get the following. +Fig. 6.14 +Region to +be tiled +Tiles +Black and white grid +Plain grid +In the black-and-white region, the problem is to tile the region with +the 2 × 1 black-and-white tiles so that each black square of a tile sits on +a black square of the grid, and each white square sits on a white square. +If the plain grid is tileable, is the black-and-white-grid tileable? +If the black-and-white grid is tileable, is the plain grid tileable? +It can be seen that the answer to both the questions is yes. +Math +Talk +Math +Talk +159 + +Ganita Prakash | Grade 7 | Part-II +Is the black-and-white region in Fig. 6.14 tileable? +Any region tiled with black-and-white-tiles must have an equal number +of black tiles and white tiles. +Since the black-and-white region in Fig. 6.14 has 8 white squares and 6 +black squares, it can never be tiled with these tiles! +Use this idea to find another unit square that, when removed from +a 5 × 3 grid, makes it non-tileable? +Isn’t it surprising how, by introducing colours and making the +problem more complicated, it becomes easier to tackle? What a creative +way of looking at the problem! +Figure it Out +Are the following tilings possible? +1. +Region to be tiled +Tile + +2. +Region to be tiled +Tile + +Tiling the Entire Plane +So far we have seen how to tile a given region. What about tiling the +entire plane? +Can you think of a shape whose copies can tile the entire plane? +Clearly, squares can. +Math +Talk +160 + +Constructions and Tilings +Are there other regular polygons that can tile the plane? +What about equilateral triangles? +Tiling with equilateral triangles shows the possibility of tiling with +another regular polygon. +A plane can be tiled using regular hexagons as well. +A plane can also be tiled using more than one shape, and using +non-regular polygons. People such as the great Dutch artist Escher +(1898 – 1972) — whose works explored mathematical themes such as +tiling — have come up with creative ways of tiling a plane with animal +shapes! Here are some examples. +(b) +(a) +161 + +Ganita Prakash | Grade 7 | Part-II +(c) +(d) +Mathematicians are still exploring various ways of tiling the plane! +Tiling (b) was found as recently as 2023. +Have you seen tilings in daily life? They are often used in buildings +and in designs. Tilings are found in nature too. The front face of +bee hives and some wasp nests are tiled using hexagonal cells! +These cells are used by the insects to keep their eggs, larvae and +pupae safe, as well as to store food. Because the region is tiled, no space +is wasted. +Scientists still wonder how bees and wasps are able to make hexagonal +cells. Next time you see any tiling, pay closer attention to it! +Tiling is still one of the most exciting and active areas of research +in geometry. +162 + +• A division of a line segment, or any geometrical quantity, into two +identical parts is called bisection. +• Any point that is of equal distance from the two endpoints of a given line +segment lies on its perpendicular bisector. This property can be used +to construct the perpendicular bisector using a ruler and compass. +• The method of constructing the perpendicular bisector can be +modified to draw a 90° angle at any point on a line using only a ruler +and compass. +• An angle can be bisected and copied using the congruence properties +of triangles. +• A 60° angle can be constructed using a ruler and compass by +constructing an equilateral triangle. +• Covering a region using a set of shapes, without gaps or overlaps, is +called tiling. +SUMMARY" +class_7,15,finding the unknown,ncert_books/class_7/gegp1dd2/gegp207.pdf,"FINDING THE +UNKNOWN +7 +7.1 Find the Unknowns +Unknown Weights +We have a weighing scale that behaves as follows. The numbers represent +same units of weight: +Find the unknown weights in the following cases: +Fig. 7.1 +Fig. 7.2 +Fig. 7.3 + +Finding the Unknown +Fig. 7.5 +Fig. 7.4 +Fig. 7.6 +Fig. 7.7 +Fig. 7.8 +Discuss the answers with your classmates. Give reasons why you +think your answer is right. +Find the unknown weight of the sack in the following cases. In Fig. 7.10, +all the sacks have the same weight. +Fig. 7.9 +Fig. 7.10 +Math +Talk +[Hint: If we remove equal weights from both the plates, will the +weighing scale still be balanced? Remove one sack from each plate for +Fig. 7.10.] +165 + +Ganita Prakash | Grade 7 | Part-II +Fig. 7.11 +Fig. 7.12 +[Hint: Can you remove objects so that the sacks are only on one plate?] +Note to the Teacher: Encourage your students to find solutions to these problems +using different strategies and methods, and ask them to compare and contrast +their methods. +Let us find an unknown value in a different setting. +Matchstick Pattern +Consider this sequence of matchstick arrangements. +1 +2 +3 +4 +Recall that we have studied this sequence in an earlier chapter. +166 + +Finding the Unknown +The figure shows the first 4 matchstick arrangements along with their +position numbers in the sequence. +Jasmine decides to make a matchstick arrangement that appears in this +sequence, using exactly 99 sticks. What will be the position number of +this arrangement in the sequence? +To answer this question, it is useful to know the number of matchsticks +in each position. +We see that +the arrangement at position 1 has 2 × (1) + 1 = 3 matchsticks, +the arrangement at position 2 has 2 × (2) + 1 = 5 matchsticks, +the arrangement at position 3 has 2 × (3) + 1 = 7 matchsticks, +and so on. +So, the nth position will have 2n+1 matchsticks. +To answer Jasmine’s question, we have to find the value of n such that +2n + 1 has the value 99, or, +2n + 1 = 99. +Can you find ways to get the value of n, such that 2n + 1 = 99? +Is it possible to make a matchstick arrangement that appears in this +sequence using exactly 200 sticks? +A statement of equality between two algebraic expressions is called +an equation. Nowadays, when using symbols, we write an equation as +two algebraic expressions with an equal sign ‘=’ between them. +Here are some more examples of equations: +3x + 4 = 7, +20 = y – 3, +a +3 = 50, +2z + 4 = 5z – 14 etc. +The process of finding the value(s) of the letter−numbers for which +the equality holds, or for which the value of the Left Hand Side (LHS) of +the equation becomes equal to the Right Hand Side (RHS) of the equation, +is called solving the equation. +As we saw in the matchstick +problem, framing an equation +using an unknown quantity as +a letter−number can help us find +its value. +Math +Talk +Left Hand Side +(LHS) +2n + 1 = 99 +Right Hand Side +(RHS) +167 + +Ganita Prakash | Grade 7 | Part-II +For the weighing scale problems in figures 7.6, 7.7, 7.8, 7.9, 7.10, +and 7.11, frame equations by using letter-numbers to denote the +unknown weight. +For the problem in Fig. 7.6, let us denote the weight of one fried egg as e. +Since each slice of bread is 2, we have 2 + 2 + 2 = 6 on one side and +e + e on the other side. Since they are equal, we have +6 = e + e, or +2e = 6. +For the problem in Fig. 7.7, + = 4, and we can denote the weight of one + as y. So, we have 16 on one side, and 4 + 2y on the other side. Thus, +we have the equation +4 + 2y = 16. +Solve the equations that you frame and check if you get the same value +for the unknown weight as you got previously. +Frame 5 equations. Find methods to solve them. +7.2 Solving Equations Systematically +How did you solve the various equations framed in the previous section? +One way to solve an equation is to substitute different values in place +of the letter−number and to check which value makes LHS = RHS. For +example, consider the equation 2n + 1 = 99. +If we substitute n = 5, we get LHS = 2 × (5) + 1 = 11. It is far away from +99, the RHS. +We can try n = 10. The LHS is 21. It is still not equal to 99. +Can we try n = 30? The LHS is now 61, still much lower than 99. +Let us try n = 40. The LHS is 81. We are getting closer to 99. +When n = 50, the LHS is 101. This is just a bit too high. +When n = 49, the LHS is 99! +Therefore, the solution to the equation 2n + 1 = 99 is n = 49. +Can this equation have any other solution? +This method is called the trial and error method. +The trial and error method can be inefficient. +Try solving 5x – 4 = 7 using trial and error. +Math +Talk +168 + +Finding the Unknown +Recall that in the case of the weighing scale, we didn’t use the +trial and error method! For some problems, finding the solution was +straightforward. For others, we used the fact that when we remove equal +weights from both the plates of a balanced weighing scale, it remains +balanced. Do equations have a similar property? +Consider an equation 15 + 8 = 23. If we add, subtract, multiply or divide +the same number on both sides, will it still preserve the equality of LHS +and RHS? +For example, you can check by adding 10 to both sides. +Since the LHS and the RHS of an equation have the same value, +performing the same operation on both sides will clearly not change +their equality. +In the weighing scale problems, we removed equal weights so that +all the unknown weights were on only one plate of the scale. This made +the problem easier to solve. We can use this same strategy to solve an +equation as well! Let us first solve some arithmetic problems. +Example 1: It is known that 14593 – 1459 + 145 – 14 + 88 = 13353. What +is the value of 14593 – 1459 + 145 – 14? +To find the value, do we need to evaluate 14593 – 1459 + 145 – 14? +No, we can get it by subtracting 88 from 13353. +Why can we do this? +We can do this because addition and subtraction are inverse +operations. So, the value of the required expression can be found by +subtracting 88 from both sides (LHS and RHS), which removes the term +88 and leaves only the expression to be evaluated on the LHS. +14593 – 1459 + 145 – 14 + 88 – 88 = 13353 – 88 +Example 2: It is known that 23 × 41 × 11 × 8 × 7 = 5,80,888. What is the +value of the expression 23 × 41 × 11 × 8? +Using the fact that multiplication and division are inverse operations, +we can simply divide 5,80,888 by 7 to get the value of the expression. +Is this the same as dividing both sides by 7, which removes the +factor 7 and leaves only the expression to be evaluated on the LHS? +Example 3: It is known that 12345 – 5432 + 135 – 24 – (–67) = 7091. +What is the value of the expression 12345 – 5432 + 135 – 24? +Math +Talk +169 + +Ganita Prakash | Grade 7 | Part-II +This problem also can be solved using the fact that addition and +subtraction are inverse operations. However, we can use the following +method that is perhaps easier to visualise. +To retain only the expression to be evaluated on the LHS, we need +to remove the term –(–67). We can remove this term by adding (–67) to +both sides of the equation. +12345 – 5432 + 132 – 24 – (–67) + (–67) = 7091 + (–67) +12345 – 5432 + 132 – 24 – (–67) + (–67) = 7091 + (–67) +Thus, 12345 – 5432 + 132 – 24 = 7024. +Example 4: It is known that ( +35 +113) × 24 × 14 × ( +8 +9) = 94080 +1017 . What is the +value of the expression ( +35 +113) × 24 × 14? +Solution: To retain only the expression to be evaluated on the LHS we +need to remove the factor ( +8 +9). We can do that by dividing both sides by ( +8 +9). +[( +35 +113) × 24 × 14 × ( +8 +9)] ÷ ( +8 +9) = 94080 +1017 ÷ ( +8 +9) +( +35 +113) × 24 × 14 × ( +8 +9) × ( +9 +8) = 94080 +1017 × ( +9 +8) +Thus, +( +35 +113) × 24 × 14 = 11760 +113 . +Let us use these ideas to solve the equation 5x – 4 = 7. +What can we do so that 5x is on one side and the equality between the +LHS and the RHS still holds? +To retain only 5x on the LHS, we need to remove the term – 4. This can +be done by adding 4 to both sides. +Thus, 5x – 4 + 4 = 7 + 4. +Hence, 5x = 11. +To retain only the unknown x on the LHS, we need to remove the +factor 5. This can be done by dividing both sides by 5. +Thus, 5x +5 = 11 +5  . +So, x = 11 +5  . +170 + +Finding the Unknown +Can we check that x = 11 +5 is the correct solution to the equation? +We can check this by substituting x with the value 11 +5 in the equation +5x – 4 = 7, and checking if the LHS = RHS. +Substituting x with 11 +5 in the LHS we get, +LHS = 5 ( +11 +5 ) – 4 += 5 ( +11 +5 ) – 4 += 11 – 4 += 7 +This is the same as the RHS. +So, 11 +5 is indeed the correct solution to the equation. +Example 5: Solve the equation 11y + (–5) = 61. +To retain only the unknown term 11y on one side, we need to remove +the term –5. This can be done by subtracting (–5) from both sides. +11y + (–5) – (–5) = 61 – (–5). +That is, +11y = 66. +We can directly find the value of y, seeing that 11 × 6 = 66. +We can also divide both sides by 11 to find y. +11y ÷11 = 66 ÷ 11. +So y = 6 is the solution to the equation 11y + (–5) = 61. +Can you check that this solution is correct? +Example 6: Solve 6y + 7 = 4y + 21. +In this equation, expressions with an unknown are on both sides. +We have seen how to solve equations when the unknown term is on one +side. What can be done to bring the unknown terms to the same side? +Subtracting 4y from both sides, we get +6y + 7 – 4y = 4y + 21 – 4y. +So, +2y + 7 = 21. +171 + +Ganita Prakash | Grade 7 | Part-II +Subtracting 7 from both sides we get +2y + 7 – 7 = 21 – 7, which gives +2y = 14. +We can directly find the value of y, seeing that 2 × (7) = 14. +We can also divide both sides by 2 to find y. +2y ÷ 2 = 14 ÷ 2, which gives +y = 7. +Remember, it is always good to check your solution. +Figure it Out +1. Solve these equations and check the solutions. +(a) 3x – 10 = 35 +(b) 5s = 3s +(c) 3u – 7 = 2u + 3 +(d) 4 (m + 6) – 8 = 2m – 4 +(e) +u +15 = 6 +2. Frame an equation that has no solution. +[Hint: 4 more than a number, and 5 more than a number can +never be equal!] +The procedure to systematically solve an equation can be made +efficient if we make an observation. Consider the equation 11y + (–5) = 61, +which we solved. +As the first step, we subtracted – 5 from both sides to remove the term –5. +11y + (– 5) – (–5) = 61 – (– 5). +Since this action removes the term – 5 from the LHS, we could have +written this step as: +11y = 61 – (–5). +Note that we could also have arrived at this step by seeing addition +and subtraction as inverse operations, as in the case of Example 1. +Similarly, when we had 11y = 66, we divided both sides by 11 to remove +the factor 11 from the LHS. +11y ÷ 11 = 66 ÷ 11. +Math +Talk +172 + +Finding the Unknown +Since this action removes the factor 11 in the LHS, we can write this +step as: +y = 66 ÷ 11. +Again, we could have arrived at this by seeing multiplication and +division as inverse operations, as in the case of Example 2. +Let us write down both these ways of solving an equation. +11y + (– 5) = 61 +11y + (– 5) – (– 5) = 61 – (– 5). +11y = 66 +11y ÷ 11 = 66 ÷ 11 + y = 6 +11y + (– 5) = 61 +11y = 61 – (– 5) +11y = 66 + y = 66 ÷ 11 + y = 6 +Let us consider another equation that we solved earlier. +6y + 7 = 4y + 21 +6y + 7 – 4y = 4y + 21 – 4y +2y + 7 = 21 +2y + 7 – 7 = 21 – 7 +2y = 14. +2y ÷ 2 = 14 ÷ 2 + y = 7 +6y + 7 = 4y + 21 +6y + 7 – 4y = 21 +2y + 7 = 21 +2y = 21 – 7 +2y = 14. + y = 14 ÷ 2 + y = 7 +What happens in cases like u +15 = 6? +Multiplying both sides by 15 leaves only u in the LHS — +u = 6 × 15 +u = 90. +Thus, we can make the following observations — +(a) When a term that is added or subtracted on one side of an equation +is removed from that side, its additive inverse should appear as a +term on the other side for the equality to hold. For example, 2y + 7 += 21 becomes 2y = 21 – 7. +(b) If one side of an equation is the product of two or more numbers +or expressions, and we remove one of the factors, then the other +side should be divided by this factor for the equality to hold. For +example, 2y = 14 becomes y = 14 ÷ 2. +(c) If one side of an equation is the quotient of two numbers or +expressions, and we remove the divisor, then the other side should +be multiplied by this divisor for the equality to hold. For example, +u +15 = 6 becomes u = 6 × 15. +173 + +Ganita Prakash | Grade 7 | Part-II +Solving Problems +Forming and solving equations gives us the ability to find solutions to +many problems in our lives. Let us see a few such examples. +Example 7: Ranjana creates a sequence of arrangements with square +tiles as shown below. Can she extend the sequence and make an +arrangement using 100 tiles? If yes, which step in the sequence will it +be? +Step 1 +Step 2 +Step 3 +She can look at the pattern in different ways. They are shown below. +Step 3 +Step 4 +Step k +Step 2 +Method 1 +Step 1 +4 + 4 + 4 + 1 += 13 +k + k + k + 1 += 3k + 1 +1 1 1 +2 +1 +2 +4 +1 +4 +3 +1 +3 +1 +2 +3 +4 +3 + 3 + 3 + 1 += 10 +1 + 1 + 1 + 1 += 4 +2 + 2 + 2 + 1 += 7 +Step 3 +Step 4 +Step k +Step 2 +Method 2 +Step 1 +3 +5 +9 = (4 × 2 + 1) +7 +1 +2 +3 +4 +1 + 3 += 4 +2 + 5 += 7 +3 + 7 += 10 +4 + 9 += 4 + (4 × 2 + 1) += 13 +k + (2k + 1) += 3k + 1 +174 + +Finding the Unknown +We have the expression 3k + 1 which gives the number of tiles needed +to make an arrangement in Step k. +To check whether an arrangement is possible using 100 tiles at some +Step k, we can solve the equation: 3k + 1 = 100. Find the value of k. +Example 8: Madhubanti wants to organise a party. She decides to buy +snacks for the party from the chaat shop in town. Each plate of snacks +costs ₹25. The shop charges an additional fixed amount of ₹50 to deliver +the snacks to Madhubanti’s house. +There are 5 members in Madhubanti’s family, including herself. Her +parents tell her she can spend ₹500 on this party. How many friends +can she invite to the party if she wants to give a plate of snacks to each +person, including her family and friends? +Fatima’s method of solving this problem is shown below. +She represented the situation using a rough diagram. +Snack +cost +Delivery +cost +25 × ++ 50 +Out of ₹500, if we subtract the fixed delivery charge, then Madhubanti +is left with ₹450. +So the question becomes “How many plates of snacks, each costing +₹25, can be bought for ₹450?”. +The answer to this is 450 ÷ 25 = 18. +18 plates of snacks can be bought for ₹450. Considering her 5 family +members, she can invite 18 – 5 = 13 friends to her party. +175 + +Ganita Prakash | Grade 7 | Part-II +Mahesh represented the unknown +quantity +of +the +total +number +of +people +who +can +attend +the +party, +including +Madhubanthi +and her family members, as p. +Cost incurred = 25p + 50. +Since this cost should be 500, we +have the equation +25p + 50 = 500. +Subtracting 50 from both sides, +we get +25p = 500 – 50. +25p = 450. +Dividing both sides by 25, we get +p = 450 ÷ 25 = 18. +18 people can be at her party, +including her 5 family members. That +means 13 friends can be invited. +Srikanth decided to represent the +unknown quantity of the total number +of friends Madhubanthi can invite as f. +What will be the cost in this case? +Cost incurred = 25 (f + 5). +Since Madhubanthi has ₹450 for +snacks, we have the equation +25 (f + 5) = 450. +Dividing both sides by 25, +f + 5 = 18. +Subtracting 5 from both sides, +f = 13. +Example 9: Two friends want to save money. Jahnavi starts with an +initial amount of ₹4000, and in addition, saves ₹650 per month. Sunita +starts with ₹5050 and saves ₹500 per month. After how many months +will they have the same amount of money? +Let m denote the number of months after which their savings are +equal. +What are Jahnavi’s savings after m +months? +Jahnavi’s savings = 4000 + 650 m. +What are Sunita’s savings after +m months? +Sunita’s savings = 5050 + 500 m. +As these amounts are equal after m +months, we get the following equation: +4000 + 650 m = 5050 + 500 m +4000 + 650 m – 500 m = 5050 (subtracting 500 m from both sides) +4000 + 150 m = 5050 +150 m = 5050 – 4000 (subtracting 4000 from both sides) +Jahnavi +Sunita +Equal +4000 +5050 +650 × m +500 × m ++ ++ +176 + +Finding the Unknown +150 m = 1050 +m = 1050 ÷150 (dividing both sides by 150) +m = 7. +So, after 7 months, both will have the same amount of money. +Check the answer. +Let us solve a few equations. +Example 10: Solve 28 (x + 4) + 300 = 1000. +Here are some ways to solve this equation. +Subtracting +300 +from +both sides, we get +28 (x + 4) = 1000 – 300 +28 (x + 4) = 700. +Dividing both sides by +28, we get +x + 4 = 700 ÷ 28 +x + 4 = 25. +So x = 25 – 4, which gives +x = 21. +28 (x + 4) + 300 = 1000 +Since 28, 300, and 1000 +are divisible by 4, we get +a simpler equation if we +divide both sides by 4. + 28 (x + 4) + 300 +4 + = 1000 +4 +Using the rules of fraction +addition, we get +28 (x + 4) +4 + + 300 +4 = 1000 +4 +7 (x + 4) + 75 = 250 +Subtracting 75 from both +sides, +7 (x + 4) = 175 +7x + 28 = 175 +Subtracting 28 from both +sides, +7x = 147 +x = 147 +7 = 21. +28 (x + 4) + 300 = 1000 +Simplifying the LHS, +28x + 112 + 300 = 1000 +28x + 412 = 1000 +Subtracting +412 +from +both sides, we get +28x = 1000 – 412 +28x = 588 +Dividing both sides by +28, +x = 588 ÷ 28 +x = 21. +177 + +Ganita Prakash | Grade 7 | Part-II +Example 11: Riyaz created a math trick, which he tries out on his friend +Akash. +Riyaz asked Akash to perform the following steps without revealing +the answer to any of the intermediate steps. +1. Think of a number. +2. Subtract 3 from the number. +3. Multiply the result by 4. +4. Add 8 to the product. +5. Reveal the final answer. +The final answer revealed by Akash was +24. Using this, Riyaz correctly figured out the +starting number that Akash had thought of. Find +this number. +Try the steps using different numbers as the +starting number. Do you see any relation between the starting number +and final answer? +The answer can be found by algebraically modeling this scenario. +In other words, we can form an equation using an unknown. Let the +unknown starting number be x. +What are the expressions we get after each step? +Steps +Expression +Think of a number. +x +Subtract 3 from the number. +x – 3 +Multiply the result by 4. +4 (x – 3) = 4x – 12 +Add 8 to the product. +4x – 12 + 8 = 4x – 4 +Since Akash’s final answer was 24, we have the equation: +4x – 4 = 24 + +4 (x – 1) = 24 + +x – 1 = 6 (Dividing both sides by 4) +Thus, Akash thought of the number 7. +Can you think of a simple rule that you can use to get the starting number +from the final answer? +178 + +Finding the Unknown +Example 12: Ramesh and Suresh have 60 marbles between them. +Ramesh has 30 more marbles than Suresh. How many marbles does +each boy have? +In this problem, we have two +unknowns. If we denote the number +of marbles with Ramesh as x and the +number of marbles with Suresh as y, +then what are the different equations +that we have? +1. The total number of marbles is 60. +x + y = 60. +2. Ramesh has 30 marbles more than Suresh. +x = y + 30. +Suresh +1 +2 +3 +30 +Ramesh +How do we find the unknowns using these equations? So far, we have +only developed a strategy to solve an equation with one unknown! +To get such an equation, we can do the following. +Denote the number of marbles with Suresh as y, and the number of +marbles with Ramesh as y + 30. +Since the total number of marbles is 60, we have the equation: +y + (y + 30) = 60 +2y + 30 = 60. +Use this to find both the unknowns. +Generating Equations +So far, we have solved a given equation to find the value of the +letter−number. Can we do the reverse — write equations using a given +value of the letter−number? +Write equations whose solution is y = 5. Share the equations you +made with each other and discuss the methods used. +Math +Talk +179 + +Ganita Prakash | Grade 7 | Part-II +Two such equations are y + 1 = 6 and 3y = 15. +Consider the following chains of equations, where one is obtained +from the previous one by performing the same operation on both sides. +y + 1 = 6 +Multiplying by – 1 +Adding y +Adding 6 +– y – 1 = – 6 +–1 = – 6 + y +5 = y +3y = 15 +Adding 6 +Dividing by 3 +Subtracting 2 +3y + 6 = 21 +y + 2 = 7 +y = 5 +Can you form a chain going from the bottom equation to the top? +Compare the operations used when going from the top to the +bottom and from the bottom to the top. +Without calculating, can you find the value of the unknown in each +equation in the chains above? +[Hint: We have seen that the value that satisfies an equation also +satisfies the new equation obtained by performing the same operation +on both sides of the original equation.] +Example 13: Can you give a real-life situation that can be modelled as +the equation, 100x + 75 = 250? +Solution: If we think of these numbers as representing money, we +can see that the total money is ₹250. +There are two terms that are adding to ₹250. The second term in the +LHS, ₹75, is fixed and does not change. The value of the first term would +change based on ‘x’. So we can think of 100 as the number of units and +‘x’ as the cost per unit. For instance, x could represent the cost of a plate +of snacks and 75 could be the delivery charge. +Math +Talk +180 + +Finding the Unknown +Figure it Out +1. Write 5 equations whose solution is x = – 2. +2. Find the value of each unknown: +(a) 2y = 60 +(b) – 8 = 5x – 3 +(c) – 53w = –15 +(d) 13 – z = 8 +(e) k + 8 = 12 – k +(f) 7m = m – 3 +(g) 3n = 10 + n +3. I am a 3-digit number. My hundred’s digit is 3 less than my ten’s digit. +My ten’s digit is 3 less than my unit’s digit. The sum of all the three +digits is 15. Who am I? +4. The weight of a brick is 1 kg more than half its weight. What is the +weight of the brick? +5. One quarter of a number increased by 9 gives the same number. +What is the number? +6. Given 4k + 1 = 13, find the values of: +(a) 8k + 2 +(b) 4k +(c) k +(d) 4k – 1 +(e) – k – 2 +7.3 Mind the Mistake, Mend the Mistake +The following are some equations along with the steps used to solve +them to find the value of the letter−number. Go through each solution +and decide whether the steps are correct. If there is a mistake, describe +the mistake, correct it and solve the equation. +4x + 6 = 10 +4x = 10 + 6 +4x = 16 +x = 4 +1 +7 – 8z = 5 +8z = 7 – 5 +8z = 2 +z = 4 +2 +2v – 4 = 6 +v – 4 = 6 – 2 +v – 4 = 4 +v = 8 +3 +5z + 2 = 3z – 4 +5z + 3z = – 4 + 2 +8z = – 2 +z = – 2 +8 +4 +15w – 4w = 26 +15w = 26 + 4w +15w = 30 +w = 2 +5 +3x + 1 = – 12 +x + 1 = – 12 +3 +x + 1 = – 4 +x = – 5 +6 +181 + +Ganita Prakash | Grade 7 | Part-II +7.4 A Pinch of History +Forming expressions using symbols and solving equations with such +expressions was an important component of mathematical explorations +in ancient India. This area of mathematics was termed bījagaṇita, also +now known as algebra. The word bīja means seed. Just as a tree is hidden +inside a seed, the answer to a problem is hidden inside an unknown +number. Solving the problem is like helping the tree grow — step by +step, we discover what is hidden. +We have seen Brahmagupta’s contributions to different areas +of mathematics like integers and fractions. In Chapter 18 of his book +Brāhmasphuṭasiddhānta (628 CE), he also explained how to add, subtract, +and multiply unknown numbers using letters — similar to how we use x +or y today. This chapter was one of the earliest known works in algebra +in history. As renowned mathematician and Fields Medalist David +Mumford remarked, ‘Brahmagupta is the key person in the creation of +algebra as we know it’. +In the 8th century, Indian mathematical ideas were translated +into Arabic. They influenced a well-known mathematician named +Al-Khwarizmi, who lived in present-day Iraq. Around 825 CE, he wrote +a book called Hisab al-jabr wal-muqabala, which means ‘calculation by +restoring and balancing’. +These ideas spread even further. By the 12th century, Al-Khwarizmi’s +book was translated into Latin and brought to Europe, where it became +very popular. The word al-jabr from his book gave us the word algebra, +which we also still use today. +Similar to how we use letters from the alphabet today to represent +unknowns, ancient Indian mathematicians from the time of +Brahmagupta used distinct symbols like yā, kā, nī, pī, lo, etc., for +different unknowns. The symbol yā was short for yāvat-tāvat (meaning +‘as much as needed’). Others like kā and nī referred to as the first letters +4 (4q + 2) = 50 + 4 (4q) = 50 – 2 + 16q = 48 + q = 3 +– 2 (3 – 4x) = 14 +–6v – 8x = 14 +– 8x = 14 + 6 +– 8x = 20 +x = – 20 +8 +3 (7y + 4) = 9 + 5y +7y + 4 = 9 +3 + 5y +7y + 4 = 3 + 5y +7y – 5y + 4 = 3 +2y = 4 – 3 +y = 1 +2 +7 +8 +9 +182 + +Finding the Unknown +of the names of colours — kālaka (black), nīlaka (blue), and so on. In +contrast to these unknowns, the known quantities in an expression had +a specific form (rūpa) and were denoted by rū. +Here are some examples of how algebraic expressions in modern +notation were written in ancient Indian notation: +Modern Notation +Ancient Indian Notation +2x + 1 +yā 2 rū 1 +(in each term the indication of +unknown/known came first) +2x – 8 +yā 2 rū 8 +. +(a dot over the number indicated +that it was negative) +3x + 4 = 2x + 8 +yā 3 rū 4 +yā 2 rū 8 +(the two sides of an equation were +presented one below the other) +Example 16: Bījgaṇita by Bhāskarāchārya (1150 CE) mentions this +problem. +One man has ₹300 rupees and 6 horses. Another man has 10 horses +and a debt of ₹100. If they are equally rich and the price of each horse is +the same, tell me the price of one horse. +Solution: +Let price of one horse = ₹ x +According to the problem +300 + 6x = 10x – 100 +300 + 6x + (100) = 10x +400 + 6x = 10x +400 = 10x – 6x +400 = 4x +400 ÷ 4 = x +100 = x +Therefore, the price of one horse = ₹ 100. +183 + +Ganita Prakash | Grade 7 | Part-II +Such problems and solutions were well understood in ancient India. +In fact, a very systematic way to solve problems with single unknowns +was first proposed by Aryabhata (499 CE) and Brahmagupta has outlined +it in his Brāhmasphuṭasiddhānta (628 CE). Let us understand his method. +Let us look at a few equations of the following form: +5x + 4 = 3x + 8, or 3x – 6 = 2x + 4. +Can we come up with a formula to solve these equations? That is, for the +first equation, can we perform some operations using 5, 4, 3, and 8 that +will directly give us the solution? Using a similar method, can you solve +the second equation using the numbers 3, – 6, 2 and 4? +To get a formula, we can generalise equations of this form by taking +the four numbers as A, B, C, and D. That is, +Ax + B = Cx + D. +Brahmagupta gives the solution to equations of this form with this +formula: +x = D – B +A – C . +Using this approach, we can find the solution to the equation: +650m + 4000 = 500m + 5050 +simply by calculating, +m = 5050 – 4000 +650 – 500 . +Using this formula can you solve this equation 2x + 3 = 4x + 5? +Ancient Indian mathematicians were excellent at converting complex +mathematical ideas into simple procedures so that everyone could use +these procedures to solve problems! +Bījagaṇita or algebra is the branch of mathematics that uses letter +symbols to solve mathematical problems. We have seen some examples in +previous pages. By studying algebra, we learn to generalise patterns — +in numbers, shapes, and situations. We express these generalisations +using the language of algebra, which is both precise and powerful. +Bījagaṇita also gives us a way to justify mathematical claims (like why +the sum of two odd numbers is always even) and to solve problems of +many kinds. +184 + +Finding the Unknown +The power of algebra was well recognised by ancient Indian +mathematicians. We hope you recognise it too — and enjoy using it! +Figure it Out +1. Fill in the blanks with integers. +(a) 5 × ___ – 8 = 37 +(b) 37 – (33 – ____ ) = 35 +(c) – 3 × (– 11 + ____ ) = 45 +2. Ranju is a daily wage labourer. She earns ₹ 750 a day. Her employer +pays her in 50 and 100 rupee notes. If Ranju gets an equal number +of 50 and 100 rupee notes, how many notes of each does she have? +3. In the given picture, each black +blob hides an equal number of +blue dots. If there are 25 dots in +total, how many dots are covered +by one blob? Write an equation +to describe this problem. +25 dots +4. Here are machines that take an input, perform an operation on it +and send out the result as an output. +(a) + +Find the inputs in the following cases: + +185 + +Ganita Prakash | Grade 7 | Part-II +(b) +Find the inputs in the following cases: +5. What are the inputs to these machines? +6. A taxi driver charges a fixed fee of ₹800 per day plus ₹20 for each +kilometer traveled. If the total cost for a taxi ride is ₹2200, determine +the number of kilometres traveled. +186 + +Finding the Unknown +7. The sum of two numbers is 76. One number is three times the other +number. What are the numbers? +8. The figure shows the diagram for a window with a grill. What is the +gap between two rods in the grill? +3 cm +2 cm +34 cm +9. In a restaurant, a fruit juice costs ₹15 less than a chocolate milkshake. +If 4 fruit juices and 7 chocolate milkshakes cost ₹600, find the cost of +the fruit juice and milkshake. +10. Given 28p – 36 = 98, find the value of 14p – 19 and 28p – 38. +11. The steps to solve three equations are shown below. Identify and +correct any mistakes. +(a) +6x + 9 = 66 +x + 9 = 11 +x = 11 – 9 +x = 2 +(b) 14y + 24 = 36 +7y + 12 = 18 +7y = 6 + y = 6 +7 +(c) +4x – 5 = 9x + 8 +4x = 9x + 8 – 5 +4x = 9x + 3 +4x – 9x = 3 +–5x = 3 +x = – 5 +3 +12. Find the measures of the angles of these triangles. +y + 15 +y +x + 10 +x – 10 +x +11 +187 + +Ganita Prakash | Grade 7 | Part-II +13. Write 4 equations whose solution is u = 6. +14. The Bakhśhāli Manuscript (300 CE) mentions the following problem. +The amount given to the first person is not known. The second person +is given twice as much as the first. The third person is given thrice +as much as the second; and the fourth person four times as much as +the third. The total amount distributed is 132. What is the amount +given to the first person? +15. The height of a giraffe is two and a half metres more than half its +height. How tall is the giraffe? +16. Two separate figures are given below. Each figure shows the first few +positions in a sequence of arrangements made with sticks. Identify +the pattern and answer the following questions for each figure: +(a) How many squares are in position number 11 of the sequence? +(b) How many sticks are needed to make the arrangement in +position number 11 of the sequence? +(c) Can an arrangement in this sequence be made using exactly +85 sticks? If yes, which position number will it correspond to? +(d) Can an arrangement in this sequence be made using exactly +150 sticks? If yes, which position number will it correspond to? +17. A number increased by 36 is equal to ten times itself. What is the +number? +18. Solve these equations: +(a) 5(r + 2) = 10 +(b) – 3(u + 2) = 2(u – 1) +(c) 2(7 – 2n) = – 6 +(d) 2(x – 4) = – 16 +(e) 6(x – 1) = 2(x – 1) – 4 +(f) 3 – 7s = 7 – 3s +(g) 2x + 1 = 6 – (2x – 3) +(h) 10 – 5x = 3(x – 4) – 2(x – 7) +19. Solve the equations to find a path from Start to the End. Show your +work in the given boxes provided and colour your path as you proceed. +188 + +Finding the Unknown +8x = 20 + 3x +8x – 3x = 20 +5x = 20 +x = 20 +5 + +x = 4 +– 7 = 11 – 3x +– 4 +– 1 +5 +4 +3 +6 +8 +5 +– 8 +2 +13 +8 +–10 +–10 +– 5 +– 4 ++ 4 +4 +–2 +21 +1 +– 44 +2x – 9 = – 3 +8m + 8 = – 72 +– 4 = 16 – 5k +2x – 9 = 3 – x +30 = 4 – 50n +– 2x = – 42 +15 = 19 – 4x +2x + 3 = x + 5 +2 (x + 1) – 10 = +18 +2x + 5 = 3 (x – 1) +20. There are some children and donkeys on a beach. Together they +have 28 heads and 80 feet. How many donkeys are there? How +many children are there? +Try +This +189 + +Ganita Prakash | Grade 7 | Part-II +• An algebraic equation is a mathematical statement that indicates the +equality of two algebraic expressions. +• When the same operation is performed on both sides of an equation, +equality is maintained. +• Finding a solution to an equation means finding the values of the +unknowns in the expressions such that the LHS is equal to the RHS. +• Equations can often be solved by performing the same operation on +both sides so that the value of the unknown becomes evident. +SUMMARY +  +Think of any number. +Now multiply it by 2. +Add 10. +Divide by 2. +Now subtract the original number you thought of. +Finally, add 3. +I predict that you now have 8. Am I correct? +Try the trick on your friends and family! +Can you explain why the trick works? +[Hint: Denote the first number thought of by x.] +Can you make your own such tricks? + +A Magic Trick +190 + +A +B +C +D +E +F +G +tANGRAM +Note: Cut each shape along the white border." +class_8,1,A square and a cube,ncert_books/class_8/hegp1dd/hegp101.pdf,"Queen Ratnamanjuri had a will written that described her fortune of +ratnas (precious stones) and also included a puzzle. Her son Khoisnam +and their 99 relatives were invited to the reading of her will. She wanted +to leave all of her ratnas to her son, but she knew that if she did so, all +their relatives would pester Khoisnam forever. She hoped that she had +taught him everything he needed to know about solving puzzles. She left +the following note in her will— +“I have created a puzzle. If all 100 of you answer it at the same time, you +will share the ratnas equally. However, if you are the first one to solve the +problem, you will get to keep the entire inheritance to yourself. Good luck.” +The minister took Khoisnam and his 99 relatives to a secret room in +the mansion containing 100 lockers. +The minister explained— “Each person is assigned a number from 1 to +100. +• Person 1 opens every locker. +• Person 2 toggles every 2nd locker (i.e., closes it if it is open, opens +it if it is closed). +• Person 3 toggles every 3rd locker (3rd, 6th, 9th, … and so on). +• Person 4 toggles every 4th locker (4th, 8th, 12th, … and so on). +This continues until all 100 get their turn. +In the end, only some lockers remain open. The open lockers reveal +the code to the fortune in the safe.” +Before the process begins, Khoisnam realises that he +already knows which lockers will be open at the end. +How did he figure out the answer? +Hint: Find out how many times each locker is toggled. +A SQUARE AND A CUBE +1 + +Ganita Prakash | Grade 8 +2 +If a locker is toggled an odd number of times, it will be open. Otherwise, +it will be closed. The number of times a locker is toggled is the same as +the number of factors of the locker number. For example, for locker #6, +Person 1 opens it, Person 2 closes it, Person 3 opens it and Person 6 closes +it. The numbers 1, 2, 3, and 6 are factors of 6. If +the number of factors is even, the locker will be +toggled by an even number of people and it will +eventually be closed. +Note that each factor of a number has a +‛partner factor’ so that the product of the pair +of factors yields the given number. Here, 1 and +6 form a pair of partner factors of 6, and 2 and 3 +form another pair. +Does every number have an even number of factors? +We see in some cases, like 2 × 2, that the numbers in the pair are the +same. +Can you use this insight to find more numbers with an odd number of +factors? +For instance, 36 has a factor pair 6 × 6 where both numbers are 6. +Does this number have an odd number of factors? If every factor of 36 +other than 6 has a different factor as its partner, then we can be sure +that 36 has an odd number of factors. Check if this is true. +Hence all the following numbers have an odd number of factors — +1 × 1, 2 × 2, 3 × 3, 4 × 4, ... +A number that can be expressed as the product of a number with +itself is called a square number, or simply a square. The only numbers +that have an odd number of factors are the squares, because they each +have one factor which, when multiplied by itself, equals the number. +Therefore, every locker whose number is a square will remain open. +6: +1 × 6 +2 × 3 +Factors are +1, 2, 3 and 6. +1: +1 × 1 +The only factor +is 1. +4: +1 × 4 +2 × 2 +Factors are +1, 2 and 4. +9: +1 × 9 +3 × 3 +Factors are +1, 3 and 9. + +A Square and A Cube +3 +Write the locker numbers that remain open. +1.1 Square Numbers +Why are the numbers, 1, 4, 9, 16, …, called squares? We know that the +number of unit squares in a square (the area of a square) is the product +of its sides. The table below gives the areas of squares with different +sides. +We use the following notation for squares. +1 × 1 = 12 = 1 +2 × 2 = 22 = 4 +3 × 3 = 32 = 9, +4 × 4 = 42 = 16 +5 × 5 = 52 = 25. +... +In general, for any number n, we write n × n = n2, which is read as ‛n squared’. +Can we have a square of sidelength 3 +5 or 2.5 units? +Yes, there area in square units are ( 3 +5) +2 = ( 3 +5) × ( 3 +5) = ( 9 +25), +and (2.5) +2 = (2.5) × (2.5) = 6.25. +Khoisnam immediately collects word clues from these 10 lockers and +reads, “The passcode consists of the first five locker numbers that +were touched exactly twice.” +Which are these five lockers? +The lockers that are toggled twice are the prime numbers, since each +prime number has 1 and the number itself as factors. So, the code is +2-3-5-7-11. +Sidelength +(in units) +Area +(in sq units) +1 +1 × 1 = 1 sq. unit +2 +2 × 2 = 4 sq. units +3 +3 × 3 = 9 sq. units +4 +4 × 4 = 16 sq. units +5 +5 × 5 = 25 sq. units +10 +10 × 10 = 100 sq. units + +Ganita Prakash | Grade 8 +4 +The squares of natural numbers are called perfect squares. For +example, 1, 4, 9, 16, 25, … are all perfect squares. +Patterns and Properties of Perfect Squares +Find the squares of the first 30 natural numbers and fill in the table below. +12 = 1 +112 = 121 +212 = 441 +22 = 4 +122 = +222 = +32 = 9 +132 = +42 = 16 +142 = +52 = 25 +152 = +62 = +162 = +72 = +172 = +82 = +182 = +92 = +192 = +102 = +202 = +What patterns do you notice? Share your observations and make +conjectures. +Study the squares in the table above. What are the digits in the +units places of these numbers? All these numbers end with 0, 1, 4, 5, 6 or +9. None of them end with 2, 3, 7 or 8. +If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square? +The numbers 16 and 36 are both squares with 6 in the units place. +However, 26, whose units digit is also 6, is not a square. Therefore, we +cannot determine if a number is a square just by looking at the digit in +the units place. But, the units digit can tell us when a number is not a +square. If a number ends with 2, 3, 7, or 8, then we can definitely say that +it is not a square. +Write 5 numbers such that you can determine by looking at their units +digit that they are not squares. +The squares, 12, 92, 112, 192, 212, and 292, all have 1 in their units place. +Write the next two squares. Notice that if a number has 1 or 9 in the +units place, then its square ends in 1. +Let us consider square numbers ending in 6: 16 = 42, 36 = 62, 196 = 142, +256 = 162, 576 = 242, and 676 = 262. +Math +Talk +Math +Talk + +A Square and A Cube +5 +Which of the following numbers have the digit 6 in the units place? + +(i) 382 + (ii) 342 (iii) 462 (iv) 562 (v) 742 (vi) 822 +Find more such patterns by observing the numbers and their squares +from the table you filled earlier. +Consider the following numbers and their squares. +If a number contains 3 zeros at the end, how many zeros will its square +have at the end? +What do you notice about the number of zeros at the end of a number +and the number of zeros at the end of its square? Will this always +happen? Can we say that squares can only have an even number of +zeros at the end? +What can you say about the parity of a number and its square? +Perfect Squares and Odd Numbers +Let us explore the differences between consecutive squares. What do +you notice? +4 – 1 = 3    9 – 4 = 5    16 – 9 = 7    25 – 16 = 9 +See if this pattern continues for the next few square numbers. +From this we observe that adding consecutive odd numbers starting +from 1 gives consecutive square numbers, as shown below. +102 = 100 +202 = 400 +402 = 800 +1002 = 10000 +2002 = 40000 +7002 = 49000 +9002 = 81000 +We have +one zero. +We have +two zeroes. +But we have +two zeroes. +But we have +four zeroes. +1 = 1 +1 + 3 = 4 +1 + 3 + 5 = 9 +1 + 3 + 5 + 7 = 16 +1 + 3 + 5 + 7 + 9 = 25 +1 + 3 + 5 + 7 + 9 + 11 = 36. + +Ganita Prakash | Grade 8 +6 +Do you remember this pattern from Grade 6? +The picture below explains why each subsequent inverted L gives the +next odd number: +We see that the sum of the first n odd numbers is n2. Alternatively, +every square is a sum of successive odd numbers starting from 1. +Also, we can find out whether a number is a perfect square by +successively subtracting odd numbers. Consider the number 25, +successively subtract 1, 3, 5, ... until you get or cross over 0, +This means 25 = 1 + 3 + 5 + 7 + 9 and is thus a perfect square. Since we +subtracted the first five odd numbers, 25 = 52. +Using the pattern above, find 362, given that 352 = 1225. +From the question we know that 1225 is the sum of the first 35 odd +numbers. To find 362, we need to add the 36th odd number to 1225. +How do we find the 36th odd number? +The 1st odd number is 1, 2nd odd number is 3, 3rd number is 5, …, 6th +odd number is 11 and so on. +What is the nth odd number? +The nth odd number is 2n–1. +Therefore, the 36th odd number is 71. +By adding 71 to 1225, we get 1296, which is 362. +Consider a number such as 38 that is not a square and subtract +consecutive odd numbers starting from 1. +This shows that 38 cannot be expressed as a sum of consecutive odd +numbers starting with 1. +1 + 3 +1 + 3 + (3 + 2) +1 + 3 + 5 +1 + 3 + 5 + (5 + 2) +1 + 3 + 5 + 7 +In mathematics, sometimes arguments and reasoning can be +presented without any words. Visual proofs can be complete by +themselves. +25 – 1 = 24    24 – 3 = 21    21 – 5 = 16    16 – 7 = 9    9 – 9 = 0 + +38 – 1 = 37 +37 – 3 = 34 +34 – 5 = 29 +29 – 7 = 22 +22 – 9 = 13 + +13 – 11 = 2 +2 – 13 = – 11 + +A Square and A Cube +7 +Thus, we can say that a natural number is not a perfect square if it +cannot be expressed as a sum of successive odd natural numbers starting +from 1. We can use this result to find out whether a natural number is a +perfect square. +Find how many numbers lie between two consecutive perfect squares. +Do you notice a pattern? +How many square numbers are there between 1 and 100? How many +are between 101 and 200? Using the table of squares you filled earlier, +enter the values below, tabulating the number of squares in each block +of 100. What is the largest square less than 1000? +Perfect Squares and Triangular Numbers +Do you remember triangular numbers? +Can you see any relation between triangular numbers and square +numbers? Extend the pattern shown and draw the next term. +Square Roots +The area of a square is 49 sq. cm. What is the length of its side? +We know that 7 × 7 = 49, or 72 = 49. +1 –100 +101 – 200 +201 – 300 +301 – 400 +401 – 500 +501 – 600  601 – 700 +701 – 800 +801 – 900 +901 – 1000 +1 +3 +6 +10 +15 +1 + 3 = 4 = 22 +3 + 6 = 9 = 32 +6 + 10 = 16 = 42 + +Ganita Prakash | Grade 8 +8 +So, the length of the side of a square with an area of 49 sq. cm is 7 cm. +We call 7 the square root of 49. +In general, if y= x2 then x is the square root of y. +What is the square root of 64? +We know that 8 × 8 is 64. So, 8 is the square root of 64. What about +– 8 × – 8 ? That is 64 too! +82 = 64, and (– 8)2 = 64. +So, the square roots of 64 are + 8 and – 8. +Every perfect square has two integer square roots. One is positive and +the other is negative. The square root of a number is denoted by +Thus, +64 +8 +=± and 100 +10 +=± +. +Note that 8 +8 +2 =± and 10 +10 +2 =± +. In general, +n +n +2 =± +. +In this chapter, we shall only consider the positive square root. +Given a number, such as 576 or 327, how do we find out if it is a +perfect square? If it is a perfect square, how can we find its square +root? +We know that perfect squares end in 1, 4, 9, 6, 5, or an even +number of zeros. But, it is not certain that a number that satisfies this +condition is a square. +We can clearly say that 327 is not a perfect square. However, we +cannot be sure that 576 is a perfect square. +1. We can list all the square numbers in sequence and find out +whether 576 occurs among them. We know that 202 = 400, we can +find squares of 21, 22, 23, … and so on until we get 576 or a number +greater than 576. +202 = 400    212 = 441    222 = 484    232 = 529    242 = 576 + +However, this process becomes inefficient for larger numbers. +2. Recall that every square can be expressed as a sum of consecutive +odd numbers starting from 1. + Consider 81. +  +From 81, we successively subtracted consecutive odd numbers +starting from 1 until we obtained 0 at the 9th step. Therefore 81 = 9. +  +Can we find the square root of 729 using this method? Yes, but it +will be time-consuming. +Math +Talk +81 – 1 = 80 +80 – 3 = 77 +77 – 5 = 72 +72 – 7 = 65 +65 – 9 = 56 +56 – 11 = 45 +45 – 13 = 32 +32 – 15 = 17 +17 – 17 = 0 + +A Square and A Cube +9 +3. We know that a perfect square is obtained by multiplying an integer +by itself. Will looking at a number’s prime factorisation help in +determining whether it is a perfect square? + +Yes, if we can divide the prime factors of a number into two +equal groups, then the product of the prime factors in either group +combine to form the square root. +Is 324 a perfect square? + + + + +324 = 2 × 2 × 3 × 3 × 3 × 3. +These can be grouped as + + + + +324 = (2 × 3 × 3) × (2 × 3 × 3). + + + + + += (2 × 3 × 3)2 = 182. +We can also write the prime factors in pairs. That is, + + + + +324 = (2 × 2) × (3 × 3) × (3 × 3), +which shows that 324 is a perfect square. Thus, + + + + +324 = (2×3×3)2 = 182. + + Therefore, 324 =18. +Is 156 a perfect square? +The prime factorisation of 156 is 2 × 2 × 3 × 13. +We cannot pair up these factors. +Therefore, 156 is not a perfect square. +Find whether 1156 and 2800 are perfect squares using prime factorisation. +We can estimate the square root of larger perfect squares by looking +at the closest perfect squares we are familiar with and then narrowing +down the interval to search. +For example, to find 1936 , we can reason as follows: + (i) +1936 is between 1600 (402) and 2500 (502), so 40 < 1936 < 50. + (ii) +The last digit of 1936 is 6. So, the last digit of the square root +must either be 4 or 6. It can be 44 or 46. +(iii) If we calculate 452, we can compare it with 1936 to halve +the interval to search from 40–50 to either 40–45 or 45–50. +We can write 452 as (40 + 5) (40 + 5) = 402 + 2 × 40 × 5 + 52 += 1600 + 400 + 25 = 2025. + (iv) 2025 > 1936. So, 40 < 1936 < 45 + (v) +From the observation in point b we can guess and then verify +that 1936 is 44. +Consider the following situations — +Aribam and Bijou play a game. One says a number and the other +replies with its square root. Aribam starts. He says 25, and Bijou quickly + +Ganita Prakash | Grade 8 +10 +responds with 5. Then Bijou says 81, and Aribam answers 9. The game +goes on till Aribam says 250. Bijou is not able to answer because 250 +is not a perfect square. Aribam asks Bijou if he can at least provide a +number that is close to the square root of 250. +For this, Bijou needs to estimate the square root of 250. +We know that 100 < 250 < 400 and 100 = 10 and +400 = 20. +So, 10 < +250 < 20. +But, we are still not very close to the number whose square is 250. +We know that 152 = 225 and 162 = 256. +Therefore, 15 < +250 < 16. Since 256 is much closer to 250 than 225, +250 is approximately 16. We also know it is less than 16. +Here is another problem that requires estimating square roots. +Akhil has a square piece of cloth of area 125 cm2. He wants to know +if he can cut out a square handkerchief of side 15 cm. If not, he wants +to know the maximum size handkerchief that can be cut out from this +piece of cloth with an integer side length. +125 is not a perfect square. The nearest perfect squares are 112 = 121 +and 122 = 144. So the largest square handkerchief with integer side length +that can be cut out from this piece of cloth has side length 11 cm. +Figure it Out +1. Which of the following numbers are not perfect squares? + +(i) 2032 +(ii) 2048 +(iii) 1027 +(iv) 1089 +2. Which one among 642, 1082, 2922, 362 has last digit 4? +3. Given 1252 = 15625, what is the value of 1262? + +(i) 15625 + 126 +(ii) 15625 + 262 +(iii) 15625 + 253 + +(iv) 15625 + 251 +(v) 15625 + 512 +4. Find the length of the side of a square whose area is 441 m2. +5. Find the smallest square number that is divisible by each of the +following numbers: 4, 9, and 10. +6. Find the smallest number by which 9408 must be multiplied so that +the product is a perfect square. Find the square root of the product. +7. How many numbers lie between the squares of the following +numbers? + (i) 16 and 17 +(ii) 99 and 100 +8. In the following pattern, fill in the missing numbers: + +A Square and A Cube +11 +12 + 22 + 22 = 32 +22 + 32 + 62 = 72 + 32 + 42 + 122 = 132 +42 + 52 + 202 = (___)2 + + + + + + +92 + 102 + (___)2 = (___)2 +9. How many tiny squares are there in the following picture? Write the +prime factorisation of the number of tiny squares. +1.2 Cubic Numbers +You know the word cube from geometry. A cube +is a solid figure all of whose all sides meet at right +angles and are equal. How many cubes of side 1 cm +make a cube of side 2 cm? +How many cubes of side 1 cm will make a cube of side 3 cm? + +Ganita Prakash | Grade 8 +12 +Consider the numbers 1, 8, 27, ... +These numbers are called perfect cubes. Can you see why they are +named so? +Each of them is obtained by multiplying a number by itself three +times. We note that +1 = 1 × 1 × 1 +8 = 2 × 2 × 2 +27 = 3 × 3 × 3 +Is 9 a cube? +We see that 2 × 2 × 2 = 8 and 3 × 3 × 3 = 27. This shows that +9 is not a perfect cube. Nor is any number from 10 to 26. +Can you estimate the number of unit cubes in a cube with an +edge length of 4 units? +It has 64 unit cubes! If you notice carefully, each layer +of this cube has 4 × 4 unit cubes. Each square layer has +16 unit cubes (4 × 4), and there are 4 such layers, so the +total number of unit cubes is 4 × 4 × 4 = 64. +Since 53 = 5 × 5 × 5 = 125, 125 is a cube. +In general, for any number n, we write the cube +n × n × n as n3. +Complete the table below. +13 = 1 +113 = 1331 +23 = 8 +123 = +33 = 27 +133 = 2197 +43 = 64 +143 = 2744 +53 = 125 +153 = +63 = +163 = +73 = +173 = 4913 +83 = +183 = 5832 +93 = +193 = 6859 +103 = +203 = +What patterns do you notice in the table above? +We know that 0, 1, 4, 5, 6, 9 are the only last digits possible for +Math +Talk + +A Square and A Cube +13 +squares. What are the possible last digits of cubes? +Similar to squares, can you find the number of cubes with 1 digit, 2 digits, +and 3 digits? What do you observe? +Can a cube end with exactly two zeroes (00)? Explain. +Just as we can take squares of fractions/decimals — ( 4 +6 )2, (13.08)2, and +(– 6)2 — we also can compute cubes of such numbers — (4 +6)3, (13.08)3, and +(– 6)3. +( 4 +6 )3 = ( 4 +6 ) × ( 4 +6 ) × ( 4 +6 ) = ( 64 +216) +(13.08)3 = 13.08 × 13.08 × 13.08 = 2237.810112 +(– 6)3 = – 6 × – 6 × – 6 = – 216. +Taxicab Numbers +Once when Srinivasa Ramanujan was working with +G. H. Hardy at the University of Cambridge, Hardy had +come to visit Ramanujan at a hospital when he was +ill. Hardy had ridden in a taxicab numbered 1729 and +he remarked that 1729 was ‘rather a dull number,’ +adding that he hoped that this was not a bad sign. +Ramanujan immediately replied, “No, Hardy, it is a +very interesting number. It is the smallest number +that can be expressed as the sum of two cubes in two +different ways”. +1729 = 13 + 123 + = 93 + 103. +Because of this story, 1729 has since been known as the Hardy– +Ramanujan Number. And numbers that can be expressed as the sum of +two cubes in two different ways are called taxicab numbers. +The next two taxicab numbers after 1729 are 4104 and 13832. Find +the two ways in which each of these can be expressed as the sum of +two positive cubes. +How did Ramanujan know this? Well, he loved numbers. All through his +life, he tinkered with numbers. During Ramanujan’s time in Cambridge, +his colleagues often marveled at his ability to see deep patterns in numbers +that seemed arbitrary to others. His colleague, John Littlewood, once said, +“Every positive integer was one of his [Ramanujan's] personal friends”. +Try +This + +Ganita Prakash | Grade 8 +14 +Perfect Cubes and Consecutive Odd Numbers +Consecutive odd numbers have a role to play with cubes too. Look at the +following pattern: +1 = 1 = 13 +3 + 5 = 8 = 23 +7 + 9 + 11 = 27 = 33 +13 + 15 + 17 + 19 = 64 = 43 +21 + 23 + 25 + 27 + 29 = 125 = 53 +31 + 33 + 35 + 37 + 39 + 41 = 216 = 63. +Later in this series, we get the following set of consecutive numbers: +91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109. +Can you tell what this sum is without doing the calculation? +Cube Roots +We know that 8 = 23. +We call 2 the cube root of 8 and denote this by 2 = 8 +3 +. +More generally, if y = x3, then x is the cube root of y. This is denoted by +x = +y +3 +. So, 8 +3 + = +23 +3 + = 2. +Similarly, +3 +27 +3 +3 +3 +3 += += and 1000 +10 +3 +3 +3 += +=3. In general, +n3 +3 += n . +How do we find out if a number is a cube? Taking inspiration from the +case of squares, let us see if we can use prime factorisations. +Let us check if 3375 is a perfect cube. +3375 = 3 × 3 × 3 × 5 × 5 × 5. +Can the factors be split into three identical groups? For 3375, we can +form three groups of (3 × 5). So, +3375 = (3 × 5) × (3 × 5) × (3 × 5) + + + + + + = (3 × 5)3 = 153. +Another way is to check if the factors can be grouped into triplet(s): +3375 = (3 × 3 × 3) × (5 × 5 × 5) = 33 × 53. +This means 3375 +15 +3 += +. +Is 500 a perfect cube? +500 = 2 × 2 × 5 × 5 × 5. We see that the factors cannot be split into three +identical groups. Therefore, 500 is not a perfect cube. + +A Square and A Cube +15 +Prime Factorisation of a +Number +Prime Factorisation of its Cube +4 = 2 × 2 +43 = 64 = 2 × 2 × 2 × 2 × 2 × 2 = 23 × 23 +6 = 2 × 3 +63 = 216 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33 +15 = 3 × 5 +153 = 3375 = 3 × 3 × 3 × 5 × 5 × 5 = 33 × 53 +12 = 2 × 2 × 3 +123 = 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 += 23 × 23 × 33 +Observe that each prime factor of a number appears three times in +the prime factorisation of its cube. +Find the cube roots of these numbers: +(i) +64 +3 + = +(ii) 512 +3 + = +(iii) +729 +3 + = +Successive Differences +We know that the differences between consecutive perfect squares gives +the sequence of odd numbers. Observe the figure below where the +differences are computed successively for perfect squares. After two +levels, all the differences are the same. +Compute successive differences over levels for perfect cubes until all the +differences at a level are the same. What do you notice? +1.3 A Pinch of History +The first known list of perfect squares and perfect +cubes was compiled by the Babylonians as far +back as 1700 BCE. These lists, found on clay tablets, +were used to quickly find square roots and cube +1 +4 +9 +16 +25 36 ... +2 +2 +2 +2 + ... +3 +5 +7 +9 +11 ... +Level 1 +Level 2 +Perfect Squares +1 +8 +27 64 +125 216 ... +Perfect Cubes + +Ganita Prakash | Grade 8 +16 +roots in problems involving land measurement, architectural design, +and other areas where geometric calculations were necessary. +In ancient Sanskrit works the term varga was used both for the square +figure or its area, as well as the square power, and the term ghana was +used both for the solid cube as well as the product of a number with +itself three times. The fourth power was called varga-varga . These terms +were used in India at least from the third century BCE. +Aryabhata (499 CE) states +“A square figure of four equal sides and the number representing its area +are called varga. The product of two equal quantities is also called varga.” +Thus, the term varga for square power has its origin in the graphical +representation of a square figure. +Why is the word ‘root’ (the root of a plant) used for the mathematical +operation √ (square root, cube root, etc.)? +It is because, in ancient India, the Sanskrit word mula, meaning +root of a plant, basis, cause, origin, etc., was used for the mathematical +operations of taking roots. +In Sanskrit, varga-mula (the basis, cause, origin of the square) was +used for square-root and ghana-mula was used for cube-root. This use of +mula for the mathematical concept of root was subsequently emulated +in Arabic and Latin through their corresponding words for the root of +a plant — jidhr and radix respectively. The term mula for root has been +used in India at least from the first century BCE. Another term used was +pada (foot, basis, cause, origin). Brahmagupta (628 CE) explains, ‘The +pada (root) of a krti (square) is that of which it is a square.’ +Figure it Out +1. Find the cube roots of 27000 and 10648. +2. What number will you multiply by 1323 to make it a cube number? +3. State true or false. Explain your reasoning. + (i) +The cube of any odd number is even. + (ii) +There is no perfect cube that ends with 8. + (iii) The cube of a 2-digit number may be a 3-digit number. + (iv) The cube of a 2-digit number may have seven or more digits. + (v) +Cube numbers have an odd number of factors. +4. You are told that 1331 is a perfect cube. Can you guess without +factorisation what its cube root is? Similarly, guess the cube roots of +4913, 12167, and 32768. + +A Square and A Cube +17 +5. Which of the following is the greatest? Explain your reasoning. + +(i) 673 – 663 +(ii) 433 – 423 +(iii) 672 – 662 +(iv) 432 – 422 + +„ +A number obtained by multiplying a number by itself is called +a square number. Squares of natural numbers are called +perfect squares. + +„ +All perfect squares end with 0, 1, 4, 5, 6 or 9. Squares can only +have an even number of zeros at the end. + +„ +Square root is the inverse operation of square. Every perfect +square has two integral square roots. The positive square root +of a number is denoted by the symbol √ . For example, √9 = 3. + +„ +A number obtained by multiplying a number by itself three +times is called a cube. For example 1, 8, 27, ... ,etc., are cubes. + +„ +A number is a perfect square if its prime factors can be split +into two identical groups. + +„ +A number is a perfect cube if its prime factors can be split into +three identical groups + +„ +The symbol 3 + denotes cube root. For example, +27 +3 + = 3. +SUMMARY + +Look at the following numbers: 3 6 10 15 1 +They are arranged such that each pair of adjacent numbers adds up to +a square. +3 + 6 = 9, 6 + 10 = 16, 10 + 15 = 25, 15 + 1 = 16. +Try arranging the numbers 1 to 17 (without repetition) in a row in a +similar way — the sum of every adjacent pair of numbers should be a +square. +Can you arrange them in more than one way? If not, can you explain +why? +Can you do the same with numbers from 1 to 32 (again, without +repetition), but this time arranging all the numbers in a circle? +Square Pairs!" +class_8,2,power play,ncert_books/class_8/hegp1dd/hegp102.pdf,"2 +POWER PLAY +2.1 Experiencing the Power Play ... +An Impossible Venture! +Take a sheet of paper, as large a sheet as you can find. Fold it once. Fold +it again, and again. +How many times can you fold it over and over? +Estu says “I heard that a sheet of paper can’t be folded more than +7 times”. +Roxie replies “What if we use a thinner paper, like a newspaper or a +tissue paper?” +Try it with different types of paper and see what happens. +Say you can fold a sheet of paper as many times as you wish. What would +its thickness be after 30 folds? Make a guess. +Let us find out how thick a sheet of paper will be after 46 folds. Assume +that the thickness of the sheet is 0.001 cm. +If you can fold a paper +46 times, it will be so +thick that it can reach +the Moon! +What! That’s crazy! +Just 46 times!? You +must have ignored +several zeros +after 46. +Well, why +don’t you find +out yourselves. + +Ganita Prakash | Grade 8 +20 +The following table lists the thickness after each fold. Observe that the +thickness doubles after each fold. +Fold +Thickness +Fold +Thickness +Fold +Thickness +1 +0.002 cm +7 +0.128 cm +13 +8.192 cm +2 +0.004 cm +8 +0.256 cm +14 +16.384 cm +3 +0.008 cm +9 +0.512 cm +15 +32.768 cm +4 +0.016 cm +10 +1.024 cm +16 +65.536 cm +5 +0.032 cm +11 +2.048 cm +17 +≈ 131 cm +6 +0.064 cm +12 +4.096 cm +(We use the sign ‘≈’ to indicate ‘approximately equal to’.) +After 10 folds, the thickness is just above 1 cm (1.024 cm). +After 17 folds, the thickness is about 131 cm (a little more than 4 feet). +Now, what do you think the thickness would be after 30 folds? +45 folds? Make a guess. +Fill the table below. +Fold +Thickness +Fold +Thickness +Fold +Thickness +18 +≈ 262 cm +21 +24 +19 +≈ 524 cm +22 +25 +20 +≈ 10.4 m +23 +26 +After 26 folds, the thickness is approximately 670 m. Burj Khalifa +in Dubai, the tallest building in the world, is 830 m tall. +Fold +Thickness +Fold +Thickness +27 +≈ 1.3 km +29 +28 +30 +After 30 folds, the thickness of the paper is about 10.7 km, the +typical height at which planes fly. The deepest point discovered in +the oceans is the Mariana Trench, with a depth of 11 km. +Fold +Thickness +Fold +Thickness +Fold +Thickness +31 +36 +41 +32 +37 +42 +33 +38 +43 +34 +39 +44 +35 +40 +45 +Math +Talk + +Power Play +21 +It might be hard to digest the fact that after just 46 folds, the thickness +is more than 7,00,000 km. This is the power of multiplicative growth, +also called exponential growth. Let us analyse the growth. +We have seen that the thickness doubles after every fold. +Notice the change in thickness after two +folds. By how much does it increase? +After any 3 folds, the thickness increases 8 +times (= 2 × 2 × 2). Check if that is true. Similarly, +from any point, the thickness after 10 folds increases by 1024 times +(= 2 multiplied by itself 10 times), as shown in the table below. +Fold +Thickness +Times increased by +0 to 10 +1.024 cm – 0.001 cm += 1.023 cm +1.024 ÷ 0.001 += 1024 +10 to 20 +10.485 m – 1.024 cm + ≈ 10.474 m +10.485 m ÷ 1.024 cm += 1024 +20 to 30 +10.737 km – 10.485 m +≈ 10.726 km +10.737 km ÷ 10.485 m += 1024 +30 to 40 +10995 km – 10.737 km + ≈ 10984.2 km +10995 km ÷ +10.737 km = 1024 +2.2 Exponential Notation and Operations +The initial thickness of the paper was 0.001 cm. +Upon folding once, its thickness became 0.001 cm × 2 = 0.002 cm. +Folding it twice, its thickness became — +0.001 cm × 2 × 2 = 0.004 cm, or 0.001 cm × 22 = 0.004 cm (in shorthand). +Upon folding it thrice, its thickness became — +0.001 cm × 2 × 2 × 2, or 0.001 cm × 23 = 0.008 cm. +When folded four times, its thickness became — +0.001 cm × 2 × 2 × 2 × 2, or 0.001 cm × 24 = 0.016 cm. +Similarly, the expression for the thickness of the paper when folded 7 +times will be 0.001 cm × 2 × 2 × 2 × 2 × 2 × 2 × 2, or 0.001 cm × 27 = 0.128 cm. +We have seen that square numbers can be expressed as n2 and cube +numbers as n3. +n × n = n2 (read as ‘n squared’ or ‘n raised to the power 2’) +Fold 4 +0.016 cm +Fold 5 +0.032 cm +Fold 9 +0.512 cm +Fold 10 +1.024 cm +Fold 4 +0.016 cm +Fold 6 +0.064 cm + +Ganita Prakash | Grade 8 +22 +n × n × n = n3 (read as ‘n cubed’ or ‘n raised to the power 3’) +n × n × n × n = n4 (read as ‘n raised to the power 4’ or ‘the 4th power of n’) +n × n × n × n × n × n × n = n7 (read as ‘n raised to the power 7’ or ‘the 7th +power of n’) and so on. +In general, we write na to denote n multiplied by itself a times. + + + 54 = 5 × 5 × 5 × 5 = 625. +54 is the exponential form of 625. Here, 4 is the +exponent/power, and 5 is the base. Exponents of +the form 5n are called powers of 5: 51, 52, 53, 54, etc. +2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 210 = 1024. +Remember the 1024 from earlier? There, it meant +that after every 10 folds, the thickness increased +1024 times. +Which expression describes the thickness of a sheet of paper after +it is folded 10 times? The initial thickness is represented by the +letter-number v. + (i) 10v +(ii) 10 + v +(iii) 2 × 10 × v +(iv) 210 +(v) 210v +(vi) 102v +Some more examples of exponential notation: +4 × 4 × 4 = 43 = 64. +(– 4) × (– 4) × (– 4) = (– 4)3= – 64. +Similarly, +a × a × a × b × b can be expressed as a3b2 (read as a cubed b squared). +a × a × b × b × b × b can be expressed as a2b4 (read as a squared b raised +to the power 4). +Remember that 4 + 4 + 4 = 3 × 4 = 12, whereas 4 × 4 × 4 = 43 = 64. +Express the number 32400 as a product of its prime factors +and represent the prime factors in their exponential form. +32400 = 2 × 2 × 2 × 2 × 5 × 5 × 3 × 3 × 3 × 3. +In exponential form, this would be +32400 = 24 × 52 × 34. +What is (– 1)5 ? Is it positive or negative? What about (– 1)56 ? +Is (– 2)4 = 16? Verify. +Figure it Out +1. Express the following in exponential form: +(i) 6 × 6 × 6 × 6 +(ii) +y × y + +(iii) b × b × b × b +(iv) +5 × 5 × 7 × 7 × 7 + +(v) 2 × 2 × a × a +(vi) +a × a × a × c × c × c × c × d +32400 +2 +16200 +2 +8100 +2 +4050 +2 +2025 +5 +405 +5 +81 +3 +27 +3 +9 +3 +3 +3 +1 +What is 02, 05 ? +What is 0n ? +54 is read as +‘5 raised to the power 4’ or +‘5 to the power 4’ or +‘5 power 4’ or +‘4th power of 5’ + +Power Play +23 +2. Express each of the following as a product of powers of their prime +factors in exponential form. +(i) 648      (ii) 405      (iii) 540      (iv) 3600 +3. Write the numerical value of each of the following: +(i) 2 × 103 +(ii) 72 × 23 +(iii) +3 × 44 + +(iv) (– 3)2 × (– 5)2 +(v) 32 × 104 +(vi) +(– 2)5 × (– 10)6 +The Stones that Shine ... +Three daughters with curious eyes, +Each got three baskets — a kingly prize. +Each basket had three silver keys, +Each opens three big rooms with ease. +Each room had tables — one, two, three, +With three bright necklaces on each, you see. +Each necklace had three diamonds so fine… +Can you count these stones that shine? +Hint: Find out the number of baskets and rooms. +How many rooms were there altogether? +The information given can be visualised as shown below. +From the diagram, the number of rooms is 34. This can be computed +by repeatedly multiplying 3 by itself, +3 × 3 = 9. +9 × 3 = 27. +27 × 3 = 81. +81 × 3 = 243. +How many diamonds were there in total? Can we find out by just one +multiplication using the products above? +The number of diamonds is 3 × 3 × 3 × 3 × 3 × 3 × 3 = 37. +King +daughters +baskets +keys +rooms + +Ganita Prakash | Grade 8 +24 +We can write +37 = (3 × 3 × 3 × 3) × (3 × 3 × 3) +We had computed till 34. To find 37, we can just multiply 34 (= 81) with +33 (= 27). + += 34 × 33 + += 81 × 27 = 2187 +37 can also be written as 32 × 35. Can you reason out why? +This can be easily extended to products where exponents are the +same letter-numbers. +Write the product p4 × p6 in exponential form. +p4 × p6 = (p × p × p × p) × (p × p × p × p × p × p) = p10. +We can generalise this to — +na × nb = na+b, where a and b are counting numbers. +Use this observation to compute the following. +(i) 29 +(ii) 57 +(iii) 46 +46 can be evaluated in these two ways, +(4 × 4 × 4) × (4 × 4 × 4) = 43 × 43 +                                        = 64 × 64 +                                        = 4096. +43 × 43 is the square of 43, i.e., +43 × 43 can also be written as +(43) +2. +(4 × 4) × (4 × 4) × (4 × 4) = 42 × 42 × 42 +                                            = 16 × 16 × 16 +                                          = 4096. +42 × 42 × 42 is the cube of 42, i.e., +42 × 42 × 42 can also be written as (42) +3. +Similarly, 74 = (7 × 7) × (7 × 7) = 72 × 72 = (72) +2, and +210 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) + = (22) × (22) × (22) × (22) × (22) + = (22) +5. +Is 2 +10 also equal to (25) +2? Write it as a product. +2 +10 = (2 × 2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2) + = (25) × (25) + = (25)2. +In general, +(na) +b =(nb) +a = n +a × b=n +ab, where a and b are counting numbers. +Write the following expressions as a power of a power in at least two +different ways: +(i) 86      (ii) 715      (iii) 914      (iv) 58 +3 × 3 × 3 × 3 × 3 × 3 × 3 +34 +33 +Math +Talk + +Power Play +25 +Magical Pond +In the middle of a beautiful, magical pond +lies a bright pink lotus. The number of +lotuses doubles every day in this pond. +After 30 days, the pond is completely +covered with lotuses. On which day was +the pond half full? +If the pond is completely covered by +lotuses on the 30th day, how much of it is +covered by lotuses on the 29th day? +Since the number of lotuses doubles every day, the pond should be +half covered on the 29th day. +Write the number of lotuses (in exponential form) when the pond was — +(i) fully covered +(ii) half covered +There is another pond in which the number of lotuses triples every day. +When both the ponds had no flowers, Damayanti placed a lotus in the +doubling pond. After 4 days, she took all the lotuses from there and put +them in the tripling pond. How many lotuses will be in the tripling pond +after 4 more days? +After the first 4 days, the number of lotuses is 1 × 2 × 2 × 2 × 2 = 24. +After the next 4 days, the number of lotuses is 24 × 3 × 3 × 3 × 3 = 24 × 34. +What if Damayanti had changed the order in which she placed the +flowers in the lakes? How many lotuses would be there? +1 × 34 × 24 = (3 × 3 × 3 × 3) × (2 × 2 × 2 × 2). +Can this product be expressed as an exponent mn, where m and n are +some counting numbers? +By regrouping the numbers, + += (3 × 2) × (3 × 2) × (3 × 2) × (3 × 2) + += (3 × 2)4 = 64. +In general form, +ma × na = (mn)a , where a is a counting number. +Use this observation to compute the value of 25 × 55. +Simplify 104 +54 and write it in exponential form. +In general, we can show that ma +na = (m +n) +a­ +. + +Ganita Prakash | Grade 8 +26 +How Many Combinations +Estu has 4 dresses and 3 caps. How many different ways can Estu combine +the dresses and caps? +For each cap, he can choose any of the 4 dresses, so for 3 caps, 4 + 4 + +4 = 4 × 3 = 12 combinations are possible. We can also look at it as — for +each dress, Estu can choose any of the 3 caps, so for 4 outfits, 3 + 3 + 3 + +3 = 3 × 4 = 12 combinations are possible. +Roxie has 7 dresses, 2 hats, and 3 pairs of shoes. How many different +ways can Roxie dress up? +Hint: Try drawing a diagram like the one above. +Estu and Roxie came across a safe containing old stamps and coins +that their great-grandfather had collected. It was secured with a +5-digit password. Since nobody knew the password, they had no +option except to try every password until it opened. They were +unlucky and the lock only opened with the last password, after +they had tried all possible combinations. How many passwords +did they end up checking? +Instead of a 5-digit lock, let us assume we have a 2-digit lock and try +to find out how many passwords are possible. +There are 10 options for the first digit (0 to 9). For each of these, there +are 10 options for the second digit (If 0 is the first digit then 00, 01, 02, +03, …, 09 are possible). Therefore the total number of combinations for +a 2-digit lock is 10 × 10 = 100. +Now, suppose we have a 3-digit lock. For each of the earlier 100 +(2-digit) passwords there are 10 choices for the third digit. So, there are +100 × 10 = 1000 combinations for a 3-digit lock. You can list them all: 000, +001, 002, ….., 997, 997, 999. +If you can’t solve a problem, try to find a simpler version of the +problem that you can solve. This technique can come in handy +many times. + +Power Play +27 +How many 5-digit passwords are possible? +Each digit has 10 choices, so a +5-digit lock will have: +10 × 10 × 10 × 10 × 10 = 105 += 1,00,000 passwords. This is +the same as writing numbers +till 99,999 with all 5 digits, i.e., +00000, 00001, 00002, …00010, +00011, …, 00100, 00101, …, 00999, +…, 30456, …, 99998, 99999. +Estu says, “Next time, I will +buy a lock that has 6 slots with +the letters A to Z. I feel it is safer.” +How many passwords are possible with such a lock? +Think about how many combinations are possible in different +contexts. Some examples are — +(i) Pincodes of places in India — The Pincode of Vidisha in +Madhya Pradesh is 464001. The Pincode of Zemabawk in +Mizoram is 796017. +(ii) Mobile numbers. +(iii) Vehicle registration numbers. +Try to find out how these numbers or codes are allotted/generated. +2.3 The Other Side of Powers +Imagine a line of length 16 units. Erasing half of it would result in +24 ÷ 2 = 2 × 2 × 2 × 2 +2 + = 2 × 2 × 2 = 23 = 8 units. +Erasing half one more time would result in, +(24 ÷ 2) ÷ 2 = 24 ÷ 22 = 2 × 2 × 2 × 2 +2 × 2 + = 2 × 2 = 22 = 4 units. +Halving 16 cm three times may be written as, +24 ÷ 23 = 2 × 2 × 2 × 2 +2 × 2 × 2 = 2 = 21 = 2 units. +From this we can see that +24 ÷ 23 = 24 – 3 = 21. +What is 2100 ÷ 225 in powers of 2? +In a generalised form, +na ÷ nb = na – b, +where n ≠ 0 and a and b are counting numbers and a > b. +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +Oh no! Maybe my +entire vacation will +be gone trying to +unlock this… +Try +This + +Ganita Prakash | Grade 8 +28 +Why can’t n be 0? +We have not covered the case when the exponent is 0; for example, +what is 20? +Let us define 20 in a way that the generalised form above holds true. +20 = 24 – 4 = 24 ÷ 24 = 2 × 2 × 2 × 2 +2 × 2 × 2 × 2 = 1. +In fact for any letter number a +20 = 2a – a = 2a ÷ 2a = 1. +In general, +xa ÷ xa = xa – a = x0, and so +1 = x0 , +where x ≠ 0 and a is a counting number. +When Zero is in Power! +When a line of length 24 units is halved 5 times, +24 ÷ 25 = +2 × 2 × 2 × 2 +2 × 2 × 2 × 2 × 2 = 1 +2 units. +Using the generalised form, we get 24 ÷ 25= 2(4 – 5) = 2– 1. +So, 2– 1 = 1 +2 . +When a line of length 24 units is halved 10 times, +we get 24 ÷ 210= 2(4 – 10) = 2– 6 units. +When expanded, 24 ÷ 210 = +2 × 2 × 2 × 2 +2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1 +26 = 1 +64, +which is also written as 2– 6. +Math +Talk + +Power Play +29 +Similarly, 10–3 = 1 +103, 7–2 = 1 +72 , etc. +Can we write 103 = 1 +10–3 ? +We can write, +1 +10–3 = +1 +1/103 = 1 ÷ +1 +10 3 = 1×103 = 103. +Similarly, 72 = +1 +7– 2 and 4a = +1 +4– a. +In a generalised form, +n– a = 1 +na and na = 1 +n–a , where n ≠ 0. +Consider the following general forms we have identified. +We had required a and b to be counting numbers. Can a and b be any +integers? Will the generalised forms still hold true? +Write equivalent forms of the following. +(i) 2– 4 +(ii) 10– 5 +(iii) (– 7)–2 + +(iv) (– 5)– 3 +(v) 10– 100 +Simplify and write the answers in exponential form. +(i) 2– 4 × 27 +(ii) 32 × 3– 5 × 36 +(iii) p3 × p–10 + +(iv) 24 × (– 4) +– 2 +(v) 8 +p × 8 +q +Power Lines +Let us arrange the powers of 4 along a line. +na × nb = na + b +(na)b = (nb)a = na × b +na ÷ nb = na – b +Math +Talk +48 +47 +46 +45 +44 +43 +42 +41 +40 +4–1 +4–2 +48 ÷ 43 +43 × 42 +47 × 4–2 +65536 ÷ 64 +64 × 16 +16384 × +÷ 4 +÷ 16 +× 16 +× 4 +65536 +16384 +4096 +1024 +256 +64 +16 +4 +1 +1 +4 +1 +16 +1 +16 + +Ganita Prakash | Grade 8 +30 +Can we say that 16384 (47) is 16 (42) times larger than 1,024 (45)? +Yes, since 47 ÷ 45 = 42. +How many times larger than 4–2 is 42? +Use the power line for 7 to answer the following questions. +2,401 × 49 = +493 = +343 × 2,401 = +16,807 +49 = +7 +343 = +16,807 +8,23,543 = +1,17,649 × +1 +343 = +1 +343 × +1 +343 = +2.4 Powers of 10 +We have used numbers like 10, 100, 1000, and so on when writing Indian +numerals in an expanded form. For example, +47561 = (4 × 10000) + (7 × 1000) + (5 × 100) + (6 × 10) + 1. +This can be written using powers of 10 as +(4 × 104) + (7 × 103) + (5 × 102) + (6 × 101) + (1 × 100). +Write these numbers in the same way: (i) 172, (ii) 5642, (iii) 6374. +How can we write 561.903? +561.903 = (5 × 100) + (6 × 10) + 1 + (9 × 1 +10) + (0 × 1 +100) + (3 × +1 +1000). +Writing it using powers of 10, we have +561.903 = (5 × 102) + (6 × 101) + (1 × 100) + (9 × 10–1) + (0 × 10–2) + (3 × 10–3). +1 +343 +1 +2401 +1 +49 +1 +7 +77 +76 +75 +74 +73 +72 +71 +70 +7–1 +7–2 +7–3 +7–4 +823543 +117649 +16807 +2401 +343 +49 +7 +1 + +Power Play +31 +Scientific Notation +Let’s look at some facts involving large numbers — +(i) The Sun is located 30,00,00,00,00,00,00,00,00,000 m from the +centre of our Milky Way galaxy. +(ii) The number of stars in our galaxy is 1,00,00,00,00,000. +(iii) The mass of the Earth is 59,76,00,00,00,00,00,00,00,00,00,000 kg. +As the number of digits increases, it becomes difficult to read the +numbers correctly. We may miscount the number of zeroes or place +commas incorrectly. We will then read the wrong value. It is like getting +₹5,000 when you were supposed to get ₹50,000. The number of zeroes is +more important than the initial digits in several cases. +Can we use the exponential notation to simplify and read these very +large numbers correctly? +For example, the number 5900 can be expressed as — +5900 = 590 × 10 = 590 × 101 +  = 59 × 100 = 59 × 102 +    = 5.9 × 1000 = 5.9 × 103 +    = 0.59 × 10000 = 0.59 × 104. +Any number can be written as the product of a number between +1 and 10 and a power of 10. For example, + +5900 = 5.9 × 103 + +20800 = 2.08 × 104 +80,00,000 = 8 × 106 +Write the large-number facts we read just before in this form. +In scientific notation or scientific form (also called standard form), +we write numbers as x × 10 y, where x ≥ 1 and x < 10 is the coefficient and +y, the exponent, is any integer. Often, the exponent y is more important +than the coefficient x. When we write the 2 crore population of Mumbai +as 2 × 107, the 7 is more important than the 2. Indeed, if the 2 is changed +to 3, the population increases by one-half, i.e., 2 crore to 3 crore, whereas +if the 7 is changed to 8, the change in population is 10 times, i.e., 2 crores +to 20 crores. Therefore, the standard form explicitly mentions the +exponent, which indicates the number of digits. +If we say that the population of Kohima is 1,42,395, then it gives the +impression that we are quite sure about this number up to the units +place. When we use large numbers, in most cases, we are more concerned +about how big a quantity or measure is, rather than the exact value. If +we are only sure that the population is around 1 lakh 42 thousand, we +can write it as 1.42 × 105. If we can only be certain that it is around 1 lakh +40 thousand, we write it as 1.4 × 105. The number of digits in the coefficient +reflects how well we know the number. The most important part of any + +Ganita Prakash | Grade 8 +32 +number written in scientific form is the exponent, and then the first +digit of the coefficient. The digits following the coefficient are small +corrections to the first digit. +These values are rounded-off estimates, averages, or approximations; +most of the time, they serve the purpose at hand. +The distance between the Sun and Saturn is 14,33,50,00,00,000 m += 1.4335 × 1012 m. +The distance between Saturn and Uranus is 14,39,00,00,00,000 m += 1.439 × 1012 m. The distance between the Sun and Earth is +1,49,60,00,00,000 m = 1.496 × 1011 m. +Can you say which of the three distances is the smallest? +The number line below shows the distance between the Sun and Saturn +(1.4335 × 1012 m). On the number line below, mark the relative position of +the Earth. The distance between the Sun and the Earth is 1.496 × 1011 m. +Express the following numbers in standard form. +(i) 59,853 +(ii) +65,950 + +(iii) 34,30,000 +(iv) +70,04,00,00,000 +How old is +this dinosaur’s +skeleton? +It is 70 million +and 15 years old… +...when I started +working here, it +was 70 million +years old. +Sun +Saturn + +Power Play +33 +2.5 Did You Ever Wonder? +Last year, we looked at interesting thought experiments in the chapter +on Large Numbers. Let us continue this journey. +Nanjundappa wants to donate jaggery equal to Roxie’s weight and +wheat equal to Estu’s weight. He is wondering how much it would cost. +What would be the worth (in rupees) of the donated jaggery? What +would be the worth (in rupees) of the donated wheat? +In order to find out, let us first describe the relationships among the +quantities present. +Worth of jaggery (in rupees) = Roxie’s weight in kg × cost of 1 kg jaggery. +Worth of wheat (in rupees) = Estu’s weight in kg × cost of 1 kg wheat. +Make necessary and reasonable assumptions for the unknowns and find +the answers. Remember, Roxie is 13 years old and Estu is 11 years old. +Assuming Roxie’s weight to be 45 kg and the cost of 1 kg of jaggery to be +₹70, the worth of donated jaggery is 45 × 70 = ₹3150. Assuming Estu’s +weight to be 50 kg and the cost of 1 kg of wheat to be ₹50, the worth of +donated wheat is 50 × 50 = ₹2500. +Roxie wonders, “Instead of jaggery if we use 1-rupee coins, how many +coins are needed to equal my weight?”. How can we find out? +For questions like these, you can consider following the steps suggested +below. +1. Guessing: Make an instinctive (quick) guess of what the answer could +be, without any calculations. +The practice of offering goods equal to the weight of a person, called +Tulābhāra or Tulābhāram, is quite old and is still followed in many +places in Southern India. It is a symbol of bhakti (surrendering oneself), +a token of gratitude; it also supports the community. + +Ganita Prakash | Grade 8 +34 +2. Calculating using estimation and approximation — +(i) Describe the relationships among the quantities that are needed +to find the answer. +(ii) Make reasonable assumptions and approximations if the required +information is not available. +(iii) Compute and find the answer (and check how close your guess +was). +Would the number of coins be in hundreds, thousands, lakhs, crores, or +even more? Make an instinctive guess. +Find the answer by making necessary and +reasonable assumptions and approximations for +the unknowns. Remember, we are not looking for +an exact answer but a reasonably close estimate. +Estu asks, “What if we use 5-rupee coins or 10-rupee notes instead? +How much money could it be?” +Make an instinctive guess first. Then find out (make necessary and +reasonable assumptions about the unknown details and find the +answers). +Estu says, “When I become an adult, I would like to donate notebooks +worth my weight every year”. Roxie says, “When I grow up, I would like +to do annadāna (offering grains or meals) worth my weight every year”. +How many people might benefit from each of these offerings in a +year? Again, guess first before finding out. +Roxie and Estu overheard someone saying — “We did pādayātra for +about 400 km to reach this place! We arrived early this morning.” +How long ago would they have started their journey? +Find answers by making necessary assumptions and approximations. +Do guess first before calculating to check how close your guess was! +How about measuring +to find out the weight +of a 1-rupee coin? +Initially, your guesses may be very far off from the answer and +it is perfectly fine! You will get better at it like as you do it often +and in different situations. Guessing and estimating can build +intuition about numbers and various quantities. +Math +Talk +Note to the Teacher: Assumptions can vary greatly at times, and as a result the +answers computed using these you can also vary. This is perfectly alright. Modelling +the situation properly is crucial, which can also be done in different ways sometimes. +The accuracy of the assumed numbers or quantities can get better with exposure +and practice. + +Power Play +35 +Before the rise of modern transport, people moved from one place to +another by walking — sometimes merchants, sages, and scholars walked +thousands of kilometres to different parts of the world across deserts, +mountains, and rivers. +How many times can a person circumnavigate (go around the world) the +Earth in their lifetime if they walk non-stop? Consider the distance around +the Earth as 40,000 km. +Linear Growth vs. Exponential Growth +Roxie tells Estu about a science-fiction novel she is reading where they +build a ladder to reach +the moon, “... I wonder if +we actually had a ladder +like that, how many steps +would it have?”. +What do you think? Make +an instinctive guess first. +Would the number of steps +be in thousands, lakhs, +crores, or even more? +To find out, we would +need to know the gap between consecutive steps of the ladder. Let’s +assume a reasonable distance of 20 cm. Visualising the problem as shown, +Pādayātra, is the traditional practice of walking long distances as part +of a religious or spiritual pursuit. People across religions in our country +observe similar forms of pilgrimage or spiritual walking, although they +may have different names or purposes. +Some of the pilgrimages are Ajmer Sharif Dargah Ziyarat, +Pandharpur Wari, Kānwar Yatra, Sabarimala Yatra, Sammed +Shikharji Yatra, Lumbini to Sarnath Yatra. + +Ganita Prakash | Grade 8 +36 +We have to find out how many 20 cm make 3,84,400 km. +If we calculate the value, we get the result as 1,92,20,00,000 steps, which +is 192 crore and 20 lakh steps or 1 billion 922 million steps. The fixed +increase in the distance from the earth with each step (a 20 cm gain after +each step) is called linear growth. +To cover the distance between the Earth and the Moon, it takes +1,92,20,00,000 steps with linear growth whereas it takes just 46 folds of +a piece of paper with exponential growth! Linear growth is additive, +whereas exponential growth is multiplicative. +20 + 20 + 20 + ………… +1,92,20,00,000 +times +0.001 × 2 × 2 × 2 ………. +46 times +Some examples of exponential growth we have seen earlier in +this chapter are ‘The Stones that Shine’, ‘Magical Pond’, ‘How Many +Combinations’. We shall explore more such interesting examples in a +later chapter and also in the next grade. +Can you come up with some examples of linear growth and of +exponential growth? +Getting a Sense for Large Numbers +Last year, we learnt about lakhs and crores, as well as millions and +billions. A lakh is 105 (1,00,000), a crore is 107 (1,00,00,000), and an arab +is 109 (1,00,00,00,000), whereas a million is 106 (1,000,000) and a billion +is 109 (1,000,000,000). +You might know the size of the world’s human population. Have you +ever wondered how many ants there might be in the world or how long ago +humans emerged? In this section, we shall explore numbers significantly +larger than arabs and billions. We shall use powers of 10 to represent and +compare these numbers in each case. +100 +As of mid-2025, there are only two northern white rhinos +remaining in the world, both females, and they reside at the +Ol Pejeta Conservancy in Kenya (= 2 × 100). + +Power Play +37 +101 +As of early 2024, the total population of Hainan gibbons is a +meagre 42 ( ≈ 4 × 101). +102 +There are just 242 Kakapo alive as of mid-2025 (≈ 2 × 102). +103 +There are fewer than 3000 Komodo dragons in the world, all +based in Indonesia (≈ 3 × 103). +104 +A 2005 estimate of the maned wolf population showed that +there are more than 17000 of them; most are located in Brazil +(1.7 × 104). +105 +As of 2018, there are around 4.15 lakh African elephants + +(≈ 4 × 105). +106 +There are an estimated 50 lakh / 5 million American alligators as +of 2025 (5 × 106). +107 +The global camel population is estimated to be over 3.5 crore/ +35 million (3.5 × 107). India has only about 2.5 lakhs of them. +The global horse population is around 5.8 crore / 58 million +(5.8 × 107), with about half of them in America. +108 +More than 20 crore / 200 million (2 × 108) water buffaloes are +estimated worldwide, with a vast majority of them in Asia. +109 +The estimated global population of starlings is around 1.3 +arab/1.3 billion (_________). The global human population as of +2025 is 8.2 arab/8.2 billion (8.2 × 109). +A picture of a starling murmuration over a farm in the UK. Starling murmuration is a +mesmerising aerial display of thousands of starlings flying in synchronised, swirling +patterns. It is often described as a ‘choreographed dance’. + +Ganita Prakash | Grade 8 +38 +With a global human population of about 8 × 109 and about 4 × 105 +African elephants, can we say that there are nearly 20,000 people for +every African elephant? +1010 +The global chicken population living at any time is estimated at +≈ 33 billion (3.3 × 1010). +1012 +The estimated number of trees (2023) globally stands at 30 +kharab/3 trillion (3 × 1012). One kharab is 100 arab, and one +trillion is 1000 billion. +1014 +The estimated mosquito population worldwide +(2023) is 11 neel/110 trillion (________). A derived +estimate of the population of the Antarctic krill +stands at 50 neel/500 trillion (5 × 1014). +1015 +An estimate of the beetle population stands +at 1 padma/1 quadrillion (1 × 1015). The +estimate of the earthworm population is +also at 1 padma/1 quadrillion. +1016 +The estimated population of ants globally +is 20 padma/20 quadrillion (2 × 1016). Ants +alone outweigh all wild birds and wild +mammals combined. +1021 is supposed to be the number of grains of sand +on all beaches and deserts on Earth. This is enough +sand to give every ant 10 little sand castles to live in. +1023 +The estimated number of stars in the observable universe is +2 × 1023. +1025 +There are an estimated 2 × 1025 drops of water on Earth (assuming +16 drops per millilitre). +Calculate and write the answer using scientific notation: +(i) How many ants are there for every human in the world? +(ii) If a flock of starlings contains 10,000 birds, how many flocks +could there be in the world? + +Power Play +39 +(iii) If each tree had about 104 leaves, find the total number of leaves +on all the trees in the world. +(iv) If you stacked sheets of paper on top of each other, how many +would you need to reach the Moon? +A different way to say your age! +“How old are you?” asked Estu. +“I completed 13 years a few weeks ago!” said Roxie. +“How old are you?” asked Estu again. +“I’m 4840 days old today!” said Roxie. +“How old are you?” asked Estu again. +“I’m ______ hours old!” said Roxie. +Make an estimate before finding this number. +Estu: “I am 4070 days old today. Can you find out my +date of birth?” +If you have lived for a million seconds, how old would +you be? +We shall look at approximate times and timelines of some events and +phenomena, and use powers of 10 to represent and compare these +quantities. +I am +69,70,710 … +old +What could this number mean? +Find out! +Time in seconds +Comparison to real-world events/phenomena +100 = 1 second +- +Time taken for a ball thrown up to fall back on the +ground (typically a few seconds). +101 = 10 seconds +- +Time blood takes to complete one full circulation +through the body: 10 – 20 seconds (1 × 101 – 2 × 101 +seconds). +- +Typical waiting time at a traffic signal. +Isn’t it quite amazing how someone is able to estimate things +like the number of ants in the world or the time blood takes +to fully circulate? You may carry this wonder whenever you +encounter such facts. You will come across such facts in subjects +like Science and Social Science, where such estimates are made +frequently. +102 seconds +≈1.6 minutes +- +Time needed to make a cup of tea: 5 – 10 minutes +(≈ 4 × 102 – 8 × 102 seconds). +- +Time for light to reach the Earth from the Sun: about +8 minutes (≈ 5 × 102 seconds). + +Ganita Prakash | Grade 8 +40 +105 seconds ≈ 1.16 days and 106 seconds ≈ 11.57 days. Think of some +events or phenomena whose time is of the order of (i) 105 seconds and +(ii) 106 seconds. Write them in scientific notation. +These liars!! +It never gets +cooked before 10 +minutes… +They should +approximate it as +order of 102 seconds +noodles  —  that way +whether it takes +120 seconds or 900 +seconds, the claim +will be true. +103 seconds +≈ 16.6 minutes +- +Satellites in low Earth orbits take +between 90 minutes (≈ 5.5 × 103  seconds) +to 2 hours to complete one full +revolution around the Earth. +104 seconds +≈ 2.7 hours +- +The time needed to digest a meal: about +2 – 4 hours to pass through the stomach. +- +Lifespan of an adult mayfly: about a day +(≈ 9 × 104 seconds). +107 seconds +≈ 115.7 days / +≈ 3.8 months +- +- +Time spent sleeping in a year: about 4 months. +Time taken by Mangalyaan mission to reach Mars: +298 days (≈ 2.65 × 107 seconds). +- +Time taken by Mars for one full revolution around +the Sun: 687 Earth-days/1.88 Earth-years +(≈ 6 × 107 seconds). +108 seconds +≈ 3.17 years +- +The typical lifespan of most dogs is 3 to 15 years. +109 seconds +≈ 31.7 years +- +The orbital period of Halley’s comet is 75 – 79 years; +the next expected return is in the year 2061 +(≈ 2.4 × 109 seconds). +- +Duration of one full revolution of Neptune around +the Sun: 60,190 Earth-days/~165 Earth-years or +89,666 Neptunian days/1 Neptunian-year (≈ 5.2 × 109 +seconds). A day on Neptune is about 16.1 hours + +Power Play +41 +Notice how rapid exponential growth is — 106 seconds is less than a +fortnight, but 109 seconds is a whopping 31 years (about half the life +expectancy of a human)! +1015 seconds +≈ 3.17 crore +years +- +Age of Himalayas: 5.5 crore years/55 million years (≈ 1.7 +× 1015 seconds); they continue to grow a few mm every +year. +- +Dinosaurs went extinct 6.6 crore years ago/66 million +years ago (≈ 2 × 1015 seconds). +- +Dinosaurs first appeared more than 20 crore/200 million +years ago (≈ 6 × 1015 seconds). +- +It takes about 23 crore years for the Sun to make one +complete trip around the Milky Way (≈ 7 × 1015 seconds). +1010 seconds +≈ 317 years +- +The Chola dynasty ruled for more than 900 years +(≈ 3 × 1010 seconds) between the 3rd Century BCE and +12th Century CE. +1011 seconds +≈ 3,170 years +- +Age of the oldest known living tree: about 5000 years +(≈ 1.57 × 1011 seconds). +- +Time since the last peak ice age: 19,000 – 26,000 years +ago (≈ 6 × 1011 seconds – 8.2 × 1011 seconds). +1012 seconds +≈ 31,700 +years +- +Early Homo sapiens first appeared 2 – 3 lakh years ago +(≈7 × 1012 – 9 × 1012 seconds). The entire population +around that time could fit in a large cricket stadium. +1013 seconds +≈ 3.17 lakh +years +- +The Steppe Mammoth is +estimated to have +appeared around +8 – 18 lakh years ago. +1014 seconds +≈ 3.17 million +years +- +A fossil of Kelenken Guillermoi, + a type of terror bird, is dated +to 15 million years ago +( ≈_______________ seconds). + +Ganita Prakash | Grade 8 +42 +Notice that 109 seconds is of the order of the lifespan of a human, +whereas 1018 seconds ago the universe did not exist according to modern +physics!! The exponential notation can capture very large quantities in +a concise manner. +Calculate and write the answer using scientific notation: +(i) If one star is counted every second, how long would it take to +count all the stars in the universe? Answer in terms of the +number of seconds using scientific notation. +(ii) If one could drink a glass of water (200 ml) every 10 seconds, +how long would it take to finish the entire volume of water +on Earth? +2.5 A Pinch of History +In the Lalitavistara, a Buddhist treatise from the first century BCE, we see +number - names for odd powers of ten up to 1053. The following occurs +as part of the dialogue between the mathematician Arjuna and Prince +Gautama, the Bodhisattva. +“Hundred kotis are called an ayuta (109), hundred ayutas a niyuta (1011), +hundred niyutas a kankara (1013), …, hundred sarva-balas a visamjna- +gati (1047), hundred visamjna-gatis a sarvajna (1049), hundred sarvajnas +a vibhutangama (1051), a hundred vibhutangamas is a tallakshana (1053).” +Mahaviracharya gives a list of 24 terms (i.e., up to 1023) in his treatise +Ganita-sara-sangraha. An anonymous Jaina treatise Amalasiddhi gives a +list with a name for each power of ten up to 1096 (dasha-ananta). A Pali +grammar treatise of Kāccāyana lists number-names up to 10140, named +asaṅkhyeya. +1016 seconds +≈ 31.7 crore years +- +Plants on land started 47 crore/470 million years +ago ( ≈ _______________ seconds). +1017 seconds +≈ 3.17 billion years +The oldest fossil evidence suggests that bacteria +first appeared about 3.7 billion years ago. +- +The Earth is 4.5 billion years old. +- +The Milky Way galaxy was formed 13.6 billion +years ago, and the Universe was formed 13.8 +billion years ago. +Try +This +Very large quantities are often beyond our experience and +comprehension. To put them into perspective, we can relate and +compare them with quantities we are familiar with. This can +give an essence of how large a number or a measure is! + +Power Play +43 +For expressing high powers of ten, Jaina and Buddhist texts use bases +like sahassa (thousand) and koṭi (ten million); for instance, prayuta (106) +would be dasa sata sahassa (ten hundred thousand). +The modern naming is similar to this, where we say, +Continuing this, a hundred kharab is a neel (1013), a hundred neel is a +padma (1015), a hundred padma is a shankh (1017) and a hundred shankh +is a maha shankh (1019). +In the American/International system, we say +A thousand thousand +is a million +1000 × 1000 = 1,000,000 +103 × 103 = 106 +A thousand million is +a billion +1000 × 1,000,000 = 1,000,000,000 +103 × 106 = 109 +A thousand billion is +an trillion +1000 × 1,000,000,000 += 1,00,000,000,000 +103 × 109 = 1012 +Continuing this, a thousand trillion is a quadrillion (1015). This pattern +continues. Observe the names million (106), billion (109), trillion (1012), +quadrillion (1015), quintillion (1018), sextillion (1021), septillion (1024), +octillion (1027), nonillion (1030), decillion (1033). +What does the first part of each name denote? +The number 10100 is also called a googol. The estimated number +of atoms in the universe is 1078 to 1082. The number 10googol is called a +googolplex. It is hard to imagine how large this number is! +The currency note with the highest denomination in India currently +is 2000 rupees. Guess what is the highest denomination of a currency +note ever, across the world. The highest numerical value banknote ever +printed was a special note valued 1 sextillion pengő (1021 or 1 milliard +bilpengő) printed in Hungary in 1946, but it was never issued. In 2009, +Zimbabwe printed a 100 trillion (1014) Zimbabwean dollar note, which at +the time of printing was worth about $30. +A hundred +thousand is a lakh +100 × 1000 = 1,00,000 +102 × 103 = 105 +A hundred lakhs +is a crore +100 × 1,00,000 = 1,00,00,000 +102 × 105 = 107 +A hundred crores +is an arab +100 × 1,00,00,000 = 1,00,00,00,000 +102 × 107 = 109 +A hundred arab is +a kharab +100 × 1,00,00,00,000 += 1,00,00,00,00,000 +102 × 109 = 1011 + +Ganita Prakash | Grade 8 +44 +Figure it Out +1. Find out the units digit in the value of 2224 ÷ 432? [Hint: 4 = 22] +2. There are 5 bottles in a container. Every day, a new container is +brought in. How many bottles would be there after 40 days? +3. Write the given number as the product of two or more powers in +three different ways. The powers can be any integers. +(i) 643 +(ii)    1928 +(iii)    32–5 +4. Examine each statement below and find out if it is ‘Always True’, +‘Only Sometimes True’, or ‘Never True’. Explain your reasoning. +(i) Cube numbers are also square numbers. +(ii) Fourth powers are also square numbers. +(iii) The fifth power of a number is divisible by the cube of that +number. +(iv) The product of two cube numbers is a cube number. +(v) q46 is both a 4th power and a 6th power (q is a prime number). +5. Simplify and write these in the exponential form. +(i) 10– 2 × 10– 5 +(ii) +57 ÷ 54 + +(iii) 9– 7 ÷ 94 +(iv) +(13– 2) +– 3 + +(v) m5n12(mn)9 +6. If 122 = 144 what is +(i) (1.2)2 +(ii) +(0.12)2 + +(iii) (0.012)2 +(iv) +1202 + +Power Play +45 +7. Circle the numbers that are the same — +24 × 36        64 × 32        610        182 × 62        624 +8. Identify the greater number in each of the following — +(i) 43 or 34 +(ii) 28 or 82 (iii) 1002 or 2100 +9. A dairy plans to produce 8.5 billion packets of milk in a year. They +want a unique ID (identifier) code for each packet. If they choose to +use the digits 0–9, how many digits should the code consist of? +10. 64 is a square number (82) and a cube number (43). Are there +other numbers that are both squares and cubes? Is there a way +to describe such numbers in general? +11. A digital locker has an alphanumeric (it can have both digits and +letters) passcode of length 5. Some example codes are G89P0, 38098, +BRJKW, and 003AZ. How many such codes are possible? +12. The worldwide population of sheep (2024) is about 109, and that of +goats is also about the same. What is the total population of sheep +and goats? +(ii) 209 +(ii) +1011 +(iii) +1010 + +(iv) 1018 +(v) +2 × 109 +(vi) +109 + 109 +13. Calculate and write the answer in scientific notation: +(i) If each person in the world had 30 pieces of clothing, find the +total number of pieces of clothing. +(ii) There are about 100 million bee colonies in the world. Find the +number of honeybees if each colony has about 50,000 bees. +(iii) The human body has about 38 trillion bacterial cells. Find the +bacterial population residing in all humans in the world. +(iv) Total time spent eating in a lifetime in seconds. +14. What was the date 1 arab/1 billion seconds ago? +Math +Talk +Try +This + +Ganita Prakash | Grade 8 +46 + +„ +We analysed some situations, asked questions, and found +answers by first guessing, then modelling the problem +statement, +followed +by +making +assumptions +and +approximations to carry out the calculations. + +„ +We experienced how rapid exponential growth, also called +multiplicative growth, can be compared to additive growth. + +„ +na is n × n × n × n ×…× n (n multiplied by itself a times) and n–a= 1 +na. + +„ +Operations with exponents satisfy + y na × nb = na+b + y (na)b = (nb)a = na × b + y na ÷ nb = na – b (n ǂ 0) + y na × ma = (n × m)a + y na ÷ ma = (n ÷ m)a (m ǂ 0) + y n0 = 1 (n ǂ 0) + +„ +The scientific notation for the number 308100000 is 3.081 × 108. +The standard form of the scientific notation of any number is +x × 10 y, where x ≥ 1 and x < 10, and y is an integer. + +„ +Engaging in interesting thought experiments can be used as +means to understand how large a number or a quantity is. +SUMMARY + +Tremendous in Ten! +Find a partner to play this game with. In 10 seconds, the person who +writes a number or an expression, using only the digits 0 - 9 and +arithmetic operations, that gives a number that is the larger between +the two wins the round. +In Round 1, Roxie wrote 10000000000000 and Estu wrote 999999 × 999999. +Between these two, Roxie’s number is greater. Can you see why? Roxie’s +number is 1013, whereas Estu’s number is less than (106) +2. +In Round 2, Roxie wrote 101000 + 101000 + 101000 + 101000 and Estu wrote +(101000000) × 9000. Can you say which is greater? +Below are some conditions that you may consider for different rounds. +(i) Exponents are not allowed. Only addition is allowed. +(ii) Exponents are not allowed. Only addition and multiplication +are allowed. +(iii) Exponents are allowed. Only addition is allowed. +(iv) Exponents are allowed. Any arithmetic operation is allowed. +You can create your own conditions and/or involve more people to play +together." +class_8,3,a story of numbers,ncert_books/class_8/hegp1dd/hegp103.pdf,"3 +A STORY OF +NUMBERS +3.1 Reema’s Curiosity +One lazy afternoon, Reema was flipping through an old book when — +whoosh! — a piece of paper slipped out and floated to the floor. She +picked it up and stared at the strange symbols all over it. “What is this?” +she wondered. +She ran to her father, holding the paper as if it were a secret treasure. +He looked at it and smiled. “Around 4000 years ago, there flourished a +civilisation in a region called Mesopotamia, in the western part of Asia, +containing a major part of the present-day Iraq and a few other +neighbouring countries. This is one of the ways they wrote their +numbers!” +Reema’s eyes lit up, “Seriously? These strange symbols were numbers?” +Her curiosity was sparked, and questions started swirling in her head. +Since +when have +humans been +counting? +What was their +need for counting? +What were they +counting? +Since when +have people +been writing +numbers in +the modern +form? +How would the +Mesopotamians +have written +20? 50? 100? + +A Story of Numbers +49 +Sensing her curiosity, her father started telling her how the idea of +number and number representation evolved over the course of time, +across geographies, to finally reach its modern efficient form. +Get ready to travel back in time with them! +Humans had the need to count even as early as the Stone +Age. They were counting to determine the quantity of food +they had, the number of animals in their livestock, details +regarding trades of goods, the number of offerings given in +rituals, etc. They also wanted to keep track of the passing +days, e.g., to know and predict when important events such +as the new moon, full moon, or onset of a season would occur. +However, when they said or wrote down such numbers, they +didn’t make use of the numbers that we use today. +The structure of the modern oral and written numbers that we use +today had its origin thousands of years ago in India. Ancient Indian texts, +such as the Yajurveda Samhita, mentioned names of numbers based on +powers of 10, almost as we say them orally today. For example, they listed +names for the numbers one (eka), ten (dasha), hundred (shata), thousand +(sahasra), ten thousand (āyuta), etc., all the way up to 1012 and beyond. +The way we write our numbers today — using the digits 0 through 9 — +also originated and were developed in India, around 2000 years ago. The +first known instance of numbers being +written using ten digits, including the digit 0 +(which was then notated as a dot), occurs in +the Bakhshali manuscript (c. 3rd century +CE). Aryabhata (c. 499 CE) was the first +mathematician to fully explain, and do +elaborate scientific computations with the +Indian system of 10 symbols. +The Indian number system was transmitted to the Arab world +by around 800 CE. It was popularised in the Arab world by the great +Persian mathematician Al-Khwārizmī (after whom the word ‘algorithm’ +is named) through his book On the Calculation with Hindu Numerals +(c. 825) and by the noted philosopher Al-Kindi through his work On the +Use of the Hindu Numerals (c. 830). +From the Arab world, the Hindu numerals were transmitted to Europe +and to parts of Africa by around 1100 CE. Though Al-Khwārizmī’s work +on calculation with Hindu numerals was translated into Latin, it was the +Italian mathematician Fibonacci who around the year 1200 really made +the case to Europe to adopt the Indian numerals. However, the Roman +numerals were so ingrained in European thinking and writing at the +time that the Indian numerals did not gain widespread use for several +more centuries. But eventually, during the European Renaissance and +by the 17th century, not adopting them became impossible or it would +impede scientific progress. +Zero in the Bakhshali manuscript + +Ganita Prakash | Grade 8 +50 +“The ingenious method of expressing every possible number using a set of ten +symbols (each symbol having a place value and an absolute value) emerged in +India. The idea seems so simple nowadays that its significance and profound +importance is no longer appreciated. Its simplicity lies in the way it facilitated +calculations and placed arithmetic foremost among useful inventions.” +— Pierre-Simon Laplace (1749 – 1827) +Their use then spread to every continent, and are now used in every +corner of the world. + Because European scholars learned the Indian numerals from the +Arab world, they called them ‛Arabic numerals’ to reflect their European +perspective. On the other hand, as noted above, Arab scholars, such as +Al-Khwārizmī and Al-Kindi, called them ‛Hindu numerals’. During the +period of European colonisation, the European term Arabic numbers +became widely used. However, in recent years, this mistake is being +corrected in many textbooks and documents around the world, including +in Europe. The most commonly used terminologies for the numbers we +use today are ‛Hindu numerals’, ‛Indian numerals’, and the transitional +‛Hindu-Arabic numerals’. It is worth noting that the word ‛Hindu’ here +does not refer to a religion, but rather a geography/people from whom +these numbers came. +The shape of the digits 0, 1, 2, ..., 9 used to write numbers in the Indian +number system today evolved over a period of time, as shown below: +Evolution of the digits used in the Indian number system +Prior to the global adoption of the Indian system of numerals, different +groups of people used different methods of representing numbers. We + +A Story of Numbers +51 +shall take a glimpse of some of them. We will not be looking at different +systems in a chronological order, but rather an order that shows us the +main stages in the development of the idea of number representation. +But first, let us explore some of the foundational ideas needed to count +and to determine the number of objects in a given collection. +The Mechanism of Counting +Imagine that we are living in the Stone Age, say, around ten thousand +years ago. Suppose we have a herd of cows. Here are some natural +questions that we might ask about our herd — +Q1. How do we ensure that all cows have returned safely after +grazing? +Q2. Do we have fewer cows than our neighbour? +Q3. If there are fewer, how many more cows would we need so +that we have the same number of cows as our neighbour? +We need to tackle these questions without the use of the +number names or written numbers of the Hindu number +system. How do we do it? +Here are some possible methods. +Method 1: We could tackle the questions by using pebbles, sticks or +any object that is available in abundance. Let us choose sticks. For every +cow in the herd, we could keep a stick. The final collection of sticks tells +us the number of cows, which can be used to check if any cows have +gone missing. +Math +Talk + +Ganita Prakash | Grade 8 +52 +This way of associating each cow with a stick, such that no two cows are +associated or mapped to the same stick is called a one-to-one mapping. +This mapping can then be used to come up with a way to represent +numbers, as shown in the table. +Number +Its representation (using sticks) +1 +2 +3 +4 +5 +. +. +. +. +. +. +How will you use such sticks to answer the other two questions (Q2 and +Q3)? +Method 2: Instead of objects, we could use a standard sequence of +sounds or names. For example, we could use the sounds of the letters +of any language. While counting, we could make a one-to-one mapping +between the objects and the letters: that is, associate each object to be +counted with a letter, following the letter-order. This mapping can then +be used to come up with a way of verbally representing numbers. +For example, we get the following number representation if we use +English letters ‘a’ to ‘z’. +Number +Its representation (using +sounds or names) +1 +2 +3 +4 +5 +. +. +. +26 +. +. +. +a +b +c +d +e +z + +A Story of Numbers +53 +An obvious limitation of using only the letters of the English alphabet +in this form is that it cannot be used to count collections having more +than 26 objects. +How many numbers can you represent in this way using the sounds +of the letters of your language? +Method 3: We could use a sequence of written symbols as follows. +Do you see a way of extending this method to represent bigger numbers +as well? How? +From the discussion above, we see that for counting and finding the +size of a collection, we need a standard sequence of objects, or names, +or written symbols, that has a fixed order. Let us call this standard +sequence a number system. A collection of objects can be counted by +making a one-to-one mapping between them and the standard sequence, +following the sequence order. +Since there is no end to numbers, the challenge is to come up with +an unending standard sequence/number system that is easy to count +with. Using sticks gives an unending standard sequence/number system. +However, it is not convenient to count larger collections, as we will +need as many sticks as the number of objects being counted. Using the +sounds of the letters of a language, as in Method 2, is convenient for +the counting process but is not an unending standard sequence/number +system. The standard sequence/number system given in Method 3 was +actually the system used in Europe before it got replaced by the Hindu +number system. It is called the Roman number system. It was widely +used in Europe for centuries, and was convenient for many purposes, +but had the similar drawback that one cannot write arbitrarily large +numbers without introducing more and more symbols. We will learn +more about this system of writing numbers later on. +Math +Talk +Number +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +Representation +using symbols +I +II +III +IV +V +VI +VII +VIII +IX +X +Number +11 +12 +13 +14 +15 +16 +17 +18 +19 +20 +Representation +using symbols +XI +XII +XIII +XIV +XV +XVI +XVII +XVIII +XIX +XX +Table 1 + +Ganita Prakash | Grade 8 +54 +As illustrated by the three methods, history gives us examples of +number systems formed using physical objects (such as sticks, pebbles, +body parts, etc.), names, and written symbols. Some groups of people +had numbers represented both by physical objects as well as by names, +while others like the Chinese had all three forms of representation. The +symbols occurring in a written number system are called numerals. +For example, 0, 1, 5, 36, 193, etc., are some of the numerals occurring in +the Hindu number system. Numerals representing ‘smaller’ numbers +always had names, and so a number system composed of written +symbols always went hand in hand with a number system composed of +names, as is the case with the modern-day Hindu system. +Figure it Out +1. Suppose you are using the number system that uses sticks to +represent numbers, as in Method 1. Without using either the +number names or the numerals of the Hindu number system, +give a method for adding, subtracting, multiplying and dividing +two numbers or two collections of sticks. +2. One way of extending the number system in Method 2 is by +using strings with more than one letter — for example, we could +use ‘aa’ for 27. How can you extend this system to represent all +the numbers? There are many ways of doing it! +3. Try making your own number system. +3.2 Some Early Number Systems +I. Use of Body Parts +Many groups of people across the world have used their hands and body +parts for counting. Here is how a group of people in Papua New Guinea +used and still use their body parts as the standard sequence/number +system. +Math +Talk +Math +Talk +Try +This + +A Story of Numbers +55 +II. Tally Marks on Bones and Other Surfaces +One of the oldest methods of number representation is by making +notches — marks cut on a surface such as a bone or a wall of a cave. +These marks are also called tally marks. +In this method, a mark is made for each object that is being counted. +So the final collection of marks represents the total number of objects. This +method is very similar to the method of using sticks to count (Method 1), +except for the fact that a mark is made instead of adding a stick. +Archaeologists have discovered bones dating back more than 20,000 +years that seem to have tally marks. The oldest known such bones with +markings that are thought to represent numbers are the Ishango bone +and the Lebombo bone. The Ishango bone, dating back 20,000 to 35,000 +years, was discovered in the Democratic Republic of Congo. It features +notches arranged in columns, possibly indicating calendrical systems. +The Lebombo bone, discovered in South Africa, is an even older tally +stick with 29 notches, estimated to be around 44,000 years old. It is +considered one of the oldest mathematical artefacts, and may have +served as a tally stick or lunar calendar. +Lebombo bone +Ishango bone + +Ganita Prakash | Grade 8 +56 +III. Number Names Obtained by Counting in Twos +A group of indigenous people in Australia called the Gumulgal had the +following words for their numbers. +Can you see how their number names are formed? +The number name for 3 is composed of number names of 2 and 1. +The number name for 4 is composed of two occurrences of the number +name for 2. +Can you see how the names of the other numbers are formed? +The numbers are counted in 2s, using which the number names are +formed: 3 = 2 + 1, 4 = 2 + 2, 5 = 2 + 2 + 1, 6 = 2 + 2 + 2. +Gumulgal called any number greater than 6 ras. +There is a very interesting and puzzling historical phenomenon +associated with this number system. Look at the following number +systems of a group of indigenous people in South America, and the +Bushmen of South Africa: +Gumulgal +(Australia) +1. urapon +2. ukasar +3. ukasar-urapon +4. ukasar-ukasar +5. ukasar-ukasar-urapon +6. ukasar-ukasar-ukasar +Map not to scale +India +Gumulgal +Bushmen +Bakairi +1 - tokale +2 - ahage +3 - ahage tokale + (or ahawao) +4 - ahage ahage tokale +5 - ahage ahage ahage +1 - xa +2 - t'oa +3 - 'quo +4 - t'oa-t'oa +5 - t'oa-t'oa-t'a +6 - t'oa-t'oa-t'oa +1 - urapon +2 - ukasar +3 - ukasar-urapon +4 - ukasar-ukasar +5 - ukasar-ukasar-urapon +6 - ukasar-ukasar-ukasar + +A Story of Numbers +57 +Despite being so far apart geographically, and with no trace of contact +between them, these three groups have developed equivalent number +systems! Historians have wondered how this happened. One theory is +that these three groups of people may have had common ancestors, who +used this number system. In course of time, their descendants migrated +to these places. +Even though the number system of Gumulgal had number names for +numbers only till 6, we can see the emergence of an idea here. Counting +in 2s is more efficient for representing numbers than, for example, a tally +system. A general form which this idea has taken in different number +systems is as follows: count in groups of a certain number (like 2 in the +case of Gumulgal’s system), and use the word or symbol associated with +this group size to represent bigger numbers. Some of the commonly +used group sizes in different number systems have been 2, 5, 10 and 20. +You can find the idea of counting by 5s in the Roman system (Table 1). +One of the phenomena that could have led people to this idea might +be the human limit for immediately knowing the size of a collection at a +glance. Let us try out the following activity. +Quickly count the number of objects in each of the following boxes: +Up to what group size could you immediately see the number of +objects without counting? Most humans find it difficult to count groups +having 5 or more objects in a single glance. +This limit of perception could have prompted people using tally +marks to replace every group of, say, 5 marks, with a new symbol, +as seen in the system shown in Table 1. +This idea of counting in a certain group size and using it to represent +numbers is an important idea in the history of the evolution of +number systems. +Math +Talk + +Ganita Prakash | Grade 8 +58 +What could be the difficulties with using a number system that counts +only in groups of a single particular size? How would you represent a +number like 1345 in a system that counts only by 5s? +Even though counting in groups of a particular size and using it for +number representation is more efficient than the tally system, this +method can still become cumbersome for larger numbers. The next +system shows a refinement of this idea. +IV. The Roman Numerals +We have already seen the Roman number system till 20 (Table 1). We +have seen that it uses I for 1, V for 5, X for 10. +To get the Roman numeral for any number till 39, it is first grouped +into as many 10s as possible, the remaining is grouped into as many 5s +as possible, and finally the remaining is grouped into 1s. +Example: Let us take the number 27. +27 = 10 + 10 + 5 + 1 + 1 +So, 27 in Roman numerals is XXVII. +Instead of representing 50 as XXXXX, a new symbol is given to it: L. +Following the way the number 4 is represented as 1 less than 5 — that +is, as IV — 40 is represented as 10 less than 50 — that is, as XL. However, +people using this system were not always consistent with this practice. +Sometimes, 40 was also represented as XXXX. +The Roman number system introduces newer symbols to represent +certain bigger numbers. Let us call all these numbers that have a new +basic symbol as landmark numbers. Here are some of the landmark +numbers of the Roman system and their associated numerals. +I +V +X +L +C +D +M +1 +5 +10 +50 +100 +500 +1,000 + +A Story of Numbers +59 +These symbols are used to denote other numbers as well. For example, +consider the number 2367. Writing it as a sum of landmark numbers +starting from 1000 such that we take as many 1000s as possible, 500s as +possible, and so on, we get +2367 = 1000 + 1000 + 100 + 100 + 100 + 50 + 10 + 5 + 1 + 1 +So in Roman numerals, this number is MMCCCLXII. +Figure it Out +1. Represent the following numbers in the Roman system. +(i) 1222      (ii) 2999      (iii) 302      (iv) 715 +We see how vastly efficient this system is compared to some of the +previous number systems that we have seen. This system seems to have +evolved out of the ancient Greek number system in around the 8th +century BCE in Rome, and evolved over time. It spread throughout +Europe with the expansion of the Roman empire. +Despite the relative efficiency of the Roman system, it doesn’t lend +itself to an easy performance of arithmetic operations, particularly +multiplication and division. +Example: Try adding the following numbers without converting them +to Hindu numerals: +(a) CCXXXII + CCCCXIII +Let us find the total number of Is, Xs, and Cs, and group them +starting from the largest landmark number. +Apparently, it looks like the largest landmark number is C, but +note that 5 Cs (100s) make a D (500). So the sum is +The efficiency of this system is due to the grouping of a given number +by not just one group size, but a sequence of group sizes that we +call landmark numbers, and then using these landmark numbers +to represent the given number. This idea is the next important +breakthrough in the history of the evolution of number systems. + +Ganita Prakash | Grade 8 +60 +Do it yourself now: +(b) LXXXVII + LXXVIII +How will you multiply two numbers given in Roman numerals, +without converting them to Hindu numerals? Try to find the product +of the following pairs of landmark numbers: V × L, L × D, V × D, +VII × IX. +Multiply +CCXXXI and +MDCCCLII +People using the Roman system made use of a calculating tool called +the abacus to perform their arithmetic operations. We will see what it +is in a later section. However, only specially trained people used this tool +for calculation. +While going through the number systems discussed above, it should +not be understood that, historically, one system developed as an +improvement over the previous system. This point should be kept in +mind when studying the upcoming number systems too. The actual story +of how each of the number systems developed is much more complex, +and many times not clearly known, and so we will not try to trace this +in the chapter. +Figure it Out +1. A group of indigenous people in a Pacific island use different +sequences of number names to count different objects. Why do +you think they do this? +2. Consider the extension of the Gumulgal number system beyond +6 in the same way of counting by 2s. Come up with ways of +performing the different arithmetic operations (+, –, ×, ÷) +for numbers occurring in this system, without using Hindu +numerals. Use this to evaluate the following: +Try +This +Math +Talk +Math +Talk + +A Story of Numbers +61 +(i) (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasar- +ukasar-urapon) +(ii) (ukasar-ukasar-ukasar-ukasar-urapon) – (ukasar-ukasar- +ukasar) +(iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar) +(iv) (ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ +(ukasar-ukasar) +3. Identify the features of the Hindu number system that make it +efficient when compared to the Roman number system. +4. Using the ideas discussed in this section, try refining the number +system you might have made earlier. +3.3 The Idea of a Base +I. The Egyptian Number System +We are now going to see a written number system that the Egyptians +developed around 3000 BCE. In this system, we see the use of landmark +numbers to group and represent a given number. However, what makes +this system special is its sequence of landmark numbers. +Imagine making collections of pebbles. The first landmark number +is 1. Group together 10 collections of the previous landmark number (1). +Its size is the second landmark number which is 10. Group together 10 +collections of the previous landmark number (10). Its size is the third +landmark number which is 10 × 10 = 100, and so on. +Math +Talk +Try +This + +Ganita Prakash | Grade 8 +62 +Each landmark number is 10 times the previous one. Since 1 is the +first landmark number, they are all powers of 10. The following are the +symbols given to these numbers — +As in the case of Roman numbers, a given number is counted in groups +of the landmark numbers, starting from the largest landmark number +less than the given number. This is then used to assign the numeral. +For example 324 which equals 100 + 100 + 100 + 10 + 10 + 4 is written +as +. +Figure it Out +1. Represent the following numbers in the Egyptian system: 10458, +1023, 2660, 784, 1111, 70707. +2. What numbers do these numerals stand for? +II. Variations on the Egyptian System and the Notion of +Base +Instead of grouping together 10 collections of size equal to the previous +landmark number (as in the case of the Egyptian system), can we get +a number system by grouping together 5 collections of size equal to +the previous landmark number? Can this 5 be replaced by any positive +integer? +Let us examine this possibility. Let 1 be the first landmark number. +Group together 5 collections of size equal to the previous landmark +number (1). Its size is the second landmark number which is 5. +Group together 5 collections of size equal to the previous landmark +number (5). Its size is the third landmark number which is 5 × 5 = 25. +Group together 5 collections of size equal to the previous landmark +number (5). Its size is the fourth landmark number which is 5 × 25 = 125. +(i) +(ii) + +A Story of Numbers +63 +Thus, we have a new number system where each landmark number +is 5 times the previous one. Since 1 is the first landmark number, they +are all powers of 5. +50 = 1            51 = 5            52 = 25          53 = 125          54 = 625          55 = 3125 +Express the number 143 in this new system. +Let us start grouping, starting with the size 53 = 125, as this is the +largest landmark number smaller than 143. We get— +143 = 125 + 5 + 5 + 5 + 1 + 1 + 1. +So the number 143 in the new system is +. +Number systems having landmark numbers in which the +(a) first landmark number is 1, and +(b) every next landmark number is obtained by multiplying the +current landmark number by some fixed number n is said to +be a base-n number system. +The Egyptian number system is a base-10 system, and the number +system that we created is a base-5 system. A base-10 number system is +also called a decimal number system. +Figure it Out +1. Write the following numbers in the above base-5 system using the +symbols in Table 2: 15, 50, 137, 293, 651. +2. Is there a number that cannot be represented in our base-5 system +above? Why or why not? +3. Compute the landmark numbers of a base-7 system. In general, what +are the landmark numbers of a base-n system? +The landmark numbers of a base-n number system are the powers of +n starting from n0 = 1, n, n2, n3,... + +Ganita Prakash | Grade 8 +64 +Advantages of a Base-n System +What is the advantage of having landmark numbers that are all the +powers of a number? To understand this, let us perform some arithmetic +operations using them. +Example: Add the following Egyptian numerals: +Let us find the total number of | and and group them starting from +the largest possible landmark number. It has a total of— +15 and 15 |. +Since 10 gives the next landmark number , the sum can be regrouped +as— +Sum = + +A Story of Numbers +65 +Since 10| gives a , we have +Sum = +Figure it Out +1. Add the following Egyptian numerals: +(i) +(ii) +2. Add the following numerals that are in the base-5 system that we +created: + Remember that in this system, 5 times a landmark number gives the +next one! ++ + +Ganita Prakash | Grade 8 +66 +Contrast the addition done in a base-n number system with that +done in the Roman system. In the Roman system, the grouping and +rearranging has to be done carefully as it is not always by the same size +that each landmark number has to be grouped to get the next one. +The advantage of a number system with a base becomes more evident +when we consider multiplication. +How to multiply two numbers in Egyptian numerals? +Let us first consider the product of two landmark numbers. +1. What is any landmark number multiplied by (that is 10)? Find the +following products — + + + Each landmark number is a power of 10 and so multiplying it with +10 increases the power by 1, which is the next landmark number. +2. What is any landmark number multiplied by (102)? Find the +following products — + +I see a similarity in the +method of adding numbers +in the Egyptian and the +Hindu system! ++ ++ + 4 7 + 5 6 +10 3 ++ +1 + +A Story of Numbers +67 +Find the following products — +Thus, the product of any two landmark numbers is another landmark +number! +Does this property hold true in the base-5 system that we created? +Does this hold for any number system with a base? +What can we conclude about the product of a number and (10), +in the Egyptian system? +(ii) +| × + + + +| is the same as + + + |. Thus, + + + +| × = ( + + + |) × + + + Will the distributive property hold here? For the same reason +that it holds for (a + b) × n, it also holds when one of the +numbers has more than 2 terms. For example, (a + b + c) × n = +an + bn + cn. So, +( + + + |) × = ( × ) + ( + × ) + (| × ) += + + + + + + + + + +    = + +Math +Talk + +Ganita Prakash | Grade 8 +68 +Now find the following products — +(i) +(ii) +What would be a simple rule to multiply a number with ? +As has been seen, a process of multiplying two numbers involves the +multiplication of landmark numbers. When the landmark numbers are +powers of a number, then their product is another landmark number. +This fact simplifies the process of multiplication. However, this is not the +case with the Roman numerals, which is why multiplication using them +is difficult. +Thus, a number system whose landmark numbers are powers of +a number, i.e., a number system with a base, is efficient not only in +number representation but also in its utility in carrying out arithmetic +operations. +The idea of a number system with a base was a turning point in the +history of the evolution of number systems. Our modern Hindu number +system is built on this structure. +Abacus that Makes Use of the Decimal System +In around the 11th century, even the people still using the Roman +numerals started using a calculating device — the abacus — constructed +using a decimal system. It was a board with lines, as shown in the Fig. 3.1. +Starting from the line that stood for 1, each successive line stood for a +successive power of 10. +Numbers were represented in it as follows: the given number was +first grouped into the landmark numbers (powers of 10), in exactly the +same way we have been grouping them so far. For each power of 10, as +many counters were placed on its line as the number of times it occurred +Fig. 3.1: Abacus + +A Story of Numbers +69 +in the grouping. The presence of a counter above a line contributed a +value of 5. +For example let us take the number 3426. It can be grouped as +3426 = 1000 + 1000 + 1000 + 100 + 100 + 100 + 100 + 10 + 10+ 1 + 1 + +1 + 1 + 1 + 1 +This number was represented as shown in Fig. 3.1. Notice how the +6 ones are represented. +To get an idea of how the abacus was used for calculations, let us +consider a simple addition problem: 2907 + 43. The two numbers were +taken on either side of the vertical partition. +How would you use this to find the sum? +The counters along each line were brought together. What is to be +done if the total in a line exceeded 10? +Hint: In this problem, the 7 ones and the 3 ones together make 10 ones +which contributes a counter to the line representing 10s. +III. Shortcomings of the Egyptian System +Despite being a number system that enabled relatively efficient number +representations for numbers till a crore (107), and relatively easy +computations, the Egyptian system had a drawback. +If larger and larger numbers needed to be represented, then there +was a need for inventing an unending sequence of symbols for higher +and higher powers of 10. Here we see the original challenge of number +representation resurfacing in a different form! +The next and the final idea in the history of the evolution of number +systems not only solves this problem but also remarkably simplifies +number representation and computations! +Figure it Out +1. Can there be a number whose representation in Egyptian +numerals has one of the symbols occurring 10 or more times? +Why not? +Math +Talk + +Ganita Prakash | Grade 8 +70 +2. Create your own number system of base 4, and represent +numbers from 1 to 16. +3. Give a simple rule to multiply a given number by 5 in the base-5 +system that we created. +3.4. Place Value Representation +I. The Mesopotamian Number System +In the beginning, the number system used in ancient Mesopotamia had +different symbols for different landmark numbers. In later times, it +became a base-60 system, also called the sexagesimal system, with a +very efficient number representation. +It has puzzled many why they chose base 60. Different theories exist to +explain this, ranging from the connection between 60 and the periods of +some important events (like the length of their lunar month which had +30 days, or the time taken for the Sun to complete one revolution around +the Earth when Earth is taken to be stationary), the ease of representing +fractions (we will not go into this here), their earlier sequence of +landmark numbers — 1, 10, 60, 600, 3600, 36000,… — getting reduced to +only the powers of 60, and so on. +The influence of the Mesopotamian sexagesimal system, also known +as the Babylonian number system, can be seen even now in our units +of time measurements — 1 hour = 60 minutes and 1 minute = 60 seconds. +This system used the symbol for 1 and + for 10. +Let us now briefly pause on the study of their number system, and +ideate on how one can build an efficient number system using the +Mesopotamian features seen so far. +Math +Talk + +A Story of Numbers +71 +Let us give our own symbols to their landmark numbers — +Note that we have actually used Indian numerals in creating these +symbols. We could have invented our own symbols but for the sake of +easy recall and use, we have chosen to take help of the familiar numerals +1, 2, 3, ... +Using and +, numbers from 1 to 59 can be represented — +Example: Let us represent the number 640 in this system. +Grouping it into landmark numbers, we see that +640 = (10) × 60 + 40. +If we use the Egyptian idea, this number would be represented using +10 +1 s, and 40 would be represented using 4 +s. +1                60              602 = 3600      603 = 216000 ... +1 +... +2 +3 +1            2             3             4             5              6              7               8              9 +10          11          12          20          30              40          50            59 +Reproduction of a Mesopotamian Tablet + +Ganita Prakash | Grade 8 +72 +Can we represent this more compactly? +We can simply represent this number as +which can be read as ten 60s and one 40, just as we have written in the +equation. +Example: Let us try another number — 7530. +7530 = (2) ×3600 + (5) × 60 + 30 +So, its representation would be +Note that when a number is grouped into powers of 60 for its +representation, no power of 60 can occur 60 or more times. If this +happens, then 60 of them can be grouped to form the next power of 60. +For example, consider the expression — +(1) × 3600 + (70) × 60 + 2 = (1) × 602 + (60 + 10) × 60 + 2 + + + + + += (1) × 602 + 602 + (10) × 60 + 2 + + + + + += (2) × 602 + (10) × 60 + 2 +Therefore, any number can be represented using the numerals from +1 – 59, along with the numerals for landmark numbers. +Now, what if we make the representation even more compact by +dropping the symbols for the different powers of 60 altogether? + +A Story of Numbers +73 +This is exactly what the Mesopotamians did! In their numeral, the +rightmost set of symbols showed the number of 1s, the set of symbols to +its left showed the number of 60s, the next showed the number of 3600s +and so on. Whenever there was no occurrence of a power of 60, a blank +space was given in that position. +It does not seem that the Mesopotamians arrived at this idea in the +same way we did. Some scholars suggest that the similarity of symbols +given to the landmark numbers 1 and 60 in their earlier number system, +and an accidental usage of them, might have made them stumble upon +this idea. +Figure it Out +1. Represent the following numbers in the Mesopotamian system — +(i) 63    (ii) 132    (iii) 200    (iv) 60    (v) 3605 +Thus, we can see how the Mesopotamian system removes the need for +generating an unending sequence of symbols for the landmark numbers +by making use of the positions where the symbols are written. Such a +number system (having a base) that makes use of the position of each +symbol in determining the landmark number that it is associated with is +called a positional number system or a place value system. +This idea of place value marks the highest point in the history of +evolution of number systems, and gives a very elegant solution to the +problem of representing the unending sequence of numbers using only +a finite number of different symbols! +The Mesopotamian system however cannot be considered a fully +developed place value system. It has certain defects that lead to confusion +while reading a number. +Look at the representation of 60. What will be the representation for +3,600? +While writing the numerals, the spacing between symbols was not +given the way we are giving it here. It was also difficult to maintain a +consistent spacing for blanks across different manuscripts written by +different people. These created ambiguities. For example, consider the +representations of the following numbers — + +Ganita Prakash | Grade 8 +74 +Because of the ambiguity in finding which symbols correspond to +which powers of 60, the same numeral can be read in different ways. +Even in our representation which uses uniform spacing between +symbols for different powers of 60, it is difficult to know the number of +blanks between two sets of symbols, as in the representation of 36002. +To address the issue arising out of blank spaces, the later +Mesopotamians used a brilliant idea of assigning a ‘placeholder’ symbol + to denote a blank space. This is like the 0 (zero) we use in our system. +Thus, zero — the symbol that shows nothingness — is indispensable as a +placeholder in a place value system in which numbers are written in an +unambiguous manner. +Even with the problem arising out of blank spaces solved, other +ambiguities still remained in the system. For example, the placeholder +symbol was primarily used in the middle of numbers and not at the +end; so they would not use it to represent a number like (what we would +write as) 3600. +II. The Mayan Number System +In Central America, there flourished a civilisation known as the Mayan +civilisation that made great intellectual and cultural progress between +the 3rd and 10th centuries CE. Among their intellectual achievements +stands their place value system designed independently of those in Asia. +They also made use of a placeholder symbol, for the modern-day ‘0’, that +looked like a seashell. + + +Ganita Prakash | Grade 8 +76 +Here we find a puzzling phenomenon. Why was their third landmark +number 360 rather than 400? Some scholars feel that this might have +something to do with their calendars. +They used a dot + for 1, and a bar + for 5. These were used to +denote numbers from 1 to 19. +The symbols associated with different landmark numbers were +written one below the other with the lowermost set of symbols +corresponding to the number of 1s, the set above corresponding to the +number of 20s, the set above to the number of 360s and so on. +Represent the following numbers using the Mayan system: +(i) 77    (ii) 100    (iii) 361    (iv) 721 +Because the Mayan system is not an actual base-20 system, it lacks the +advantages that a system with a base has for computations. Nevertheless, +their place value notation and their use of a placeholder symbol for zero +is considered an important advance in the history of number systems. +A curious fact is that we can still find the use of base-20 in the number +names of some European languages. +III. The Chinese Number System +The Chinese used two number systems — a written system for recording +quantities, and a system making use of rods for performing computations. +The numerals in the rod-based number system are called rod numerals. +Here we discuss the rod numerals, which are more efficient in writing +and computing with numbers than the written system of the Chinese. +The rod numerals developed in China by at least by 3rd century AD +and were used till the 17th century. It was a decimal system (base-10). +The symbols for 1 to 9 were as follows: + +A Story of Numbers +77 +CHINESE NUMBER SYSTEM +Base-10 or Decimal +Note: The zongs represent units, hundreds, tens of thousands, etc., and the hengs tens, +thousands, hundreds of thousands, etc. +How is this to be read? +2 (Heng) +6 (Zong) +3 (Heng) +4 (Zong) +Landmark number +positions +103 +102 +10 +1 += +(2) × 103 ++ (6) × 10² + +(3) × 10 ++ +(4) × 1 += +2634 +A numeral +Like the Mesopotamians, the rod numerals used a blank space to +indicate the skipping of a place value. However, because of the slightly +more uniform sizes of the symbols for one through nine, the blank +spaces were easier to locate than in the Mesopotamian system. + +Ganita Prakash | Grade 8 +78 +Notice how similar the rod numerals are to the Hindu system. +The Chinese system, with a symbol for zero, would be a fully developed +place value system. +IV. The Hindu Number System +Where does the Hindu/Indian number system figure in the evolution of +ideas of number representation? What are its landmark numbers? And +does it use a place value system? +Hindu Number System +Base-10 or Decimal +Ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 +A numeral 375 +How is this to be read? +3 +7 +5 +Landmark number positions += (3) × 10² ++ +(7) × 10 + +(5) × 1 +102 +10 +1 += +375 +As can be seen, the Hindu number system is a place value system. +The Hindu number system has had a symbol for 0 at least as early as +200 BCE. Because of the use of 0 as a digit, and the use of a single digit in +each position, this system does not lead to any kind of ambiguity when + +A Story of Numbers +79 +reading or writing numerals. It is for this reason that the Hindu number +system is now used throughout the world. +The use of 0 as a digit, and indeed as a number, was a breakthrough +that truly changed the world of mathematics and science. In Indian +mathematics, indeed, zero was not just used as a placeholder in the place +value system, but was also given the status of a number in its own right, +on par with other numbers. The arithmetic properties of the number 0 +(e.g., that 0 plus any number is the same number, and that 0 times any +number is zero) were explicitly used by Aryabhata in his Āryabhaṭīya in +499 CE to compute with and do elaborate scientific computations using +Hindu numerals. The use of 0 as a number like any other number, on +which one can perform the basic arithmetic operations, was codified +by Brahmagupta in his work Brāhmasphuṭasiddhānta in 628 CE, as we +learned in an earlier grade. +By introducing 0 as a number, along with the negative numbers, +Brahmagupta created what in modern terms is called a ring, i.e., a set of +numbers that is closed under addition, subtraction, and multiplication +(i.e., any two numbers in the set can be added, subtracted, or multiplied +to get another number in that set). These new ideas laid the foundations +for modern mathematics, and particularly for the areas of algebra and +analysis. +Hopefully, this gives you a sense of all the ideas that went into writing +and computing with numbers in the way that we do today. The discovery +of 0 and the resulting Indian number system is truly one of the greatest, +most creative, and most influential inventions of all time — appearing +constantly in our daily lives and forming the basis of much of modern +science, technology, computing, accounting, surveying, and more. The +next time you are writing numbers, think about the incredible history +behind them and all the deep ideas that went into their discovery! +Evolution of Ideas in Number +Representation +1. Count in groups of a single number. + +ukasar-ukasar-urapon +2. Group using landmark numbers. + +I V X L C M +3. Choosing powers +of +a number as landmark +numbers — the idea of a base. + 1 101 102 103 104 … +4. Using positions to denote the landmark numbers — the +idea of place value system. + +1 7 2 9 +5. The idea of 0 as a positional digit and as a number. + +Ganita Prakash | Grade 8 +80 +Figure it Out +1. Why do you think the Chinese alternated between the Zong and +Heng symbols? If only the Zong symbols were to be used, how +would 41 be represented? Could this numeral be interpreted in +any other way if there is no significant space between two successive +positions? +2. Form a base-2 place value system using ‘ukasar’ and ‘urapon’ as the +digits. Compare this system with that of the Gumulgal’s. +3. Where in your daily lives, and in which professions, do the Hindu +numerals, and 0, play an important role? How might our lives have +been different if our number system and 0 hadn’t been invented or +conceived of? +4. The ancient Indians likely used base 10 for the Hindu number +system because humans have 10 fingers, and so we can use our +fingers to count. But what if we had only 8 fingers? How would +we be writing numbers then? What would the Hindu numerals +look like if we were using base 8 instead? Base 5? Try writing the +base-10 Hindu numeral 25 as base-8 and base-5 Hindu numerals, +respectively. Can you write it in base-2? +Math +Talk +Math +Talk +Map not to scale +The map shows the locations of the different civilisations. They existed in different time periods. + +A Story of Numbers +81 + +„ +To represent numbers, we need a standard sequence of objects, +names, or written symbols that have a fixed order. This standard +sequence is called a number system. + +„ +The symbols representing numbers in a written number system +are called numerals. + +„ +In a number system, landmark numbers are numbers that +are easily recognisable and used as reference points for +understanding and working with other numbers. They serve +as anchors within the number system, helping people to orient +themselves and make sense of quantities, particularly larger +ones. + +„ +A number system whose landmark numbers are the powers of +a number n is referred to as a base-n number system. + +„ +Number systems having a base that make use of the position +of a symbol in determining the landmark number that it is +associated with are called positional number systems or +place value systems. + +„ +Place value representations were used in the Mesopotamian +(Babylonian), Mayan, Chinese and Indian civilisations. + +„ +The system of numerals that we use throughout the world today +is the Hindu number system (also sometimes called the Indian +number system, or the Hindu-Arabic number system). It is a +place value system with (usually) 10 digits, including the digit +0 which is treated on par with other digits. Due to its use of 0 +as a number, the system enables the writing of all numbers +unambiguously using just finitely many symbols, and also +enables efficient computation. The system originated in India +around 2000 years ago, and then spread across the world, and +is considered one of human history’s greatest inventions. +SUMMARY" +class_8,4,quadrilaterals,ncert_books/class_8/hegp1dd/hegp104.pdf,"In this chapter, we will study some interesting types of four-sided +figures and solve problems based on them. Such figures are commonly +known as quadrilaterals. The word ‘quadrilateral’ is derived from Latin +words — quadri meaning four, and latus referring to sides. +Observe the following figures. +Figs. (i), (ii), and (iii) are quadrilaterals, and the others are not. Why? +The angles of a quadrilateral are the angles between its sides, as +marked in Figs. (i), (ii), and (iii). +We will start with the most familiar quadrilaterals — rectangles and +squares. +(i) +(ii) +(iii) +(iv) +(v) +QUADRILATERALS +4 + +Quadrilaterals +83 +4.1 Rectangles and Squares +We know what rectangles are. Let us define them. +Rectangle: A rectangle is a quadrilateral in which — +(i) The angles are all right angles (90°), and +(ii) The opposite sides are of equal length. +The definition precisely states the conditions a quadrilateral has to +satisfy to be called a rectangle. +Are there other ways to define a rectangle? +Let us consider the following problem related to the construction of +rectangles. +A Carpenter’s Problem +A carpenter needs to put together +two thin strips of wood, as shown +in Fig. 1, so that when a thread is +passed through their endpoints, it +forms a rectangle. +She already has one 8 cm long +strip. What should be the length of +the other strip? Where should they +both be joined? +Let us first model the structure +that the carpenter has to make. The +strips can be modelled as line segments. They are the diagonals of the +quadrilateral formed by their endpoints. For the quadrilateral to be a +rectangle, we need to answer the following questions — +1. What is the length of the other diagonal? +2. What is the point of intersection of the two diagonals? +3. What should the angle be between the diagonals? +Let us answer these questions using +geometric reasoning (deduction). If +that is challenging, try to construct/ +measure some rectangles. +To find the answers to these +questions, let us suppose that we +have placed the diagonals such that +their endpoints form the vertices of +a rectangle, as shown in Fig. 2. +A +B +D +C +O +Fig. 1 +A +D +B +C +Fig. 2 +AC = 8 cm + +Ganita Prakash | Grade 8 +84 +Deduction 1 — What is the length of the other diagonal? +This can be deduced using congruence as follows — +Since ABCD is a rectangle, we have +AB = CD +∠BAD = ∠CDA = 90° +AD is common to both triangles. +So, ∆ADC ≅ ∆DAB by the SAS congruence +condition. +Therefore, AC = BD, since they are corresponding parts of congruent +triangles. This shows that the diagonals of a rectangle always have the +same length. +So the other diagonal must also be 8 cm long. You can verify this +property by constructing/measuring some rectangles. +Deduction 2 — What is the point of intersection of the two diagonals? +This can also be found using congruence. Since we need to know the +relation between OA and OC, and OB and OD, which two triangles of the +rectangle ABCD should we consider? +Common side +A +B +D +C +The blue angles are equal since +they are vertically opposite +angles. +In order to show congruence, +consider ∠1 and ∠2. Are they +equal? +O +1 +2 +A +B +D +C +A +B +D +C +O +Since ∠B = 90°, +∠3 + ∠1 = 90°. +In ∆BCD, since +∠3 + ∠2 + 90 = 180, +we have ∠3 + ∠2 = 90°. +1 +3 +O +A +B +D +C +2 +3 +O +A +B +D +C + +Quadrilaterals +85 +So, ∠1 = ∠2 (= 90° – ∠3). +Thus, by the AAS condition for congruence, ∆AOB ≅ ∆COD. +Hence OA = OC and OB = OD, since they are corresponding parts of +congruent triangles. So, O is the midpoint of AC and BD. +This shows that the diagonals of a rectangle always intersect at +their midpoints. +Therefore, to get a rectangle, the diagonals must be drawn so that +they are equal and intersect at their midpoints. +When the diagonals cross at their midpoints, we say that the diagonals bisect +each other. Bisecting a quantity means dividing it into two equal parts. +Verify this property by constructing some rectangles and measuring +their diagonals and the points of intersection. +Can the following equalities be used to establish that ∆AOD ≅ ∆COB? +AO = CO (proved above) +∠AOB = ∠COD (vertically opposite angles) +AD = CB +Deduction 3 — What are the angles between the diagonals? +Let us check what quadrilateral we get if we draw the +two diagonals such that their lengths are equal, they +bisect each other and have an arbitrary angle, say +60°, between them as shown in the figure to the right. +Can you find all the remaining angles? +Math +Talk +60° +A +B +D +C +O +We can find the remaining angles between the +diagonals using our understanding of vertically +opposite angles and linear pairs. +O +60° +60° +120° +120° +A +B +D +C +In ∆AOB, since OA = OB, the angles opposite them +are equal, say a. +Can you find the value of a? +O +a +a +A +B +D +C +60° +60° +120° +120° + +Ganita Prakash | Grade 8 +86 +In ∆AOB, we have, +a + a + 60 = 180 (interior angles of a triangle). +Therefore 2a = 120. +Thus a = 60. +Similarly, we can find the values of all the other angles. +Can we now identify what type of quadrilateral ABCD is? +Notice that its angles all add up to 90° (30° + 60°). +What can we say about its sides? +We can see that ∆AOB ≅ ∆COD and ∆AOD ≅ ∆COB. Hence, AB = CD, and +AD = CB, since they are corresponding parts of congruent triangles. +Therefore, ABCD is a rectangle since it satisfies the definition of a +rectangle. +Will ABCD remain a rectangle if the angles between the +diagonals are changed? Can we generalise this? +Take one of the angles between the diagonals as x. +O +60° +60° +120° +120° +60° +60° +60° +60° +30° +30° +30° +30° +A +B +D +C +60° +60° +O +A +B +D +C +120° +120° +O +A +B +D +C +x +B +D +C +A +O + +Quadrilaterals +87 +We can compute the four angles between the +diagonals to be x, x, 180 – x, and 180 – x. +Can you find the other angles? +Since we know that ∆AOB is isosceles, we can +denote the measures of both of its base angles by a. +What is the value of a (in degrees) in terms of x? +We have, +a + a + x = 180 +(sum of the interior angles of a triangle) +2a = 180 – x +a = (180 – x) +2 + = 90 – x +2. +Similarly, in the isosceles ∆AOD, let the base angles be b. +b + b + 180 – x = 180 +2b = 180 – (180 – x) +2b = 180 – 180 + x +2b = x +b = x +2 . +All the angles of the quadrilateral are a + b, which is +90 – x +2 + x +2 = 90. +Thus, all four angles of the quadrilateral ABCD are 90°. +What can we say about AB and CD, and AD and BC? +We have ∆AOB ≅ ∆COD and ∆AOD ≅ ∆COB. Hence, AB = CD, and AD = CB, +since they are the corresponding parts of congruent triangles. +O +A +B +D +C +x +x +180° – x +180° – x + 90°– x +2 + 90°– x +2 + 90°– x +2 + 90°– x +2 +x +2 + x +2 +x +2 + x +2 +A +B +D +C +x +a +a +b +b +x +180° – x +180° – x +O +A +B +D +C +x +x +180° – x +180° – x + +Ganita Prakash | Grade 8 +88 +Hence, no matter what the angles between the diagonals are, if the +diagonals are equal and they bisect each other, then the angles of the +quadrilateral formed are 90° each, and the opposite sides are equal. +Thus, the quadrilateral is a rectangle. +Now we know how the wooden strips have to be put together to form +the vertices of a rectangle! They should be equal and connected at their +midpoints. +This method is actually used in practice to +make rectangles. Carpenters in Europe use +this method to get a rectangular frame. It is +also known that farmers in Mozambique, +a country in Africa, use this method while +constructing houses to get the base of the +house in a rectangular shape. +The Process of Finding Properties +As we have been seeing from lower grades, +properties of geometric objects such as parallel +lines, angles, and triangles can be deduced through geometric reasoning. +We will continue to deduce properties of special types of quadrilaterals +in this chapter. +Once you have deduced a property of a quadrilateral, it is good to verify +it with a real-world quadrilateral, either the quadrilateral constructed +on paper or simply a surface having the shape of the quadrilateral. +If you are not able to figure out the property using deduction, you +could experiment by taking real-world quadrilaterals and observing the +property through measurement. Note that these observations give +useful insights about the property, but with them, we can only form a +conjecture, that is, a statement about which we are highly confident, +but not yet sure if it always holds true. For example, constructing a +few rectangles and observing through measurement that their diagonals +bisect each other does not necessarily mean that this will always be the +case — can we be sure that the 1000th rectangle we construct will also +have this property? The only way we can be sure of this property is by +justifying or proving the statement, just as we did in Deduction 2. +A +B +D +C +A +B +D +C +A +C +O +O +O +4 cm +4 cm +4 cm +4 cm +4 cm +4 cm +4 cm +4 cm +4 cm +4 cm +Note to the Teacher: Gently encourage students to deduce or justify properties. +Whenever students face challenges in doing it, encourage them to experiment and +observe, and use their intuition to figure out the properties. + +Quadrilaterals +89 +The Carpenter’s Problem shows that rectangles can also be defined as +follows — +Rectangle: A rectangle is a quadrilateral whose diagonals are equal +and bisect each other. +Observe how different this definition is from the earlier one. Yet, both +capture the same class of quadrilaterals. Further, it turns out that the +first definition can be simplified. +In the earlier definition, we stated that a rectangle has (a) opposite sides +of equal length, and (b) all angles equal to 90°. Would we be wrong if we +just define a rectangle as a quadrilateral in which all the angles are 90°? +If you think that this definition is incomplete, try constructing a +quadrilateral in which the angles are all 90° but the opposite sides are +not equal. +Are you able to construct such a quadrilateral? +Let us prove why this is impossible. +Deduction 4 — What is the shape of a quadrilateral with all the +angles equal to 90°? +Consider a quadrilateral ABCD with all angles measuring 90°. What can +we say about the opposite sides of such a quadrilateral? +Join BD. ∆BAD and ∆DCB seem congruent. Can we justify this claim? +Two equalities can be directly +seen in the triangles. What can we +say about ∠1 and ∠2? +Recall that we tackled a very similar problem in Deduction 2. We can +use the same reasoning here. +1 +A +B +D +C +2 +Common side +Since ∠B = 90°, +∠3 + ∠1 = 90°. +In ∆BCD, since +∠3 + ∠2 + 90° = 180°, +∠3 + ∠2 = 90°. +1 +3 +A +B +D +C +2 +3 +A +B +D +C + +Ganita Prakash | Grade 8 +90 +So, ∠1 = ∠2. +Thus, by the AAS congruence condition, ∆BAD ≅ ∆DCB. +Therefore, AD = CB, and DC = BA, since these are corresponding sides +of congruent triangles. +Is it wrong to write ∆BAD ≅ ∆CDB? Why? +Thus, we have established that if all the angles of a quadrilateral are +right angles, then the opposite sides have equal lengths. Therefore, the +quadrilateral is a rectangle. Thus, a rectangle can simply be defined as +follows — +Rectangle: A rectangle is a quadrilateral in which the angles are all 90°. +Let us list the properties of a rectangle. +Property 1: All the angles of a rectangle are 90°. +Property 2: The opposite sides of a rectangle are equal. +Are the opposite sides of a rectangle parallel? +They definitely seem so. This fact can be justified using one of the +transversal properties. +Notice that AB acts as a transversal to AD +and BC, and that ∠A + ∠B = 90° + 90° = 180°. +When the sum of the internal angles on +the same side of the transversal is 180°, +the lines are parallel. We can use this fact +to conclude that the lines AD and BC are +parallel, which we represent as AD || BC. +Can you similarly show that AB is +parallel to DC (AB||DC)? +Property 3: The opposite sides of a rectangle are parallel to each other. +Property 4: The diagonals of a rectangle are of equal length and they +bisect each other. +A Special Rectangle +In the quadrilaterals below, are there any non-rectangles? +A +B +D +C +(i) +(ii) +(iii) +(iv) +5 cm +5 cm +2 cm +6 cm +3.6 cm +5 cm +5 cm +1 cm +2 cm +6 cm +3.6 cm +1 cm +4 cm +4 cm +4 cm +4 cm + +Quadrilaterals +91 +All these quadrilaterals are rectangles, including (iv). Quadrilateral +(iv) is a rectangle because all its angles are 90°. However, it is a special +kind of rectangle with all sides of equal length. We know that this +quadrilateral is also called a square. +Square: A square is a quadrilateral in which all the angles are equal +to 90°, and all the sides are of equal length. +Thus, every square is also a rectangle, but clearly every rectangle is +not a square. +I am an +Indian +and I am a +Malayali. +Wait, which one of +them are you — an +Indian or a Malayali? +How can you be both? +This relation can be pictorially represented using a Venn diagram. +We have seen these diagrams before. In a Venn diagram, a set of objects +is represented as points inside a closed curve. Typically, these closed +curves are ovals or circles. +For example, the set of all squares is represented as +Each point in the region represents a square, thereby covering all the +possible squares. +Since every square is a rectangle, the Venn diagram representation of +these two sets would be as follows — +Square +Rectangle +Square + +Ganita Prakash | Grade 8 +92 +Let us consider the Carpenter’s Problem again. If the wooden strips +have to be placed such that the thread passing through their endpoints +forms a square, what must be done? +As in the previous case, let us try to +construct a square, one of whose diagonals is +of length 8 cm. +While solving the Carpenter’s Problem for +the case of a rectangle, we have seen that to +get a quadrilateral with all angles 90° (and +opposite sides of equal length), the diagonals +have to be drawn such that — +(i) they are of equal lengths, and +(ii) they bisect each other. +What more needs to be done to get equal sidelengths as well? Can this be +achieved by properly choosing the angle between the diagonals? +See if you can reason and/or experiment to figure this out! +Deduction 5 — What should be the angle formed by the diagonals? +The angle between the diagonals can be found using the notion of congruence! +Suppose we join the equal diagonals such that they bisect each other +and result in a square. Let us label the square ABCD. +To find the angle formed by the diagonals, what are the two triangles +we should consider for congruence? +By the SSS condition for congruence, +∆BOA ≅ ∆BOC +Can this be used to find the angles ∠BOA and ∠BOC formed by the +diagonals? +Since these angles are corresponding parts of congruent triangles, +they are equal. Further, these angles together form a straight angle. So +∠BOA + ∠BOC = 180°. Thus, these angles have to be 90° each. +A +B +D +C +O +D +A +C +B +O +Common Side + +Quadrilaterals +93 +This shows that the diagonals of a square bisect each other at right +angles. This means that the diagonals have to be drawn such that they are +of equal lengths and bisect each other at right angles. Since the endpoints +of the diagonals uniquely determine the vertices of a quadrilateral, we +will get a square when the diagonals are joined this way. +Using this fact, construct a square with a diagonal of length 8 cm. +Properties of a Square +Since a square is a special type of rectangle, all the properties of a +rectangle hold true for a square. +Verify if this is true by going through geometric reasoning in Deduction +1 and Deduction 2, and see if they apply to a square as well. +Property 1: All the sides of a square are equal to each other. +Property 2: The opposite sides of a square are parallel to each other. +Property 3: The angles of a square are all 90°. +Property 4: The diagonals of a square are of equal length and they bisect +each other at 90°. +There is one more special property of a square. +What are the measures of ∠1, ∠2, ∠3, and ∠4? See if you can reason +and/or experiment to figure this out! +In ∆ADC, we have, +∠1 + ∠3 + 90 = 180 +Since AD = DC, we have ∠1 = ∠3. +Thus, ∠1 = ∠3 = 45°. +Similarly, find ∠2 and ∠4. +Thus, we have another property of a square — +Property 5: The diagonals of a square divide the angles of the square into +equal halves. This can also be expressed as — The diagonals +of a square bisect the angles of the square. +A +B +D +C +1 +2 +3 +4 + +Ganita Prakash | Grade 8 +94 +Figure it Out +1. Find all the other angles inside the following rectangles. +2. Draw a quadrilateral whose diagonals have equal lengths of 8 cm +that bisect each other, and intersect at an angle of + +(i) 30° +(ii) 40° +(iii) 90° +(iv) 140° +3. Consider a circle with centre O. Line segments PL and AM are two +perpendicular diameters of the circle. What is the figure APML? +Reason and/or experiment to figure this out. +4. We have seen how to get 90° using paper folding. Now, suppose +we do not have any paper but two sticks of equal length, and a +thread. How do we make an exact 90° using these? +5. We saw that one of the properties of a rectangle is that its opposite +sides are parallel. Can this be chosen as a definition of a rectangle? +In other words, is every quadrilateral that has opposite sides +parallel and equal, a rectangle? +4.2 Angles in a Quadrilateral +Is it possible to construct a quadrilateral with three angles equal to 90° +and the fourth angle not equal to 90°? +You might have observed through constructions that this may not be +possible. +But why not? +This is due to a general property of quadrilaterals related to their +angles. +We have seen that the sum of the angles of a triangle is 180°. There is +a similar regularity in the sum of the angles of a quadrilateral. +30° +110° +(i) +(ii) +S +R +P +Q +D +C +A +B +Math +Talk +Try +This + +Quadrilaterals +95 +Consider a quadrilateral SOME. +Draw a diagonal SM. We get two +triangles ∆SEM and ∆SOM. +In ∆SEM, we have ∠1 + ∠2 + ∠3 = 180°. +And in ∆SOM, ∠4 + ∠5 + ∠6 = 180°. +What do we get when we add all six +angles? +We will have +∠1+ ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 180° + 180° += 360°. +Or, (∠1+ ∠4) + (∠3 + ∠6) +∠2 + ∠5 = 360°. +Since (∠1 + ∠4), (∠3 + ∠6), ∠2 and ∠5 are the angles of this quadrilateral, +we have the following result — +The sum of all angles in any quadrilateral is 360°. +This explains why it is impossible for a quadrilateral to have three +right angles, with the fourth angle not right angle. +4.3 More Quadrilaterals with Parallel Opposite +Sides +Rectangles (and therefore squares) have parallel opposite sides. +Are there quadrilaterals that have parallel opposite sides that are not +rectangles? +Let us try constructing one. +This can be easily done by drawing two pairs of parallel lines, ensuring +that they do not meet at right angles. +Construct such a figure by recalling how parallel lines can be +constructed using a ruler and a set-square, +or a compass and a ruler. +Observe the quadrilateral ABCD. It has +parallel opposite sides but is not a rectangle. +Thus, a larger set of quadrilaterals exists +in which the opposite sides are parallel. Such +quadrilaterals are called parallelograms. +A +B +D +C +1 +2 +3 +4 +5 +6 +S +E +M +O + +Ganita Prakash | Grade 8 +96 +Is a rectangle a parallelogram? +A rectangle has opposite sides parallel. So, it satisfies the +parallelogram’s definition. Hence, it is indeed a parallelogram. More +specifically, a rectangle is a special kind of parallelogram with all its +angles equal to 90°. +Let us represent this relation using a Venn diagram. +To understand the relations between the sides and the angles of a +parallelogram, let us construct the following figure. +Rectangle +Square +Parallelogram +Draw a parallelogram with adjacent +sides of lengths 4 cm and 5 cm, and +an angle of 30° between them. + 30° +5 cm +4 cm +A +B +D +C +ABCD is the required parallelogram. +What are the remaining angles of the parallelogram? What are the +lengths of the remaining sides? See if you can reason out and/or +experiment to figure these out. +Step 1: Draw line segments AB = 4 +cm and AD = 5 cm with an angle of +30° between them. +30° +5 cm +4 cm +A +B +D +C +30° +5 cm +4 cm +B +D +A +Step 2: Draw a line parallel to AB +through the point D, and a line +parallel to AD through B. Mark the +point at which these lines intersect +as C. + +Quadrilaterals +97 +Deduction 6 — What can we say about the angles of a parallelogram? +In the parallelogram ABCD, AB||CD, and AD is a transversal to them. +∠A + ∠D = 180° (sum of the +internal angles on the same side of +a transversal). +Therefore, +∠D = 180 – ∠A = 180 – 30 = 150°. +Similarly, AD||BC, and AB and +CD are transversals to them. +So, ∠A + ∠B = 180°. +So, ∠C + ∠D = 180°. +Using these equations, we get ∠B = 150° and ∠C = 30°. +We see that in this parallelogram, the adjacent pairs of angles add +up to 180° and opposite pairs of angles are equal. +Thus, +∠A + ∠B = 180°, ∠A + ∠D = 180°, ∠C + ∠D = 180°, and ∠B + ∠C = 180°. +And, +∠A = ∠C, and ∠B = ∠D. +Since the adjacent angles are the interior angles on the same side of +a transversal to a pair of parallel lines, they must add up to 180°. +What about the opposite angles? Will they be equal in all parallelograms? +If yes, how can we be sure? +Let us take one of the angles to be x. +What are the other angles? +Since ∠P + ∠R = 180°, +∠R = 180 – ∠P = 180 – x. +Similarly, since ∠A + ∠R = 180°, +∠A = 180 – ∠R = 180 – (180 – x) = 180 – 180 + x = x. +Thus, ∠P = ∠A = x. +Similarly, we can deduce that ∠ R = ∠E = 180 – x. +Therefore, this shows that the opposite angles of a parallelogram +are always equal. +Deduction 7 — What can we say about the sides of a parallelogram? +By looking at a parallelogram, it appears that the opposite sides are +equal. +Can we again use congruence to show this? Which two triangles can +be considered for this? + 30° +5 cm +4 cm +A +B +D +C + x +P +E +R +A + +Ganita Prakash | Grade 8 +98 +In ∆ABD and ∆CDB, the angles +marked with a single arc are equal +as they are the opposite angles of a +parallelogram. +Since AD||BC, and BD is a transversal +to it, the angles marked with double +arcs are equal as they are alternate +angles. +So, by the AAS condition, the +triangles are congruent, that is, ∆ABD ≅ ∆CDB. +Therefore, AD = CB, and AB = CD. +Thus, the opposite sides of a parallelogram are equal. +Is it wrong to write ∆ABD ≅ ∆CBD? Why? +From these deductions we can find the remaining sides and angles of +the parallelogram. + 30° +5 cm +4 cm +A +B +D +C +5 cm +4 cm + 30° + 150° + 150° +Let us list the properties of a parallelogram. +Property 1: The opposite sides of a parallelogram are equal. +Property 2: The opposite sides of a parallelogram are parallel. +Property 3: In a parallelogram, the adjacent angles add up to 180°, and +the opposite angles are equal. +Are the diagonals of a parallelogram always equal? Check with the +parallelogram that you have constructed. +We see that the diagonals of a parallelogram need not be equal. +Do they bisect each other (do they intersect at their midpoints)? Reason +and/or experiment to figure this out. +Deduction 8 — What is the point of intersection of the two diagonals +in a parallelogram? +As in the case of a rectangle, we can find out if the diagonals bisect +each other by examining the congruence of ∆AOE and ∆YOS in the +parallelogram EASY. +A +B +D +C +Common +Side + +Quadrilaterals +99 +AE = YS (as they are the opposite sides +of the parallelogram) +The angles marked using a single arc +are equal, and so are the angles marked +using a double arc, since they are alternate +angles of parallel lines. +Thus, by the ASA condition, the triangles +are congruent, that is, +∆AOE ≅ ∆YOS. +Therefore, OA = OY, and OE = OS, since they are corresponding parts +of congruent triangles. +Thus, O is the midpoint of both diagonals. +Is it wrong to write ∆AOE ≅ ∆SOY? Why? +Property 4: The diagonals of a parallelogram bisect each other. +Do the diagonals of a parallelogram intersect at a particular angle? +4.4 Quadrilaterals with Equal Sidelengths +Are squares the only quadrilaterals that have +equal sidelengths? Let us explore this question +through construction. +Draw two equal sides AD and AB, that are +not perpendicular to each other. +Can we complete this quadrilateral so that all its sides are of the same +length? +Mark a point C whose distance from +B and D is equal to AB (or AD). To do this, +measure AB using a compass. Keeping +this length as the radius, cut arcs from +B and D. +Now we have a quadrilateral with +equal sidelengths and one of its angles +50°. Note that we could have constructed +such a quadrilateral by taking any angle less than 180° (in place of 50°). +A quadrilateral in which all the sides have the same length is a +rhombus. + 50° + A + B + D +50° +A +B +D +C +E +Y +A +S +O + +Ganita Prakash | Grade 8 +100 +What are the other angles of the rhombus ABCD that we have +constructed? Reason and/or experiment to figure this out. +Deduction 9 — What can we say about the angles in a rhombus? +Consider a rhombus GAME. +In ∆GAE, since GE = GA, a = d. +Similarly, in ∆MAE, since ME = MA, +b = c. +It can be seen that ∆GAE ≅ ∆MAE +(How?) +So, a = b, c = d and ∠G = ∠M (since +they are corresponding parts of congruent triangles). +Thus, we have, a = b = c = d. +These facts hold for any rhombus. Let us apply them to the rhombus +ABCD that we constructed earlier. Let the four equal angles formed by +the diagonal be a, as shown in the figure +In ∆ADB, we have +a + a + 50 = 180°. +So, a = 65°. +Thus, the angles of the rhombus ABCD +are 50°, 130°, 50°, and 130°. +So, in a rhombus opposite angles are +equal to each other. +Interestingly, there is one more way by +which we could have figured out the other +angles of the rhombus ABCD. We have +shown that in a general rhombus GAME, the four angles formed by a +diagonal are equal to each other. +G +A +E +M + a + b + d + c +B +A +D +C + a +a +a +a + 50° +50° +A +B +D +C + 65° + 50° + 50° + 65° + 65° + 65° +G +A +E +M + a + a +G +A +E +M +a + a +Consider the lines EM and GA and its +transversal AE. Since the alternate +angles are equal, EM||GA. +Similarly consider the lines GE +and AM and its transversal AE. +Since the alternate angles are +equal, GE||AM. + +Quadrilaterals +101 +As opposite sides are parallel, GAME +is also a parallelogram. Thus, every +rhombus is a parallelogram, and the +properties of a parallelogram hold +true for a rhombus as well. Thus, the +adjacent angles of a rhombus add up to +180°, and the opposite angles are equal +(verify that the arguments in Deduction 6 can be applied to a rhombus +as well!). +Thus, in rhombus ABCD, +∠A = ∠C = 50°, and +∠D = ∠B = 180  –  50 = 130°. +So a rhombus is a parallelogram, and a rectangle is also a parallelogram. +How can this be represented using a Venn diagram? +Where will the set of squares occur in this diagram? +We know that a square is a rectangle. Since the opposite sides of a +square are parallel, a square is also a parallelogram. Further, since all +the sides of a square have the same length, a square is also a rhombus. +Thus, the Venn diagram will be as follows. +Rectangle +Square +Rhombus +Rectangle +Rhombus +Parallelogram +Square +Let us list the properties of a rhombus. +Property 1: All the sides of a rhombus are equal to each other. +Property 2: The opposite sides of a rhombus are parallel to each other. +Property 3: In a rhombus, the adjacent angles add up to 180°, and the +opposite angles are equal. +Are the diagonals of a rhombus equal? +Property 4: The diagonals of a rhombus bisect each other. +Property 5: The diagonals of a rhombus bisect its angles. +50° +A +B +D +C + +Ganita Prakash | Grade 8 +102 +Do the diagonals of a rhombus intersect at any particular angle? Reason +out and/or experiment to figure this out! +Deduction 10 — What can we say about the angles formed by the +diagonals of a rhombus at their point of intersection? +In the rhombus GAME, we have +∆GEO ≅ ∆MEO (why?). +So, ∠GOE = ∠MOE, as they are +corresponding parts of congruent +triangles. As they add up to 180°, +they should be 90° each. +Property 6: Diagonals of a rhombus intersect each other at an angle of +90°. +Figure it Out +1. Find the remaining angles in the following quadrilaterals. +2. Using the diagonal properties, construct a parallelogram whose +diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of +140°. +3. Using the diagonal properties, construct a rhombus whose diagonals +are of lengths 4 cm and 5 cm. +G +A +E +M +O +S +R +P +Q +110° +R +A +P +E +40° +(iii) +U +V +X +W +30° +A +O +I +E +20° +(i) +(ii) +(iv) + +Quadrilaterals +103 +4.5 Playing with Quadrilaterals +Geoboard Activity +Take a geoboard and some rubber bands. If you do not have these, you +could just use the dot grid papers given at the end of the book for this +activity. +Place two rubber bands perpendicular to each other, forming +diagonals of equal length. Join the ends. +What is the quadrilateral that you get? Justify your answer. +Extend one of the diagonals on both sides by 2 cm. +What quadrilateral will you get now? Justify your answer. + +Ganita Prakash | Grade 8 +104 +Joining Triangles +1. Take two cardboard cutouts of an equilateral triangle of sidelength +8 cm. +8 cm +8 cm +8 cm +8 cm +8 cm +8 cm +Can you join them to get a quadrilateral? +8 cm +8 cm +8 cm +8 cm +8 cm +What type of a quadrilateral is this? Justify your answer. +2. Take two cardboard cutouts of an isosceles triangle with sidelengths +8 cm, 8 cm, and 6 cm. +6 cm +8 cm +8 cm +6 cm +8 cm +8 cm +What are the different ways they can be joined to get a quadrilateral? +6 cm +8 cm +8 cm +8 cm +8 cm +8 cm +6 cm +6 cm +8 cm +8 cm +Joining them in this +way you get +Joining them in this +way you get +What quadrilaterals are these? Justify your answers. + +Quadrilaterals +105 +3. Take two cardboard cutouts of a scalene triangle with sides 6 cm, +9 cm, and 12 cm. +What are the different ways they can be joined to get a quadrilateral? +Are you able to identify the different quadrilaterals that are +obtained by joining the triangles? Justify your answer whenever +you identify a quadrilateral. +4.6 Kite and Trapezium +Kite +One of the ways the two triangles of sides 6 cm, 9 cm and 12 cm can be +joined together is as follows — +This quadrilateral looks like a kite. Observe that +the adjacent sides are of the same length. +Kite: A kite is a quadrilateral that can be labelled +ABCD such that AB = BC, and CD = DA. +Property 1: In the kite, show that the diagonal BD +(i) bisects ∠ABC and ∠ADC, +(ii) bisects the diagonal AC, that is, +AO = OC, and is perpendicular to it. +Hint: Is ∆AOB ≅ ∆COB? +6 cm +9 cm +12 cm +6 cm +9 cm +12 cm +6 cm +9 cm +12 cm +6 cm +9 cm +A +B +C +D +O + +Ganita Prakash | Grade 8 +106 +Trapezium +Parallelograms are quadrilaterals that have +parallel opposite sides. We get a new type +of quadrilateral if we relax this condition. +Trapezium: A trapezium is a quadrilateral +with at least one pair of parallel opposite +sides. +Construct a trapezium. Measure the base angles (marked in the figure). +Can you find the remaining angles without measuring them? +Since PQ||SR, we have +Property 1: ∠ S + ∠P = 180° and ∠R + ∠Q = 180°. +Using these facts, the remaining angles can easily be found. Verify +your answer after finding them. +When the non-parallel sides of a trapezium have the same lengths, +the trapezium is called an isosceles trapezium. +How do we construct an isosceles trapezium? +Construct an isosceles trapezium UVWX, with UV||XW. Measure ∠U. +Mark X and W such +that UX = VW +X +W +V +U +X +W +V +U +Can you find the remaining angles without measuring them? +Does it appear that the angles opposite to the equal sides — ∠U and +∠V — are also equal? Can we find congruent triangles here? +Consider line segments XY and WZ perpendicular to UV. +What type of quadrilateral is XWZY? +Since XW||UV, +a = 180° – ∠XYZ = 90°, and +b = 180° – ∠WZY = 90° (since the internal +angles on the same side of a transversal +add up to 180°) +Hence, XWZY is a rectangle. +P +Q +R +S +V +X +W +U +Y +Z +a +b + +Quadrilaterals +107 +Now, it can be shown that ∆UXY ≅ ∆VWZ. (How?) +Thus, ∠U = ∠V. +Using this fact, the remaining angles of the isosceles trapezium can be +determined. Verify the angles by measurement. +Property 2: In an isosceles trapezium, the angles opposite to the equal +sides are equal. +Figure it Out +1. Find all the sides and the angles of the quadrilateral obtained by +joining two equilateral triangles with sides 4 cm. +2. Construct a kite whose diagonals are of lengths 6 cm and 8 cm. +3. Find the remaining angles in the following trapeziums — +105° +135° + 100° +4. Draw a Venn diagram showing the set of parallelograms, kites, +rhombuses, rectangles, and squares. Then, answer the following +questions — +(i) What is the quadrilateral that is both a kite and a +parallelogram? +(ii) Can there be a quadrilateral that is both a kite and a rectangle? +(iii) Is every kite a rhombus? If not, what is the correct relationship +between these two types of quadrilaterals? +5. If PAIR and RODS are two rectangles, find ∠IOD. +30° +A +P +I +D +R +O +S +5 cm +5 cm + +Ganita Prakash | Grade 8 +108 +6. Construct a square with diagonal 6 cm without using a protractor. +7. CASE is a square. The points U, V, W and X are the midpoints of the +sides of the square. What type of quadrilateral is UVWX? Find this +by using geometric reasoning, as well as by construction and +measurement. Find other ways of constructing a square within a +square such that the vertices of the inner square lie on the sides of +the outer square, as shown in Figure (b). +8. If a quadrilateral has four equal sides and one angle of 90°, will it be +a square? Find the answer using geometric reasoning as well as by +construction and measurement. +9. What type of a quadrilateral is one in which the opposite sides are +equal? Justify your answer. +Hint: Draw a diagonal and check for congruent triangles. +10. Will the sum of the angles in a quadrilateral such as the following +one also be 360°? Find the answer using geometric reasoning as +well as by constructing this figure and measuring. +D +A +B +C +11. State whether the following statements are true or false. Justify your +answers. + (i) A quadrilateral whose diagonals are equal and bisect each +other must be a square. +E +S +C +A +U +V +W +X +(a) +(b) +Try +This + +Quadrilaterals +109 + (ii) A quadrilateral having three right angles must be a rectangle. + (iii) A quadrilateral whose diagonals bisect each other must be a +parallelogram. + (iv) A quadrilateral whose diagonals are perpendicular to each +other must be a rhombus. + (v) A quadrilateral in which the opposite angles are equal must +be a parallelogram. + (vi) A quadrilateral in which all the angles are equal is a rectangle. + (vii) Isosceles trapeziums are parallelograms. + +„ +A rectangle is a quadrilateral in which the angles are all 90°. +Properties of a rectangle — + y Opposite sides of a rectangle are equal. + y Opposite sides of a rectangle are parallel to each other. + y Diagonals of a rectangle are of equal length and they bisect +each other. + +„ +A square is a quadrilateral in which all the angles are 90°, and +all the sides are of equal length. +Properties of a square — + y The opposite sides of a square are parallel to each other. + y The diagonals of a square are of equal lengths and they +bisect each other at 90°. + y The diagonals of a square bisect the angles of the square. + +„ +A parallelogram is a quadrilateral in which opposite sides are +parallel. +Properties of a parallelogram — + y The opposite sides of a parallelogram are equal. + y In a parallelogram, the adjacent angles add up to 180°, and +the opposite angles are equal. + y The diagonals of a parallelogram bisect each other. + +„ +A rhombus is a quadrilateral in which all the sides have the +same length. +SUMMARY + +Ganita Prakash | Grade 8 +110 +Properties of a rhombus — + y The opposite sides of a rhombus are parallel to each other. + y In a rhombus, the adjacent angles add up to 180°, and the +opposite angles are equal. + y The diagonals of a rhombus bisect each other at right +angles. + y The diagonals of a rhombus bisect its angles. + +„ +A kite is a quadrilateral with two non-overlapping adjacent +pairs of sides having the same length. + +„ +A trapezium is a quadrilateral having at least one pair of +parallel opposite sides. + +„ +The sum of the angle measures in a quadrilateral is 360°. +Parallelogram +Kite +Trapezium +Rectangle +Square +Rhombus + +Quadrilaterals +111 +Which Quad? +Gameplay +1. Fold a sheet into half. +2. Now, fold it once more into a quarter. +3. Make a triangular crease at the corner +that is at the middle of the paper. +4. Open the sheet. What is the shape +formed by the creases? +5. How would you fold the quarter paper to get the kinds of creases +shown in the following image. +6. How would you fold the quarter paper such that a square is formed?" +class_8,5,number play,ncert_books/class_8/hegp1dd/hegp105.pdf,"5.1 Is This a Multiple Of? +Sum of Consecutive Numbers +Anshu is exploring sums of consecutive numbers. He has written the +following — +Now, he is wondering — +• “Can I write every natural number as a sum of consecutive +numbers?” +• “Which numbers can I write as the sum of consecutive numbers in +more than one way?” +• “Ohh, I know all odd numbers can be written as a sum of two +consecutive numbers. Can we write all even numbers as a sum of +consecutive numbers?” +• “Can I write 0 as a sum of consecutive numbers? Maybe I should +use negative numbers.” +Explore these questions and any others that may occur to you. +Discuss them with the class. +Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place ‘+’ and +‘–’ signs in between the numbers. How many different possibilities exist? +Write all of them. +7 = 3 + 4 +10 = 1 + 2 + 3 + 4 +12 = 3 + 4 + 5 +15 = 7 + 8 + = 4 + 5 + 6 + = 1 + 2 + 3 + 4 + 5 +Math +Talk +3 + 4 – 5 + 6 +3 – 4 – 5 – 6 +NUMBER PLAY +5 + +Number Play +113 +Eight such expressions are possible. You can use the diagram below to +systematically list all the possibilities. +Evaluate each expression and write the result next to it. Do you notice +anything interesting? +Now, take four other consecutive numbers. Place the ‘+’ and ‘–’ signs +as you have done before. Find out the results of each expression. +What do you observe? +Repeat this for one more set of 4 consecutive numbers. Share your +findings. +Some sums appear always no matter which 4 consecutive +numbers are chosen. Isn’t that interesting? +Do these patterns occur no matter which 4 consecutive numbers are +chosen? Is there a way to find out through reasoning? +Hint: Use algebra and describe the 8 expressions in a general form. +You might have noticed that the results of all expressions are even +numbers. Even numbers have a factor of 2. Negative numbers having +a factor 2 are also even numbers, for example, – 2, – 4, – 6, and so on. +Check if anyone in your class got an odd number. +When 4 consecutive numbers are chosen, no matter how the ‘+’ and +‘–’ signs are placed between them, the resulting expressions always have +even parity. +3 +4 +4 +5 +5 +5 +5 +6 +6 +6 +6 +6 +6 +6 +6 ++ +– +– ++ ++ +– +– +– +– +– ++ ++ ++ ++ +3 + 4 + 5 + 6 +3 + 4 + 5 – 6 +Math +Talk +3 + 4 – 5 + 6 = 8 +3 – 4 – 5 – 6 = – 12 +. +. +. +5 + 6 – 7 + 8 = 12 +5 – 6 – 7 – 8 = – 16 +. +. +. +__ + __ – __ + __ = __ +__ – __ – __ – __ = __ +. +. +. + +Ganita Prakash | Grade 8 +114 +Now take any 4 numbers, place ‘+’ and ‘–’ signs in the eight different +ways, and evaluate the resulting expression. What do you observe about +their parities? +Repeat this with other sets of 4 numbers. +Is there a way to explain why this happens? +Hint: Think of the rules for parity of the sum or difference of two +numbers. +Explanation 1: Let us consider any of the 8 expressions formed by four +numbers a, b, c, and d. When one of its signs is switched, its value always +increases or decreases by an even number! Let us see why. +Consider one of the expressions: a + b – c – d. +Replacing +b by – b, we get +a – b – c – d. +By how much has the number changed? It has changed by +(a + b – c – d) – (a – b – c – d) += a + b – c – d – a + b + c + d (notice how the signs changed when we +opened the second set of brackets) += 2b (this is an even number). +If the difference between two numbers is even, can they have different +parities? No! So either both are even or both are odd. +Now, let us see what happens when a negative sign is switched to a +positive sign. +Replace any negative sign in the expression a + b – c – d with a positive +sign and find the difference between the two numbers. +What do you conclude from this observation? +Starting from any expression, we can get 7 expressions by switching one +or more ‘+’ and ‘–’ signs. Thus, all the expressions have the same parity! +Explanation 2: We know that +odd ± odd = even +even ± even = even +odd ± even = odd. +We have seen that the parity of a + b and a – b is the same, regardless +of the parities of a and b. +In short, a ± b have the same parity. By the same argument, a ± b + c +and a ± b – c have the same parity. Extending this further, we can say that +all the expressions a ± b ± c ± d have the same parity. +Math +Talk + +Number Play +115 +Explanation 3: This can also be explained +using the positive and negative token +model you studied in the chapter on +Integers. Try to think how. +The number of ways to choose 4 +numbers a, b, c, d and combine them using +‘+’ and ‘–’ signs is infinite. Mathematical +reasoning allows us to prove that all the +combinations a ± b ± c ± d always have the +same parity, without having to go through +them one by one. +Is the phenomenon of all the expressions having the same parity limited +to taking 4 numbers? What do you think? +Breaking Even +We know how to identify even numbers. Without computing them, find +out which of the following arithmetic expressions are even. +Using our understanding of how parity behaves under different +operations, identify which of the following algebraic expressions give +an even number for any integer values for the letter-numbers. +3g + 5h +2a + 2b +2u – 4v +4m + 2n +6m – 3n +13k – 5k +b +2 + 1 +4k × 3j +x +2 + 2 +Several problems in mathematics can be thought about and +solved in different ways. While the method you came up with +may be dear to you, it can be amusing and enriching to know +how others thought about it. Two tidbits: ‘share’ and ‘listen’. +‘What if …?’, ‘Will it always happen?’— Wondering and posing +questions and conjectures is as much a part of mathematics as +problem solving. +672 – 348 +43 + 37 +708 – 477 +4 × 347 × 3 +119 × 303 +809 + 214 +513 +3 +543 – 479 +a +b +b +c +c +c +c +d +d +d +d +d +d +d +d ++ +– +– ++ ++ +– +– +– +– +– ++ ++ ++ ++ + +Ganita Prakash | Grade 8 +116 +The expression 4m + 2q will always evaluate to an even number for +any integer values of m and q. We can justify this in two different ways — +• We know 4m is even and 2q is even for any integers m and q. +Therefore, their sum will also be even. +• The expression 4m + 2q is equal to the expression 2(2m + q). Here, +the expression 2(2m + q) means 2 times 2m + q. In other words, 2 is +a factor of this expression. Therefore, this expression will always +give an even number for any integers m and q. +For example, if m = 4 and q = – 9, the expression 4m + 2q becomes +4 × 4 + 2 × (–9) = – 2, which is an even number. +In the expression x +2 + 2, x +2 is even if x is even, and x +2 is odd if x is odd. +Therefore, the expression x +2 + 2 will not always give an even number. +An example and a non-example for when the expression evaluates to an +even number — (i) if x = 6, then x +2 + 2 = 38, and (ii) if x = 3, then x +2 + 2 + = 11. +Similarly, determine and explain which of the other expressions always +give even numbers. Write a couple of examples and non-examples, as +appropriate, for each expression. +Write a few algebraic expressions which always give an even number. +Pairs to Make Fours +Take a pair of even numbers. Add them. Is the sum divisible by 4? +Try this with different pairs of even numbers. +When is the sum a multiple of 4, and when is it not? +Is there a general rule or a pattern? +Even numbers can be of two types based on the remainders they leave +when divided by 4. +. +. +. +. +. +. +. +Even numbers that are multiples +of 4 leave a remainder of 0 when +divided by 4. +Even +numbers +that +are +not +multiples of 4 leave a remainder 2 +when divided by 4. + +Number Play +117 +When will two even numbers add up to give a multiple of 4? +This problem is similar to the question of identifying when adding +two numbers will result in an even number. Can you see this? +There are three cases to examine: +What happens when we add a multiple of 4 to an even number that is +not a multiple of 4? Is it similar to the case of the parity of the sum of an +even and an odd number? +Look at the following expressions and the visualisation. Write the +corresponding explanation and examples. +Explanation with Algebra and Visualisation +Examples +Adding +two (even) +numbers that +are multiples +of 4 will always +give a multiple +of 4. +4p and 4q. +4p + 4q += 4 (p + q). +p rows +. +. +. +. +q rows +. +. +. +. ++ += +. +. +. +. +. +. +. +. +(p+q) rows +4, 12, 16, 24, +36. +12 + 16 += 4 (3 + 4) += 28. +16 + 28 += 4 (4 + 7) += 44. +Adding two +even numbers +that are not +multiples +of 4 will +always give a +multiple of 4 +because their +remainders of +2 add up to 4. +(4p + 2) and (4q + 2). +(4p + 2) + (4q + 2) += 4p + 4q + 4 += 4 (p + q + 1). +p rows +q rows ++ += +(p + q + 1) rows +. +. +. +. +. +. +. +. +. +. +. +. +p rows +1 row +q rows +2, 6, 10, 18, +22, 42. +2 + 6 = 8. +6 + 10 = 16. +22 + 6 = 28. + +Ganita Prakash | Grade 8 +118 +Notice how we are able to generalise and prove properties of +arithmetic using algebra and also using visualisation. +Always, Sometimes, or Never +We examine different statements about factors and multiples and +determine whether a statement is ‘Always True’, ‘Sometimes True’, or +‘Never True’. +We know that the sum of any two multiples of 2 is also a multiple of 2. +1. If 8 exactly divides two numbers separately, it must exactly divide +their sum. +Statement 1 is always true. Determine if it is true with subtraction. +Explanation with Algebra and Visualisation +Examples +4p and (4q + 2) += 4p + (4q + 2) += 4p + 4q + 2 += 4 (p + q) + 2. +p rows +q rows ++ += +(p + q) rows with +a remainder 2 +. +. +. +. +. +. +. +. +. +. +. +. +. +. +Explanation with Algebra and Visualisation +Examples +The two numbers +have 8 as a factor; +in other words, the +two numbers are +multiples of 8. +8a and 8b. +8 and 16. +16 and 56. +80 and 120. +As multiples of 8 +are obtained by +repeatedly adding +8, the sum of two +multiples of 8 will +also be a multiple +of 8. +8a + 8b += 8 (a + b). +a rows +b rows +. +. +. +. +. +. +. +. +(a+b) rows +8 + 16 = 8(1 + 2) += 24. +16 + 56 = 72. +80 + 120 = 200. + +Number Play +119 +In general, if a divides M and a divides N, then a divides M + N and +a divides M – N. In other words, if M and N are multiples of a, then +M + N and M – N will also be multiples of a. +2. If a number is divisible by 8, then 8 also divides any two numbers +(separately) that add up to the number. +So, statement 2 is sometimes true. +3. If a number is divisible by 7, then all multiples of that number will +be divisible by 7. +The number 7jm or (7 × j × m) has a factor of 7. We can see that +Statement 3 is always true. +Explanation with Algebra and Visualisation +Examples +A number +divisible by 8 is +a multiple of 8. +8m +8, 16, 56, 72. +A number +divisible +by 8 can be +expressed as +a sum of two +multiples of 8 or +sum of two non- +multiples of 8. +8m = 8a + 8b +8m = p + q +(p, q not +multiples of +8) +a rows +b rows +. +. +. +. +. +. +. +. +m = (a+b) rows +. +. +p +. +. +. +. +. q += +72 = 48 + 24 +(8×9 += 8×6 + 8×3). +72 = 50 + 22 +Explanation with Algebra and Visualisation +Examples +Numbers divisible by 7 +will have 7 as a factor. +7j +14 = 7 × 2 (j = 2). +42 = 7 × 6 (j = 6). +98 = 7 × 14 (j = 14). +This contains a total of +mj rows. So this is also a +multiple of 7. +(7j) × m +. +. +. +. +. +j rows +. +. +. +. +. +j rows +. +. +. +. +. +j rows +. +. +. +. +m times +Some multiples of +14: +28 = (7 × 2) × 2. +70 = (7 × 2) × 5. +154 = (7 × 2) × 11 + +Ganita Prakash | Grade 8 +120 +In general, if A is divisible by k, then all multiples of A are divisible by k. +4. If a number is divisible by 12, then the number is also divisible by all +the factors of 12. +In general, if A is divisible by k, then A is divisible by all the factors +of k. Hence, Statement 4 is always true. +5. If a number is divisible by 7, then it is also divisible by any multiple +of 7. +We can see that this statement is only sometimes true. +Explanation with Algebra and Visualisation +Examples +A number divisible by +12 is a multiple of 12. +12m +12, 24, 36, 48, +108, 132. +Factors of multiples of +12 will include factors +of 12. +12m += 2 × 6 × m += 3 × 4 × m +. +. +. +. +. +m rows +. . . . +. . . . +12 +A factor of 12 covers a row fully. +Hence, it covers all multiples of 12 +fully. +Factors of 24: 1, +2, 3, 4, 6, 8, 12, +24. +Explanation with Algebra and Visualisation +Examples +Numbers divisible by +7 are multiples of 7. +7k +. +. +. +. +. +k rows +Multiples of 7. +7k will be divisible by +7m if and only if m is a +factor of k. +7m +If k = ym +then +7k ÷ 7m = +7ym ÷ 7m + = y +. +. +. +. +. +m rows +. +. +. +. +. +m rows +. +. +. +. +. +m rows +. +. +. +. +k rows +42 (7 × 6) is divisible +by 7 but it is not +divisible by 28 (7 × 4). +42 (7 × 6) is divisible +by 7 and it is divisible +by 14 (7 × 2). + +Number Play +121 +Examine each of the following statements, and determine whether +it is ‛Always true’, ‛Sometimes true’, ‛Never true’. +6. If a number is divisible by both 9 and 4, it must be divisible by 36. +7. If a number is divisible by both 6 and 4, it must be divisible by 24. +In general, if A is divisible by k and A is also divisible by m, then +A is divisible by the LCM of k and m. This is because A is a multiple of +k and also a multiple of m, so A’s prime factorisation should contain the +prime factorisation of LCM (k, m). +8. When you add an odd number to an even number we get a multiple +of 6. +We know that multiples of 6 are all even +numbers. The sum of an odd number and an +even number will be an odd number. Therefore, +this statement is never true. We can also explain +this algebraically. Suppose, +(2n) + (2m + 1) = 6j, +where 2n is an even number, 2m + 1 is an odd +number, and 6j is a multiple of 6. Then +2n + 2m = 6j – 1 +2 (n + m) = 6j – 1 +which means 2(n + m), which is an even number, should be equal to +6j – 1, which is an odd number. This is never true. +What Remains? +Find a number that has a remainder of 3 when divided by 5. Write more +such numbers. +Which algebraic expression(s) capture all such numbers? +(i) 3k + 5    (ii) 3k – 5    (iii) 3k +5     (iv) 5k + 3    (v) 5k – 2    (vi) 5k – 3 +The numbers that leave a remainder of 0 when divided by 5 are the +multiples of 5. But we want numbers that leave a remainder of 3 when +divided by 5. These numbers are 3 more than multiples of 5. Multiples of +5 are of the form 5k. So, numbers that leave a remainder of 3 when +divided by 5 are those of the form 5k + 3 +Math +Talk +Can I write an even and +an odd number as 2n and +2n+1 instead? +. +. +. +k rows +5 +k = 0 1 2 3 4 +5k + 3 = 3 8 13 18 23 + +Ganita Prakash | Grade 8 +122 +Let us consider another expression, 5k – 2, and see the values it takes for +different values of k. +Numbers that leave a remainder of 3 when +divided by 5 can also be seen as 2 less than +multiples of 5; 5k – 2, where k ≥ 1. +Are there other expressions that generate +numbers that are 3 more than a multiple of 5? +Figure it Out +1. The sum of four consecutive numbers is 34. What are these numbers? +2. Suppose p is the greatest of five consecutive numbers. Describe the +other four numbers in terms of p. +3. For each statement below, determine whether it is always true, +sometimes true, or never true. Explain your answer. Mention +examples and non-examples as appropriate. Justify your claim using +algebra. +(i) +The sum of two even numbers is a multiple of 3. +(ii) If a number is not divisible by 18, then it is also not divisible +by 9. +(iii) If two numbers are not divisible by 6, then their sum is not +divisible by 6. +(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3. +(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9. +4. Find a few numbers that leave a remainder of 2 when divided by 3 +and a remainder of 2 when divided by 4. Write an algebraic +expression to describe all such numbers. +5. “I hold some pebbles, not too many, +When I group them in 3’s, one stays with me. +Try pairing them up — it simply won’t do, +A stubborn odd pebble remains in my view. +Group them by 5, yet one’s still around, +But grouping by seven, perfection is found. +More than one hundred would be far too bold, +Can you tell me the number of pebbles I hold?” +6. Tathagat has written several numbers that leave a remainder of 2 +when divided by 6. He claims, “If you add any three such numbers, +the sum will always be a multiple of 6.” Is Tathagat’s claim true? +k = 1 2 3 4 5 +5k – 2 = 3 8 13 18 23 + +Number Play +123 +7. When divided by 7, the number 661 leaves a remainder of 3, and +4779 leaves a remainder of 5. Without calculating, can you say what +remainders the following expressions will leave when divided by 7? +Show the solution both algebraically and visually. +(i)  4779 + 661 +(ii)  4779 – 661 +8. Find a number that leaves a remainder of 2 when divided by 3, +a remainder of 3 when divided by 4, and a remainder of 4 when +divided by 5. What is the smallest such number? Can you give a +simple explanation of why it is the smallest? +5.2 Checking Divisibility Quickly +Earlier, you have learnt shortcuts to check whether a given number, +written in the Indian number system is divisible by 2, 4, 5, 8, and 10. Let +us revisit them. +Divisibility by 10, 5, and 2: If the units digit of a number is ‘0’, then +it is divisible by 10. Let us understand why this works through algebra. +We can write the general form of a number in the Indian system +using a set of letter-numbers. For example, a 5-digit number can be +expressed as, edcba denoting e × 10000 + d × 1000 + c × 100 + b × 10 + a. The +letter-numbers e, d, c, b, and a denote each digit of a 5-digit number. +Any number can be written in general as … dcba, where the letter- +numbers a, b, c and d represent the units, tens, hundreds and thousands +digit, respectively, and so on. As a sum of place values, this number is — +… +1000d + 100c + 10b + a. +(For example, in the number 4075, d = 4, c = 0, b = 7, and a = 5.) +We know that each place value, with the exception of the units place, +is a multiple of 10. So, 10b, 100c, … all will be multiples of 10. Hence, the +number will be divisible by 10 if and only if the units digit a is 0. +Similarly, explain using algebra why the divisibility shortcuts for 5, 2, 4, +and 8 work. +Let us now examine shortcuts to check divisibility by some other +numbers and explain why they work! +A Shortcut for Divisibility by 9 +Can you say, without actually calculating, which of these numbers are +divisible by 9: 999, 909, 900, 90, 990? +All of them. + +Ganita Prakash | Grade 8 +124 +Can we say that any number made up of only the digits ‘0’ and ‘9’, in any +order, will always be divisible by 9? +Yes, if each digit is either 0 or 9, then each term in its expanded form +will be 9 × or 0 × (the ‘   ’ denotes a place value). This means each +term will be a multiple of 9, for example, +99009 = 9 × 10000 + 9 × 1000 + 0 × 100 + 0 × 10 + 9 × 1. +But this shortcut alone cannot identify all the multiples +of 9. Unlike the numbers 2, 5, and 10, we cannot identify +the multiples of 9 by just looking at the unit’s digit. 99 and +109 are two numbers with 9 as the units digit; but 99 is +divisible by 9, while 109 is not. +Is 10 divisible by 9? If not, what is the remainder? +Check the divisibility of other multiples of 10 (10, 20, 30, ...) by 9. +You will notice that for any multiple of 10, the remainder is the same +as the number of tens. +Similarly, look at the remainder when the multiples of 100 (100, 200, +300, … ) are divided by 9. What do you notice? +The remainder is the same as the number of hundreds for any multiple +of 100. +Using this observation, find the remainder when 427 is divided by 9. +We see that 427 has 4 hundreds; thus, its corresponding remainder +(upon division by 9) would be 4. 427 has 2 tens, and its corresponding +remainder would be 2. We have 7 units also remaining. Adding all the +remainders, we get 4 + 2 + 7 = 13. We can make one more group of 9 with +13, leaving a remainder of 4. Therefore, 427 ÷ 9 gives a remainder of 4. +99 +99 +1 +1 +9 +1 +9 +1 +9 +1 +99 +1 +9 +1 +9 +1 +99 +1 +99 +1 +99 +1 +1 +1 +1 +1 +1 +1 +1 +400 +20 +7 +99 +1 +9 +1 +9 +1 +99 +1 +99 +1 +99 +1 +1 +1 +1 +1 +1 +1 +1 +4 +9 +(Remainder) + +Number Play +125 +Will this work with bigger numbers? +You can see that this is true for any place value: +1 = 0 + 1 +10 = 9 + 1 +100 = 99 + 1 +1000 = 999 + 1 +10000 = 9999 + 1, and so on. Each digit thus denotes the remainder when +the corresponding place value is divided by 9. +For example, to find the remainder of 7309 when divided by 9, we can +just add all the digits — 7 + 3 + 0 + 9 — to get 19. This can be seen as +follows: + 7 × 1000 + 3 × 100 + 0 × 10 + 9 × 1 += 7 × (999 + 1) + 3 × (99 + 1) + 0 × (9 + 1) + 9 × (0+1) += (7 × 999 + 3 × 99 + 0 × 9 + 9 × 0) + (7 × 1 + 3 × 1 + 0 × 1 + 9 × 1) += (7 × 999 + 3 × 99 + 0 × 9 + 9 × 0) + (7 + 3 + 0 + 9). +This means that the number 7309 is 19 more than some multiple of 9. +The digits 1 and 9 can further be added to get 1 + 9 = 10. Now, we can say +that 7309 is 10 more than a multiple of 9. And repeating this step for the +number 10, we get the remainder to be 1 + 0 = 1, meaning 7309 is 1 more +than a multiple of 9. Therefore, 7309 ÷ 9 gives a remainder of 1. +A number is divisible by 9 if and only if the sum of its digits is divisible +by 9. Also, we can add the digits of a number repeatedly till a single +digit is obtained. This single digit is the remainder when the number is +divided by 9. +Look at each of the following statements. Which are correct and why? +(i) +If a number is divisible by 9, then the sum of its digits is +divisible by 9. +9 +1 +99 +1 +999 +1 +9999 +1 +7 × 1000 +3 × 100 +9 × 1 +999 +1 +999 +1 +999 +1 +999 +1 +999 +1 +999 +1 +999 +1 +99 +1 +99 +1 +99 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +7 × 999 +3 × 99 +9 +3 +7 +999 +999 +999 +999 +999 +999 +999 +99 +99 +99 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +This is a +multiple of 9 +So, we need to just +consider this part +(i.e., 7 + 3 + 9 = 19) +This is a multiple of 9 +So, we need to just consider this part + +Ganita Prakash | Grade 8 +126 +(ii) If the sum of the digits of a number is divisible by 9, then the +number is divisible by 9. +(iii) If a number is not divisible by 9, then the sum of its digits is +not divisible by 9. +(iv) If the sum of the digits of a number is not divisible by 9, then +the number is not divisible by 9. +Figure it Out +1. Find, without dividing, whether the following numbers are divisible +by 9. +(i) 123    (ii) 405    (iii) 8888    (iv) 93547    (v) 358095 +2. Find the smallest multiple of 9 with no odd digits. +3. Find the multiple of 9 that is closest to the number 6000. +4. How many multiples of 9 are there between the numbers 4300 and +4400? +A Shortcut for Divisibility by 3 +We know that all the multiples of 9 are also multiples of 3. That is, if a +number is divisible by 9, it will also be divisible by 3. However, there are +other multiples of 3 that are not multiples of 9 for example — 15, 33, +and 87. +The shortcut to find the divisibility by 3 is similar to the method for 9. +A number is divisible by 3 if the sum of its digits is divisible by 3. Explore +the remainders when powers of 10 are divided by 3. Explain why this +method works. +A Shortcut for Divisibility by 11 +Interestingly, the shortcut for 11 is also based on checking the +remainders with place value. Let us see how. +Learning maths is not just about knowing some shortcuts and +following procedures but about understanding ‘why’ something +works. + +Number Play +127 +Units place +(1) +11 × 0 = 0 +1 = 11 × 0 + 1 +1 is one more +than a multiple +of 11. +Tens place +(10) +11 × 1 = 11 +10 = 11×1 – 1 +10 is one less +than a multiple +of 11. +11 +Hundreds +place (100) +11 × 9 = 99 +100 = 11×9 + 1 +100 is one more +than a multiple +of 11. +11 +... +9 +Thousands +place +(1000) +11 × 91 = 1001 +1000 = 11×91 +– 1 +1000 is one less +than a multiple +of 11. +11 +...... +91 +. +. +. +. +. +. +. +. +. +. +. +. +. +. +. +. +This alternating pattern of one more than 11 and one less than 11 +continues for higher place values. +Since 400 contains 4 hundreds, 400 is 4 more than a multiple of +11 (396 + 4). Since 60 contains 6 tens, 60 is 6 less than a multiple of +11 (66 – 6). Since 2 contains 2 units, 2 is 2 more than a multiple of 11, +i.e., 2 = (0 + 2). +Using these observations, can you tell whether the number 462 is +divisible by 11? +What could be a general method or shortcut to check divisibility by +11? +Math +Talk +Math +Talk + +Ganita Prakash | Grade 8 +128 +We saw that the place values alternate as 1 more and 1 less than a +multiple of 11. Using this observation, +Steps +Purpose +Example for the Number 320185 +1. Add the digits of place +values which are 1 +more (than a multiple +of 11), i.e., place values +corresponding to 1, 100, +10000, and so on. +To know +how much in +excess we are +with respect +to a multiple +of 11 for +these place +values. +11 +... +9 +320185 +11 +...... +909 +11 +...... +909 +2 × 10,000 +1 × 100 +5 × 1 +Total excess, 2 + 1 + 5 = 8. +2. Add the digits of place +values which are 1 +less (than a multiple of +11), i.e., place values +corresponding to 10, +1000, 100000, and so on. +To know how +short we are +with respect +to a multiple +of 11 for +these place +values. +320185 +11 +........ +9091 +3 × 10,000 +0 × 100 +8 × 10 +11 +........ +9091 +11 +........ +9091 +11 +11 +11 +11 +11 +11 +11 +11 +Total short, 3 + 0 + 8 = 11. +3. Compute the difference +between these two +sums, i.e., (number +in excess) – (number +short). +To know the +remainder +obtained +when divided +by 11. +8 – 11 = – 3. +(3 short of a multiple of 11) + +The difference between these two sums 8 – 11 = – 3, indicating that the +number 3,28,105 is 3 short of or 8 more than a multiple of 11. +If this difference is 11 or a multiple of 11, what does that say about the +remainder obtained when the number is divisible by 11? +Using this shortcut, find out whether the following numbers are divisible +by 11. Further, find the remainder if the number is not divisible by 11. +(i)  158    (ii)  841    (iii)  481    (iv)  5529    (v)  90904    (vi)  857076 + +Number Play +129 +Look at the following procedure — +Is this method similar to or different from the method we saw just +before? +Fill in the following table. Find a quick way to do this? +Number +Divisible by +2 +3 +4 +5 +6 +8 +9 +10 +11 +128 +Yes +No +No +No +No +Yes +No +No +No +990 +1586 +275 +6686 +639210 +429714 +2856 +3060 +406839 +More on Divisibility Shortcuts +Divisibility Shortcuts for Other Numbers +How can we find out if a number is divisible by 6? +Will checking its divisibility by its factors 2 and 3 work? Use the shortcuts +for 2 and 3 on these numbers and divide each number by 6 to verify — +38, 225, 186, 64. +Steps to follow +Example for the number 328105 +1. Place alternating ‘+’ and +‘–’ signs before every digit +starting from the unit’s digit. +–3 + 2 – 8 + 1 – 0 + 5 +2. Evaluate the expression. +–3 + 2 – 8 + 1 – 0 + 5 = – 3 +3. The result denotes the +remainder obtained when the +number is divided by 11. +328105 is 3 less than or 8 more +than a multiple of 11 +Math +Talk + +Ganita Prakash | Grade 8 +130 +How about checking divisibility by 24? Will checking the divisibility by +its factors, 4 and 6, work? Why or why not? +Determining divisibility by 24 by checking divisibility by 4 and by 6 +does not work. For example, the number 12 is divisible by both 4 and 6, +but not by 24. +To check for the divisibility by 24, we can instead check for the +divisibility by 3 and divisibility by 8. +Explain using prime factorisation why checking divisibility by 3 and +8 works for checking divisibility by 24, but checking divisibility by 4 and +6 is not sufficient for checking divisibility by 24. +There are such shortcuts to check divisibility by every number until +100, and for some numbers beyond 100. You may try to understand how +these work after learning certain concepts in higher grades. +Digital Roots +Take a number. Add its digits repeatedly till you get a single-digit +number. This single-digit number is called the digital root of the number. +For example, the digital root of the number 489710 will be +2 (4 + 8 + 9 + 7 + 1 + 0 = 29, 2 + 9 = 11, 1 + 1 = 2). +What property do you think this digital root will have? Recall that we +did this while finding the divisibility shortcut for 9. +Between the numbers 600 and 700, which numbers have the digital +root: (i) 5, (ii) 7, (iii) 3? +Write the digital roots of any 12 consecutive numbers. What do you +observe? +We saw that the digital root of multiples of 9 is always 9. +Now, find the digital roots of some consecutive multiples of (i) 3, (ii) 4, +and (iii) 6. +What are the digital roots of numbers that are 1 more than a multiple of +6? What do you notice? +Try to explain the patterns noticed. +I’m made of digits, each tiniest and odd, +No shared ground with root #1 — how odd! +My digits count, their sum, my root — +All point to one bold number’s pursuit — +The largest odd single-digit I proudly claim. +What’s my number? What’s my name? +Math +Talk + +Number Play +131 +Aryabhata II’s (c. 950 CE) work Mahāsiddhānta, mentions the method +of computing the digital root of a number by repeatedly adding the digits +till a single-digit number is obtained. This method is known to have been +used to perform checks on calculations of arithmetic operations. +Figure it Out +1. The digital root of an 8-digit number is 5. What will be the digital +root of 10 more than that number? +2. Write any number. Generate a sequence of numbers by repeatedly +adding 11. What would be the digital roots of this sequence of +numbers? Share your observations. +3. What will be the digital root of the number 9a + 36b + 13? +4. Make conjectures by examining if there are any patterns or +relations between +(i) +the parity of a number and its digital root. +(ii) +the digital root of a number and the remainder obtained +when the number is divided by 3 or 9. +5.3 Digits in Disguise +Last year, we saw cryptarithms — puzzles where each letter stands for a +digit, each digit is represented by at most one letter, and the first digit of +a number is never 0. +Solve the cryptarithms given below. +(i) + A1 ++ 1B +B0 +    (ii) + AB + + 37 +6A +    (iii) + ON +  ON + + ON +PO +    (iv) + QR + QR + + QR +PRR +    +Let us now try solving some cryptarithms involving multiplication. +(v) PQ × 8 = RS. +Guna says, “Oh, this means a 2-digit number +multiplied by 8 should give another 2-digit number. +I know that 10 × 8 = 80. But the units digits of 10 and +80 are the same, which we don’t want. For the same +reason PQ cannot be 11 as P and Q correspond to +different digits. 12 × 8 = 96 fits all the conditions”. Can +PQ be 13? Think. +It is not possible because 13 × 8 = 104. For all 2-digit +numbers greater than 12, the product with 8 is a 3-digit number. +Math +Talk + +Ganita Prakash | Grade 8 +132 +(vi) Try this now: GH × H = 9K. +This means a 2-digit number multiplied by a 1-digit number gives +another 2-digit number in the 90s. Observe the letters corresponding to +the units digits in this cryptarithm. Pick the solution to this question +from the options given below: +11 × 9 = 99, 12 × 8 = 96, 46 × 2 = 92, 24 × 4 = 96, 47 × 2 = 94, 31 × 3 = 93, +16 × 6 = 96. +(vii) Here is one more: BYE × 6 = RAY. +Anshu says, “Since the product is a 3-digit number, B can’t +be 2 or more. If B = 2, i.e., 2 hundreds, the product will be +more than 1200. So, B = 1.” +What can you say about ‘Y’? What digits are possible/not +possible? +“Y cannot be 7 or more because, if Y = 7, then 170 × 6 = 1020; but we +want a 3-digit product. Also, Y will be even”, Anshu explains. +We can solve cryptarithms using patterns, properties, and reasoning +related to numbers and operations. +Solve the following: +  (i) UT × 3 = PUT +(ii) +AB × 5 = BC +(iii) L2N × 2 = 2NP +(iv) XY × 4 = ZX + (v) +PP × QQ = PRP + (vi) JK × 6 = KKK +Figure it Out + 1. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? +Explain why there are two answers to this problem. + 2. “I take a number that leaves a remainder of 8 when divided by 12. I +take another number which is 4 short of a multiple of 12. Their sum +will always be a multiple of 8”, claims Snehal. Examine his claim +and justify your conclusion. + 3. When is the sum of two multiples of 3, a multiple of 6 and when +is it not? Explain the different possible cases, and generalise the +pattern. + 4. Sreelatha says, “I have a number that is divisible by 9. If I reverse +its digits, it will still be divisible by 9”. +  (i) Examine if her conjecture is true for any multiple of 9. +(ii) Are any other digit shuffles possible such that the number +formed is still a multiple of 9? + 5. If 48a23b is a multiple of 18, list all possible pairs of values for a +and b. + +Number Play +133 + 6. If 3p7q8 is divisible by 44, list all possible pairs of values for p +and q. + 7. Find three consecutive numbers such that the first number is a +multiple of 2, the second number is a multiple of 3, and the third +number is a multiple of 4. + + Are there more such numbers? How often do they occur? + 8. Write five multiples of 36 between 45,000 and 47,000. +Share your approach with the class. + 9. The middle number in the sequence of 5 consecutive even numbers +is 5p. Express the other four numbers in sequence in terms of p. + 10. Write a 6-digit number that it is divisible by 15, such that when the +digits are reversed, it is divisible by 6. + 11. Deepak claims, “There are some multiples of 11 which, when +doubled, are still multiples of 11. But other multiples of 11 don’t +remain multiples of 11 when doubled”. Examine if his conjecture is +true; explain your conclusion. + 12. Determine whether the statements below are ‘Always True’, +‘Sometimes True’, or ‘Never True’. Explain your reasoning. + (i) The product of a multiple of 6 and a multiple of 3 is a multiple +of 9. + (ii) The sum of three consecutive even numbers will be divisible +by 6. + (iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6. + (iv) 8 (7b – 3) – 4 (11b + 1) is a multiple of 12. + 13. Choose any 3 numbers. When is their sum divisible by 3? Explore +all possible cases and generalise. + 14. Is the product of two consecutive integers always multiple of 2? +Why? What about the product of these consecutive integers? Is it +always a multiple of 6? Why or why not? What can you say about +the product of 4 consecutive integers? What about the product of +five consecutive integers? + 15. Solve the cryptarithms — +(i) EF × E = GGG +(ii) WOW × 5 = MEOW + 16. Which of the following Venn diagrams captures the relationship +between the multiples of 4, 8, and 32? +Try +This +Math +Talk + +Ganita Prakash | Grade 8 +134 +(i) +(iii) +(ii) +(iv) +Multiples +of 4 +Multiples +of 32 +Multiples +of 8 +Multiples +of 4 +Multiples +of 32 +Multiples +of 8 +Multiples +of 32 +Multiples +of 4 +Multiples +of 8 +Multiples +of 4 +Multiples +of 32 + +„ We explored and learnt various properties of divisibility— +• If a is divisible by b, then all multiples of a are divisible by b. +• If a is divisible by b, then a is divisible by all the factors of b. +• If a divides m and a divides n, then a divides m + n and m – n. +• If a is divisible by b and is also divisible by c, then a is divisible by +the LCM of b and c. + +„ We learnt shortcuts to check divisibility by 3, 9 and 11, and why they +work. + +„ Through all this we were exposed to the power of mathematical +thinking and reasoning, using algebra, visualisation, examples and +counterexamples. +SUMMARY +Multiples +of 8 + +135 +Navakankari +Navakankari, also known as Sālu Mane Āṭa, Chār-Pār, or Navkakri, is a +traditional Indian board game that is the same as ‛Nine Men’s Morris’ or +‛Mills in the West’. It is a strategy game for two players where the goal is +to form lines of three pawns to eliminate the opponent’s pawns or block +their movement. +Gameplay +1. Each player starts with 9 pawns. The players take turns in placing +their pawns on the marked intersections. An intersection can have +at most one pawn. +2. Once all the pawns are placed, the players take turns to move one of +their pawns to adjacent empty intersections to form lines of three. +The line can be horizontal or vertical. +3. Once a player makes a line with their pawns they can remove any +one of the opponent’s pawns as long as it is not a part of one of their +lines. +A player wins if the opponent has less than 3 pawns or is unable to make +a move." +class_8,6,"we distribute, yet things multiply",ncert_books/class_8/hegp1dd/hegp106.pdf,"6 +WE DISTRIBUTE, +YET THINGS +MULTIPLY +We have seen how algebra makes use of letter symbols to write general +statements about patterns and relations in a compact manner. Algebra +can also be used to justify or prove claims and conjectures (like the many +properties you saw in the previous chapter) and to solve problems of +various kinds. +Distributivity is a property relating multiplication and addition that +is captured concisely using algebra. In this chapter, we explore different +types of multiplication patterns and show how they can be described in +the language of algebra by making use of distributivity. +6.1 Some Properties of Multiplication +Increments in Products +Consider the multiplication of two numbers, say, 23 × 27. +1. By how much does the product increase if the first number (23) is +increased by 1? +2. What if the second number (27) is increased by 1? +3. How about when both numbers are increased by 1? +Do you see a pattern that could help generalise our observations to the +product of any two numbers? +Let us first consider a simpler problem — find the increase in the +product when 27 is increased by 1. From the definition of multiplication +(and the commutative property), it is clear that the product increases by +23. This can be seen from the distributive property of multiplication as +well. If a, b and c are three numbers, then — + +We Distribute, Yet Things Multiply +137 +This is called the distributive property of multiplication over +addition. Using the identity a (b + c) = ab + ac with a = 23, b = 27, and +c = 1, we have +Remember that here, a (b + c) and 23 (27 + 1) mean a × (b + c), and +23 × (27 + 1), respectively. We usually skip writing the ‘×’ symbol before +or after brackets, just as in the case of expressions like 5a, xy, etc. +We can also similarly expand (a + b) c using the distributive property +as follows — + + + +(a + b) c = c (a + b) (commutativity of multiplication) + + + + +     = ca + cb (distributivity) + + + + +     = ac + bc (commutativity of multiplication) +We can use the distributive property to find, in general, how much a +product increases if one or both the numbers in the product are increased +by 1. Suppose the initial two numbers are a and b. If one of the numbers, +say b, is increased by 1, then we have — +Now let us see what happens if both numbers in a product are +increased by 1. If in a product ab, both a and b are increased by 1, then +we obtain (a + 1) (b + 1). +a (b + c) = ab + ac +a (b + c) + . . . . . + . . . . . + . . . . . +. . . . . +. . . . . +. . . . . +. . . . . + . . . . . + . . . . . + . . . . . . + . . . . . . + . . . . . . + . . . + . . . + . . . + . . . + . . . . . . + . . . . . . +ab +ac +b columns +c columns +a rows +This property can be visualised nicely using a diagram: +Increase +23 ( 27 + 1) = 23 × 27 + 23 +Increase +a ( b + 1) = ab × a + +Ganita Prakash | Grade 8 +138 +How do we expand this? +Let us consider (a + 1) as a single term. Then, by the distributive +property, we have +Thus, the product ab increases by a + b + 1 when each of a and b are +increased by 1. +What would we get if we had expanded (a + 1) (b + 1) by first taking (b + 1) +as a single term? Try it? +What happens when one of the numbers in a product is increased by 1 +and the other is decreased by 1? Will there be any change in the product? +Let us again take the product ab of two numbers a and b. If a is +increased by 1 and b is decreased by 1, then their product will be (a + 1) +(b – 1). Expanding this, we get +Will the product always increase? Find 3 examples where the product +decreases. +What happens when a and b are negative integers? +Check by substituting different values for a and b in each of the above +cases. For example, a = –5, b = 8; a = –4, b = –5; etc. +We have seen that integers also satisfy the distributive property, that +is, if x, y and z are any three integers, then x (y + z) = xy + xz. +Thus, the expressions we have for increase of products hold when the +letter-numbers take on negative integer values as well. +Recall that two algebraic expressions are equal if they take on the +same values when their letter-numbers are replaced by numbers. These +(a + 1) (b + 1) = (a + 1) b + (a + 1) 1 + + + += ab + (b + a + 1) +Increase +(a + 1) (b + 1) = (a + 1) b + (a + 1) 1 + Again applying the distributive +property, we obtain +(23 + 1) (27 + 1) = (23 + 1) 27 + (23 + 1) 1 + + + +  = 23 × 27 + (27 + 23 + 1) +Increase +If a = 23, and, b = 27, we get +(a + 1) (b – 1) = (a + 1) b – (a + 1) 1 + + + += ab + b – (a + 1) + + + += ab + b – a – 1 +Increase +(23 + 1) (27 – 1) = (23 + 1) 27 – (23 + 1) 1 + + + + = 23 × 27 + 27 – (23 + 1) + + + + = 23 × 27 + 27 – 23 – 1 +Increase +If a = 23, and b = 27, we get + +We Distribute, Yet Things Multiply +139 +numbers could be any integers. Mathematical statements that express +the equality of two algebraic expressions, such as +a (b + 8) = ab + 8a, +(a + 1) (b – 1) = ab + b – a – 1, etc., +are called identities. +By how much will the product of two numbers change if one of the +numbers is increased by m and the other by n? +If a and b are the initial numbers being multiplied, they become +a + m and b + n. +(a + m) (b + n) = (a + m)b + (a + m)n + = ab + mb + an + mn +The increase is an + bm + mn. +Notice that the product is the sum of the product of each term of +(a + m) with each term of (b + n). +This identity can be visualised as follows — +(a + m) (b + n) = ab + mb + an + mn +Identity 1 +b columns +n columns +a rows +m rows +(a + m) (b + n) + . . . . . + . . . . . + . . . . . +. . . . . +. . . . . +. . . . . +. . . . . + . . . . . + . . . . . +ab + . . . . . . + . . . . . . + . . . . . . + . . . + . . . + . . . + . . . + . . . . . . + . . . . . . +an + . . . . . . . + . . . . . . . +. . . . +. . . . +. . . . + . . . . . . +mb + . . . . . . . + . . . . . . . + . . . + . . . + . . . + . . . . . . . +mn + +Ganita Prakash | Grade 8 +140 +This identity can be used to find how products change when the numbers +being multiplied are increased or decreased by any amount. Can you see +how this identity can be used when one or both numbers are decreased? +For example, let us reconsider the case when one number is increased +by 1 and the other decreased by 1. Let us write the product (a + 1) (b – 1) +as (a + 1) (b + (–1). Taking m = 1 and n = –1 in Identity 1, we have +ab + (1) × b + a × (–1) + (1) × (–1) = ab + b – a – 1, +which is the same expression that we obtain earlier. +Use Identity 1 to find how the product changes when +(i) one number is decreased by 2 and the other increased by 3; +(ii) both numbers are decreased, one by 3 and the other by 4. +Verify the answers by finding the products without converting the +subtractions to additions. +Generalising this, we can find the product (a + u) (b – v) as follows. +(a + u) (b – v) = (a + u) b – (a + u) v + = ab + ub – (av + uv) + = ab + ub – av – uv. +Check that this is the same as taking m = u and n = –v in Identity 1. +As in Identity 1, the product (a + u) (b – v) is the sum of the product of +each term of a + u (a and u) with each term of b – v (b and (–v)). Notice +that the signs of the terms in the products can be determined using the +usual rules of integer multiplication. +Expand (i) (a – u) (b + v), (ii) (a – u) (b – v). +We get + (a – u) (b + v) = ab – ub + av – uv, and +(a – u) (b – v) = ab – ub – av + uv. +The distributive property is not restricted to two terms within a bracket. +Example 1: Expand 3a +2 (a – b + 1 +5). +3a +2 (a – b + 1 +5) = (3a +2 × a) – (3a +2 × b) + (3a +2 × 1 +5 ). +The terms can be simplified as follows — +3a +2 × a = 3 +2 × (a × a). +See how the rules of integer multiplication allows us to handle +multiple cases using a single identity! + +We Distribute, Yet Things Multiply +141 +Using exponent notation, we can write 3 +2 × (a × a) = 3 +2 a2. +3a +2 × b = 3 +2 × (a × b) = 3 +2 ab. +3a +2 × 1 +5 = (3 +2 × 1 +5) a = 3 +10 a +So we get +3a +2 (a – b + 1 +5) = 3 +2a2 – 3 +2 ab + 3 +10 a. +Can any two terms be added to get a single term? +For example, can 3 +2 a2 and 3 +10 a be added to get a single term? +We see that no two terms have exactly the same letter-numbers, +which would have allowed them to be simplified into a single term. So, +a further simplification of the expression is not possible. +Recall that we call terms having the same letter-numbers like terms. +Example 2: Expand (a + b) (a + b). +We have (a + b) (a + b) = (a+b) a + (a + b)b = a × a + b × a + ab + b × b + + + + + + + = a2 + ba + ab + b2 +Since ba = ab, we have two terms having the same letter-numbers ab +(or, that are like terms), and so can be added — +ba + ab = ab + ab = 2ab +So we get +(a + b) (a + b) = a2 + 2ab + b2. +Example 3: Expand (a + b) (a2 + 2ab + b2). +(a + b) (a2 + 2ab + b2) = (a + b)a2 + (a + b) × 2ab + (a + b)b2 + + + + + = (a × a2) + ba2 + (a × 2ab) + (b × 2ab) + ab2 + (b × b2) +The terms can be simplified as follows — +a × a2 = a3 (why?) +ba2 = a2b +a × 2ab = 2 × a × a × b = 2a2b + b × 2ab = 2 × a × b × b = 2ab2 +b × b2 = b3 +So, (a + b)(a2 + 2ab + b2) = a3 + a2b + 2a2b + 2ab2 + ab2 + b3. +We see that a2b and 2a2b have the same letter-numbers (or, are like +terms) and so can be added — + +Ganita Prakash | Grade 8 +142 +a2b + 2a2b = (1 + 2) a2 b = 3a2 b. +Similarly, ab2 and 2ab2 are like terms and so can be added — +ab2 + 2ab2 = (1 + 2)ab2 = 3ab2. +Thus, we have +(a + b) × (a2 + 2ab +b2 ) = a3 + 3a2b + 3ab2 + b3. +A Pinch of History +The distributive property of multiplication over addition was implicit +in the calculations of mathematicians in many ancient civilisations, +particularly in ancient Egypt, Mesopotamia, Greece, China, and India. For +example, the mathematicians Euclid (in geometric form) and Āryabhaṭa +(in algebraic form) used the distributive law in an implicit manner +extensively in their mathematical and scientific works. The first explicit +statement of the distributive property was given by Brahmagupta in his +work Brahmasphuṭasiddhānta (Verse 12.55), who referred to the use of +the property for multiplication as khaṇḍa-guṇanam (multiplication by +parts). His verse states, “The multiplier is broken up into two or more +parts whose sum is equal to it; the multiplicand is then multiplied by +each of these and the results added”. That is, if there are two parts, then +using letter symbols this is equivalent to the identity (a + b) c = ac + bc. In +the next verse (Verse 12.56), Brahmagupta further describes a method +for doing fast multiplication using this distributive property, which we +explore further in the next section. +Figure it Out +1. Observe the multiplication grid below. Each number inside the grid +is formed by multiplying two numbers. If the middle number of a +3 × 3 frame is given by the expression pq, as shown in the figure, +write the expressions for the other numbers in the grid. +x +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +1 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +2 +2 +4 +6 +8 +10 +12 +14 +16 +18 +20 +3 +3 +6 +9 +12 +15 +18 +21 +24 +27 +30 +4 +4 +8 +12 +16 +20 +24 +28 +32 +36 +40 +5 +5 +10 +15 +20 +25 +30 +35 +40 +45 +50 +6 +6 +12 +18 +24 +30 +36 +42 +48 +54 +60 +7 +7 +14 +21 +28 +35 +42 +49 +56 +63 +70 +8 +8 +16 +24 +32 +40 +48 +56 +64 +72 +80 +9 +9 +18 +27 +36 +45 +54 +63 +72 +81 +90 +10 +10 +20 +30 +40 +50 +60 +70 +80 +90 +100 +3 × 5 +3 × 6 +3 × 7 +4 × 5 +4 × 6 +4 × 7 +5 × 5 +5 × 6 +5 × 7 +pq + +We Distribute, Yet Things Multiply +143 +2. Expand the following products. + +(i) (3 + u) (v – 3) +(ii) +2 +3 (15 + 6a) + +(iii) (10a + b) (10c + d) +(iv) (3 – x) (x – 6) + +(v) (–5a + b) (c + d) +(vi) (5 + z) (y + 9) +3. Find 3 examples where the product of two numbers remains +unchanged when one of them is increased by 2 and the other is +decreased by 4. +4. Expand (i) (a + ab – 3b2) (4 + b), and (ii) (4y + 7) (y + 11z – 3). +5. Expand (i) (a – b) (a + b), (ii) (a – b) (a2 + ab + b2) and +(iii) (a – b)(a3 + a2b + ab2 + b3), Do you see a pattern? What would be +the next identity in the pattern that you see? Can you check it by +expanding? +Fast Multiplications Using the Distributive Property +The distributive property can be used to come up with quick methods of +multiplication when certain types of numbers are multiplied. +When one of the numbers is 11, 101, 1001, ... +Use the following multiplications to find the product of a number with +11 in a single step. +(a) 3874 × 11 + + + +(b) 5678 × 11 +Let us take the first multiplication. + 3874 × 11 = 3874 (10 + 1) = 38740 + 3874 +Notice how the digits are getting added. + Let us take a 4-digit number dcba, that is, the number that has d in the +thousands place, c in the hundreds place, b in the tens place and a in the +units place. +dcba × (10 + 1) = dcba × 10 + dcba. +This becomes + + d  c    b    a    o ++    d      c      b      a + d (c + d)  (b + c)  (a + b)  a +Math +Talk + 38740 ++ 3874 + +Ganita Prakash | Grade 8 +144 +This can be used to obtain the product in one line. +3874 × 11 + 614 +3874 × 11 + 2614 +3874 × 11 + 42614 ++ ++ +Step 4 +Step 5 +3874 × 11 + 4 +Step 1 +Step 3 +1 +1 +1 +1 +1 +1 +1 +1 +1 +3874 × 11 + 14 ++ +Step 2 +Describe a general rule to multiply a number (of any number of digits) +by 11 and write the product in one line. +Evaluate (i) 94 × 11,  (ii) 495 × 11,  (iii) 3279 × 11,  (iv) 4791256 × 11. +Can we come up with a similar rule for multiplying a number by +101? +Multiply 3874 by 101. +Let us take a 4-digit number dcba. +dcba × 101 = dcba × (100 + 1) = dcba × 100 + dcba. +This becomes + d  c    b    a    o    o ++           d      c      b      a + d  c     (b + d)  (a + c)  b      a +Use this to multiply 3874 × 101 in one line. +What could be a general rule to multiply a number by 101 and write the +product in one line? Extend this rule for multiplication by 1001, 10001, … +Use this to find (i) 89 × 101, (ii) 949 × 101, (iii) 265831 × 1001, +(iv) 1111 × 1001, (v) 9734 × 99 and (vi) 23478 × 999. +Such methods of applying the distributive property to easily multiply +two +numbers +were +discussed +extensively +in +the +ancient +mathematical works of Brahmagupta (628 CE), Sridharacharya +(750 CE) and Bhaskaracharya (Lilavati, 1150 CE). In his work +Brahmasphuṭasiddhānta (Verse 12.56), Brahmagupta refers to such +methods for fast multiplication using the distributive property as +ista-gunana. +Math +Talk +Math +Talk + +We Distribute, Yet Things Multiply +145 +6.2 Special Cases of the Distributive Property +Square of the Sum/Difference of Two Numbers +The area of a square of sidelength 60 units is 3600 sq. units (602) and that +of a square of sidelength 5 units is 25 sq. units (52). Can we use this to +find the area of a square of sidelength 65 units? +A square of sidelength 65 can be split into 4 regions as shown in the +figure — a square of sidelength 60, a square of sidelength 5, and two +rectangles of sidelengths 60 and 5. The area of the square of sidelength 65 +is the sum of the areas of all its constituent parts. Can you find the areas +of the four parts in the figure above? +We get +652 = (60 + 5)2 = 602 + 52 + 2 × (60 × 5). + + + + = 3600 + 25 + 600 = 4225 sq. units. +Let us multiply (60 + 5) × (60 + 5) using the +distributive property. +(60 + 5) × (60 + 5) = 60 × 60 + 5 × 60 + 60 × 5 + 5 × 5 + +        = 602 + 2 × (60 × 5) + 52. +What if we write 652 as (30 + 35)2 or (52 + 13)2? +Draw the figures and check the area that you get. +Let us look at the general expression for the +square of sum of two numbers, (a + b)2. +If a and b are any two integers, is (a + b)2 always greater than +a2 + b2? If not, when is it greater? +Use Identity 1A to find the values of 1042, 372. (Hint: Decompose 104 and 37 +into sums or differences of numbers whose squares are easy to compute.) +60 +60 +5 +5 +602 +52 +5 × 60 +60 × 5 +a +a +b +b +a2 +b2 +b × a +a × b + +Using the distributive property, +(a+b)2 can be expanded as +(a+b) × (a+b) = a×a + a×b + b×a + b×b + = a2 + 2ab + b2, +as we had already seen in Example 2. + Identity 1A +(a+b)2 = a2 + 2ab + b2 +Math +Talk + +Ganita Prakash | Grade 8 +146 +Use Identity 1A to write the expressions for the following. +(i) (m + 3)2 +(ii) (6 + p)2 +Expand (6x + 5)2. +Expand (3j + 2k)2 using both the identity and by applying the distributive +property. +Can we use 602 (=3600) and 52 (=25) to find the value of (60 – 5)2 or 552? +Let us approach this through geometry by drawing a square of side +length 55 sitting inside a square of sidelength 60. +Area of a square of sidelength 55 is (60 – 5)2 = 552. +Using the Distributive Property +Using the Identity +(6x + 5)2 = (6x + 5) (6x + 5) + + += (6x × 6x) + (5 × 6x) +(6x × 5) + 5 × 5 + + += (6x)2 + 2 (6x × 5) + 52 + + += 36x2 + 60x + 25. +(6x + 5)2 = (6x)2 + 52 + 2 × (6x × 5) + + += 36x2 + 25 + 60x. +If you have difficulty remembering or using the general rule, you +can just apply the distributive property to multiply and get the +desired result. +55 +60 +55 +5 +60 × 60 +– 5 × 60 +– 60 × 5 ++5 × 5 + +We can get the area of +the square of sidelength +55 by taking the area of +the square of sidelength +60 and removing the areas +of the two rectangles of +sidelengths 60 and 5, i.e., +602 – (60 × 5) – (5 × 60). +By doing this, we remove the +area of the small square of +sidelength 5 twice. What can +we do with this expression to +get the actual area? + +We can add back the area +of the square of sidelength 5 +to this expression. That way, +we are only subtracting this +area once. + +We Distribute, Yet Things Multiply +147 +So, +(60 – 5)2 = 602 – (60 × 5) – (5 × 60) + 52 += 3600 – 300 – 300 + 25 += 3025. +The area of the square of sidelength 55 is 3025 sq. units. +We have seen what (a + b)2 gives when expanded. What is the +expansion of (a – b)2? +Using the distributive property, +(a – b)2 = (a – b) × (a – b) += (a)2 – ba – ab + (b) +2 += a2 – 2ab + b2. +We can also use the expansion of (a + b)2 to find the expansion of +(a – b)2. Think how. +Hint: (a – b)2 = (a + (–b))2. +We can now directly use the expansion of (a + b)2. +(a + (–b)2 = (a)2 + (–b)2 + 2 × (a) × (–b) +Find the general expansion of (a – b)2 using geometry, as we did for 552. +Use the identity (a – b)2 to find the values of (a) 992 and (b) 582. +Expand the following using both Identity 1B and by applying the +distributive property +(i) (b – 6)2 +(ii) (–2a + 3)2 +(iii) (7y – 3 +4z) +2 +Investigating Patterns +Pattern 1 +Look at the following pattern. +2 (22 + 12) = 32 + 12 +2 (32 + 12) = 42 + 22 + 2 (62 + 52) = 112 + 12 + 2 (52 + 32) = 82 + 22. +Identity 1B (a – b)2 = a2 + b2 – 2ab + +Ganita Prakash | Grade 8 +148 +Take a pair of natural numbers. Calculate the sum of their squares. +Can you write twice this sum as a sum of two squares? +Try this with other pairs of numbers. Have you figured out a pattern? +Notice that 2 (52 + 62) = (6 + 5)2 + (6 – 5)2. +Do the identities below help in explaining the observed pattern? +(a + b)2 = a2 + 2ab + b2 +(a – b)2 = a2 – 2ab + b2 +(a + b)2 + (a – b)2 = (a2 + 2ab + b2) + (a2 – 2ab + b2) +Adding the like terms a2 + a2 = 2a2 , b2 + b2 = 2b2 and 2ab – 2ab = 0, +we get +2 (a2 + b2) = (a + b)2 + (a – b)2. +Pattern 2 +Here is a related pattern. Try to describe the pattern using algebra to +determine if the pattern always holds. +9 × 9 – 1 × 1= 10 × 8 +8 × 8 – 6 × 6 = 14 × 2 +7 × 7 – 2 × 2 = 9 × 5 +10 × 10 – 4 × 4 = 14 × 6 +The pattern here appears to be a2 – b2 = (a + b) × (a – b). +Is this a true identity? Using the distributive property, we get +(a + b) × (a – b) = a2 – ab + ba – a2. +Adding the like terms, ab + (–ab) = 0, we see that indeed +You had seen this identity earlier in Figure it Out 5 (i). +Use Identity 1C to calculate 98 × 102, and 45 × 55. +Show that (a + b) × (a – b) = a2 – b2 geometrically. +Identity 1C (a + b) × (a – b) = a2 – b2. +Try +This + +We Distribute, Yet Things Multiply +149 +Hint: +Sridharacharya (750 CE) gave an interesting method to quickly +compute the squares of numbers using Identity 1C! Consider the +following modified form of this identity — +a2 = (a + b) (a – b) + b2 +Why is this identity true? +Now, for example, 312 can be found by taking a = 31 and b = 1. + 312 = (31 + 1) (31 – 1) + 12 + = 32 × 30 + 1 += 961. + 1972 can be found by taking a = 197, and b = 3. + 1972 = (197 + 3) (197 – 3) + 32 + = 200 × 194 + 9 + = 38809. +Figure it Out +1. Which is greater: (a – b)2 or (b – a)2? Justify your answer. +2. Express 100 as the difference of two squares. +3. Find 4062, 722, 1452, 10972, and 1242 using the identities you have +learnt so far. +4. Do Patterns 1 and 2 hold only for counting numbers? Do they +hold for negative integers as well? What about fractions? Justify +your answer. +a +b +What do we get +when this piece +is moved? +a +b +Math +Talk + +Ganita Prakash | Grade 8 +150 +6.3 Mind the Mistake, Mend the Mistake +We have expanded and simplified some algebraic expressions below to +their simplest forms. +(i) Check each of the simplifications and see if there is a mistake. +(ii) If there is a mistake, try to explain what could have gone +wrong. +(iii) Then write the correct expression. +6.4 This Way or That Way, All Ways Lead to the Bay +Observe the pattern in the figure below. Draw the next figure in the +sequence. How many circles does it have? How many total circles +are there in Step 10? Write an expression for the number of circles +in Step k. +There are many ways of interpreting this pattern. Here are some +possibilities: + –3p (–5p + 2q) += –3p + 5p – 2q += p – 2q +1 + 2(x – 1) + 3 (x + 4) += 2x – 1 + 3x + 4 += 5x + 3 +2 + y + 2 (y + 2) += (y + 2)2 += y2 + 4y + 4 +3 + (5m + 6n)2 += 25m2 + 36n2 +4 + (– q + 2)2 += q2 – 4q + 4 +5 + 3a (2b × 3c) += 6ab × 9ac += 54a2bc +6 + 1 +2 (10s – 6) + 3 += 5s – 3 + 3 += 5s +7 + 5w2 + 6w += 11w2 +8 + 2a3 + 3a3 + 6a2b + + 6ab2 += 5a3 + 12a2b2 +9 + (a + 2) (b + 4) += ab + 8 + ab2 + a2b + a2b2 += ab (a + b + ab) +  (x + 2)(x + 5) += (x + 2)x + (x + 2)5 += x2 + 2x + 5x +10 += x2 + 7x + 10 +10 +11 +12 +Math +Talk +1 2 3 +. . . . . . + +We Distribute, Yet Things Multiply +151 +Step 1 Step 2 Step 3 Step 4 . . . Step k + 22 – 1 += (1 + 1)2 – 1 + 32 – 1 += (2 + 1)2 – 1 + 42 – 1 += (3 + 1)2 – 1 + 52 – 1 += (4 + 1)2 – 1 +(k + 1)2 – 1 +. . . +Method 1 +. . . +Step 1 Step 2 Step 3 Step 4 . . . Step k + 1 + 2 × 1 += 12 + 2 × 1 + 22 + 2 × 2 += 22 + 2 × 2 +. . . +Method 2 + 32 + 2 × 3 += 32 + 2 × 3 + 42 + 2 × 4 += 42 + 2 × 4 +k2 + 2 × k +. . . + 1 × 2 + 1 += 1 × (1 + 1) + 1 +Method 3 + 2 × 3 + 2 += 2 × (2 + 1) + 2 + 3 × 4 + 3 += 3 × (3 + 1) + 3 + 4 × 5 + 4 += 4 × (4 + 1) + 4 +k × (k + 1) + k +Step 1 Step 2 Step 3 Step 4 . . . Step k +. . . +. . . + +Ganita Prakash | Grade 8 +152 + 1 × 3 += 1 × (1 + 2) +. . . +Method 4 + 2 × 4 += 2 × (2 + 2) + 3 × 5 += 3 × (3 + 2) + 4 × 6 += 4 × (4 + 2) +k × (k+2) +. . . +Step 1 Step 2 Step 3 Step 4 . . . Step k +Does your method match any of these, or is it different? Each +expression that we have identified appears different, but are they really +different? Since they describe the same pattern, they should all be the +same. Let us simplify each expression and find out. +When carried out correctly, all methods lead to the same answer; +k2 + 2k. The expression k2 + 2k gives the number of circles at Step k of this +pattern. +Use this formula to find the number of circles in Step 15. +Consider the pattern made of square tiles in the picture below. + (k + 1)2 – 1 += k2 + 1 + 2k – 1 += k2 + 2k +k2 + 2 × k += k2 + 2k +k × (k + 1) + k + = k2 + k + k + = k2 + 2k + k × (k + 2) += k2 + 2k +In Mathematics, there are often multiple ways of looking at +a pattern, and different ways of approaching and solving the +same problem. Finding such ways often requires a great deal +of creativity and imagination! While one or two of the ways +might be your favourite(s), it can be amusing and enriching to +explore other ways as well. +1 2 3 +. . . . . . + +We Distribute, Yet Things Multiply +153 +How many square tiles are there in each figure? +How many are there in Step 4 of the sequence? What about Step 10? +Write an algebraic expression for the number of tiles in Step n. +Share your methods with the class. Can you find more than one +method to arrive at the answer? +Find the area of the (interior) shaded region in the figure below. All four +rectangles have the same dimensions. +Tadang’s method: +The total region is a square of side (m + n) with an area +(m + n)2. +Subtracting the area of four rectangles from the total +area will give the area of the interior shaded region. +That is, (m + n)2 – 4mn . +Yusuf’s method: +The shaded region is a square with sidelength (n – m). So, +its area is (n – m)2. +By expanding both expressions, +check that (m + n)2– 4mn = (n – m)2. +Find out the area of the region with slanting lines in the +figure. All three rectangles have the same dimensions +(Fig. 1). +Anusha’s method: +Required area = Area (ABCD) – Area (EFGH) +Area of ABCD = x2. +Area of EFGH = xy. +Required area = x2 – xy. +Vaishnavi’s method: +QS = y + x + y + += x + 2y. +Area of PQSR = x (x + 2y) +Required area = Area of PQSR – (area of the three +rectangles) + = x (x + 2y) – 3xy. +Math +Talk +m +n +Fig. 1 +x +y + +Ganita Prakash | Grade 8 +154 +Aditya’s method: +The required area is 2 times the area of JKLM. +JK = +x – y +2 , KM = x +Area (JKML) = x ( +x – y +2 ) +Required area = 2 × Area of JKML + = 2x ( +x – y +2 ) + = x (x – y). +By expanding the expressions, verify that all three expressions are +equivalent. If x = 8 and y = 3, find the area of the shaded region. +Write an expression for the area of the dashed region in the figure +below. Use more than one method to arrive at the answer. Substitute +p = 6, r = 3.5, and s = 9, and calculate the area. +Figure it Out +1. Compute these products using the suggested identity. +(i) 462 using Identity 1A for (a + b)2 +(ii) 397 × 403 using Identity 1C for (a + b) (a – b) +(iii) 912 using Identity 1B for (a – b)2 +(iv) 43 × 45 using Identity 1C for (a + b) (a – b) +2. Use either a suitable identity or the distributive property to find each +of the following products. +(i) (p – 1) (p + 11) +(ii) (3a – 9b) (3a + 9b) +(iii) –(2y + 5) (3y + 4) +(iv) (6x + 5y)2 +(v) (2x – 1 +2)2 +(vi) (7p) × (3r) × (p + 2) +Math +Talk +r +p +r +s + +We Distribute, Yet Things Multiply +155 +3. For each statement identify the appropriate algebraic expression(s). +(i) Two more than a square number. +  2 + s      (s + 2)2      s2 + 2      s2 + 4      2s2      22s +(ii) The sum of the squares of two consecutive numbers +  m2 + n2 +  (m + n)2 + +m2 + 1 +m2 + (m + 1)2 +  m2 + (m – 1)2   (m + (m + 1))2 +(2m)2 + (2m + 1)2 +4. Consider any 2 by 2 square of numbers in a calendar, as shown in the +figure. +Find products of numbers lying along each diagonal — 4 × 12 = 48, +5 × 11 = 55. Do this for the other 2 by 2 squares. What do you +observe about the diagonal products? Explain why this happens. +Hint: Label the numbers in each 2 by 2 square as +a +(a + 1) +a + 7 +(a + 8) +5. Verify which of the following statements are true. +(i) (k + 1) (k + 2) – (k + 3) is always 2. +(ii) (2q + 1) (2q – 3) is a multiple of 4. +(iii) Squares of even numbers are multiples of 4, and squares of +odd numbers are 1 more than multiples of 8. +(iv) (6n + 2)2 – (4n + 3)2 is 5 less than a square number. +6. A number leaves a remainder of 3 when divided by 7, and another +number leaves a remainder of 5 when divided by 7. What is the +remainder when their sum, difference, and product are divided by 7? +7. Choose three consecutive numbers, square the middle one, and +subtract the product of the other two. Repeat the same with other +Math +Talk +Su +M +Tu +W +Th +F +Sa +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +13 +14 +15 +16 +17 +18 +19 +20 +21 +22 +23 +24 +25 +26 +27 +28 +February + +Ganita Prakash | Grade 8 +156 + + sets of numbers. What pattern do you notice? How do we write this +as an algebraic equation? Expand both sides of the equation to +check that it is a true identity. + 8. What is the algebraic expression describing the following +steps — add any two numbers. Multiply this by half of the sum of +the two numbers? Prove that this result will be half of the square of +the sum of the two numbers. + 9. Which is larger? Find out without fully computing the product. +(i) 14 × 26 or 16 × 24 +(ii) 25 × 75 or 26 × 74 + 10. A tiny park is coming up in Dhauli. +The plan is shown in the figure. The +two square plots, each of area g2 sq. +ft., will have a green cover. All the +remaining area is a walking path w +ft. wide that needs to be tiled. Write +an expression for the area that needs to be tiled. + 11. For each pattern shown below, +(i) Draw the next figure in the sequence. +(ii) How many basic units are there in Step 10? +(iii) Write an expression to describe the number of basic units in +Step y. +g2 sq. ft. +g2 sq. ft. +w ft. +w ft. +Step 1 +Step 2 +Step 3 +Step 1 +Step 2 +Step 3 + +We Distribute, Yet Things Multiply +157 +„ We extended the distributive property to find the product of +two expressions each of which has two terms. The general form +for the same is (a + b) × (c + d) = ac + ad + bc + bd. +„ We saw some special cases of this identity. + y (a + b)2 = a2 + 2ab + b2 + y (a – b)2 = a2 – 2ab + b2 + y (a + b) (a – b) = a2 – b2. +„ We considered different patterns, and explored how to +understand them using algebra. We saw that, often, there are +multiple ways to solve a problem and arrive at the same correct +answer. Finding different methods to approach and solve the +same problem is a creative process. +SUMMARY + +158 +Coin Conjoin +Arrange 10 coins in a triangle as shown in the figure below on the left. +The task is to turn the triangle upside down by moving one coin at a +time. How many moves are needed? What is the minimum number of +moves? +A triangle of 3 coins can be inverted (turned upside down) with a +single move, and a triangle of 6 coins can be inverted by moving 2 coins. +The 10-coin triangle can be flipped with just 3 moves; did you figure +out how? Find out the minimum possible moves needed to flip the +next bigger triangle having 15 coins. Try the same for bigger triangular +numbers. +Is there a simple way to calculate the minimum number of coin moves +needed for any such triangular arrangement?" +class_8,7,propositional reasoning - 1,ncert_books/class_8/hegp1dd/hegp107.pdf,"7.1 Observing Similarity in Change +We are all familiar with digital images. We often change the size and +orientation of these images to suit our needs. Observe the set of images +below — +We can see that all the images are of different sizes. +Which images look similar and which ones look different? +Images (A, C, and D) look similar, even though they have different sizes. +PROPORTIONAL +REASONING-1 +7 +Image A +Image D +Image E +Image B +Image C + +Ganita Prakash | Grade 8 +160 +Do images B and E look like the other three images? +No, they are slightly distorted. The tiger appears elongated in B, and +compressed and fatter in E! +Why? +You may notice that images A, C, and D are rectangular, but E is +square. Maybe that is why E looks different. But B is also a rectangle! +Why does it look different from the other rectangular images? +Can we observe any pattern to answer this question? Perhaps by +measuring the rectangles? +Image +Width (in mm) +Height (in mm) +Image A +60 +40 +Image B +40 +20 +Image C +30 +20 +Image D +90 +60 +Image E +60 +60 +What makes images A, C, and D appear similar, and B and E different? +When we compare image A with C, we notice that the width of C is +half that of A. The height is also half of A. Both the width and height +have changed by the same factor (through multiplication), 1 +2 in this +case. Since the widths and heights have changed by the same factor, the +images look similar. +When we compare image A with image B, we notice that the width of +B is 20 millimetre (mm) less than that of A. The height too is 20 mm less +than the height of A. Even though the difference (through subtraction) is +the same, the images look different. Have the width and height changed +by the same factor? The height of B is half the height of A. But the width of +B is not half the width of A. Since the width and height have not changed +by the same factor, the images look different. +Can you check by what factors the width and height of image D change +as compared to image A? Are the factors the same? +Images A, C, and D look similar because their widths and heights have +changed by the same factor. We say that the changes to their widths and +heights are proportional. +Math +Talk + +Proportional Reasoning-1 +161 +7.2 Ratios +We use the notion of a ratio to represent such proportional relationships +in mathematics. +We can say that the ratio of width to height of image A is +60 : 40. +The numbers 60 and 40 are called the terms of the ratio. +The ratio of width to height of image C is 30 : 20, and that of image D +is 90 : 60. +In a ratio of the form a : b, we can say that for every ‘a’ units of the +first quantity, there are ‘b’ units of the second quantity. +So, in image A, we can say that for every 60 mm of width, there are 40 +mm of height. +We can say that the ratios of width to height of images A, C, and D +are proportional because the terms of these ratios change by the same +factor. Let us see how. +Image A — 60 : 40 +Multiplying both the terms by 1 +2, we get +60 × 1 +2 : 40 × 1 +2 +which is 30 : 20, the ratio of width to height in image C. +By what factor should we multiply the ratio 60 : 40 (image A) to get 90 : 60 +(image D)? +A more systematic way to compare whether the ratios are proportional +is to reduce them to their simplest form and see if these simplest forms +are the same. +7.3 Ratios in their Simplest Form +We can reduce ratios to their simplest form by dividing the terms by +their HCF. +In image A, the terms are 60 and 40. What is the HCF of 60 and 40? It +is 20. Dividing the terms by 20, we get the ratio of image A to be 3 : 2 in +its simplest form. +The ratio of image D is 90 : 60. Dividing both terms by 30 (HCF +of 90 and 60), we get the simplest form to be 3 : 2. So the ratios of +images A and D are proportional as well. +What is the simplest form of the ratios of images B and E? +The ratio of image B is 40 : 20; in its simplest form, it is 2 : 1. +The ratio of image E is 60 : 60; in its simplest form, it is 1 : 1. + +Ganita Prakash | Grade 8 +162 +These ratios are not the same as 3 : 2. So, we can say that the ratios of +width to height of images B and E are not proportional to the ratios of +images A, C, and D. +When two ratios are the same in their simplest forms, we say that the +ratios are in proportion, or that the ratios are proportional. We use +the ‘::’ symbol to indicate that they are proportional. +So a : b :: c : d indicates that the ratios a : b and c : d are proportional. +Thus, +60 : 40 :: 30 : 20 and 60 : 40 :: 90 : 60. +7.4 Problem Solving with Proportional Reasoning +Example 1: Are the ratios 3 : 4 and 72 : 96 proportional? +3 : 4 is already in its simplest form. +To find the simplest form of 72 : 96, we need to divide both terms by +their HCF. +What is the HCF of 72 and 96? +The HCF of 72 and 96 is 24. Dividing both terms by 24, we get 3 : 4. +Since both ratios in their simplest form are the same, they are proportional. +Example 2: Kesang wanted to make lemonade for a celebration. She +made 6 glasses of lemonade in a vessel and added 10 spoons of sugar to +the drink. Her father expected more people to join the celebration. So he +asked her to make 18 more glasses of lemonade. +To make the lemonade with the same sweetness, how many spoons of +sugar should she add? +To maintain the same sweetness, the ratio of +the number of glasses of lemonade to the number +of spoons of sugar should be proportional. For 6 +glasses of lemonade, she added 10 spoons of sugar. +The ratio of glasses of lemonade to spoons of +sugar is 6 : 10. If she needs to make 18 more glasses +of lemonade, how many spoons of sugar should +she use? We can model this problem as — +6 : 10 :: 18 : ? +We know that each term in the ratio must change +by the same factor, for the ratios to be proportional. +How can we find the factor of change in the ratio? +The first term has increased from 6 to 18. To find the factor of change, +we can divide 18 by 6 to get 3. + +Proportional Reasoning-1 +163 +The second term should also change by the same factor. When 10 +increases by a factor of 3, it becomes 30. Thus, +6 : 10 :: 18 : 30. +So, she should use 30 spoons of sugar to make 18 glasses of lemonade +with the same sweetness as earlier. +Example 3: Nitin and Hari were constructing a compound wall around +their house. Nitin was building the longer side, 60 ft in length, and Hari was +building the shorter side, 40 ft in length. Nitin used 3 bags of cement but +Hari used only 2 bags of cement. Nitin was worried that the wall Hari built +would not be as strong as the wall he built because she used less cement. +Is Nitin correct in his thinking? +In Nitin and Hari’s case, we should compare the ratio of the length +of the wall to the bags of cement used by each of them and see whether +they are proportional. +The ratio in Nitin’s case is 60 : 3, i.e., 20 : 1 (in its simplest form). +The ratio in Hari’s case is 40 : 2, i.e., 20 : 1 (in its simplest form). +Since both ratios are proportional, the walls are equally strong. Nitin +should not worry! +Example 4: In my school, there are 5 teachers and 170 students. The +ratio of teachers to students in my school is 5 : 170. Count the number +of teachers and students in your school. What is the ratio of teachers to +students in your school? Write it below. +______ : ______ +Is the teacher-to-student ratio in your school proportional to the one in +my school? +Example 5: Measure the width and height (to the nearest cm) of the +blackboard in your classroom. What is the ratio of width to height of the +blackboard? +______ : ______ +Can you draw a rectangle in your notebook whose width and height are +proportional to the ratio of the blackboard? +Compare the rectangle you have drawn to those drawn by your +classmates. Do they all look the same? +Math +Talk +Note to the Teacher: Give more such examples that students can relate to and ask +them to give reasons why they think they are correct. Engaging with these problems +and finding solutions through a process of proportional reasoning should go along +with learning procedures and methods to solve the problems. + +Ganita Prakash | Grade 8 +164 +Example 6: When Neelima was 3 years old, her mother’s age was 10 +times her age. What is the ratio of Neelima’s age to her mother’s age? +What would be the ratio of their ages when Neelima is 12 years old? +Would it remain the same? +The ratio of Neelima’s age to her mother’s age when Neelima is 3 years +old is 3 : 30 (her mother’s age is 10 times Neelima’s age). In the simplest +form, it is 1 : 10. +When Neelima is 12 years old (i.e., 9 years later), the ratio of their ages +will be 12 : 39 (9 years later, her mother would be 39 years old). In the +simplest form, it is 4 : 13. +When we add (or subtract) the same number from the terms of a ratio, +the ratio changes and is not necessarily proportional to the original ratio. +Example 7: Fill in the missing numbers for the following ratios that are +proportional to 14 : 21. +______ : 42      6 : ______       2 : ______ +In the first ratio, we don’t know the first term. But the +second term is 42. It is 2 times the second term of the +ratio 14 : 21. So, the first term should also be 2 times 14 +(the first term). Hence the proportional ratio is 28 : 42. +For the second ratio, the first term is 6. +What factor should we multiply 14 by to get 6? Can it +be an integer? Or should it be a fraction? +We can model this as 14y = 6. So, y = 6 +14 = 3 +7. +So, we need to multiply 21 (the second term of 14 : 21) also by the +same factor 3 +7. +21 × 3 +7 is 9. So, the ratio is 6 : 9. +In the third ratio, the first term is 2. +We can see that when we divide 14 (the first term of 14 : 21) by 7 +(HCF of 14 and 21) we get 2. +If we divide 21 also by 7, we get 3. So, the ratio is 2 : 3. +Filter Coffee! +Filter coffee is a beverage made by mixing coffee +decoction with milk. Manjunath usually mixes 15 +mL of coffee decoction with 35 mL of milk to make +one cup of filter coffee in his coffee shop. +In this case, we can say that the ratio of coffee +decoction to milk is 15 : 35. +If +customers +want +‘stronger’ +filter +coffee, +Manjunath mixes 20 mL of decoction with 30 mL of +milk. The ratio here is 20 : 30. +14:21 +× 2=? +× 2 +:42 +______ + +Proportional Reasoning-1 +165 +Why is this coffee stronger? +And when they want ‘lighter’ filter coffee, he mixes 10 mL of +coffee and 40 mL of milk, making the ratio 10 : 40. +Why is this coffee lighter? +The following table shows the different ratios in which Manjunath +mixes coffee decoction with milk. Write in the last column if the coffee +is stronger or lighter than the regular coffee. +Coffee Decoction +(in mL) +Milk (in mL) +Regular/Strong/ +Light +300 +600 +150 +500 +200 +400 +24 +56 +100 +300 +Figure it Out +1. Circle the following statements of proportion that are true. + +(i) +4 : 7 :: 12 : 21 +(ii) +8 : 3 :: 24 : 6 + +(iii) 7 : 12 :: 12 : 7 +(iv) +21 : 6 :: 35 : 10 + +(v) 12 : 18 :: 28 : 12 +(vi) +24 : 8 :: 9 : 3 +2. Give 3 ratios that are proportional to 4 : 9. + +______ : ______     ______ : ______     ______ : ______   +3. Fill in the missing numbers for these ratios that are proportional to +18 : 24. +3 : ______    12 : ______    20 : ______    27 : ______ +Math +Talk +Math +Talk + +Ganita Prakash | Grade 8 +166 +4. Look at the following rectangles. Which rectangles are similar to +each other? You can verify this by measuring the width and height +using a scale and comparing their ratios. +5. Look at the following rectangle. Can you draw a smaller rectangle +and a bigger rectangle with the same width to height ratio in your +notebooks? Compare your rectangles with your classmates’ drawings. + +Are all of them the same? If they are different from yours, +can you think why? Are they wrong? +6. The following figure shows a small portion of a long brick wall with +patterns made using coloured bricks. Each wall continues this +pattern throughout the wall. What is the ratio of grey bricks to +coloured bricks? Try to give the ratios in their simplest form. +D +A +B +C +E +Math +Talk +(a) +(b) + +Proportional Reasoning-1 +167 +7. Let us draw some human figures. Measure your friend’s body — the +lengths of their head, torso, arms, and legs. Write the ratios as +mentioned below— + + + + +head : torso +______ : ______ +torso : arms +______ : ______ +torso : legs +______ : ______ +  +Now, draw a figure with head, torso, arms, and legs with equivalent +ratios as above. +Does the drawing look more realistic if the ratios are proportional? +Why? Why not? +Trairasika — The Rule of Three +Example 8: For the mid-day meal in a school with 120 students, the cook +usually makes 15 kg of rice. On a rainy day, only 80 students came to +school. How many kilograms of rice should the cook make so that the +food is not wasted? +The ratio of the number of students to the amount of rice needs to be +proportional. +So, 120 : 15 :: 80 : ? +What is the factor of change in the first term? +We can find that by dividing the terms 80 +120 = 2 +3. +The number of students is reduced by a factor of 2 +3. +On multiplying the weight of rice by the same factor, we get, +15 × 2 +3 = 10. +So, the cook should make 10 kg of rice on that day. +The situation above is a typical example of a problem where we need +to use proportional reasoning to find a solution. Four quantities are +Math +Talk +Note to the Teacher: In all these activities, encourage students to reason why their +drawings are proportional. + +Ganita Prakash | Grade 8 +168 +linked proportionally, out of which three are known and we must find +the fourth, unknown, quantity. +To solve such problems, we can model two proportional ratios using +algebraic notation as — +a : b :: c : d. +For these two ratios to be proportional, we know that term c should +be a multiple of term a by a factor, say f, and term d should be a multiple +of term b by the same factor f. So, +c = f a ...(1) +d = f b ...(2) +From (1) and (2), we can say that, +f = c +a and f = d +b . +Therefore, c +a = d +b. +Multiplying both sides by ab, we get, +ab × c +a = ab × d +b +ab × c +a = ab × d +b +bc = ad or ad = bc +Thus, when a : b :: c : d, then ad = bc. This is known as cross multiplication +of terms. +Since ad = bc, we can show that +d = bc +a . +Two ratios are proportional if their terms are equal when cross +multiplied. The fourth unknown quantity can be found through such +cross multiplication. + +In ancient India, Āryabhaṭa (199 CE) and others called such problems +of proportionality Rule of Three problems. There were 3 numbers +given — the pramāṇa (measure — ‘a’ in our case), the phala (fruit — ‘b’ +in our case), and the ichchhā (requisition — ‘c’ in our case). To find the +ichchhāphala (yield — ‘d’ in our case), Āryabhaṭa says, +“Multiply the phala by the ichchhā and divide the resulting product +by the pramāṇa.” +In other words, Āryabhaṭa says, +“pramāṇa : phala :: ichchhā : ichchhāphala,” therefore, +pramāṇa × ichchhāphala = phala × ichchhā. + +Proportional Reasoning-1 +169 +Thus, +ichchhāphala = phala × ichchhā +pramāṇa +. +Using the cross multiplication method proposed by Āryabhaṭa, ancient +Indians solved complex problems that involved proportionality. +Example 9: A car travels 90 km in 150 minutes. If it continues at the +same speed, what distance will it cover in 4 hours? +If it continues at the same speed, the ratio of the time taken should be +proportional to the ratio of the distance covered. +150 : 90 :: 4 : ? +Is this the right way to formulate the question? +No, because 150 is in minutes, but 4 is in hours. The second ratio +should use the same units for time as the first ratio. Since 4 hours is 240 +minutes, the right form is +150 : 90 :: 240 : ? +How can you find the distance covered in 240 minutes? +Discuss with your classmates and find the answer using different +strategies. +We can model this proportion as +150 : 90 :: 240 : x. +By cross multiplication, we get +150 × x = 240 × 90 +Therefore, + + +x = 240 × 90 +150 + . + + + + + + +48 3 + + + + + + = 240 × 90 +150 + = 144. + + + + + + + 5 + + + + + + + 1 +The distance covered by the car in 4 hours is 144 km. +Example 10: A small farmer in Himachal Pradesh sells each 200 g packet +of tea for ₹200. A large estate in Meghalaya sells each 1 kg packet of +tea for ₹800. Are the weight-to-price ratios in both places proportional? +Which tea is more expensive? +Note to the Teacher: Instead of giving one ‘method’ to solve the problem of the +distance, encourage students to reason out the answer through different strategies. +They can use their understanding of equivalent fractions and equivalent ratios to +find the answer. + +Ganita Prakash | Grade 8 +170 +The ratio of weight to price of the Himachal tea is 200 : 200. +What is the weight to price ratio of the Meghalaya tea? Is it 1 : 800? +This would not be appropriate, because we considered the weight in +grams in the case of Himachal. So, the weight to price ratio is 1000 : 800 +in Meghalaya after we convert the weight to grams. +To check if the ratios are proportional, we need to see if both ratios +are the same in their simplest forms. +The Himachal tea ratio in its simplest form is 1 : 1. +The Meghalaya tea ratio in its simplest form is 5 : 4. +So, the ratios are not proportional. +Which tea is more expensive? Why? +To answer the question as to which tea is more expensive, we should +compare the price of tea for the same weight in both places. +What is the price of 1 kg of tea from Meghalaya? It is ₹800. +In Himachal, if 200 g of tea costs rupees 200, what is the cost of 1 kg +of tea? +Let us say that the price of 1 kg of tea is x rupees. 200 g is 1 +5 of 1 kg. +So, + + + + +1 +5 × x = 200. +Multiplying both sides by 5, we get +1 +5 × x × 5 = 200 × 5 +1 +5 × x × 5 = 1000 +x = 1000. +So, the cost of 1 kg of tea is ₹800 in Meghalaya and ₹1,000 in Himachal +Pradesh. +Therefore, the tea from Himachal Pradesh is more expensive. +Activity 1: Take your favourite dish. Find out all the ingredients and +their respective quantities needed to make the dish for your family. +Suppose you are celebrating a festival and you want to invite 15 guests. +Find out the quantities of the ingredients required to cook the same dish +for them. +Figure it Out +1. The Earth travels approximately 940 million kilometres around the +Sun in a year. How many kilometres will it travel in a week? +2. A mason is building a house in the shape shown in the diagram. He +needs to construct both the outer walls and the inner wall that +Note to the Teacher: Encourage a discussion on the more expensive tea, how they +came to their conclusions and what the reasons for that tea being more expensive +could be. + +Proportional Reasoning-1 +171 +separates two rooms. To build a wall of 10-feet, he requires +approximately 1450 bricks. How many bricks would he need to build +the house? Assume all walls are of the same height and thickness. +Puneeth’s father went from Lucknow to Kanpur in 2 hours by riding +his motorcycle at a speed of 50 km/h. If he drives at 75 km/h, how +long will it take him to reach Kanpur? Can we form this problem as +a proportion — +50 : 2 :: 75 : __ + +Would it take Puneeth’s father more time or less time to reach +Kanpur? Think about it. + +Even though this problem looks similar to the previous +problems, it cannot be solved using the Rule of Three! + +The time of travel would actually decrease when the speed +increases. So this problem cannot be modelled as 50 : 2 :: 75 : __. +Activity 2: Go to the market and collect the prices of different +sizes of shampoo containers of the same shampoo and create a table like +the one given below. See if the volume of shampoo is proportional to the +price. +Container +Volume +Price +Sachet +6 mL +₹2 +Small Bottle +180 mL +₹154 +Medium Bottle +340 mL +₹276 +Large Bottle +1000 mL +₹540 +Math +Talk +12 ft +15 ft +9 ft +9 ft +6 ft + +Ganita Prakash | Grade 8 +172 +Let us compare the ratios for the sample table above. +The ratio of the volume of a sachet to a small bottle is 6 : 180. The ratio +of their prices is 2 : 154. Are these ratios proportional? +Why do you think that the ratio of the prices is not proportional to +the ratio of the volumes? +Discuss the pros and cons of different size bottles for the company +and for customers. For reducing ecological footprint, what would +you recommend to the company and to the customer? +Does the same occur for other products? +Make similar tables for other products in the market, capturing +different prices for different measures of the same product, e.g., +rice or atta (flour). +Observe the products for which the prices are proportional to +the different measures. +Discuss in class the proportionality of prices to measures of the +same product. +7.5 Sharing, but Not Equally! +Activity 3: Form a pair. Collect 12 +countable objects or counters (it can +be coins, seeds, or pebbles). Now, +share them between the two of you +in different ways. +If you divide them equally, what is +the ratio of the number of counters +with each of you? +Each of you will get 6 counters. +So, the ratio is 6 : 6, or 1 : 1 in its +simplest form. +Now let us not share equally. +If your partner gets 5 counters, how +many objects will you get? What is +the ratio of the counters? +Math +Talk +Note to the Teacher: Give a project to students. Start by dividing the class into +groups. Each group should go to one shop and collect prices for different measures +of the same product. For example, they should note down the prices of 500 g of +rice, 1 kg of rice, and 10 kg of rice. They should make tables with measure sizes and +prices, and present them to the rest of the class. They should discuss if the prices are +proportional or not, and why. + +Proportional Reasoning-1 +173 +The ratio of the counters of your partner to yours is 5 : 7. +Now, if you want to share the counters between the two of you in the +ratio of 3 : 1, how many counters would each of you get? +Share the counters in different ways and see which combination is in +the ratio 3 : 1. +One way to share the counters in the ratio of 3 : 1 is as follows — +1. Your partner takes 3 counters and you take 1 counter. There are now +8 counters left. +2. Your partner takes 3 more counters and you take 1 more counter. +There are now 4 counters left. +3. Your partner takes 3 more counters and you take 1 more counter. +There are no more counters left. +So, your partner gets 9 counters in total and you get 3 counters. +When we divide 12 counters in the ratio of 3 : 1 between two people, +one gets 9 counters and the other gets 3 counters. +Now, if you want to share 42 counters between the two of you in the +ratio of 4 : 3, how will you do it? +Using the same procedure would take a long time! There is a simpler +way to share the whole with parts in a specified ratio. +You need to divide 42 into groups such that your partner gets 4 groups +and you get 3 groups. +What is the size of each group? +If your partner gets 4 groups and you get 3 groups, the total number +of groups is 7. So, the size of each group is 42 ÷ 7 = 6. +Multiplying the number of groups by the size of each group, your +partner gets 24 counters and you get 18 counters, when you share 42 +counters in the ratio of 4 : 3. +12 +3 +1 +: +: +9 +3 +Sharing 12 +in the ratio 3 : 1 +The ‛whole’ is 12 +The ‛whole’ is 4 (= 3 + 1) +and 12 is 3 times 4 +The ‛parts’ are +3 times the ratio 3 : 1 +× 3 +× 3 + +Ganita Prakash | Grade 8 +174 +In general, when we want to divide a quantity, say x, in the ratio m: n, +we do the following: +1. We need to split x into groups such that it can be divided into two +parts where the first part has m groups and the second part has +n groups. +2. But what is the size of each group? This can be found out by dividing +x by the number of groups. The number of groups are m + n. So, the +size of each group is +x +m + n . +3. So, the first part has m × +x +m + n objects and the second part has n × +x +m + n +objects. +Thus, if we want to divide a quantity x in the ratio of m: n, then the +parts will be m × +x +m + n and n × +x +m + n. We see that +m × +x +m + n : n × +x +m + n :: m : n. +Example 11: Prashanti and Bhuvan started a food cart business near +their school. Prashanti invested ₹75,000 and Bhuvan invested ₹25,000. +At the end of the first month, they gained a profit of ₹4,000. They +decided that they would share the profit in the same ratio as that of their +investment. What is each person’s share of the profit? +The ratio of their investment is 75000 : 25000. +Reducing this ratio to its simplest form, we get 3 : 1. +3 + 1 is 4 and dividing the profit of 4000 by 4, we get 1000. +So, Prashanti’s share is 3 × 1000 and Bhuvan’s share is 1 × 1000. +So, Prashanti would get ₹3,000 and Bhuvan would get ₹1,000 of the profit. +Example 12: A mixture of 40 kg contains sand and cement in the ratio +of 3 : 1. How much cement should be added to the mixture to make the +ratio of sand to cement 5 : 2? +× 6 +× 6 +42 +4 +3 +: +: +24 +18 +Sharing 42 +in the ratio 4 : 3 +The ‛whole’ is 42 +The ‛whole’ is 7 (= 4 + 3) +and 42 is 6 times 7 +The ‛parts’ are +6 times the ratio 4 : 3 + +Proportional Reasoning-1 +175 +Let us find the quantity of sand and cement in the original mixture. +The ratio is 3 : 1 and the total weight is 40 kg. +So, the weight of sand is +3 +(3 + 1) × 40 = 30 kg. +The weight of cement is +1 +(3 + 1) × 40 = 10 kg. +The weight of sand is the same in the new mixture. It remains 30. But +the new ratio of sand to cement is 5 : 2. So the question is, +5 : 2 :: 30 : ? +If the ratio is 5 : 2, then the second term is 2 +5 times the first term. Since +the new ratio is equivalent to 5 : 2, the second term in the new ratio +should also be 2 +5 times of 30. + 2 +5 × 30 = 12. +The new mixture should have 12 kg of cement if the ratio of sand to +cement is to be 5 : 2. +There is 10 kg of cement already. So, we need to add 2 kg of cement to +the original mixture. +Figure it Out +1. Divide ₹4,500 into two parts in the ratio 2 : 3. +2. In a science lab, acid and water are mixed in the ratio of 1 : 5 to make +a solution. In a bottle that has 240 mL of the solution, how much acid +and water does the solution contain? +3. Blue and yellow paints are mixed in the ratio of 3 : 5 to produce +green paint. To produce 40 mL of green paint, how much of these +two colours are needed? To make the paint a lighter shade of green, +I added 20 mL of yellow to the mixture. What is the new ratio of blue +and yellow in the paint? +4. To make soft idlis, you need to mix rice and urad dal in the ratio +of 2 : 1. If you need 6 cups of this mixture to make idlis tomorrow +morning, how many cups of rice and urad dal will you need? +5. I have one bucket of orange paint that I made by mixing red and +yellow paints in the ratio of 3 : 5. I added another bucket of yellow +paint to this mixture. What is the ratio of red paint to yellow paint in +the new mixture? +7.6 Unit Conversions +We have noticed earlier that solving problems with proportionality +often requires us to convert units from one system to another. Here are + +Ganita Prakash | Grade 8 +176 +a few important unit conversions for your reference. +Length +1 metre = 3.281 feet +Area +1 square metre = 10.764 square feet +1 acre = 43,560 square feet +1 hectare = 10,000 square metres +1 hectare = 2.471 acres +Volume +1 millilitre (mL) = 1 cubic centimetre (cc) +1 litre = 1,000 mL or 1,000 cc +Temperature +Temperature conversion between Fahrenheit and Celsius is a bit more +complicated. 0 +oC = 32 +oF, and +Fahrenheit = 9 +5 × Celsius + 32 +and +Celsius = 5 +9 × (Fahrenheit–32) +For example, 25 +oC is 77 +oF. +Figure it Out + 1. Anagh mixes 600 mL of orange juice with 900 mL of apple juice to +make a fruit drink. Write the ratio of orange juice to apple juice in +its simplest form. + 2. Last year, we hired 3 buses for the school trip. We had a total of +162 students and teachers who went on that trip and all the buses +were full. This year we have 204 students. How many buses will +we need? Will all the buses be full? + 3. The area of Delhi is 1,484 sq. km and the area of Mumbai +is 550 sq. km. The population of Delhi is approximately +30 million and that of Mumbai is 20 million people. +Which city is more crowded? Why do you say so? + 4. A crane of height 155 cm has its neck and the rest of +its body in the ratio 4 : 6. For your height, if your neck +and the rest of the body also had this ratio, how tall +would your neck be? + 5. Let us try an ancient problem from Lilavati. At that +time weights were measured in a unit named palas +and niskas was a unit of money. “If 2 1 +2 palas of saffron + +Proportional Reasoning-1 +177 +costs 3 +7 niskas, O expert businessman! tell me quickly what quantity +of saffron can be bought for 9 niskas?” + 6. Harmain is a 1-year-old girl. Her elder brother is 5 years old. What +will be Harmain’s age when the ratio of her age to her brother’s age +is 1 : 2? + 7. The mass of equal volumes of gold and water are in the ratio 37 : 2. +If 1 litre of water is 1 kg in mass, what is the mass of 1 litre of gold? + 8. It is good farming practice to apply 10 tonnes of cow manure for 1 +acre of land. A farmer is planning to grow tomatoes in a plot of size +200 ft by 500 ft. How much manure should he buy? (Please refer to +the section on Unit Conversions earlier in this chapter). + 9. A tap takes 15 seconds to fill a mug of water. The volume of the mug +is 500 mL. How much time does the same tap take to fill a bucket of +water if the bucket has a 10-litre capacity? + 10. One acre of land costs ₹15,00,000. What is the cost of 2,400 square +feet of the same land? + 11. A tractor can plough the same area of a field 4 times faster than a +pair of oxen. A farmer wants to plough his 20-acre field. A pair of +oxen takes 6 hours to plough an acre of land. How much time would +it take if the farmer used a pair of oxen to plough the field? How +much time would it take him if he decides to use a tractor instead? + 12. The ₹10 coin is an alloy of copper and nickel called ‘cupro-nickel’. +Copper and nickel are mixed in a 3 : 1 ratio to get this alloy. The +mass of the coin is 7.74 grams. If the cost of copper is ₹906 per +kg and the cost of nickel is ₹1,341 per kg, what is the cost of +these metals in a ₹10 coin? + +„ Ratios in the form of a : b indicate that for every ‘a’ unit of the first +quantity, there are ‘b’ units of the second quantity. ‘a’ and ‘b’ are the +terms in the ratio. + +„ Two ratios — a : b and c : d — are proportional (written a : b :: c : d) if +their terms change by the same factor, i.e., if ad = bc. + +„ If x is divided into two parts in the ratio m : n, then the quantity of the +first part is m × +x +m + n and the quantity of the second part is n × +x +m + n. +SUMMARY +Try +This + +Binairo +Binairo, also known as Takuzu, is a logic puzzle with simple rules. Binairo +is generally played on a square grid with no particular size. Some cells +start out filled with two symbols: here horizontal and vertical lines. The +rest of the cells are empty. The task is to fill cells in such a way that: +1. Each row and each column must contain an equal number of +horizontal and vertical lines. +2. More than two horizontal or vertical lines can’t be adjacent. +3. Each row is unique. Each column is unique. +Solve the following Binairo puzzles: +Puzzle +Solution + +Dot Grid + +Dot Grid" +class_8,8,fractions in disguise,ncert_books/class_8/hegp2dd/hegp201.pdf,"1.1 Fractions as Percentages +You might have heard statements like, “Mega Sale — up to 50% off!” or +“Hiya scored 83% in her board exams”. +Do you know what the symbol ʻ%ʼ means? +This symbol is read as per cent. +The word ‘per cent’ is derived from the Latin phrase ‘per centum’, +meaning ‘by the hundred’ or ‘out of hundred’. +So, 25 per cent (25%) means 25 out of every 100 — like 25 people out +of 100, 25 rupees out of 100 rupees, or 25 marks out of 100 marks. +If we say 50% of some quantity s, it means +50% = 50 × 1 +100 × s (50 times the unit fraction of s) += 50 +100 × s = 1 +2s. +Thus, percentages are simply fractions where the denominator +is 100. Examples: +20% = 20 +100 = 2 +10 = 1 +5, +33% = 33 +100. +We saw that percentages are just fractions. Given any fraction, can we +express it as a percentage? Yes, let us see how. +Expressing Fractions as Percentages +Example 1: Surya wants to use a deep orange colour to capture the +sunset. He mixes some red paint and yellow paint to make this colour. +The red paint makes up 3 +4 of this mixture. What percentage of the colour +is made with red? +3 +4 is 3 out of every 4. +That is, 6 out of every 8 (equivalent fraction). +That is, 30 out of every 40. +That is, 75 out of every 100. +This means 75%. +3 +4 = 6 +8 = 30 +40 = 75 +100 +FRACTIONS IN +DISGUISE +1 + +Ganita Prakash | Grade 8 | Part-II +2 +Explaining this in a different way: To express 3 +4 as a percentage, we +need to find its equivalent fraction with 100 as the denominator. Two +ways of going ahead are shown below. +Method 1 +Method 2 +3 +4 = 3 × 25 +4 × 25 += 75 +100 = 75% +3 +4 = x +100 . +3 +4 × 100 = x +100 × 100 +[multiply both sides by 100] +x = 3 +4 × 100 += 75. +So, 3 +4 can be expressed as 75%. +Observe the following bar model diagram showing the equivalence +between 3 +4 and 75%. +0 +(25%) +(50%) +(75%) +(100%) +1 +4 +25 +100 +50 +100 +75 +100 +100 +100 +3 +4 +2 +4 = 1 +2 +4 +4 = 1 +0 +Can you tell what percentage of the colour was made using yellow? +Example 2: Surya won some prize money in a contest. He wants to save +2 +5 of the money to purchase a new canvas. Express this quantity as a +percentage. +Try to understand the different methods for solving this problem, as +shown below. +Method 1 +Method 2 +2 +5 = 20 +50 = 40 +100 += 40%. +2 +5 = x +100 . +x = 2 +5 × 100 = 40. + +Fractions in Disguise +3 +Try completing Method 3 by filling the boxes. +Method 3 +100% +Total prize +money +0% +Isn’t it the same +as finding +2 +5 + th of 100? +1 +5 +2 +5 +3 +5 +4 +5 +5 +5 = 1 +0 +Savings for canvas +Several problems in mathematics can be approached and solved +in different ways. While the method you came up with may be +dear to you, it can be amusing and enriching to know how others +thought about it. +A fraction is of a unit, while a percentage is per 100. Therefore, to +express a fraction as a percentage, we can just multiply the fraction +by 100. +Example 3: Given a percentage, can you express it as a fraction? For +example, express 24% as a fraction. +Since a percentage is a fraction, 24% is the same as 24 +100. +We can find other equivalent forms of 24 +100 = 12 +50 = 6 +25 = 48 +200. +In general, we can say that a percentage, z%, can be expressed by any +of the fractions that are equivalent to z +100. +Figure it Out +1. Express the following fractions as percentages. + (i) 3 +5 +(ii) 7 +14 +(iii) 9 +20 + (iv) 72 +150 +(v) 1 +3 +(vi) 5 +11 +2. Nandini has 25 marbles, of which 15 are white. What percentage of +her marbles are white? + (i) 10% +(ii) 15% +(iii) 25% + (iv) 60% +(v) 40% +(vi) None of these + +Ganita Prakash | Grade 8 | Part-II +4 +3. In a school, 15 of the 80 students come to school by walking. What +percentage of the students come by walking? +4. A group of friends is participating in a long-distance run. The positions +of each of them after 15 minutes are shown in the following picture. +Match (among the given options) what percentage of the race each +of them has approximately completed. +Finish +Start +A +B +C +D +55% +20% +38% +72% +84% +93% +5. Pairs of quantities are shown below. Identify and write appropriate +symbols ‘>’, ‘<’, ‘=’ in the blanks. Try to do it without calculations. + (i) 50% ____ 5% +(ii) 5 +10 ____ 50% + (iii) 3 +11 _____ 61% +(iv) 30% ____ 1 +3 +Well, if percentages are just a particular type of fraction, why do we +need them? Why can’t we just continue using fractions? +Let us consider an example. +A biscuit-making factory is experimenting with 2 new varieties of +biscuits. Sugar makes up 9 +34 of Variety 1 and 13 +45 of Variety 2. Which variety +is more sugary? It may not be clear at first glance and we may have +to do some calculations. But, when the same information is presented +as — Sugar makes up 26.47% of Variety 1 and 28.88% of Variety 2, it is +immediately clear which variety is more sugary. +If we want to have the same denominator, why choose 100 in +particular? Why not 10, 50, 1000, or 43? Think. +In principle, we could choose any number as the denominator. But +with 100, there are some advantages. Since our number system is base 10, +numbers like 10, 100, and 1000 fit easily with decimals. For example, +31% = 31 +100 = 0.31. +Converting between fractions, decimals, and percentages becomes +quick and intuitive. +The number 100 is round, and easy to understand +and work with. We could say “per 1000” or +“per 100,000” (this usage is present in statistics +like “per thousand people” or “per lakh”), but +100 hits the sweet spot — it’s large enough to give +detail, yet simple enough to grasp mentally. Per 10 +would be too small for many purposes. +9 +34 = 26.47 percent (per 100) +9 +34 = 2.647 per decem (per 10) +9 +34 = 264.7 per mille (per 1000) + +Fractions in Disguise +5 +Long before the decimal fraction was introduced, the need for it was +felt in computations by tenths, twentieths, and hundredths. The idea of +‘per hundred’ can be found as early as the 4th century BCE in Kautilya’s +Arthaśhāstra, “An interest of a pana and a quarter per month per cent +is just. Five panas per month per cent is commercial interest. Ten panas +per month per cent prevails among forests. Twenty panas per month +per cent prevails among sea traders”. +Around the same time, the Romans used taxes of 1 +20, 1 +100 in transactions +related to trade and auctions. In the Italian manuscripts of the 15th +century, expressions such as ‘xx p cento’, ‘x p cento’, ‘vii p cento’ can be +found (equivalent to our 20%, 10%, and 7%). +Percentages Around Us +Percentages are widely used in a variety of contexts. Here are some +interesting findings that involve percentages. +The human body, on +average, is about 60% +water by weight. +Ice cream is about +30 – 50% air by +volume. +45% of the world̛s +population watched at +least part of the 2022 +FIFA World Cup. +Over 80% of the teenagers +globally fail to meet the +recommendation of at +least one hour of daily +physical actvity. +About 99.86% of the +Solar System̛s mass is +contained in the Sun. +An estimated 52% of +the agricultural land +worldwide is degraded. + +Ganita Prakash | Grade 8 | Part-II +6 +1.2 Percentage of Some Quantity +Example 1: Madhu and Madhav each ate +biscuits of a different variety. Madhu’s biscuits +had 25% sugar, while Madhav’s had 35% sugar. +Can you tell who ate more sugar? +As we just saw, percentages represent +fractional quantities or proportions. It would +be +inappropriate +to +compare +just +the +percentages when they are referring to +different quantities or wholes. That is, if they +both had 100 g of biscuits, then clearly Madhav ate more sugar — 35 g +(35% of 100 g is 35 g per 100 g) vs. Madhu’s 25 g (25% of 100 g is 25 g +per 100 g). +Suppose Madhu ate 120 g of biscuits and Madhav ate 95 g of biscuits. +Who consumed more sugar? Try to find out. +We know that the weight of sugar is proportional to the weight of the +biscuits consumed. Madhu ate 120 g of biscuits having 25% sugar. The +amount of sugar he ate is the value in the blank — +25 : 100 :: ____ : 120. +A few ways of going forward are shown. +25% sugar means — +25 g sugar per 100 g of +biscuits, which means +5 g sugar per 20 g +biscuits, so +30 g sugar per 120 g +biscuits. +The proportional +relationship can be +written as + 25 +100 = s +120. +s = 25 +100 × 120 = 30 . +0 +0 +25% +50% +75% +100% +30 g +60 g +90 g +120 g +Sugar +Do any of the methods match your thinking? Were you able to +understand all the methods? +Now, let us find out how much sugar Madhav ate. He ate 95 g of +biscuits with 35% sugar. Sometimes the numbers may not be convenient +to calculate using different methods as we did just before. +100 g of the biscuits he ate has 35 g of sugar. +This means, 1 g of the biscuits has 35 +100 g of sugar. +We can say that 95 g of the biscuit will have 35 +100 g × 95 = 33.25 g. +This can also be solved by finding the value of s in the proportional +relationship 35 +100 = s +95. +Therefore, Madhav ate more sugar. +35 +100 = x +1 → x = 35 +100 + +Fractions in Disguise +7 +More generally, y% of some value, say 80, is given by y +100 × 80. +We can also say that 45% of some value, say z, is given by 45 +100 × z. +Free-hand Computations +We just calculated 25% of 120. Is 25% the same as 1 +4th (a quarter)? +Suppose we want to find 25% of 40. Is it the same as 1 +4th of 40? +Yes, since 25 is 1 +4th (a quarter) of 100 ( +25 +100 = 1 +4). +Therefore 25% of 40 = 25 +100 × 40 is the same as 1 +4 × 40. +Try to calculate (without using pen and paper) the indicated percentages +of the values shown in the table below. Write your answers in the table. +100 +200 +50 +80 +10 +35 +287 +25% +25 +10% +20% +5% +How did you find these values? Discuss the methods with the class. +Do you find anything interesting in the table? +You may have noticed that 20% of a value is double that of 10% +of the same value. This will always happen as 20% (20 parts out of +100) is twice that of 10% (10 parts out of 100). +Using this understanding, mentally calculate how much 40% of the +values in the table above would be. +What relationship do you observe among 20%, 5% and 25% of a value? +It appears that (20% of y) + (5% of y) = 25% of y. We can verify that this +property always holds: +( +20 +100 × y) + ( +5 +100 × y) = ( +25 +100 × y). +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +8 +Using this observation, mentally calculate how much 15% of the values +in the table would be. +Suppose you have to mentally calculate the following percentages +of some value: 75%, 90%, 70%, 55%. How would you do it? Discuss. +The FDP Trio — Fractions, Decimals, and Percentages +Example 2: We can find 50% of a value by multiplying 1 +2 with the value. +Will multiplying the value by 0.5 also give the answer for 50% of the +value? +Yes, since 1 +2 = 0.5. +50% = 50 +100 = 1 +2 = 0.5 +1 = 0.5. +Similarly, to find 10% of a quantity, what decimal value should be +multiplied? +Complete the following table: +Per cent +50% +100% +25% +75% +10% +1% +5% +43% +Fraction +50 +100 +Decimal +0.5 +Activity: How Close Can You Get? +Make a pair. Each of you choose a number. Suppose, the numbers +chosen are a and b. Share your numbers with each other. Both of +you should estimate the percentage equivalent to the fraction a +b +(where a < b) and announce your answers by a fixed time, say, 5 +seconds. The one whose estimate is the closest wins this round. Play +this for 10 rounds. +Example 3: The maximum marks in a test are 75. If students score 80% +or above in the test, they get an A grade. How much should Zubin score +at least to get an A grade? +We can find 80% of 75 in different ways, using our understanding of +fraction and decimal multiplication, as well as of proportionality. +Math +Talk +50% of 24 = 12 +0.5 × 24 = 12 + +Fractions in Disguise +9 +Fraction Multiplication → 80 +100 × 75 += 4 +5 × 75 = 60. +Decimal Multiplication → 0.8 × 75 = 60. +Proportional Reasoning → +Out of 100, the minimum mark is 80. +Out of 75, it is +75 × 80 +100 + = 60. +Example 4: To prepare a particular millet kanji (porridge), suppose the +ratio of millet to water to be mixed for boiling is 2:7. What percentage +does the millet constitute in this mixture? If 500 ml of the mixture is to +be made, how much millet should be used? +This situation can be modelled as shown +in the bar model on the right side. +The ratio of millet to the volume of the +mixture is 2:9. In other words, in one unit +of the mixture, millet occupies 2 +9 units and +water occupies 7 +9 units. +Estimate first what percentage 2 +9 would be. +The percentage (i.e., in 100 such units) of +millet in the mixture is +2 +9 × 100 = 22.22%. +The percentage of water in the mixture will +be 100 – 22.22 = 77.78%. +A mixture with 22.22% millet means +100 ml mixture will have 22.22 ml millet. +Therefore, 500 ml with 22.22% millet will have 5 × 22.22 = 111.1 ml of +millet. +A given ratio can be converted to a fraction, and then to a percentage. +The practice of estimating first before calculating can improve +number sense and help reduce mistakes. Often, we may not need +exact values. The ability to make quick estimates is useful at +these times. +Note to the Teacher: Incorporate the practice of estimating before calculating or +solving as part of the problem-solving process. You may remind and encourage +students as needed. +0 +0 +15 +30 +45 +60 +0.2 +(20%) +0.4 +(40%) +0.6 +(60%) +0.8 +(80%) +75 +1.0 +(100%) +Total +marks +Minimum marks for A grade +2 +7 +9 +Half of 9 is 4.5. +So, 2 +9 is clearly less +than 50%. Half of +4.5 is 2.25. So, 2 +9 is +less than 25%. +10% of 9 is 0.9. +20% of 9 is 1.8. +So, 2 +9 could +be between +20% – 25%. + +Ganita Prakash | Grade 8 | Part-II +10 +Example 5: A cyclist cycles from Delhi to Agra and completes 40% of the +journey. If he has covered 92 km, how many more kilometres does he +have to travel to reach Agra? +Let us first try to model +this situation by a bar model. +40% +100% +92 km +Delhi +Agra +? +Estimate first before solving further. +A few ways of solving this problem are shown below. Does your method +match any of the given ones? Do you like any of the other methods? +It is given that 40% of the distance is 92 km. We have to find out how +much the rest of the 60% distance is. +Method 1 +Method 3 +Method 4 +40% is 92 km, therefore +20% is 46 km. +This makes 60% to be +92 + 46 = 138 km. +40 +100 = 92 +d +(d is the total distance) +d = 92 × 100 +40 = 230. +Remaining distance = +230 km – 92 km += 138 km. +If x is the remaining +distance then the total +distance from Delhi to +Agra is x + 92. Since, we +know that 92 is 40% of +this total distance, +40 +100 × (x + 92) = 92. +(x + 92) = 92 × 100 +40 +x = 230 – 92 = 138. +Method 2 +If 40% is 92, 60% is ? +40 : 92 :: 60: ? +40 +92 = 60 +r +(r is the remaining +distance) +r = 60 × 92 +40 = 138. +Drawing rough diagrams can help understand the given situation +better and make it easier to think further about the problem. +Percentages Greater than 100 +Till now, we saw percentages with a value 100 or less than 100. Can there +be percentages with a value more than 100? What could it mean when a +percentage is greater than 100? Let us explore. +Example 6: Kishanlal recently opened a garment shop. He aims to +achieve a daily sales of at least ₹5000. The sales on the first 2 days were +₹2000 and ₹3500. What percentage of his target did he achieve? + +Fractions in Disguise +11 +The percentage target achieved is visualised below. +0 +Day 1: 2000 +0 +100% +5000 +% of target: + 2000 +5000 × 100 = 40% +40% = 40 +100 = 2 +5 = 0.4 +0 +0 +100% +5000 +Day 2: 3500 +% of target: +3500 +5000 × 100 = 70% +70% = 70 +100 = 7 +10 = 0.7 +It is 40% on Day 1 and 70% on Day 2. +Another way of saying it is — he was 60% short of his target on Day 1 +and 30% short of his target on Day 2. +In the next two days, he made ₹5000 and ₹6000 respectively. What +percentage of his target are these values? +His target is ₹5000, and he made ₹5000 on Day 3 — this is 100%. On +Day 4, he made ₹6000, which is 1000 more than his target. +What percentage of the target was achieved on Day 4? +0% +Day 3: 5000 +0 +100% +5000 +% of target: +5000 +5000 × 100 = 100% +100% = 100 +100 = 1 +1 = 1.0 +0% +0 +100% +Day 4: 6000 +5000 +20% +% of target: +6000 +5000 × 100 = 120% +120% = 120 +100 = 6 +5 = 1.2 +1000 is 20% of 5000. Therefore, 6000, (5000 + 1000) is 100% + 20% += 120% of 5000. It can also be computed as 6000 +5000 × 100 = 6 +5 × 100 = 120%. +This means he achieved 120% of his target, i.e., 20% more than his target. +On Days 5 and 6 his sales were ₹7800 and ₹9550 respectively. Calculate +the percentage of the target achieved on these days. +0% +Day 5: 7800 +0 +100% +0% +0 +100% +Day 6: 9550 +5000 +5000 +200% +10,000 +2800 +4550 +200% +10,000 + +Ganita Prakash | Grade 8 | Part-II +12 +On Day 7, he achieved 150% of his target. On Day 8, he achieved 210% of +his target. Find the sales made on these days. +Suppose on some day, he made ₹2500. This can be expressed as “He +achieved 1 +2 of his target” or “He achieved 50% of his target” or “He +achieved 0.5 of his target”. On some other day, he made ₹10,000. We +can say “He achieved twice/double/2 times his target” or “He achieved +200% of his target”. +Complete the table below. Mark the approximate locations in the +following diagram. +Percent +90% +110% +200% +250% +15% +173% +358% +28.9% +305% +Fraction +Decimal +(0) +(1) +(2) +(3) +(4) +0% +100% +200% +300% +400% +90% +Example 7: A farmer harvested 260 kg of wheat last year. This year, they +harvested 650 kg of wheat. What percentage of last year’s harvest is this +year’s harvest? +This year’s harvest = 650 +260 × 100 = 250% of last year’s harvest. +250% indicates that it is 2.5 times the original value. +Figure it Out +Estimate first before making any computations to solve the following +questions. Try different methods including mental computations. +1. Find the missing numbers. The first problem has been worked out. +100% +(i) +20% +75 +100% +60 +100% +(ii) +? +90 +100% +? + +Fractions in Disguise +13 +100% +(iii) +? +140 +100% +? +2. Find the value of the following and also draw their bar models. + (i) 25% of 160 +(ii) 16% of 250 +(iii) 62% of 360 + (iv) 140% of 40 +(v) 1% of 1 hour +(vi) 7% of 10 kg +3. Surya made 60 ml of deep orange paint, how much red paint did he +use if red paint made up 3 +4 of the deep orange paint? +4. Pairs of quantities are shown below. Identify and write appropriate +symbols ‘>’, ‘<’, ‘=’ in the boxes. Visualising or estimating can help. +Compute only if necessary or for verification. + (i) 50% of 510 + 50% of 515 +(ii) 37% of 148 + 73% of 148 + (iii) 29% of 43 + 92% of 110 +(iv) 30% of 40 + 40% of 50 + (v) 45% of 200 + 10% of 490 +(vi) 30% of 80 + 24% of 64 +5. Fill in the blanks appropriately: +(i) 30% of k is 70, 60% of k is _____, 90% of k is _____, 120% of k +is ______. +(ii) 100% of m is 215, 10% of m is _____, 1% of m is ______, 6% of m +is ______. +(iii) 90% of n is 270, 9% of n is ______, 18% of n is _____, 100% of n +is ______. +(iv) Make 2 more such questions and challenge your peers. +6. Fill in the blanks: +(i) 3 is ____ % of 300. +(ii) _____ is 40% of 4. +(iii) 40 is 80% of _____. +7. Is 10% of a day longer than 1% of a week? Create such questions +and challenge your peers. +8. Mariam’s farm has a peculiar bull. One day she gave the bull +2 units of fodder and the bull ate 1 unit. The next day, she gave +the bull 3 units of fodder and the bull ate 2 units. The day after, +she gave the bull 4 units and the bull ate 3 units. This continued, +and on the 99th day she gave the bull 100 units and the bull ate +99 units. Represent these quantities as percentages. This task +can be distributed among the class. What do you observe? +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +14 +9. Workers in a coffee plantation take 18 days to pick coffee berries in +20% of the plantation. How many days will they take to complete +the picking work for the entire plantation, assuming the rate of +work stays the same? Why is this assumption necessary? +10. The badminton coach has planned the training +sessions such that the ratio of warm up : play : cool +down is 10% : 80% : 10%. If he wants to conduct a +training of 90 minutes. How long should each activity +be done? +11. An estimated 90% of the world’s population lives in the Northern +Hemisphere. Find the (approximate) number of people living in the +Northern Hemisphere based on this year’s worldwide population. +12. A recipe for the dish, halwa, for 4 people has the following +ingredients in the given proportions — Rava: 40%, Sugar: 40%, and +Ghee: 20%. +(i) If you want to make halwa for 8 people, what is the proportion +of each of the above ingredients? +(ii) If the total weight of the ingredients is 2 kg, how much rava, +sugar and ghee are present? +1.3 Using Percentages +To Compare Proportions +Example 1: Eesha scored 42 marks out of 50 on an English test and 70 +marks out of 80 in a Science test. Since she lost only 8 marks in English +but 10 marks in Science, she thinks she has done better at English. Reema +does not agree! She argues that since Eesha has scored more marks in +Science, she has done better at Science. Vishu thinks we cannot compare +the scores because the maximum marks are different. Who do you think +is correct? +If the maximum marks are the same, the comparison becomes easier, +isn’t it? For these kinds of comparisons, we need to convert both values +to percentages. +English score as a percentage = 42 +50 × 100 = 84% +Science score as a percentage = 70 +80 × 100 = 87.5%. +The Science score (as a percentage) is higher than the English score +(as a percentage). So, we can conclude that Eesha has scored better on +the Science test. + +Fractions in Disguise +15 +Know Your Contents (KYC) +Example +2: +Madhu +and +Madhav recently learnt about +the importance of reading +labels on processed food +before purchase. They are at +a shop to buy badam drink +mix. They are looking at two +products +and +wondering +which has a larger share of +badam. Can you figure it out? +Which product uses a smaller +proportion of food chemicals? +It is easier to compare the proportions of the ingredients if we convert +them into percentages. For example, +DEF’s sugar content as a percentage of total weight = 99 +150 × 100 = 66%. +Complete this table by calculating the percentages to answer the +questions: +Maybe they +should call it +ʻSugar drink +mixʼ +Ya! +It is very important to +Know Your Contents. +Do +Do KYC +KYC! + +Sugar +Milk +Solids +Badam +Powder +Food +Chemicals +DEF +66% +Zacni +Check if the percentages of each product add up to 100. +Percentage Increase or Decrease +Percentages are often used to describe the rate of change of quantities. +For example, +1. Suppose the price of 1 kg tomatoes 3 years ago was ₹30, and the price +now is ₹42. The increase in the price is ₹12. +Percentage increase = +amount of increase +original amount or base × 100 += 12 +30 × 100 = 40%. +We say the price of tomatoes increased by 40% over the last 3 years. + +Ganita Prakash | Grade 8 | Part-II +16 +2. The average footfall in this theater before COVID was 160. Now it is +just 100. The decrease in the footfall is 60. +Percentage decrease = +amount of decrease +original amount or base × 100 += 60 +160 × 100 = 37.5%. +We say the footfall in this theater post-COVID has decreased by 37.5%. +Example 3: Do the following two statements mean the same thing? +(i) The population of this state in 1991 is 165% of that in 1961. +(ii) The population of this state has increased by 65% from 1961 to 1991. +Yes, both mean the same. Suppose p is the population of the state in +1961 and q is the population of the state in 1991. +Statement A implies, +q = 165% of p +q = 165 +100 × p = 1.65p +Statement B implies, +q = p + 65% of p +q = p + 0.65 × p = 1.65p +In other words, the population of the state in 1991 is 1.65 times that +in 1961. +Profit and Loss +You may have the experience of buying something — snacks, groceries, +clothes, toys, etc. Very often, the shopkeeper quotes a price and after +some bargaining the customer pays the negotiated amount and buys the +item(s). We call the price quoted by the shopkeeper the marked price +(sometimes this can be the MRP of an item). The price the customer pays +after a discount is called the selling price. Also, the price the shopkeeper +paid to purchase that item is called the cost price. Let us see how these +labels are relative to the context through an example. +The picture below shows the journey of a sweater from the +manufacturer to the customer and what the cost price (CP), the marked +price (MP), and the selling price (SP) mean in each step. + +Fractions in Disguise +17 +Kishanlal (retailer) buys sweaters from a wholesaler at a price of ₹300 +per sweater. The marked price he quotes his customers is ₹480. After +bargaining, he sells this sweater at ₹430. Notice that the selling price is +greater than the cost price, resulting in a profit of ₹430 – ₹300 = ₹130. If +the selling price is less than the cost price, it will result in a loss. +Example 4: Find out the percentage profit Kishanlal made on this +sweater. +We shall consider the cost price to +be 100% to find out the percentage +profit made with reference to the cost +price. The following rough diagram +describes this situation. +The profit amount is ₹130. +The percentage profit is 130 +300 × 100 = 43.3%. +Find the profit percentage of the wholesaler and the manufacturer. +Shambhavi owns a stationery shop. She procures 200 page notebooks at +₹36 per book. She sells them with a profit margin of 20%. Find the selling +price. +She sells crayon boxes at ₹50 per box with a profit margin of 25%. How +much did Shambhavi buy them from the wholesaler? +Example 5: The rice stock in Raghu’s provision store is getting old. He +had purchased the rice at ₹35 per kg. To clear his stock, he sells 10 kg rice +for ₹300. Find out the percentage loss. +The amount Raghu had paid towards +buying the 10 kg rice is ₹350. He sold it +for ₹300. +The loss is ₹350 – ₹300 = ₹50. +The percentage loss is 50 +350 × 100 = 14.28%. +Could we have just calculated the loss percentage per kg instead? Would +it be the same? +Example 6: Shyamala had procured decorative vases at ₹2650 per piece. +One of the pieces was slightly damaged. She decides to sell it at a loss of +18%. How much will she get by selling this piece? +Try making an estimate. Draw a rough diagram depicting the given +situation. +0 +0 +100% +300 +430 +? +Selling price +130 +Kishanlal’s buying price +0 +0 +100% +350 +50 +Selling price +Cost price + +Ganita Prakash | Grade 8 | Part-II +18 +Two methods of solving this are shown. +With respect to the buying price +being 100%, the selling price is +18% less than the buying price. +That is, the selling price would +be 82%. +82% of 2650 = 0.82 × 2650 = 2173. +The loss amount is 18% of 2650. +That is, 18 +100 × 2650 = 477. +Reducing this from the +buying price, 2650 – 477 = 2173. +The sale amount of the damaged vase would be ₹2173. +Due to heavy rains, Snehal could not transport +strawberries to Hyderabad from his farm in Panchgani. +He sells some of his stock at ₹80 per kg with a 12% loss. +What is the cost price? +You have probably seen percentages mentioned when +shops offer discounts! Do you know what 30% off (or +30% discount) means? It means that the shop is willing to +reduce the price of the item by 30%. +A utensil store is offering a 35% discount on the cooker +with an MRP ₹1800. What is the selling price? If the cost +price was ₹900, what is the percentage profit made after the sale? +Suppose Kishanlal achieved a sales of ₹80,000 last month. Out of this, +the amount he spent on buying these goods he sold last month was +₹48,000. The difference amount, ₹80,000 – ₹48,000 = ₹32,000, is called +gross profit. +After deducting the other expenses he incurred (such as +transport cost, employee’s salary, electricity bill, etc.) amounting +to ₹8,000, the amount remaining is called net profit, which is +₹32,000 – ₹8,000 = ₹24,000. +Note: In this chapter, the term profit refers to gross profit. +Manisha sells fertiliser. She buys 50 kg bags +at ₹500 per bag. She sells it at ₹750 per bag +making a profit of 250 +500 × 100 = 50%. +Although, with respect to the ₹750 amount +she has earned by selling a bag, the profit +percentage is 250 +750 × 100 = 33.33%. +Profit percentage is calculated based on +the price the goods bought when we want +to know “How much profit did I make +compared to what I invested in buying the +goods?”. +0 +0 +100% +500 +50% +250 +0 +0 +100% +500 +33.33% +250 +750 + +Fractions in Disguise +19 +Profit percentage is calculated based on the revenue (sales amount) +when we want to know “How much (net) profit did I make on my +overall revenue?”. +Suppose, in a month she made sales of ₹1,50,000. The cost of +buying the goods was ₹1,00,000. So, the gross profit is ₹50,000. The +monthly expenses amounted to ₹5,000. The net profit is ₹45,000. +Out of monthly revenue (₹1,50,000), the net profit percentage is +45000 +150000 × 100 = 3 +10 × 100 = 30%. +Taxes +Tax rates, like the GST (Goods and Services Tax) +rate or the Income Tax rates, are also specified as +percentages. You may have noticed GST mentioned +as part of bills. This means that the tax is part of +the amount we pay and this amount goes to the +government. +Check if the calculations are correct in the bill +shown. +You may share any bills you have at home with the class. Observe +the different elements present in the bills. Are there any similarities +or differences in these bills? +Figure it Out +1. If a shopkeeper buys a geometry box for ₹75 and sells it for ₹110, +what is his profit margin with respect to the cost? +2. I am a carpenter and I make chairs. The cost of materials for a chair +is ₹475 and I want to have a profit margin of 50%. At what price +should I sell a chair? +3. The total sales of a company (also called revenue) was ₹2.5 crore +last year. They had a healthy profit margin of 25%. What was the +total expenditure (costs) of the company last year? +4. A clothing shop offers a 25% discount on all shirts. If the original +price of a shirt is ₹300, how much will Anwar have to pay to buy +this shirt? +5. The petrol price in 2015 was ₹60 and ₹100 in 2025. What is the +percentage increase in the price of petrol? +(i) 50% + + +(ii) 40% + + +(iii) 60% +(iv) 66.66% + +(v) 140% + + +(vi) 160.66% +Try +This + +Ganita Prakash | Grade 8 | Part-II +20 +3. Samson bought a car for ₹4,40,000 after getting a 15% discount from +the car dealer. What was the original price of the car? +4. 1600 people voted in an election and the winner got 500 votes. What +percent of the total votes did the winner get? Can you guess the +minimum number of candidates who stood for the election? +5. The price of 1 kg of rice was ₹38 in 2024. It is ₹42 in 2025. What is the +rate of inflation? (Inflation is the percentage increase in prices.) +6. A number increased by 20% becomes 90. What is the number? +7. A milkman sold two buffaloes for ₹80,000 each. On one of them, he +made a profit of 5% and on the other a loss of 10%. Find his overall +profit or loss. +8. The population of elephants in a national park increased by 5% in +the last decade. If the population of the elephants last decade is p, +the population now is + (i) + +p × 0.5 + +(ii) p × 0.05 +(iii) p × 1.5 +(iv) + +p × 1.05 + +(v) p + 1.50 +9. Which of the following statement(s) mean the same as — “The +demand for cameras has fallen by 85% in the last decade”? +(i) The demand now is 85% of the demand a decade ago. +(ii) The demand a decade ago was 85% of the demand now. +(iii) The demand now is 15% of the demand a decade ago. +(iv) The demand a decade ago was 15% of the demand now. +(v) The demand a decade ago was 185% of the demand now. +(vi) The demand now is 185% of the demand a decade ago. +Growth and Compounding +You might have come across statements like, “1 year interest for FD +(Fixed Deposit) in the bank @ 6% per annum” or “Savings account +with interest @ 2.5% per annum”. Interest is the extra money paid by +institutions like banks or post offices on money deposited (kept) with them. +Interest is also paid by people or institutions when they borrow money. +In a Fixed Deposit (FD), you deposit a specific amount of money for +a fixed period at a predetermined interest rate. The money remains +locked for the chosen duration, and the bank pays you interest on it. +You cannot withdraw the amount before the maturity date without +incurring a penalty. At the end of the term, you receive both your +original deposit and the interest earned. + +Fractions in Disguise +21 +For example, a bank says that the interest rate for fixed deposits is 10% +p.a. What do you understand by this statement? What is the ‘p.a.’ next to +the percentage? +‘p.a.’ is the short form of per annum, which means for every year. The +specified interest rate indicates that if you invest ₹6000 as a deposit for +a year with the bank, they will give you ₹600 as interest on this deposit. +The 10% is the rate of interest and the ₹6000 on which the interest is +calculated is the principal. The amount after 1 year in your bank account +will be +6000 + (0.10 × 6000) +(principal) + (10% interest on principal) += 6000 + 600 = 6600. +In other words, the amount deposited, +₹6000, will become 110% (or increase by 10%), +6000 × 110% = 6000 × 110 +100 = 6000 × 1.1 = 6600. +This can also be expressed as +Amount after 1 year = Principal (P) + rate of interest (r) of P +Example 7: If one deposits ₹6000 in the bank, what is the amount after +3 years? +That depends on the choice of FD. There are two possibilities: +1. Option 1: The interest is paid out regularly (for example, every year). +The principal amount is returned after the maturity period. +Interest returned +(10% p.a.) +Amount in +the FD +Year 1 +Beginning +₹6000 +Ending +₹600 +returned +₹6000 +Year 2 +Beginning +₹6000 +Ending +₹600 +returned +₹6000 +Year 3 +Beginning +₹6000 +Ending +₹600 +returned +₹6000 +Total amount received +₹1800 + ₹6000 = ₹7800 +× 0.10 +× 0.10 +× 0.10 +0 +0 +100% +1000 ? +Amount after 1 year +Principal (initial deposit) +10% + +Ganita Prakash | Grade 8 | Part-II +22 +2. Option 2: The interest gained every time (say after each year) is +added back to the FD, thus increasing the principal amount for the +subsequent period. After the maturity period, the entire amount is +returned. This phenomenon is called compounding. +Interest added +back (10% p.a.) +Amount in +the FD +Year 1 +Beginning +₹6000 +Ending +₹600 +added back +₹6600 +Year 2 +Beginning +₹6600 +Ending +₹660 +added back +₹7260 +Year 3 +Beginning +₹7260 +Ending +₹726 +₹7986 +₹7986 is returned +Total amount received +₹7986 +× 0.10 +× 1.10 +× 1.10 +× 1.10 +× 0.10 +× 0.10 +We can see that with compounding, the final amount is more. +Example 8: What percent is the total amount received with respect to +the amount deposited in both the options? +This can be calculated by finding total amount received +amount deposited × 100. +Without Compounding +With Compounding +7800 +6000 × 100 = 130% = 1.3. +In other words, the total amount +received = 6000 × (1 + 0.1 + 0.1 + 0.1) += 6000 × 1.3. +The percentage gain over 3 years is 30%. +7986 +6000 × 100 = 133.1% = 1.331. +In other words, the total amount received += 6000 × 1.1 × 1.1 × 1.1. += 6000 × 1.331 +The percentage gain over 3 years is 33.1%. +Figure it Out +1. Bank of Yahapur offers an interest of 10% p.a. Compare how +much one gets if they deposit ₹20,000 for a period of 2 years with +compounding and without compounding annually. + +Fractions in Disguise +23 +2. Bank of Wahapur offers an interest of 5% p.a. Compare how +much one gets if one deposits ₹20,000 for a period of 4 years with +compounding and without compounding annually. +3. Do you observe anything interesting in the solutions of the two +questions above? Share and discuss. +Let us try to generalise the pattern observed in each of the options. +Example 9: What is the amount we get back if we invest ₹6000 at an +interest rate of 10% p.a. for ‘t’ years? +No Compounding +Here, the interest gained every term is paid back. Therefore, the +principal amount for every term shall remain the same, and as a result, +the interest gained every term also shall be the same. +The interest gained in 1 term is 6000 × 0.1 +The interest gained in 3 terms is +6000 × 0.1 × 3 +The total amount at the end of an FD of 3 +years is +6000 + (6000 × 0.1 × 3). +The interest gained in 1 term is p × r +(p is the principal, r is the rate of interest in +percentage) +The interest gained in t terms is +p × r × t +The total amount at the end of an FD of t +years is +p + (p × r × t )= p + prt += p (1 + rt). +With Compounding +Here, the interest gained every term/year is added back to the FD. +Therefore, the principal amount increases after every term, and as a +result, the interest gained every term also increases proportionately. +The total amount in the FD after Year 1 is +(6000) × 1.1 +(principal for Year 1) +The total amount in the FD after Year 2 is +(6000 × 1.1) × 1.1 +(principal for Year 2) +The total amount in the FD after Year 3 is +(6000 × 1.1 × 1.1) × 1.1 +(principal for Year 3) +The total amount in the FD after t years is +6000 × (1.1 × 1.1 × 1.1…..× 1.1) +t times +6000 × (1.1)t. +The total amount in the FD after Year 1 is +(p) × (1 + r) +(principal for Year 1) +The total amount in the FD after Year 2 is +p × (1 + r) × (1 + r) +(principal for Year 2) +The total amount in the FD after Year 3 is +p × (1 + r) × (1 + r) × (1 + r) +(principal for Year 3) +The total amount in the FD after t years is +p × (1 + r) × (1 + r) × …..× (1 + r) +t times +p × (1 + r)t. + +Ganita Prakash | Grade 8 | Part-II +24 +Suppose we want to know the expression/formula to find the total +interest amount gained at the end of the maturity period. What +would be the formula for each of the two options? +Figure it Out +4. Jasmine invests amount ‘p’ for 4 years at an interest of 6% p.a. Which +of the following expression(s) describe the total amount she will get +after 4 years when compounding is not done? + (i) p × 6 × 4 +(ii) p × 0.6 × 4 +(iii) p × 0.6 +100 × 4 + (iv) p × 0.06 +100 × 4 +(v) p × 1.6 × 4 +(vi) p × 1.06 × 4 +(vii) p + (p × 0.06 × 4) +5. The post office offers an interest of 7% p.a. How much interest would +one get if one invests ₹50,000 for 3 years without compounding? +How much more would one get if it was compounded? +6. Giridhar borrows a loan of ₹12,500 at 12% per annum for 3 years +without compounding and Raghava borrows the same amount for +the same time period at 10% per annum, compounded annually. +Who pays more interest and by how much? +7. Consider an amount ₹1000. If this grows at 10% p.a., how long +will it take to double when compounding is done vs. when +compounding is not done? Is compounding an example of +exponential growth and not-compounding an example of linear +growth? +8. The population of a city is rising by about 3% every year. If the +current population is 1.5 crore, what is the expected population +after 3 years? +9. In a laboratory, the number of bacteria in a certain experiment +increases at the rate of 2.5% per hour. Find the number of bacteria +at the end of 2 hours if the initial count is 5,06,000. +Decline +Several items or materials lose financial value over time. Suppose +someone buys a bike at your home for ₹1,00,000 and after a few years +they want to sell it. The value of the bike at that time will be less than +₹1,00,000. It could depend on various factors, including how many years +have passed since the purchase, how many kilometres the vehicle has +been used for, if there has been any damage, or if any parts have been +replaced. This is called depreciation — reduction of value due to use +and age of the item. +Math +Talk +Math +Talk + +Fractions in Disguise +25 +Example 10: A TV is bought at a price of ₹21,000. After 1 year, the value +of the TV depreciates by 5%. Find the value of the TV after one year. +The amount of reduction in the +value is 5% of 21,000 = 0.05 × 21,000 += 1050. +The current value is 21,000 – 1050 += 19,950. +The value of the TV after 1 year will +be 95% of the current value += 95% of 21,000 = 0.95 × 21,000 += 19,950. +The value of the TV after 1 year will be ₹19,950. +Example 11: The population of a village was observed to be reducing by +about 10% every decade. If the current population is 1250, what is the +expected population after 3 decades? +The population 1 decade later will +be 0.9 times the population of the +current decade. +Therefore, the population after +1 decade will be 1250 × 0.9. +The population after 2 decades will +be 1250 × 0.9 × 0.9. +The population after 3 decades will +be 1250 × 0.9 × 0.9 × 0.9 += 911.25. +First decade’s decrease = 0.1 × 1250 += 125. +Population after 1 decade += 1250 – 125 = 1125. +Second decade’s decrease += 0.1 × 1125 = 112.5 ≅ 112. +Population after 2 decades += 1125 – 112 = 1013 +Third decade’s population decrease += 0.1 × 1013 = 101.3 ≅ 101. +Population after 3 decades += 1013 – 101 = 912. +Rounding off, we can say that the expected population after 3 decades +will be around 910. +Tricky Percentages +Would You Rather? +You have won a contest. The organisers offer you two options +to choose from: +Option A: You deposit ₹100 and you get back ₹300. +Option B: You deposit ₹1000 and you get back ₹1500. +What is the percentage gain each option gives? You can choose any +option only once. Which option would you choose? Why? +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +26 +While comparing percentages, we have to be mindful that we are comparing +fractions or proportions and not absolute values. +A provision store is offering a stock clearance sale. Customers can +choose one of the two options — 20% discount or ₹50 discount—for any +purchase above ₹150. Which option would you choose if you want to: +(i) buy items worth ₹180 +(ii) buy items worth ₹225 +(iii) buy items worth ₹300 +Example 12: A bakery called Cakely is +offering a 30% + 20% discount on all cakes. +Another bakery called Cakify is offering a +50% discount on all cakes. Would you rather +choose Cakely or Cakify if you want the +cheaper cost? +It seems that both the options should give +the same benefit. Although mathematically +30% + 20% is the same as 50%, the usage of +30% + 20% in shopping means compounding. +Suppose you want to buy a cake worth ₹200. +Cakely’s 30% + 20% → +Applying the 30% discount → the price of cake +is ₹200 – ₹60 = ₹140. +Applying the 20% discount on ₹140 → the +price of cake is ₹140 – ₹28 = ₹112. +Cakify’s 50% → +The 50% discount makes the price of the cake +₹100. + +Fractions in Disguise +27 +A Mishap +Example 13: After Surbhi bought cookware from the wholesaler, she +kept a profit margin of 50% on all the products. To clear off the remaining +stock, she thought she would offer a 50% discount and come out without +any loss. +(i) Do you think she didn’t make any loss? +(ii) If she had sold goods (originally) for ₹12,000 after discount, how +much loss did she incur? What is the percentage loss? +(iii) What should have been the percentage discount offered so that she +sold the goods at the price she had bought (i.e., no profit +or loss)? +Let us model the situation first. +Suppose the worth of the products +she bought from the wholesaler is x. +The worth corresponding to the +selling price (with a 50% margin) is 1.5x. +A 50% discount on this price will +make the worth 0.75x. +(i) This means the selling price is 3 +4 of the price the goods were bought +at, i.e., a 25% loss. +(ii) If she had sold goods worth ₹12,000, +0.75x = 12,000 +x = 16,000. +She lost ₹4000. +(iii) To sell the goods at the same price, the discount offered should be +1.5x – d × (1.5x) = x +d = 1 +3 = 0.33. +The discount offered should have been 33.33%. +Ariba and Arun have some marbles. Ariba says, “The number of +marbles with me is 120% of the marbles Arun has”. What would be +an appropriate statement Arun could make comparing the number +of marbles he has with Ariba’s? +x +Price bought at +1.5x +Selling price including profit +0.75x +After discount +Try +This + +Ganita Prakash | Grade 8 | Part-II +28 +Figure it Out +1. The population of Bengaluru in 2025 is about 250% of its population +in 2000. If the population in 2000 was 50 lakhs, what is the population +in 2025? +2. The population of the world in 2025 is about 8.2 billion. The +populations of some countries in 2025 are given. Match them with +their approximate percentage share of the worldwide population. +[Hint: Writing these numbers in the standard form and estimating +can help]. +3. The price of a mobile phone is ₹8,250. A GST of 18% is added to +the price. Which of the following gives the final price of the phone +including the GST? + (i) 8250 + 18 +(ii) 8250 + 1800 + (iii) 8250 + 18 +100 +(iv) 8250 × 18 + (v) 8250 × 1.18 +(vi) 8250 + 8250 × 0.18 + (vii) 1.8 × 8250 +4. The monthly percentage change in population (compared to the +previous month) of mice in a lab is given: Month 1 change was +5%, +Month 2 change was –2%, and Month 3 change was –3%. Which of +the following statement(s) are true? The initial population is p . +(i) The population after three months was p × 0.05 × 0.02 × 0.03. +(ii) The population after three months was p × 1.05 × 0.98 × 0.97. +(iii) The population after three months was p + 0.05 – 0.02 – 0.03. +(iv) The population after three months was p. +(v) The population after three months was more than p. +(vi) The population after three months was less than p. +5. A shopkeeper initially set the price of a product with a 35% profit +margin. Due to poor sales, he decided to offer a 30% discount on the +selling price. Will he make a profit or a loss? Give reasons for your +answer. +6. What percentage of area is occupied by the region marked ‘E’ in the +figure? +13% +8% +18% +10% +1% +35% +2% +2% +0.1% +Germany +83 million +India +1.46 billion +Bangladesh +175 million +USA +347 million + +Fractions in Disguise +29 +7. What is 5% of 40? What is 40% of 5? + What is 25% of 12? What is 12% of 25? + What is 15% of 60? What is 60% of 15? + What do you notice? + Can you make a general statement and +justify it using algebra, comparing x% +of y and y% of x? +8. A school is organising an excursion for +its students. 40% of them are Grade +8 students and the rest are Grade 9 +students. Among these Grade 8 students, 60% are girls. [Hint: +Drawing a rough diagram can help]. +(i) What percentage of the students going to the excursion are +Grade 8 girls? +(ii) If the total number of students going to the excursion is 160, +how many of them are Grade 8 girls? +9. A shopkeeper sells pencils at a price such that the selling price +of 3 pencils is equal to the cost of 5 pencils. Does he make a +profit or a loss? What is his profit or loss percentage? +10. The bus fares were increased by 3% last year and by 4% this year. +What is the overall percentage price increase in the last 2 years? +11. If the length of a rectangle is increased by 10% and the area is +unchanged, by what percentage (exactly) does the breadth decrease +by? +12. The percentage of ingredients in a 65 g chips +packet is shown in the picture. Find out the +weight each ingredient makes up in this +packet. +13. Three shops sell the same items at the same +price. The shops offer deals as follows: +Shop A: “Buy 1 and get 1 free” +Shop B: “Buy 2 and get 1 free” +Shop C: “Buy 3 and get 1 free” +Answer the following: +(i) If the price of one item is ₹100, what is the effective price per +item in each shop? Arrange the shops from cheapest to costliest. +(ii) For each shop, calculate the percentage discount on the items. +[Hint: Compare the free items to the total items you receive.] +(iii) Suppose you need 4 items. Which shop would you choose? Why? +Try +This +A +B +C +D +E + +Ganita Prakash | Grade 8 | Part-II +30 +14. In a room of 100 people, 99% are left-handed. How many +left-handed people have to leave the room to bring that +percentage down to 98%? +15. Look at the following graph. +The ability to use computers is highest among those in their twenties and teenagers. +Ability to use computer by age and gender (2023) +Female +Male +Children +Teenage +Twenties +Thirties +Forties +Fifties +Seniors +0% +4% +24% +29% +26% +37% +14% +7% +4% +2% +25% +14% +9% +4% +5% +10% +15% +20% +25% +30% +35% +40% +Source: NSS Round 79, Comprehensive Annual Modular Survey, National Statistics Office +Based on the graph, which of the following statement(s) are valid? +(i) People in their twenties are the most computer-literate among +all age groups. +(ii) Women lag behind in the ability to use computers across age +groups. +(iii) There are more people in their twenties than teenagers. +(iv) More than a quarter of people in their thirties can use computers. +(v) Less than 1 in 10 aged 60 and above can use computers. +(vi) Half of the people in their twenties can use computers. +Try +This + +Fractions in Disguise +31 + +y +Percentages are widely used in our daily life. + y +Percentages are fractions with denominator 100. Percentages are +denoted using the symbol ʻ%ʼ, pronounced ʻper centʼ. x% = x +100. + +y +Fractions can be converted +to +percentages +and +vice versa. Decimals too can +be converted to percentages +and vice versa. For example, +4 +10 = 0.4 = 40%. + +y +We have learnt to find the exact number when a certain +percentage of the total quantity is given. + +y +When parts of a quantity are given to us as ratios, we have seen +how to convert them to percentages. + +y +The increase or decrease in a certain quantity can also be +expressed as a percentage. + +y +The profits or losses incurred in transactions, and tax rates, can +be expressed in terms of percentages. + +y +We have seen how a quantity or a number grows when +compounded. Interest rates are a common example of +compounding. If p is the principal, r is the rate of interest, +and t is the number of terms, then the total amount after the +maturity period is +Without +compounding, +p (1 + rt) +p remains the +principal for all +the terms. +With compounding, +p × (1 + r) × (1 + r) × … × (1 + r) = p (1 + r)t +Principal for term 1 +Principal for term 2 +Principal for term 3 +Principal for term t + +y +A situation or a problem can often be solved by describing +it using a rough diagram. We have learnt to estimate and do +mental computations to solve problems related to percentages. +0 +0% +100% +40% +40 +100 = 0.4 +100 +100 = 1 +SUMMARY + +Peaceful Knights +Place 8 knights on the chess board so that no knight attacks another. +A knight moves in an ‘L-shape’. It can move either (a) two steps vertically +and one step horizontally, or (b) two steps horizontally and one step +vertically. Possible moves of a knight are shown below." +class_8,9,the baudhayana-pythagoras theorem,ncert_books/class_8/hegp2dd/hegp202.pdf,"2.1 Doubling a Square +In Baudhāyana’s Śulba-Sūtra (c. 800 BCE), Baudhāyana considers the +following question: +How can one construct a square having double the area +of a given square? +A first guess might be to simply double the length of +each side of the square. Will this new square have double +the area of the original square? +It’s not hard to see that the new square will have area 2 × 2 = 4 times +the area of the original square: +So how can one make a square that has double the area? Baudhāyana +in his Śulba-Sūtra (Verse 1.9) gave an elegant answer to this question: +The diagonal of a square produces a square of double the area of the +original square. +As Baudhāyana says, the key is to construct a square on the diagonal +of the original square: +THE BAUDHĀYANA- +PYTHAGORAS THEOREM +2 + +Ganita Prakash | Grade 8 | Part-II +34 +Why does the new dotted square have double the +area of the original square? +In many of the constructions in the Śulba-Sūtra, +it is desirable to construct, where needed, what +Baudhāyana calls ‘east-west’ and ‘north-south’ +lines, i.e., horizontal and vertical lines that are +perpendicular to each other. Can you draw some +horizontal and vertical lines to see why the new +square has double the area of the original square? +You could draw some horizontal and vertical +lines as shown on the right. +Why should the extension of the vertical and horizontal sides of +the original square pass through the vertices of the dotted square? +[Hint: From the diagonal property of a square, the line that bisects +an angle passes through the opposite vertex. Argue why the vertical and +horizontal sides of the original square bisect the two angles of the dotted +square.] +So, the new square has double the area of the original square, because +the original square is made up of two small triangles, while the new +square is made up of four small triangles. +Moreover, all these small triangles are congruent to each other. Can you +explain why? +Adding some more horizontal and vertical lines can make the situation +even clearer: +Math +Talk + +The Baudhāyana-Pythagoras Theorem +35 +So we can make the following sequence of squares: +Each square has double the area of the previous one, as they are made +up of 2, 4, and 8 small triangles, respectively. +Doubling a Square Using Paper +Cut out two identical squares of paper. Draw, label, and cut as follows: +Square 1 +Identical Square 2 +1 +3 +2 +4 +5 +7 +6 +8 +Now place the pieces 5, 6, 7, and 8 around Square 1 to get a square +with double the area. + +Ganita Prakash | Grade 8 | Part-II +36 +2.2 Halving a Square +Now suppose we are given a square, and we want to +construct a square whose area is half that of the original +square. How would you do it? +One way to do it is to reverse the construction of the +previous section. We draw a tilted smaller square inside +the larger one: +Why is the smaller inside square half the area of the larger square? +Again, adding some east-west and north-south lines can explain it: +Halving a Square Using Paper +Cut out a square from a piece of paper. Now make a square whose area +is half the area of the first square. +Will the square having half the sidelength have half the area? Why not? +How many such squares will fill the original square? +Fold the square paper inward, as shown, such that the crease lines +pass through the midpoints of the sides. PQRS is the required square +with half the area. +S +P +Q +R + +The Baudhāyana-Pythagoras Theorem +37 +Why is PQRS a square? Why is its area half that of the original paper? +Explain by connecting QS and PR, finding the different angles formed, +and then using tringle congruence. +2.3 Hypotenuse of an Isosceles Right Triangle +Recall that in a right triangle, the side opposite to the right angle is called +the hypotenuse. +Find the hypotenuse of this isosceles right triangle. +We know that a square of side 1 unit is made of two +such isosceles right triangles. We also know that the +square constructed on the diagonal of this square has +twice the area of the original square. +T +S +R +1 +1 +1 +P +E +1 +A +We do not yet know the length of the hypotenuse, but we know the +area of the square REST on it! +Area of REST = 2 × Area of PEAR + += 2 × 1 = 2 sq. units. +We know the relation between the side and the area of a square. If c +is the hypotenuse ER, then + +Area of REST = c × c = c2. + +So, c2 = 2. + +Therefore, c = 2. +Thus, the hypotenuse is of length 2 units. +1 +1 +2 +1 +1 +2 +1 +1 +? + +Ganita Prakash | Grade 8 | Part-II +38 +In the rest of this chapter, we will assume that all the lengths are of a +fixed unit unless specified otherwise. +What is the value of 2? +Decimal Representation of 2 +The decimal representation of +2 can be found using the following +argument. +Is 2 less than or greater than 1? +A square of sidelength 1 unit has an area of 1 sq. unit. A square of +sidelength 2 has an area of 2 sq. units. So, 1 is less than 2. +In other words, 12 = 1, and 22 = 2. +Therefore, 1 < 2. +Is 2 less than or greater than 2? +A square of sidelength 2 units has an area of 4 sq. units. A square of +sidelength 2 has an area of 2 sq. units. So, 2 is greater than 2. +In other words, 22 = 2, and 22 = 4. +Therefore, 2 < 2. +Thus, 1< 2 < 2. +We call 1 a lower bound on 2 and 2 an upper bound. +Can we find closer bounds for 2? +Will we ever get a number with a terminating decimal representation +whose square is 2? +If there is such a terminating decimal starting with 1.414… whose +square is 2, then it must have a non-zero last digit. If this is the case, then +the decimal representation of its square will also have a non-zero last +digit after the decimal point. For example, if 2 is of the form 1.414…4, +then its square will be of the form — +6 +. . . +. +1.12 = 1.21 +1.22 = 1.44 +1.32 = 1.69 +1.42 = 1.96 +1.52 = 2.25 +So, 1.4 < 2 < 1.5 +1.412 = 1.9881 +1.422 = 2.0164 +So, 1.41 < 2 < 1.42 +1.4112 = 1.990921 +1.4122 = 1.993744 +1.4132 = 1.996569 +1.4142 = 1.999396 +1.4152 = 2.002225 +So, 1.414 < 2 < 1.415 + +The Baudhāyana-Pythagoras Theorem +39 +So a terminating decimal cannot have 2 or 2.000... as its square. +Thus, the decimal expansion of +2 must go on forever, i.e., it has a +non-terminating decimal representation. +Can +2 be expressed as a fraction m +n , where m and n are counting +numbers? +If 2 could be expressed as m +n , then we would have + +2 = m +n +2 = +m2 +n2 + +2n2 = m2. +Recall that in the prime factorization of a square number, each prime +occurs an even number of times. So in the equation 2n2 = m2, the prime +2 would occur an odd number of times on the left side and an even +number of times on the right side. This is impossible. Thus 2 cannot be +expressed as a fraction m +n . +This beautiful proof that 2 cannot be expressed as m +n where m, n are +counting numbers was given by Euclid in his book Elements (c. 300 BCE). +We will discuss it in more detail in a later class. +Thus the number 2 cannot be expressed as a terminating decimal or +a fraction. But we can think of it as a certain non-terminating decimal: +2 = 1.41421356… +Figure it Out +1. Earlier, we saw a method to create a square with double the +area of a given square paper. There is another method to do this +in which two identical square papers are cut in the following +way. +1 +2 +3 +4 +Can you arrange these pieces to create a square with double the area +of either square? +2. The length of the two equal sides of an isosceles right triangle is +given. Find the length of the hypotenuse. Find bounds on the length +of the hypotenuse such that they have at least one digit after the +decimal point. +(i)  3 + +(ii)  4 +(iii)  6 +(iv)  8 +(v) 9 +Try +This +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +40 +3. The hypotenuse of an isosceles right triangle is 10. What are its other +two sidelengths? [Hint: Find the area of the square composed of two +such right triangles.] +General Solution +The relation between the areas of a square and the square on its diagonal +can be used to find a general relation between the hypotenuse and the +other two sides of an isosceles right triangle. +a +a +P +Q +S +R +V +U +c +Let a be the length of the equal sides and c the length of the hypotenuse. +Area of SQVU = 2 × Area of PQRS + +So, c2 = 2a2. +This formula can be used to find c when a is known, or to find a when +c is known. +Example 1: Find the hypotenuse of an isosceles right triangle whose +equal sides have length 12. +We have a = 12. Using the formula, we get +c =  2×122 =  288 . +We have 162 = 256, and 172 = 289. +So,  288 lies between 16 and 17. +The length of the hypotenuse of an isosceles right triangle, whose +length of the equal sides is 12 units, is between 16 and 17 units. +Example 2: If the hypotenuse of an isosceles right triangle is 72, find +its other two sides. +We have c = 72. Using the formula, we get +c2 = 2a2 + +The Baudhāyana-Pythagoras Theorem +41 +So, ( 72)2 = 2a2 +72 = 2a2. +Thus, a2 = 72 +2 = 36. +So, a = 36 = 6. +Therefore, each of the other two sides has length 6. +Use this formula to check your answers in the Figure it Out on page 39. +2.4 Combining Two Different Squares +We have seen in the previous sections how to combine two copies of the +same square to make a larger square whose area is the sum of the areas +of the two smaller squares. ++ +→ +The sidelength of the larger square is the length of the diagonal of +either of the smaller squares. +What if we wish to combine two squares of ‘different’ sizes to make a +large square whose area is the sum of the two smaller squares? ++ +→ +? +In his Śulba-Sūtra (Verse 1.12), Baudhāyana also gives a truly +remarkable solution to this more general problem of combining two +different sized squares. He writes: +The area of the square produced by the diagonal is the sum of the +areas of the squares produced by the two sides. + +Ganita Prakash | Grade 8 | Part-II +42 +That is, to combine two different squares, make a right-angled triangle +whose perpendicular sides are the sidelengths of the two squares. The +square whose sidelength is the hypotenuse of this right-angled triangle +has an area that is the sum of the areas of the original two squares. ++ +→ +Area A +Area B +Area A + Area B +Why does Baudhāyana’s method work? +Can you see why the method works in the case where the two squares +are the same size? Does it agree with the method we used earlier to +combine two same sized squares into a bigger square? +Subsequently in his Śulba-Sūtra (Verse 2.1), Baudhāyana provides +another verse that helps in explaining why the method works in general: +To combine different squares, mark a rectangular portion of the +larger square using a side of the smaller square. The diagonal of this +rectangle is the side of a square that has area equal to the sum of the +areas of the smaller squares. +Let us follow Baudhayan᾽s instructions. +• Join the two squares. + +b +b +a +a +b +b +a +a + +The Baudhāyana-Pythagoras Theorem +43 +• Mark a rectangular portion of the larger square using a side of the +smaller one, and draw its diagonal. By doing this, we get a right +triangle with perpendicular sides a and b. +1. +a +b +b +a +a +b – a + +a +b +b +a +a +b – a +• Make a 4-sided figure over the hypotenuse by drawing three more +of such right triangles: +2. +a +b +b +a +a +b – a +3. +a +b +b +a +a +a +b – a +b – a +a +a +b +b +a +a +a +b – a +b – a +a + +Ganita Prakash | Grade 8 | Part-II +44 +4. +b +a +b – a +T +U +V +X +W +a +a +a +b +a +a +a +b – a +b – a +a +b +• The 4-sided figure obtained (T + U + V) is in fact a square with an +area equal to the sum of the areas of the two smaller squares! +Why? +• As T, U, X and W are all congruent, the sides of this new 4-sided +figure all have the same length. Notice the angles. +Explain why all the angles of this new 4-sided figure are right angles and +so it is a square. +Notice that this new square has as its sidelength, the hypotenuse of the +right triangle of sides a and b. +• Baudhāyana’s assertion is now clear: +The area of the square on the hypotenuse = the sum of the areas of T, +U, and V = the sum of the areas of V, W, and X = the sum of the areas of +the two given squares. +a +b +a +a +b – a +b – a +a +x +x +x +x +90 – x +90 – x +90 – x +90 – x +b + +The Baudhāyana-Pythagoras Theorem +45 +T +U +V +V +X +W +Combining Two Squares Using Paper +Cut out and join two different sized squares as follows: +a +a +b +b +Now make two cuts to make three pieces: +a +a +b +c +c +b – a +Rearrange the three pieces into a larger square: +c +c +You have now combined two smaller squares into a larger square! +Now make a right triangle using the two smaller squares. Draw a +square on the hypotenuse. + +Ganita Prakash | Grade 8 | Part-II +46 +c +a +b +Cover the square on the hypotenuse using your pieces. +a +b +c +This shows that if a right triangle has shorter sides of length a and b, +and hypotenuse of length c, then the areas of the two smaller squares, a2 +and b2, add up to the area of the larger square, c2 : +a2 + b2 = c2. +This is the famous and fundamental theorem of Baudhāyana on +right-angled triangles: +Baudhāyana’s Theorem on Right-angled triangles: If a right-angled +triangle has sidelengths a, b, and c, where c is the length of the +hypotenuse, then a2 + b2 = c2. + +The Baudhāyana-Pythagoras Theorem +47 +Baudhāyana was the first person in history to state this theorem in this +generality and essentially modern form. The theorem is also known as +the Pythagorean Theorem, after the Greek philosopher-mathematician +Pythagoras (c. 500 BCE) who also admired and studied this theorem, and +lived a couple hundred years after Baudhāyana. It is also often called the +transitional name ‘Baudhāyana-Pythagoras Theorem’ so that everyone +knows what theorem is being referred to. +Using Baudhāyana’s Theorem +Make a right-angled triangle in your notebook +whose shorter sidelengths are 3 cm and 4 cm. +Now, measure the length of the hypotenuse. +It should read about 5cm. +In fact, we could have used Baudhāyana’s +Theorem to predict that the hypotenuse is +exactly 5cm! Let a = 3 and b = 4, the lengths in cm of the two shorter +sides. Then, by Baudhāyana’s Theorem, the length c of the hypotenuse +satisfies the equation, + +a2 + b2 = c2 +32 + 42 = c2 + +9 + 16 = c2 + +25 = c2 + +So, c = 5 cm. +Figure it Out +1. If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm, +then what is the length of its hypotenuse? First draw the right-angled +triangle with these sidelengths and measure the hypotenuse, then +check your answer using Baudhāyana’s Theorem. +2. If a right-angled triangle has a short side of length 8 cm and +hypotenuse of length 17 cm, what is the length of the third side? +Again, try drawing the triangle and measuring, and then check your +answer using Baudhāyana’s Theorem. +3. Using the constructions you have now seen, how would you construct +a square whose area is triple the area of a given square? Five times +the area of a given square? (Baudhāyana’s Śulba-Sūtra, Verse 1.10) +4. Let a, b and c denote the length of the sides of a right triangle, with +c being the length of the hypotenuse. Find the missing sidelength in +each of the following cases: + (i) a = 5, b = 7 +(ii) a = 8, b = 12 + +(iii) a = 9, c = 15 +(iv) a = 7, b = 12 +(v) a = 1.5, b = 3.5 +4 cm +3 cm +? + +Ganita Prakash | Grade 8 | Part-II +48 +2.5 Right–Triangles Having Integer Sidelengths +In his Śulba-Sūtra (Verse 1.13) Baudhāyana lists a number of integer +triples (a, b, c) that form the sidelengths of a right-triangle and therefore +satisfy a2 + b2 = c2. These include (3, 4, 5), (5, 12, 13) (8, 15, 17), (7, 24, 25), +(12, 35, 37), and (15, 36, 39). +For this reason, such triples (a, b, c) that form the sidelengths of a +right triangle (equivalently, satisfy a2 + b2 = c2) are called Baudhāyana +triples. They are also called Baudhāyana-Pythagoras triples, +Pythagorean triples, and right-angled triangle triples. +List down all the Baudhāyana triples with numbers less than or +equal to 20. +Is there an unending sequence of Baudhāyana triples? +Mathematicians have answered this question and have found a +method to generate all such triples! Let us take a few steps in this +direction. +We have seen that (3, 4, 5) is a Baudhāyana triple. +Is (30, 40, 50) a Baudhāyana triple? +Is (300, 400, 500) a Baudhāyana triple? +The list of Baudhāyana triples having numbers less than or equal to +20 contains the following triples — +(3, 4, 5), (6, 8, 10), (9, 12, 15), (12, 16, 20). +Do you see any pattern among them? +All these triples can be obtained by multiplying each term of (3, 4, 5) +by a certain positive integer. +Can we form a conjecture on Baudhāyana triples based on this +observation? +Conjecture: (3k, 4k, 5k) is a Baudhāyana triple, where k is any positive +integer. +Is this true? +We have to check if (3k)2 + (4k)2 = (5k)2. +We have (3k)2 + (4k)2 = 9k2 + 16k2 = 25k2, which is equal to (5k)2. +So, (3k, 4k, 5k) is indeed a Baudhāyana triple. +This shows that there are infinitely many Baudhāyana triples. +Can we further generalise the conjecture? +Math +Talk + +The Baudhāyana-Pythagoras Theorem +49 +If (a, b, c) is a Baudhāyana triple, then (ka, kb, kc) is also a Baudhāyana +triple where k is any positive integer. Is this statement true? +This statement can be shown to be true in the same way. Since (a, b, +c) is a Baudhāyana triple, we have a2 + b2 = c2. We need to check whether +(ka)2 + (kb)2 = (kc)2. +(ka)2 = ka × ka = k2 a2, and (kb)2 = kb × kb = k2 b2. +So, (ka)2 + (kb)2 = k2 a2 + k2 b2. +Taking out the common factor, we have +(ka)2+(kb)2 = k2 (a2 + b2). +Since a2 + b2 = c2, we have +(ka)2 + (kb)2 = k2 c2 = (kc)2. +Thus, (ka, kb, kc) is a Baudhāyana triple if (a, b, c) is a Baudhāyana +triple. We call (ka, kb, kc) a scaled version of (a, b, c). +A Baudhāyana triple that does not have any common factor greater +than 1 is called a primitive Baudhāyana triple. So, (3, 4, 5) is primitive, +whereas (9,12,15) is not. +Is (5, 12, 13) a primitive Baudhāyana triple? What are the other primitive +Baudhāyana triples with numbers less than or equal to 20? +Generate 5 scaled versions of each of these primitive triples. Are these +scaled versions primitive? +If (a, b, c) is non-primitive, and the integers have f — greater than 1 — +as a common factor, then is ( +a +f , b +f , c +f ) a Baudhāyana triple? Check this +statement for (9, 12, 15). Justify this statement. +If we can find all the primitive triples, we can find all the Baudhāyana +triples. +How do we generate more primitive triples? +We know the relation between the sum of consecutive odd numbers +and square numbers. +1 = 12 +1 + 3 = 22 +1 + 3 + 5 = 32 +The sum of the first n odd numbers is n2. Let us express this algebraically. +For this, we need to know the nth odd number. What is it? +The nth odd number is 2n – 1. So, +1 + 3 + 5 + … + (2n – 3) + (2n – 1) = n2 +Sum of first (n – 1) odd numbers + +Ganita Prakash | Grade 8 | Part-II +50 +What is the sum of the first (n – 1) odd numbers? +Thus, +(n – 1)2 + (2n – 1) = n2. +Note that this equation could also have been directly obtained by +expanding (n – 1)2 and adding 2n – 1 to it. +If the nth odd number, 2n – 1, is also a square number, then we have +a sum of two square numbers equal to another square number. We will +use this idea to generate Baudhāyana triples. +1. 9 is an odd square. It is the 5th odd number (9 = 2 × 5 – 1). So, we have +1 + 3 + 5 + 7 + 9 = 52 +42 + 32 = 52 +Could we have obtained this triple using the equation (n – 1)2 + (2n – 1) = n2? +Since we took the 5th odd number, the value of n is 5. Substituting n = 5 +into the equation, we get +(5 – 1)2 + 9 = 52 +42 + 32 = 52. +2. 25 is an odd square. It is the 13th odd number (25 = 2 × 13 – 1). So, +Figure it Out +1. Find 5 more Baudhāyana triples using this idea. +2. Does this method yield non-primitive Baudhāyana triples? +[Hint: Observe that among the triples generated, one of the smaller +sidelengths is one less than the hypotenuse.] +3. Are there primitive triples that cannot be obtained through this +method? If yes, give examples. +2.6 A Long-Standing Open Problem +The study of Baudhāyana triples inspired the great French +mathematician Fermat — who lived during the 17th century — to make +a general statement about the sum of powers of positive integers. +1 + 3 + 5 + … + 23 + 25 = 132 +122 + 52 = 132 +We have n = 13. + Substituting this value in +the equation, we get +(13 – 1)2 + 25 = 132 +122 + 52 = 132 +Math +Talk + +The Baudhāyana-Pythagoras Theorem +51 +We have seen that there are an infinite number of square numbers +that can be written as a sum of two square numbers. This made Fermat +wonder if there is a perfect cube that can be written as a sum of two +perfect cubes, a fourth power that can be written as a sum of two fourth +powers, and so on. In other words, he wondered if there is a solution to +the equation +xn + yn = zn, +where x, y, and z are natural numbers, and n > 2. +In the margin of a book that dealt with properties and patterns of +positive integers (like that of Baudhāyana triples), Fermat wrote that +amongst the unending sequence of numbers, one cannot find a single +perfect cube that is a sum of two perfect cubes, a fourth power that is a +sum of two fourth powers, and so on. So, the equation has no solution +for powers greater than 2. In addition to stating this, Fermat wrote, +“I have found a truly marvellous proof of this statement, but the margin +is too small to contain it”. +No one could ever find Fermat’s proof of this statement, which is +called Fermat’s Last Theorem. +After his death, many great mathematicians tried their hand at proving +this theorem. There followed more than 300 years of failed attempts in +proving it. +In 1963, a 10-year-old boy named Andrew Wiles read a book (The +Last Problem by Eric Bell) about Fermat’s Last Theorem and its history. +Despite reading about the failures of so many great mathematicians, +this young boy resolved to prove this theorem. +He eventually did prove this theorem in 1994! + +Ganita Prakash | Grade 8 | Part-II +52 +2.7 Further Applications of the Baudhāyana - +Pythagoras Theorem +The Baudhāyana-Pythagoras theorem is one of the fundamental +theorems of geometry. Let us see some of its applications. +A Problem from Bhāskarāchārya’s Līlāvatī +The following is a translation of a problem from Bhāskarāchārya’s +(Bhāskara II) Līlāvatī. Try to visualise what you read. +“In a lake surrounded by chakra +and krauñcha birds, there is a lotus +flower peeping out of the water, +with the tip of its stem 1 unit above +the water. On being swayed by a +gentle breeze, the tip touches the +water 3 units away from its original +position. Quickly tell the depth of +the lake.” +At first glance, it seems like +there is insufficient data to find the +solution. But the solution exists! +Let x be the length of the stem inside the water. This is the required +depth of the lake. Since the stem of the lotus is sticking 1 unit above the +water, the total length of the stem is x + 1. +We can make a (very reasonable) assumption that the lotus stem is +perpendicular to the surface of the water in the first position. With this +assumption, we get a right triangle of sides 3, x and x + 1. As this satisfies +the Baudhāyana-Pythagoras theorem, we have +32 + x2 = (x + 1)2 +9 + x2 = x2 + 2x + 1. +Subtracting x2 from both sides, we get +9 = 2x + 1. +x = 4. +Thus, the depth of the lake is 4 units. +Figure it Out +1. Find the diagonal of a square with sidelength 5 cm. +2. Find the missing sidelengths in the following right triangles: + +The Baudhāyana-Pythagoras Theorem +53 +3. Find the sidelength of a rhombus whose diagonals are of length 24 +units and 70 units. +4. Is the hypotenuse the longest side of a right triangle? Justify your +answer. +5. True or False — Every Baudhāyana triple is either a primitive triple +or a scaled version of a primitive triple. +6. Give 5 examples of rectangles whose sidelengths and diagonals are +all integers. +7. Construct a square whose area is equal to the difference of the areas +of squares of sidelengths 5 units and 7 units. +8. (i) +Using the dots of a grid as the vertices, +can you create a square that has an +area of (a) 2 sq. units, (b) 3 sq. units, +(c) 4 sq.units, and (d) 5 sq. unit? + +(ii) Suppose the grid extends indefinitely. +What are the possible integer-valued +areas of squares you can create in this +manner? +7 +9 +4 +10 +40 +41 +27 +45 +200 +10 +150 +10 +Math +Talk +Try +This + +9. Find the area of an equilateral triangle with sidelength 6 units. +[Hint: Show that an altitude bisects the opposite side. Use this to +find the height.] + +y +The Baudhāyana-Pythagoras Theorem is one of the most +fundamental theorems in geometry. It expresses the +relationship among the three sides of a right-angled triangle. + +y +If a, b, c, are the sidelengths of a right-angled triangle, where c +is the length of hypotenuse, then a2 + b2 = c2. + +y +In an isosceles triangle with sidelengths a, a, c, we have the +relation a2 + a2 = 2a2 = c2, i.e., c = a +2 . + +y +The number +2 lies between 1.414 and 1.415. However, it +cannot be expressed as a terminating decimal. It also cannot +be expressed as a fraction m +n with m, n positive integers. + +y +A triple (a, b, c) of positive integers satisfying a2 + b2 = c2 is called +a Baudhāyana-Pythagoras triple. Examples include (3, 4, 5), +(6, 8, 10), and (5, 12, 13). Infinitely many such triples can be +constructed. + +y +The equation an + bn = cn has no solution in positive integers +when n > 2. This is known as ‘Fermat’s Last Theorem’. It was +proven by Andrew Wiles in 1994. +SUMMARY +Try +This +Find the Colours! +There are 3 closed boxes — one containing only red balls, the second +containing only blue balls and the third containing only green balls. The +boxes are labelled RED, BLUE and GREEN such that ‘no’ box has the +correct label. We need to find which label goes with which box. How can +this be done if we are allowed to open only one box? +Ganita Prakash | Grade 8 | Part-II" +class_8,10,proportional reasoning-2,ncert_books/class_8/hegp2dd/hegp203.pdf,"3.1 Proportionality — A Quick Recap +In an earlier chapter, we explored proportional relationships between +quantities and we used the ratio notation to represent such relationships. +When two or more related quantities change by the same factor, we call +that relationship a proportional relationship. For example, idli batter is +made by mixing rice and urad dal. The proportion of these two can have +regional variations. One of the proportions used is: for 2 cups of rice, +we add 1 cup of urad dal. We represent this relationship using the ratio +notation 2 : 1. +Viswanath made idlis by mixing 6 cups of rice with 3 cups of urad dal, +while Puneet made idlis by mixing 4 cups of rice with 2 cups of urad dal. +If cooked in the same way, would their idlis taste the same? +Viswanath’s mixture can be represented as 6 : 3 and Puneet’s mixture +as 4 : 2. +Recall that, to verify that these two ratios are proportional, we can +use the cross-multiplication method: +× +× +6 : 3 +4 : 2 +The two products are the same (12). So, the two ratios are proportional. +It is likely that the idlis would taste the same, if all the other ingredients +are proportional too! +In general, we can say that two ratios a : b and c : d are proportional if +a × d = b × c, or +a +c = b +d. +PROPORTIONAL +REASONING–2 +3 + +Ganita Prakash | Grade 8 | Part-II +56 +3.2 Ratios in Maps +Have you noticed that in many maps there is a ratio given, usually in +the lower right corner of the map? It usually contains 1 and a very large +number, such as 1 : 60,00,000. What does RF 1 : 60,00,000 mean? What +does it indicate? Can you guess? +A Representative Fraction (RF) is an expression that shows the ratio +between a distance on the map and the corresponding actual distance +on the ground. +For example, if the ratio on a map is 1 : 60,00,000, that means a distance +of 1 cm on the map is equivalent to a geographical distance of 60,00,000 cm. +Remember, this is geographical distance and not road distance! +Convert 60,00,000 cm to kilometres. +It is 60 km. Verify this. +Using the map above, can you find the geographical distance between +Bengaluru and Chennai? Also, find the geographical distance between +Mangaluru and Chennai. +[Hint: Use a ruler to find the distance between the cities on the map. +Then, use the ratio given on the map to find the actual geographical +distance.] +Map not to scale + +Proportional Reasoning–2 +57 +Try to find the distances between the same two pairs of cities with +different maps that have different scales (ratios). Do they all give the +same geographical distance, approximately? +Note to the Teacher: Bring maps and atlases to the classroom and encourage +students to observe the scale given as a ratio (usually in the lower right corner) +on the map. Use the maps and atlases in the library and ask students to find the +geographical distances between two locations on the map that are of local interest. +Ask them to verify with each other if they get similar distances and to find the +reasons if the distances are very different. +Map Making Activity: Guide students to make a sketch of their +classroom with an accurate scale (ratio of 1: 50). They should mark +the location of various objects in the classroom like the teacher᾿s +desk, blackboard, fans and lights, according to scale. Students can use +appropriate symbols to represent different objects like fans, lights, +tables, chairs, and so on. +3.3 Ratios with More than 2 Terms +Viswanath is experimenting with a spice +mix powder. He makes the powder by +grinding 8 spoons of coriander seeds, +4 red chillies, 2 spoons of toor dal, and +1 spoon of fenugreek (methi) seeds. +For his spice mix powder, the ratio of +coriander seeds to red chillies to toor +dal to fenugreek seeds is +8 : 4 : 2 : 1. +Notice that the ratio has 4 terms. +Ratios can have many terms if each of the quantities change by the same +factor to maintain the proportional relationship. +Puneet has only 2 red chillies in his kitchen. But he wants to make spice +mix powder that tastes the same as Viswanath’s spice mix powder. How +much of the other ingredients should Puneet use to make his spice mix +powder? +For Puneet’s spice mix powder to be similar to Viswanath’s, the ratio of +all the ingredients should be the same as Viswanath’s spice mix powder. +Puneet has only 2 red chillies. He has half the number of chillies that +Viswanath used in his mixture. So, the quantity of the other ingredients +should also be reduced to half. +Thus, Puneet should add 4 spoons of coriander seeds, 2 red chillies, +1 spoon of toor dal, and half a spoon of fenugreek seeds. The ratio is +4 : 2 : 1 : 0.5. + +Ganita Prakash | Grade 8 | Part-II +58 +Both the ratios are proportional to each other. We denote this by +8 : 4 : 2 : 1 :: 4 : 2 : 1 : 0.5. +In general, when two ratios with multiple terms are proportional +a : b : c : d :: p : q : r : s +then, +a +p = b +q = c +r = d +s . +Example 1: To make a special shade of purple, paint must be mixed in +the ratio, Red : Blue : White :: 2 : 3 : 5. If Yasmin has 10 litres of white +paint, how many litres of red and blue paint should she add to get the +same shade of purple? +In the ratio 2 : 3 : 5, the white paint corresponds to 5 parts. +If 5 parts is 10 litres, 1 part is 10 ÷ 5 = 2 litres. +Red = 2 parts = 2 × 2 = 4 litres. +Blue = 3 parts = 3 × 2 = 6 litres. +So, the purple paint will have 4 litres of red, 6 litres of blue, and 10 +litres of white paint. +What is the total volume of this purple paint? +The total volume of purple paint is 4 + 6 + 10 = 20 litres. +Example 2: Cement concrete is a mixture of cement, sand, and gravel, +and is widely used in construction. The ratio of the components in the +mixture varies depending on how strong the structure needs to be. For +structures that need greater strength like pillars, beams, and roofs, the +ratio is 1 : 1.5 : 3, and the construction is also reinforced with steel rods. +Using this ratio, if we have 3 bags of cement, how many bags of concrete +mixture can we make? +The concrete mixture is in the ratio +Bags of cement : bags of sand : bags of gravel :: 1 : 1.5 : 3. +If we have 3 bags of cement, we have to multiply the other terms by +3. So, the ratio is +cement : sand : gravel :: 3 : 4.5 : 9. +In total, we have 3 + 4.5 + 9 = 16.5 bags of concrete. +3.4 Dividing a Whole in a Given Ratio +In an earlier chapter, we learnt how to divide a whole in a ratio, e.g., 12 +in the ratio 2 : 1. To do this, we add the terms (2 + 1 = 3), and divide the +whole by this sum (12 ÷ 3 = 4). We multiply each term by this quotient : +2 × 4 = 8 and 1 × 4 = 4 . So, 12 divided in the ratio 2 : 1 is 8 : 4. + +Proportional Reasoning–2 +59 +We can extend this to ratios with multiple terms. +Let us look at the earlier example of making a concrete mixture. We need +to mix cement, sand, and gravel in the ratio of 1 : 1.5 : 3 to get the concrete. +Example 3: For some construction, 110 units of concrete are needed. +How many units of cement, sand, and gravel are needed if these are to +be mixed in the ratio 1 : 1.5 : 3? +For 1 unit of cement, we need to add 1.5 units of sand and 3 units of +gravel. Together, they add up to 5.5 units of concrete. We need to do this +20 times (110 ÷ 5.5 = 20) to get 110 units of concrete. So each term has to +be multiplied by 20. +1 × 20 = 20 units of cement, +1.5 × 20 = 30 units of sand, and +3 × 20 = 60 units of gravel. +So, we need 20 units of cement, 30 units of sand, and 60 units of gravel +to make the concrete. +When we divide a quantity x in the ratio a : b : c : …, the terms in the +ratio are — +x × +a +(a + b + c + …) , x × +b +(a + b + c + …) , x × +c +(a + b + c + …) , and so on. +Example 4: You get a particular shade of purple paint by mixing red, +blue, and white paint in the ratio 2 : 3 : 5. If you need 50 ml of purple +paint, how many ml of red, blue, and white paint will you mix together? +Red paint = 50 × +2 +(2 + 3 + 5) = 50 × 2 +10 = 10 ml. +Blue paint = 50 × +3 +(2 + 3 + 5) = 50 × 3 +10 = 15 ml. +White paint = 50 × +5 +(2 + 3 + 5) = 50 × 5 +10 = 25 ml. + +Ganita Prakash | Grade 8 | Part-II +60 +Example 5: Construct a triangle with angles in the ratio 1 : 3 : 5. +We know that the sum of the angles in a triangle is 180°. So the angles +are +∠A = 180° × +1 +(1 + 3 + 5) = 180° × 1 +9 = 20°. +∠B = 180° × +3 +(1 + 3 + 5) = 180° × 3 +9 = 60°. +∠C = 180° × +5 +(1 + 3 + 5) = 180° × 5 +9 = 100°. +Figure it Out +1. A cricket coach schedules practice sessions that include different +activities in a specific ratio — time for warm-up/cool-down : time +for batting : time for bowling : time for fielding :: 3 : 4 : 3 : 5. If each +session is 150 minutes long, how much time is spent on each activity? +2. A school library has books in different languages in the following ratio — +no. of Odiya books : no. of Hindi books : no. of English books :: 3 : 2 : 1. +If the library has 288 Odiya books, how many Hindi and English books +does it have? +3. I have 100 coins in the ratio — no. of ₹10 coins : no. of ₹5 coins : no. of +₹2 coins : no. of ₹1 coins :: 4 : 3 : 2 : 1. How much money do I have in +coins? +4. Construct a triangle with sidelengths in the ratio 3 : 4 : 5. Will all +the triangles drawn with this ratio of sidelengths be congruent +to each other? Why or why not? +5. Can you construct a triangle with sidelengths in the ratio 1 : 3 : 5? +Why or why not? +3.5 A Slice of the Pie +Have you seen pie charts like the one shown in the +figure? Pie charts show different proportions of a +whole. This one shows the proportions of students +that have scored each grade in an assessment. +These kinds of visualisations help us quickly +interpret data. +Let us try to create this pie chart. Here is a table +showing the grades scored by students: +Grade +A +B +C +D +E +Students +12 +10 +8 +6 +4 +20° +100° +C +A +B +60° +Math +Talk +B +C +A +D +E +Grade Distribution (40 students) +12 +4 +6 +10 +8 + +Proportional Reasoning–2 +61 +How do we mark the different slices of the pie chart? +To mark a slice in the pie chart, the angle corresponding to a grade +should be proportional to the number of students who have scored that +grade. The total angle in a circle is 360°. So, we need to divide 360 in the +ratio of 12 : 10 : 8 : 6 : 4. +Can we reduce this ratio to its simplest form? +We can use the same procedure we used to reduce a ratio with two +terms to its simplest form. We divide all the terms by their HCF to get the +simplest form. +In this example, 2 is the HCF of the terms. We get the simplest form by +dividing all the terms by 2. The ratio becomes 6 : 5 : 4 : 3 : 2. +So, the angles are: +Grade A = +6 +(6 + 5 + 4 + 3 + 2) × 360° = 6 +20 × 360° = 6 × 18 = 108°. +Grade B = +5 +(6 + 5 + 4 + 3 + 2) × 360° = 5 × 18 = 90°. +Grade C = 4 × 18 = 72°. +Grade D = 3 × 18 = 54°. +Grade E = 2 × 18 = 36°. +Now let us construct a pie chart using these angles. +Step 1: Draw a circle and mark +the radius AB as shown below: +Step 2: To draw the slice to +represent +the +proportion +of +students who got an A grade, +measure 108° from the segment +AB on A (anti-clockwise), and +mark the new radius AC. +A +B +108° +A +B +C + +Ganita Prakash | Grade 8 | Part-II +62 +Step 3: Measure 90° from AC and +draw AD. +Step 4: Measure 72° from AD and +draw AE. +108° +90° +A +B +C +D +108° +90° +A +B +C +D +E +72° +Steps 5,6: Similarly, we can +complete the rest of the pie +chart. +Step 7: You can colour and label +the different slices of the pie +chart appropriately. +108° +90° +A +B +C +D +E +72° +F +54° +36° +Grade A +Grade B +Figure it Out +1. A group of 360 people were asked to vote for their favourite season +from the three seasons — rainy, winter and summer. 90 liked the +summer season, 120 liked the rainy season, and the rest liked the +winter. Draw a pie chart to show this information. +2. Draw a pie chart based on the following information about viewers᾿ +favourite type of TV channel: Entertainment — 50%, Sports — 25%, +News — 15%, Information — 10%. +3. Prepare a pie chart that shows the favourite subjects of the students +in your class. You can collect the data of the number of students for + +Proportional Reasoning–2 +63 +each subject shown in the table (each student should choose only one +subject). Then write these numbers in the table and construct a pie +chart: +Subject +Language +Arts +Education +Vocational +Education +Social +Science +Physical +Education +Maths +Science +Number of +Students +3.6 Inverse Proportions +Do you recall the rule of three? When two ratios are proportional, i.e., +when +a : b :: c : d, then d = bc +a . We call such proportions direct proportions. +We use this understanding to find the value of the fourth quantity (d), +when the value of three quantities (a, b, and c) are given. +Example 1: If 5 workers can move 4500 bricks in a day, how many +workers are needed to move 18000 bricks in a day? +This can be represented as a statement of proportionality —  +4500 : 18000 :: 5 : x. We can find the value of x by +x = 18000 × 5 +4500 + = 20 +Thus, the number of workers needed are 20. +Example 2: Puneeth’s father went from Lucknow to Kanpur in 3 hours +by riding his motorcycle at a speed of 30 km/h. If he takes a car instead +and drives at 60 km/h, how long will it take him to reach Kanpur? +Can we represent this problem with the following statement of +proportionality — 30 : 60 :: 3 : x ? Will the travel time increase or +decrease as the speed of the motorcycle increases? +The following table shows the time taken to travel from Lucknow +to Kanpur using different modes of transport: +Walk +Bicycle +Motorcycle +Car +Speed (km/h) +5 +15 +30 +60 +Time (in hours) +18 +6 +3 +1.5 +Math +Talk +From this table, we notice that +when the speed increases, the time +taken to travel the same distance +decreases. + +Ganita Prakash | Grade 8 | Part-II +64 +Does it decrease by the same rate (or factor)? +Going by bicycle is 3 times faster than walking (15 ÷ 5). The speed has +increased 3 times. The travel time has decreased 3 times too (18 ÷ 6). +The speed has increased by the same factor by which the travel time has +decreased. +Check if this is the case for the other modes of transport. +Since both quantities, speed and time, change by the same factor, they +are proportional. But they change in opposite directions, or inversely. +Such proportions are called inverse proportions. +From the table, we can see that the product of speed and time are the +same for all modes of transport, namely 90 km. +Two quantities x and y vary in inverse proportion if there exists a +relation of the type xy = k, where k is a constant. +In the previous example, if x represents the speed and y represents +the time taken, then k is the distance between Lucknow and Kanpur, +which remains constant. +Thus, if quantities x and y are inversely proportional, and x1 and x2 +are values of x that have corresponding y values y1 and y2, then +x1 y1 = x2 y2= k (some constant). +From these we can see that +x1 +x2 + = +y2 +y1 +. +Let us check if this is true for other modes of transport, such as walking +and by car. Let us use x to represent speed and y to represent time. +x1 = 5, x2 = 60, y1 = 18, y2 = 1.5. +x1 +x2 + = 5 +60 = 0.083333... +What is the value of 1.5 +18 +y2 +y1 + ? It is 0.083333... too! +Speed (km/h) +5 +15 +30 +60 +Time (in hours) +18 +6 +3 +1.5 +× 4 +× 3 +× 2 +× 1 +3 +× 1 +2 +× 1 +4 + +Proportional Reasoning–2 +65 +Figure it Out +1. Which of these are in inverse proportion? +(i) +x +40 +80 +25 +16 +y +20 +10 +32 +50 +(ii) +x +40 +80 +25 +16 +y +20 +10 +12.5 +8 +(iii) +x +30 +90 +150 +10 +y +15 +5 +3 +45 +2. Fill in the empty cells if x and y are in inverse proportion. +x +16 +12 +36 +y +9 +48 +Example 3: 20 workers take 4 days to complete laying a road. How many +days will 10 workers take to complete laying the same length of road? +If we decrease the number of workers, then the number of days to +complete the work will increase by the same factor. So these quantities +are inversely proportional. So, x1 y1 = x2 y2. +Thus, +20 × 4 = 10 × y2 +y2 = 20 × 4 +10 = 8. +It will take 8 days for 10 workers to complete the work. +We notice that when the number of workers halved, the number of +days to complete the work doubled. Quantities are inversely proportional +if, when one quantity changes by a factor n, the other quantity changes +by the inverse 1 +n. + +Ganita Prakash | Grade 8 | Part-II +66 +Example 4: 2 pumps can fill a tank in 18 hours. How much time will it +take to fill the tank if we add 2 more pumps of the same kind? +If we add 2 more pumps, we will have 4 pumps. Let us denote the +time taken to fill the tank with 4 pumps by x. If we increase the number +of pumps by n, then the time taken to fill the tank will decrease by the +same factor n, so the quantities are inversely proportional. Since the +quantities are inversely proportional, the product remains constant. So, +2 × 18 = 4 × x. +x = 2 × 18 +4 + = 9. +It will take 9 hours to fill the tank. +Example 5: A school has food provisions to feed 80 students for 15 +days. If 20 more students join the school, for how many days will the +provisions last? +More students → fewer days the provisions will last. The quantities are +inversely proportional. If x is the number of days, +80 × 15 = 100 × x. +x = 80 × 15 +100 + = 12. +The provisions will last for only 12 days. +Example 6: If Ram takes 1 hour to cut a given quantity of vegetables +and Shyam takes 1.5 hours to cut the same quantity of vegetables, how +much time will they take to cut the vegetables if they do it together? +Consider the work done to cut the given quantity of vegetables as 1 +unit of work. Let us figure out the work done by each person in 1 hour. +• +Ram finishes the work in 1 hour, so in 1 hour he does 1 unit of work. +• +Shyam finishes the work in 1.5 hours, so in 1 hour he does 1 +1.5 = 2 +3 +units of work. +So, the work done by both in 1 hour is 1 + 2 +3 = 5 +3 units of work. +Therefore, to complete 5 +3 units of work, it takes them 1 hour if they +work together. How much time will it take them to complete 1 unit +of work? +Is the quantity of work and time taken to complete it directly or +inversely proportional? +It is directly proportional. So, this can be represented as +5 +3 : 1 :: 1 : x where x is the time taken. +Math +Talk + +Proportional Reasoning–2 +67 +Since it is a direct proportion, we know that +5 +3 +1×1 +1×1 +5 +3 +1 +5 +3 +3 +5 += += += += +x +x +. +× +If they cut the vegetables together, they will finish in 3 +5 hours. +Figure it Out +1. Which of the following pairs of quantities are in inverse proportion? +(i) The number of taps filling a water tank and the time taken to +fill it. +(ii) The number of painters hired and the days needed to paint a +wall of fixed size. +(iii) The distance a car can travel and the amount of petrol in the +tank. +(iv) The speed of a cyclist and the time taken to cover a fixed route. +(v) The length of cloth bought and the price paid at a fixed rate +per metre. +(vi) The number of pages in a book and the time required to read +it at a fixed reading speed. +2. If 24 pencils cost ₹120, how much will 20 such pencils cost? +3. A tank on a building has enough water to supply 20 families +living there for 6 days. If 10 more families move in there, how +long will the water last? What assumptions do you need to +make to work out this problem? +4. Fill in the average number of hours each living being sleeps in a +day by looking at the charts. Select the appropriate hours from this +list : 15, 2.5, 20, 8, 3.5, 13, 10.5, 18. +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +68 +5. The pie chart on the right shows the +result of a survey carried out to find the +modes of transport used by children to go +to school. Study the pie chart and answer +the following questions. +(i) What is the most common mode of +transport? +(ii) What fraction of children travel by +car? +(iii) If 18 children travel by car, how many children took part in +the survey? How many children use taxis to travel to school? +(iv) By which two modes of transport are equal numbers of +children travelling? +6. Three workers can paint a fence in 4 days. If one more worker joins +the team, how many days will it take them to finish the work? What +are the assumptions you need to make? +7. It takes 6 hours to fill 2 tanks of the same size with a pump. How +long will it take to fill 5 such tanks with the same pump? +8. A given set of chairs are arranged in 25 rows, with 12 chairs in each +row. If the chairs are rearranged with 20 chairs in each row, how +many rows does this new arrangement have? +9. A school has 8 periods a day, each of 45 minutes duration. How long +is each period, if the school has 9 periods a day, assuming that the +number of school hours per day stays the same? +10. A small pump can fill a tank in 3 hours, +while a large pump can fill the same tank in +2 hours. If both pumps are used together, +how long will the tank take to fill? +11. A factory requires 42 machines to +produce a given number of toys +in 63 days. How many machines +are required to produce the +same number of toys in 54 days? +12. A car takes 2 hours to reach a +destination, travelling at a speed +of 60 km/h. How long will the +car take if it travels at a speed of +80 km/h? +Walk +Cycle +Bus +Two- +wheeler +Car +120° +90° +60° +60° + +Proportional Reasoning–2 +69 + +y +Ratios in the form a : b : c : d : … indicate that for every a units +of the first quantity, there are b units of the second quantity, +c units of the third quantity, and so on. + +y +If x is divided into many parts in the ratio p : q : r : s : …, then the +quantity of the first part is x × +p +( p + q + r + s + …) , the quantity of +the second part is x × +q +( p + q + r + s + …) , and so on. + +y +Two quantities are directly proportional when they both +change by the same factor, and their quotient remains the +same. For example, if x and y are two quantities that are +directly proportional, and (x1, x2, x3, …) and ( y1, y2, y3, …) are the +corresponding values of x and y, then +x1 +y1 + = +x2 +y2 + = +x3 +y3 + = … = k, where +k is a constant. + +y +Quantities are inversely proportional if, when one quantity +changes by a factor n, the other quantity changes by the inverse +1 +n. For example, if x and y are two quantities that are inversely +proportional, and ( x1, x2, x3, …) and ( y1, y2, y3, …) are the +corresponding values of x and y, then x1 y1 = x2 y2 = x3 y3 = … = n, +where n is a constant. +SUMMARY" +class_8,11,exploring some geometrics themes,ncert_books/class_8/hegp2dd/hegp204.pdf,"In this chapter, we will explore two geometric themes. We will study +fractals which are self-similar shapes. They exhibit the same or similar +pattern over and over again — but at smaller and smaller scales. We will +then look at different ways of visualising solids. +4.1 Fractals +One of the most beautiful examples of a fractal that also +occurs in nature is the fern. The fern is seen to have smaller +copies of itself as its leaves, and these in turn have even +smaller copies of themselves in their sub-leaves, and so on! +Similar phenomena of self-similarity occur in trees +(where a trunk has limbs, and a limb has branches, and the +branches have branchlets, and so on), clouds, coastlines, +mountains, lightning, and many other objects in nature. +Other mathematical fractals can also be very beautiful. +We will explore some of them here. +Sierpinski Carpet +The Polish mathematician Sierpinski discovered a type of fractal known +as the Sierpinski Carpet. It is made by taking a square, breaking it into +9 smaller squares, and then removing the central square (see the figure +below); the same procedure is then repeated on the remaining 8 squares, +and so on. One then sees the same pattern at smaller and smaller scales. +Step 0 +Step 1 +Step 2 +. . . +EXPLORING SOME +GEOMETRIC THEMES +4 +Fern + +71 +Exploring Some Geometric Themes +Sierpinski Carpet +Draw the initial few steps (at least till Step 2) of the shape sequence that +leads to the Sierpinski Carpet. +By its construction, each step in the sequence has +(i) squares of the same size that remain in the figure, and the size of +these squares becomes smaller and smaller as the step number +increases, and +(ii) square holes that are formed by removing square pieces. +Do you see any pattern in the number of holes and squares that remain +at each step? +Let Rn represent the number of remaining squares at the nth step, and +Hn represent the number of holes at the nth step. +Let us understand how these numbers grow by analysing how the +holes and squares that remain are generated from the previous step. +Every square that remains at a given step, say Step n, gives rise to +8 squares that remain at the (n + 1)th step. Thus, we have +Rn + 1 = 8 Rn. +Can this be used to get a formula for Rn? +We have +R0 = 1 +R1 = 8 × 1 = 8 +R2 = 8 × 8 = 82. +In general, Rn = 8n. +Similarly, how do we find the number of holes at a given step? +Every square that remains at the nth step gives rise to a hole in the +(n + 1)th step. All the holes present at the nth step remain in the (n + 1)th +step as well. Thus, +Hn + 1 = Hn + Rn. +So we have +R0 = 1 +H0 = 0 +R1 = 8 +H1 = 1 +R2 = 82 +H2 = 1 + 8 +R3 = 83 +H3 = 1 + 8 + 82 +... + +Ganita Prakash | Grade 8 | Part-II +72 +Sierpinski Gasket +Sierpinski came up with another fractal made in a similar way. An +equilateral triangle is broken up into 4 identical equilateral triangles by +joining the midpoints of the bigger triangle, and then the central triangle +is removed. This procedure is repeated on the 3 remaining triangles, +and so on. +Step 0 +Step 1 +Step 2 +Sierpinski Triangle +. . . +Show that by joining the midpoints of an equilateral triangle, we divide +it into 4 identical equilateral triangles. +[Hint: Note that the corner triangles are isosceles.] +This fractal is called the Sierpinski Triangle/Gasket. +Figure it Out +1. Draw the initial few steps (at least till Step 2) of the shape sequence +that leads to the Sierpinski Triangle. +2. Find the number of holes, and the triangles that remain at each step +of the shape sequence that leads to the Sierpinski Triangle. +3. Find the area of the region remaining at the nth step in each of the +shape sequences that lead to the Sierpinski fractals. Take the area of +the starting square/triangle to be 1 sq. unit. + +73 +Exploring Some Geometric Themes +Koch Snowflake +The Koch Snowflake is another fractal, named after the Swedish +mathematician Von Koch, who first described it in 1904. We have already +encountered this fractal in Grade 6, Ganita Prakash. +Step 0 +Step 1 +Step 2 +. . . +Koch Snowflake +To generate it, we start with an equilateral triangle. Each side is +(i) divided into 3 equal parts, and +(ii) an equilateral triangle is raised over the middle part, and then +the middle part is removed. +Effectively, each side gets replaced by a ‘bump’ - shaped structure +. +This procedure is repeated on the sides of the new resulting shape, and +so on. +Figure it Out +1. Draw the initial few steps (at least till Step 2) of the shape sequence +that leads to the Koch Snowflake. +2. Find the number of sides in the nth step of the shape sequence that +leads to the Koch Snowflake. +3. Find the perimeter of the shape at the nth step of the sequence. +Take the starting equilateral triangle to have a sidelength of 1 unit. + +Ganita Prakash | Grade 8 | Part-II +74 +Fractals in Art +Fractals have also long been used in human-made art! Perhaps the +oldest such fractals appear in the temples of India. An example occurs +in the Kandariya Mahadev Temple in Khajuraho, Madhya Pradesh +which was completed in around 1025 C.E.; there one sees a tall temple +structure, which is made up of smaller copies of the full structure, +on which there are even smaller copies of the same structure, and +so on. Fractal-like patterns also occur in temples in Madurai, Hampi, +Rameswaram, Varanasi, among many others. +Kandariya Mahadev Temple +Fractals are also common in traditional African cultures. For example, +patterns on Nigerian Fulani wedding blankets often exhibit fractal +structures. +Nigerian Fulani Wedding Blanket + +75 +Exploring Some Geometric Themes +In the blanket shown in the figure, there are diamond shaped patterns, +inside of which there are smaller diamond shaped patterns, and so on. +The modern maestro of fractal art is undoubtedly the Dutch artist +M.C. Escher. We have already seen that some of his prints explore the +mathematical theme of tiling. One famous example involving fractals +is his work ‘Smaller and Smaller’, which exhibits the identical pattern +of lizards but at smaller and smaller scales. Here is a print inspired by +Escher’s ‘Smaller and Smaller’. +4.2 Visualising Solids +Build it in Your Imagination +We will start this section by practising visualisation. For each prompt, +feel free to talk to your partner, gesture, draw it in the air — but do not +actually draw on paper! +My method is different. I do not rush into +actual work. When I get an idea I start at once +building it up in my imagination. I change the +construction, make improvements, and operate +the device entirely in my mind. +— Nikolas Tesla (1856 – 1943), the great +Serbian-American engineer and inventor, who +made fundamental contributions to electrical +engineering and other fields. +1. Picture your name, then read off the letters backwards. Make sure +to do this by sight, not by sound — really see your name! Now try +with your friend’s name. + +Ganita Prakash | Grade 8 | Part-II +76 +2. Cut off the four corners of an imaginary square, with each cut going +between midpoints of adjacent edges. What shape is left over? How +can you reassemble the four corners to make another square? +3. Mark the sides of an equilateral triangle into thirds. Cut off each +corner of the triangle, as far as the marks. What shape do you get? +4. Mark the sides of a square into thirds and cut off each of its corners +as far as the marks. What shape is left? +When we see a solid object, we are really seeing its profile from a +specific viewpoint. Depending on our viewpoint, the outline of this +profile can vary dramatically! +Perhaps you’ve noticed interesting profiles of objects yourself, while +taking a photograph or observing the shadow of an object? Or, maybe +you’ve seen a cartoon like the following, in which a character bursts +through a wall and leaves a hole shaped like their outline! + +77 +Exploring Some Geometric Themes +In previous classes, you’ve seen solids that are much simpler than +an elephant or cat, such as cubes, spheres, cylinders, and cones. What +would the profiles of these look like, from different viewpoints? +Can you describe a solid and a viewpoint that would result in each +of the following cases? If it helps, you can imagine the solid passing +through a wall like Tom did, and leaving a hole of the appropriate shape. +5. A solid whose profile has a square outline +6. A solid whose profile has a circular outline +7. A solid whose profile has a triangular outline +As we saw with the elephant, a given solid might have very different +profiles from different viewpoints. Can you visualise solids that have +the following contrasting profiles? +Spend some time on this, and if you are finding it difficult to visualise, +you may look around and use objects that are around you, or that you +will make in the next section. Feel free to consider viewpoints from any +direction, including directly above the object. +8. A solid with a rectangular profile from one viewpoint and a circular +profile from another viewpoint +9. A solid with a circular profile from one viewpoint and a triangular +one from another viewpoint +10. A solid with a rectangular profile from one viewpoint and a +triangular one from another viewpoint +11. A solid with a trapezium shaped profile from one viewpoint and a +circular one from another viewpoint +12. A solid with a pentagonal profile from one viewpoint and a +rectangular one from another viewpoint +Are there unique solids for each of the conditions, or can you come up +with multiple possibilities? +Making Solids +Many basic solid shapes such as a cuboid, parallelopiped, cylinder, +cone, prism, and pyramid can be made using foldable flat materials like +paper, cardboard, and even metallic sheets! This is one of the methods +for manufacturing hollow solids. + +Ganita Prakash | Grade 8 | Part-II +78 +Cuboid +Parallelopiped +Cylinder +Cone +Triangular prism +Triangular +pyramid +While discussing solids that contain plane surfaces in their boundaries, +the notion of faces, edges and vertices serves a useful purpose. +Vertex +Edge +Face +Vertex +Edge +Face +Faces are the plane/flat surfaces of a solid that form its boundary. +Edges are the line segments that form the sides of the faces, and vertices +are the points at which the edges meet. For example, a cuboid or cube +has 6 faces, 12 edges and 8 vertices. + +79 +Exploring Some Geometric Themes +There are multiple types of prisms and pyramids, whose names +depend on the shapes of their faces. +Triangular +Prism +Pentagonal +Prism +Hexagonal +Prism +Prism: A prism, in general, has two congruent polygons as opposite +faces, with edges connecting the corresponding vertices of these +polygons. All the other faces are parallelograms. Based on the shape +of the congruent polygons, prisms may be called triangular prisms, +pentagonal prisms, and so on. +Pyramids: A pyramid, in general, has a polygonal base and a +point outside it, and edges connect the point with each of the vertices +of the base. Based on the shape of the base, pyramids may be called +triangular pyramids, square pyramids, pentagonal pyramids, and so on. +A triangular pyramid is also known as a tetrahedron. +Triangular +Pyramid +Pentagonal +Pyramid +Hexagonal +Pyramid +If the congruent polygons of a prism have 10 sides, how many faces, +edges and vertices does the prism have? What if the polygons have n +sides? +If the base of a pyramid has 10 sides, how many faces, edges and vertices +does the pyramid have? What if the base is an n-sided polygon? +Now we will see how to make different solids using a foldable flat +surface such as paper, cardboard, etc. +The basic idea is to create a shape on a flat surface that can be folded +into the solid. Such a shape is called a net. In other words, a net is +obtained by ‘unfolding’ a solid onto a plane. + +Ganita Prakash | Grade 8 | Part-II +80 +What is a net of a cube? +Fig. 4.1 is a net of a cube. +Visualise how it can be folded to form a cube. +Practical Aspects of Using a Net +To make an actual cube from its net, certain practical +aspects have to be kept in mind. One is that the material +used must have sufficient sturdiness for the cube to +stand. The second aspect is that there should be some +mechanism for attaching the faces together. If cardboard or a similar +material is used, cello tape is one way to join the adjacent faces. For less +sturdy materials, such as chart paper, it is useful to have extra ‘flaps’ on +some of the faces that can be stuck to the adjacent faces. You might have +observed this strategy in packaging boxes. +(i) +(ii) +(iii) +However, when we talk about the net of a solid, we consider only the +shape formed by unfolding the solid, and not the other supporting flaps +that we may use to actually make the solid. +Apart from using a net, there are other ways to make a cube +from paper. +There are multiple possible nets one can use to build a cube. +Figure it Out +1. Which of the following are the nets of a cube? First, try to answer by +visualisation. Then, you may use cutouts and try. +(i) +(ii) +(iii) +Fig. 4.1 + +81 +Exploring Some Geometric Themes +(iv) +(v) +(vi) +2. A cube has 11 possible net structures in total. In this count, two nets +are considered the same if one can be obtained from the other by a +rotation or a flip. For example, the following nets are all considered +the same — + Find all the 11 nets of a cube. +3. Draw a net of a cuboid having sidelengths: +(i) 5 cm, 3 cm, and 1 cm +(ii) 6 cm, 3 cm, and 2 cm +Let us find out the nets of other solids. +A tetrahedron whose faces are equilateral triangles is called a regular +tetrahedron. +What is a net of a regular tetrahedron? Which of the following are nets +of a regular tetrahedron? +Are there any other possible nets? +A regular tetrahedron has only 2 possible nets. +Draw a net with appropriate measurements that can be folded into +a regular tetrahedron. Verify if it works by making an actual cutout. +Draw a net with appropriate measurements that can be folded into +a square pyramid. Verify if it works by making an actual cutout. +Try +This +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +82 +What is the net of a cylinder? +If the circular faces of a cylinder are unfolded, and if a cut is made +along the height of the cylinder, as shown in the figure below, then +we get +height +What are the sidelengths of the rectangle obtained? +How will the net of a cone look? +O +l +If the cone is slit open along the line l and then unrolled, what will we +get? +Observe that all the points on the boundary of the base +circle are at equal distances from ‘O’. So after unrolling the +cone, the boundary of the net will be a portion of a circle +with centre O. +What surface do you construct by using the above net, in which O +is not the centre of the boundary circle? Make a physical model to +help you answer this question! +Draw a net with appropriate measurements that can be folded into +a triangular prism. Verify that it works by making an actual cutout. +O +Math +Talk + +83 +Exploring Some Geometric Themes +Here is a solid called the octahedron, made by joining two square +pyramids at their square bases. +Octahedron +Can you visualise its net? This is one of its nets. +Taking all the triangles in the net to be equilateral, make a cutout +of the net and fold it to form an octahedron. +As in the case of cubes, an octahedron too has 11 different nets. +So far, we have seen solids whose faces are all equilateral triangles +(regular tetrahedrons), and all squares (cubes). Does there exist a solid +whose faces are all pentagons? Interestingly, the answer is yes! +This solid is called a dodecahedron. This solid too can be made from +a net! Mathematicians have been even able to determine exactly how +many nets a dodecahedron has — 43,380. +Dodecahedron +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +84 +Net of a sphere? Experiment and see if you can make a paper cutout +that can perfectly wrap around a ball without leaving any wrinkles, gaps +or overlaps. +Shortest Paths on a Cube +Let us consider an interesting problem related to the discussion so far. +We know that on a plane, the shortest path between two points is the +straight line between them. Now, what is the shortest path between two +points on the surface of a cuboid, if we are allowed to travel only along +its surface? +Let us imagine a hungry ant living on the surface of a cuboid. To its +good fortune, there is a laddu on the surface. +Laddu (at the centre of the top face) +Ant (at the centre of the side face) +What is the shortest path for the ant to reach the laddu? +What about in the following case? +Laddu (at the centre of the edge) +If we think that a certain path is the shortest, how can we be sure that it +truly is, among all the infinite possibilities? +For example, are either of these the shortest path? +Math +Talk + +85 +Exploring Some Geometric Themes +[Hint: Draw the net and see how these paths appear on the net.] +What does this show? +A path on the surface of the cuboid can be transformed to a path of the +same length on the net. Conversely, every path on the net transforms to +a path of the same length on the cuboid. +In the first case, the path marked on the cuboid is the shortest since +it transforms to a straight line between the ant and laddu on the net, +which is the shortest path between them. +In the second case, the path marked on the cuboid does not transform +to a straight line path on the net, and hence is not the shortest path for +the ant to take. +Laddu (at the centre of the edge) +The red path is the shortest +Thus, by using a net, we convert the problem of finding the shortest +path on a cuboid to the problem of finding the shortest path on the net. +Have we now completely analysed the problem of finding the shortest +path between two points on a cuboid? + +Ganita Prakash | Grade 8 | Part-II +86 +Find the shortest path between the ant and the laddu in the following +case: +Ant (at the centre of the face) +8 cm +2 cm +4 cm +4 cm +If we unfold the cuboid as before, we get +8 cm +4 cm +4 cm +4 cm +4 cm +4 cm +4 cm +4 cm +Notice the line segment between the ant and laddu going outside the +net! Obviously, this does not correspond to any path on the cuboid. +So what do we do now? +If the net is unfolded in the following way, then we get a path on the +cuboid. +The shortest path +Thus, the way a cuboid is unfolded matters! + +87 +Exploring Some Geometric Themes +Now we will look at a trickier case. +What is the length of the shortest path between the ant and the laddu? +6 cm +30 cm +12 cm +1 +cm +1 +cm +laddu (stuck to the +back of the box) +12 cm +12 cm +6 cm +12 cm +Here are some of the ways of unfolding the cuboid. +42 cm +24 cm +32 cm +In the second case, the distance d between the ant and the laddu can +be calculated using the Baudhayana Theorem, since we have a right +triangle here. +d2 = 242 + 322 +d = 1600 = 40 cm +We see that in each of these unfoldings, the lengths of the line segments +between the ant and the laddu are different! So we have to carefully list +all the possible different unfoldings to find the answer! +Representation of Solids on a Plane Surface +Drawing is one of the oldest human activities. People have been +doing it for thousands of years for both aesthetic as well as practical +and engineering purposes. Regarding the latter, making a drawing +of an object is a useful way to record or convey information about it. +Visual representation also helps us in thinking about the objects being +represented. Such needs often arise in engineering — while constructing +buildings, designing machines, etc. Further, engineering drawings need +Try +This + +Ganita Prakash | Grade 8 | Part-II +88 +to be detailed enough so that one can physically construct the objects +they depict. +We will now explore some ways of drawing solids on a plane. +Projections +The net of a solid (when it exists) is one way of representing a solid +on the plane. But it can be difficult to visualise the solid from the net, +without physically cutting out the net and folding it up. +Another way to represent a solid is by looking at the projections of all +its points on a plane. This idea is closely related to the profile of a solid +from a specific viewpoint, which we discussed earlier. Let us see how +this technique works. +O +X +P +M +O is the projection of P +Consider a point P in space, and a plane M. Imagine a line from P +intersecting the plane M at a point O. +We say that OP is perpendicular to the plane if for any line OX on +the plane, ∠POX = 90°. In this case, we say that point O is the projection +of point P on the plane. The projections of all the points of an object +together form the projection of the object on the plane. +Let us visualise the projection of a line. +Fig. 4.2 +Projection +Projection + +89 +Exploring Some Geometric Themes +What happens to the length of a line in its projection? +Let us consider projections in Fig. 4.3. +For example let l be the actual length of the +line and p be the length of its projection. +Draw AE ⊥ BC. AECD is a rectangle (why?). +So, AE = DC = p. Also, ∠AEB = 90°. +Can you now compare the lengths p and l? +When is the length of the projected line equal +to its actual length? +What do you think are the different possible projections of a square +that we get based on its orientation? +What do you think is the projection of a parallelogram under +different orientations? +Can this ever be a quadrilateral that is not a parallelogram? +As a starting point, you could think about the projection of a pair of +parallel lines. +What can you say about the projection of an n-sided regular polygon? +[Hint: Projection of a polygon is composed of the projections of its +sides.] +Let us consider projections of solids now. +How would the projections of a cube and a cone look? +Fig. 4.4 +Fig. 4.5 +D +C +p +A +E +p +B +l +Fig. 4.3 +Math +Talk +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +90 +If an object were to pass perpendicularly through a plane and form +a hole, the shape of the hole would be the same as the shape of the +projection of the object. +See Figures 4.2 – 4.5. In each case, see if you can visualise another object +that gives the same projection. +Fig. 4.6 +Different lines and different cuboids giving the same projections +We see that a given projection is not made by a unique object. + +91 +Exploring Some Geometric Themes +Find another object that makes the same projection as that of a given +cone. +For this reason, we often take three mutually perpendicular +projections of an object. As shown in the figure, we consider a plane in +front of the object called the vertical plane, a plane below the object +called the horizontal plane and a plane to the side of the object called +the side plane. +Front +View +Side +View +Top +View +Vertical +Plane +Side +Plane +Horizontal +Plane +We give names to the projections on these three planes based on the +directions from which the object is viewed. The projection on the vertical +plane is called the front view, on the horizontal plane the top view and +on the side plane the side view. These projections formalise the notion +of profiles that we explored in an earlier section. +Let us see what these projections are for the objects shown in Fig. 4.6, +when the planes shown are taken to be vertical planes. +Projections of the different lines in Fig. 4.6 — + +Ganita Prakash | Grade 8 | Part-II +92 +Front View +Top View +Side View +Projections of the different cuboids in Fig. 4.6 — +Front View +Top View +Side View +Figure it Out +1. Observe the front view, top view and side view of the different +lines in Fig. 4.6. Is there any relation between their lengths? +2. Find the front view, top view and side view of each of +the following solids, fixing its orientation with respect +to the vertical, horizontal and side planes: cube, cuboid, +parallelepiped, cylinder, cone, prism, and pyramid. If needed, +see the next problem for clues. +Math +Talk + +Exploring Some Geometric Themes +3. Match each of the following objects with its projections. +FRONT +TOP +SIDE + +Ganita Prakash | Grade 8 | Part-II +94 +Shadows +Place an object in front of a plane, such as a wall of your room. Shine a +torch light on the object in a direction perpendicular to the wall. +What do you see? +We will see that the shape of the shadow on a plane is quite similar +to the shape of the projection on that plane! However the shadow may +be scaled up, stretched, or even distorted slightly, depending on how the +object is held. +Observe what happens to the size of the shadow as you vary the distance +between your torch and your object. +Why does this happen? +Now, imagine your torch is extremely powerful and will continue +casting a shadow of the object on the wall, no matter how far back you +move it. +If this imaginary torch always points in a direction perpendicular +to the wall, then, as the distance between the torch and the object +increases, the shadow of the object will become indistinguishable from +the projection of the object to the wall plane! +Although this experiment may sound fanciful, you have encountered +one such extremely powerful and distant torch before … namely, the +Sun! Indeed, when sunlight is perpendicular to a plane, the shadows it +casts on that plane are indistinguishable from projections! +Earlier we asked what the projection of a parallelogram might look +like. You can now physically answer this question by making a cutout +of a parallelogram and viewing its shadows under sunlight. No matter +how you orient the parallelogram, you will see that the shadow remains +a parallelogram! + +More generally, the projection of a pair of parallel lines will always +remain parallel. +Try +This + +95 +Exploring Some Geometric Themes +This property is useful for drawing projections of objects. +Figure it Out +1. Draw the top view, front view and the side view of each of the +following combinations of identical cubes. +Front +Top +Side +Front +Top +Side +Top +Front +Top +Side +Front +Top +Side +Front +Side +Top +Side +Front +2. Imagine eight identical cubes, glued together along faces to form +the letter ‘ +’. +(i) This looks like a ‘ +’ from the front. What does it look like from +the side? From the top? +(ii) Glue additional cubes to make a shape that looks like ‘ +’ +from the front and ‘ +’ from the top. +(iii) Now, can you glue even more cubes to make it look like ‘ +’ +from the front, ‘ +’ from the top, and ‘ +’ from the side? +(iv) Can you think of other letter combinations to make with +a single combination of cubes in this manner? +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +96 +3. Which solid corresponds to the given top view, +front view, and side view? +Front View +Top View +Side View +Front +Top +Side +Front +Top +Side +Top +Side +Front +Side +Front +Top +(i) +(ii) +(iii) +(iv) +Front +Front +Front +Side +Side +Side +Top +Top +Top +(v) +(vi) +(vii) +4. Using identical cubes, make a solid that gives the following +projections. +Top View +Front View +Side View +Top View +Front View +Side View +Top View +Front View +Side View +(i) +(iv) +(vii) +(ii) +(v) +(viii) +(iii) +(vi) +(ix) + +97 +Exploring Some Geometric Themes +5. Find the number of cubes in this stack of identical cubes. +6. What are the different shapes the projection of a cube can make +under different orientations? +Isometric Projections +In general we lose information when projecting a solid to a plane. But +depending on the orientation of the solid, we can sometimes recover +much of the information we’ve lost. +Let us consider an orientation of a cube in which the lengths of the +projections of all the edges of the cube are equal. Such a projection is +called the isometric projection of the cube. ‘Isometric’ means ‘equal +measure’ in Greek. +Imagine balancing a cube perfectly on one of its corner vertices, and +then projecting it down to the floor plane. This projection is isometric +and appears as shown. +Construct a model of a cube and use your hands to keep it balanced +on one corner vertex. Can you try to understand why all the +projected edges have equal length? +The isometric representation of a cube is a regular +hexagon. If the cube is made of glass, then all the edges +would be visible. +This structure is the basis for isometric drawing! Tile +the plane with hexagons and we get an isometric grid. +Math +Talk +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +98 +Isometric grids are widely used in engineering. They make it easy to +draw projections of solids and to measure lengths along all the 3 primary +directions — length, depth and height, as shown in Fig. 4.7. +Vertical +Plane +Side Plane +Horizontal +Plane +height +depth +length +Fig. 4.7 +Drawing on Isometric Grids +Have you played Tetris? There are five basic shapes in Tetris, +corresponding to the different ways of arranging four squares. +(i) +(ii) +Fig. 4.8 +(iii) +(iv) +(v) + +99 +Exploring Some Geometric Themes +Imagine these are cubes, not squares. Draw each of these on your +isometric paper (you can find it at the end of the book). +As seen in Fig. 4.7, there are three principal axes of the solid shape, +which we will call the depth axis, length axis, and height axis. +The edges of the grid appear in three different orientations: +, and |. We will associate height to | , depth to , and length to . +length +depth +height +While drawing on the grid, it may be useful to draw edge by edge, +counting the number of units you want to go along a given axis. +For example, you can draw a 1 × 1 × 1 cube as follows. How would you +draw a 2 × 2 × 2 cube? Feel free to add shading, if it helps you visualise +the solid. +The first tetris shape is a row of four cubes joined face-to-face. In +order to draw this solid, you will need to choose an orientation for the +row of four cubes. +Let’s draw it oriented along the depth axis. You could draw it +cube-by-cube, but in that case you may need to erase the lines that get +hidden by additional cubes. (If you don’t have an eraser, you can also +draw faint lines initially, and then darken the visible ones later.) +length +depth +height + +Ganita Prakash | Grade 8 | Part-II +100 +You could also draw it all at once, by counting edges as you go: +length +depth +height +Similarly, you could draw the shape oriented along the height axis or +the length axis. +In your drawing, moving in the vertical direction on the isometric +paper corresponds to moving along the height axis of the solid. Similarly, +moving along the or direction on the paper corresponds to moving +along the length or depth axis respectively of the solid. +Why is this correspondence between directions on isometric paper and +axes of the solid so effective for communicating the shape of the solid? +Earlier we observed that parallel lines project to parallel lines. This +geometric fact ensures that the three families of parallel edges on the +solid (corresponding to height, the length, and depth) project to three +families of parallel edges on the isometric paper: |, , and . Moreover, +the isometric nature of the projection ensures that the projections of +unit distances along the three axes are equal. +Can you try drawing the other tetris shapes on isometric paper? +Figure it Out +1. In addition to the 5 ways shown in Fig. 4.8, are there any +additional ways of gluing four cubes together along faces? Can +you visualise and draw these as well? +length +depth +height +Math +Talk + +101 +Exploring Some Geometric Themes +2. Draw the following figures on the isometric grid. +[Hint: It may be useful to determine whether the edge to be currently +drawn — say, along the height — goes from down to up or up to down. +Accordingly, draw the line segment on the grid either in the direction +of the height axis or opposite to it.] +3. Is there anything strange about the path of this ball? Recreate +it on the isometric grid. +[Hint: Consider a portion of this figure that is physically realisable +and identify the 3 primary directions.] +4. Observe this triangle. +(i) Would it be possible to build a model out of actual cubes? What +are the front, top, and side profiles of this impossible triangle? +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +102 +(ii) Recreate this on an isometric grid. +(iii) Why does the illusion work? + +y +Fractals are self-similar geometric objects found in nature and +in art. + +y +The Sierpinski Carpet, Sierpinski Gasket, and Koch Snowflake +are some examples of mathematical fractals. They can be +obtained by repeatedly applying certain geometric operations +that generate a sequence of shapes approaching the fractal. + +y +Cuboids, tetrahedrons, cylinders, cones, prisms, pyramids, and +octahedrons are some of the solids that can be obtained by +folding suitable nets. + +y +The shortest path between two points on the surface of a cuboid +can be found by using a suitable net of the cuboid. + +y +Any object can be represented on a plane surface by using its +projections on plane surfaces. For this purpose, we generally +use the front view (projection on the vertical plane), top view +(projection on the horizontal plane), and side view (projection +on the side plane) of the object. + +y +A cube can be oriented such that the lengths of all its edges +in the projection are equal. Such a projection is called the +isometric projection. Isometric projections of different solids +can be drawn using isometric grid paper. +SUMMARY +Math +Talk" +class_8,12,tales by dots and lines,ncert_books/class_8/hegp2dd/hegp205.pdf,"5.1 The Balancing Act +Last year, we learnt about the mean and median. Recall that the mean of +some data is the sum of all the values divided by the number of values in +the data. The median is the middle value when the data is sorted. +We shall try to understand the mean and median from a different +perspective and see how the mean behaves with changing data. +Consider any 2 numbers. Find their average/arithmetic mean. Repeat +this by taking other pairs. What do you observe? +For example, let the two numbers be 3 and 7. Their average is +3 + 7 +2 = 5. Taking another pair of numbers, say 8 and 9, their average is +8 + 9 +2 = 8.5. Visualising these as dot plots we get +Notice that the mean is exactly halfway between the two numbers. +We have learnt earlier that the arithmetic mean is a measure of +central tendency and represents the ‘centre’ of the data. Let us see how +the mean represents the ‘centre’ in the case of 3 numbers. +Calculate and mark the mean of each collection of data below. +TALES BY DOTS +AND LINES +5 + +Ganita Prakash | Grade 8 | Part-II +104 +Can you explain how the mean is the centre of each collection? +Mark the mean for the collections below. +Can you explain how the mean is the centre of each collection? +Is the mean the midpoint of the two endpoints/extremes of the +data? It is not always so. Instead, the total distances are equal on +both the sides of the mean. This is illustrated through the following +dot plots. +Verify that this holds for all the collections of data shown earlier. +Can there be more than one such ‘centre’? In other words, is there any +other value such that the sum of the distances to the values lower than +it and the values higher than it will still be equal? +In the case of the collection 10, 10, 11, and 17 whose mean is 12, suppose +there is a different centre larger than 12. +Clearly, all the distances on the LHS will increase and the distances +on the RHS will decrease. Thus, it is no longer the ‘centre’. Similarly, for +any value smaller than 12, the distances on the LHS will decrease while +those on the RHS will increase. +Math +Talk +Math +Talk + +Tales by Dots and Lines +105 +Both these cases are illustrated in the following diagram. Therefore, +there is only one centre. +Will including a new value in the data increase or decrease the mean? +When a new value greater than the mean is included, the mean +increases to maintain the balance between the sum of distances on the +LHS and RHS, as illustrated below. +Similarly, if a value smaller than the mean is included, then the new +mean will be less than before. +What happens to the mean when an existing value is removed? +When will the mean increase, decrease, or stay the same? +What happens to the mean if a value equal to the mean is included +or removed? +Try to explain this using the fair-share interpretation of mean that we +studied last year. +Unchanging Mean! +We saw earlier how the mean varies when a value is included or +removed. +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +106 +Explore if it is possible to include or remove 2 values such that the mean +is unchanged. +You may use the following data to experiment with. +How about including or removing 3 values without changing the mean? +Is it possible? +Can we include 2 values less than the mean and 1 value greater than the +mean, so that the mean remains the same? +One of the possibilities is shown here. +Try to include 2 values greater than the mean and 1 value less than the +mean, so that the mean stays the same. +Relatively Unchanged! +We saw what happens to the mean when values are included or removed +from the collection. What happens to the mean if every value in the +collection increases by some fixed number? +Consider the data: 8, 3, 10, 13, 4, 6, 7, 7, 8, 8, 5. Calculate its mean. +Now, consider this data with every value increased by 10: 18, 13, 20, 23, 14, +16, 17, 17, 18, 18, 15. What is its mean? Is there a quicker way to find out? +[Hint: Observe the following dot plots corresponding to the two data +collections.] + +Tales by Dots and Lines +107 +The mean of the new collection also increases by 10. Notice that the +relative position of the mean stays the same. +We get the following dot plot if we reduce every value by 1. +This can be explained using algebra — +Suppose there are n values in the collection. Let these values be +represented by x1, x2, x3, … xn. Their average is given by — +x1 + x2 + x3 + … + xn +n + = a. +When a fixed number, for example, 3 is added to every value in the +collection, the new average becomes — +(x1 + 3) + (x2 + 3) + (x3 + 3) + … + (xn + 3) +n += +x1 + x2 + x3 + … + xn + 3n +n += +x1 + x2 + x3 + … + xn +n + + 3n +n += a + 3. +That is, the new average is 3 more than the previous average. +Try to explain, using algebra, what the average is when a fixed +number, e.g., 2 is subtracted from every value in the collection. +Try to explain this using the fair-share interpretation of average +that you learnt last year. +What happens to the average if every value in the collection is +doubled? +You may have guessed that the average also doubles. The following is +an example with the data we saw earlier — +Try +This + +Ganita Prakash | Grade 8 | Part-II +108 +We can see that the average has doubled. +Let us prove it using algebra. +Suppose there are n values in the collection. Let these values be +represented by x1, x2, x3, … xn. Their average is — +x1 + x2 + x3 + … + xn +n + = a. +When a fixed number, for example, 5 is multiplied to every value in +the collection, the new average becomes — +(5x1) + (5x2) + (5x3) + … + (5xn) +n += (x1 + x2 + x3 + … + xn) × 5 +n + (using the distributive property) += (x1 + x2 + x3 + … + xn) +n + × 5. += 5a. +That is, the new average is 5 times more than the previous average. +Tinkering with Median +We know that the median is the middle value in the sorted data — there +are an equal number of values less than it and greater than it. +Will including a new value to the data increase or decrease the median? +Let us consider the following data. The median of this data is 8. +Median = 9.5 +Median = 8 + +Tales by Dots and Lines +109 +Suppose we include a value 11. The new value included is greater +than the (earlier) median — the median can no longer be 8 as there are +more values greater than it. Therefore, the median will also increase. +The median, now, will be the average of the two middle values 8 and 11, +which is 9.5. +We can similarly argue that when a value less than the median is +included, the median will decrease. +Finding the Unknown +Coach Balwan noted down the weights of the kushti +players (wrestlers) and the mean as shown. But one +value that was written down got smudged. Can you find +out the missing value? +Average weight of the players = +Sum of weight of all players +Number of players +. +Let the unknown weight be w kg. += 42 + 40 + 39 + 33 + 48 + 38 + 42 + 35 + 32 + w +10 + = 39.2. +Simplifying this we get, +349 + w = 392, +w = 392 – 349 = 43. +The missing value is 43 kg. +Venkayya keeps track of the coconut harvest +in his farm. He calculates the average harvest +per tree as 25.6. His son verifies the counts +and finds that one tree’s harvest count is +incorrectly noted as 3 more than the actual +number. Can you find the correct average if +the number of trees is 15? +The average harvest per tree = +Total number of coconuts harvested +Number of trees +. +The data of harvest per tree is not given. +Can we still find out the number of coconuts harvested? +Let the initial number of coconuts harvested be z. Based on what is given, +25.6 = z +15, +Simplifying, +z = 25.6 × 15 = 384. +The initial count of coconuts harvested is 384. +We know that one tree’s count is 3 more than the actual. Therefore, +the actual total harvest count is 384 – 3 = 381. +The correct average harvest is 381 +15 = 25.4. + +Ganita Prakash | Grade 8 | Part-II +110 +Mean and Median with Frequencies +What is the average family size of students in your class? How would +you find this out? +We can collect the data of how many family +members each student has, add them up, and divide +it by the number of students. The family size data of +students in a class is shown in the table here. +What is the average family size of this class? +Some of you may have thought, “Easy! It will be +3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 +8 + = 52 +8 = 6.5.” +Remember that finding the average involves +adding all the values in the data. The number 3 occurs +three times, the number 4 occurs eleven times, and +so on. Do these reflect in your calculation? +We know that the average = Sum of all the values in the data +Number of value in the data . +Accounting for the frequencies of each value, the average will be += (3 × 3) + (4 × 11) + (5 × 9) + (6 × 7) + (7 × 3) + (8 × 1) + (9 × 1) + (10 × 1) +3 + 11 + 9 + 7 + 3 + 1 + 1 + 1 += 9 + 44 + 45 + 42 + 21 + 8 + 9 + 10 +36 += 188 +36 = 5.22. +The average family size of this class is 5.22. +What is the median family size of this class? +We know that there are 36 values in the +data. The median would be the average of +the 18th and 19th value in the sequence when the data is ordered. +Do we need to write all the 36 numbers in order? Is there a quicker way +to find out? +We can make use of the table where the frequencies are listed. We +successively add the frequencies starting from the smallest value until +we reach 18 and 19. +Number +3 +4 +5 +6 +7 +8 +9 +10 +Frequency +3 +11 +9 +7 +3 +1 +1 +1 +Adding the frequencies of 3 and 4, we get 3 + 11 = 14. This means the +value in the 14th position when the data is sorted is 4. Adding the +frequencies of 3, 4, and 5, we get 3 + 11 + 9 = 23. This means the value in +the 23rd position when the data is sorted is 5. We can see that all the values +from positions 15 to 23 are 5. Therefore, the median of this data is 5. +Number +Frequency +3 +3 +4 +11 +5 +9 +6 +7 +7 +3 +8 +1 +9 +1 +10 +1 + +Tales by Dots and Lines +111 +Spreadsheets +Sudhakar has collected the mid-term exam marks obtained by his +Grade 8 students in the following table: +Name +Odia +Telugu +English +Maths +Social +Science +Science +Ratna +25 +39 +29 +36 +34 +37 +Nagesh +41 +43 +48 +39 +40 +39 +Ashwin +29 +31 +33 +34 +30 +28 +Farooq +47 +46 +38 +42 +49 +44 +Mrinal +33 +35 +28 +32 +30 +36 +Gowri +27 +29 +34 +31 +32 +30 +Pankaj +16 +19 +22 +17 +18 +20 +Jaya +31 +38 +40 +50 +43 +46 +Ganesh +39 +37 +35 +38 +36 +40 +Shravan +12 +17 +21 +20 +14 +18 +Aishwarya +48 +45 +46 +47 +44 +43 +Hari +25 +28 +24 +21 +23 +26 +Trupti +29 +36 +30 +33 +27 +33 +Veeresh +23 +25 +28 +31 +19 +22 +Vidhya +34 +36 +37 +40 +32 +34 +Sanskruti +35 +42 +41 +46 +38 +40 +Shanker +42 +45 +39 +36 +31 +39 +Vyshnavi +37 +32 +29 +33 +31 +35 +Govind +15 +18 +12 +20 +20 +18 +Shiva +29 +24 +32 +34 +28 +30 +Tarun +41 +44 +39 +43 +37 +42 +Jyothi +29 +30 +33 +28 +34 +29 + +Ganita Prakash | Grade 8 | Part-II +112 +Sudhakar has to calculate the total marks scored by each student. +He is also interested in knowing the average marks scored in each +subject. But there are so many numbers! He will have to spend too much +time and effort to complete this task. Of course, he can use a calculator +to speed up the task. +Is there any way to further quicken this process? +One way is to use a computer. Computers have different applications/ +tools that can be used to perform tasks. One such application is a +spreadsheet — it is like a digital notebook with rows and columns of +small boxes called cells. In each cell, we can type text or numbers. The +picture below shows a snapshot of a spreadsheet where Sudhakar has +entered the marks data of his class. Try to get access to a computer +before proceeding. +Before learning how to calculate the average marks per subject +and the total marks scored by each student, let us first understand the +structure of spreadsheets and how to read them. +Can you tell which cell has the marks obtained by Farooq in Mathematics? +Cells are named and referred to using the column headers labelled A, +B, C, … , and row headers labelled 1, 2, 3, … Farooq’s score in Mathematics +is in cell E5. +Can you tell what data is in column B7? +In which subjects has Ashwin scored more than 30 marks? +How can we use spreadsheets to quickly calculate totals and averages? +In addition to text and numbers, we can also enter formulae in a cell. +We can write a formula that computes the sum or average of a row, +or column of cells. +We can describe a row of cells by an expression of the form Start:End, +indicating the first and last cell in the row. For instance, Nagesh’s marks +are described by the expression B3:G3, while Gowri’s marks in Odiya, +Telugu and English are described by the expression B7:D7. We can use +similar expressions to describe columns. For instance, D2:D6 describes +the marks in English for the first five students. +We can then write a formula to compute the sum or the average of a +row or column. For instance =SUM(B3:G3) calculates the total marks for +Nagesh across all subjects, while =AVERAGE(B7:D7) calculates Gowri’s +average marks across Odia, Telugu and English. + +Tales by Dots and Lines +113 +What formula would you type to find out the class average marks in +Science? +Find out if the class average marks in Odia is greater than the class +average marks in Telugu. +Show the average marks in other subjects after the last row by typing +the appropriate formulae. +Try these out on a computer. You can download the +tabular data from this QR code. +Get the total scores of each student by typing the +appropriate formulae. +Note to the Teacher: You can use any spreadsheet software application such as +Microsoft Excel, Google Sheets, LibreOffice Calc, etc. If there are not sufficient +computers for each student, students can share a computer in groups. If that’s not +possible, the computer screen can be projected for the whole class if there is only +one computer available. +Figure it Out +1. Find the mean of the following data and share your observations: +(i) The first 50 natural numbers. +(ii) The first 50 odd numbers. +(iii) The first 50 multiples of 4. +2. The dot plot below shows a collection of data and its average; but +one dot is missing. Mark the missing value so that the mean is 9 (as +shown below). +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +114 +3. Sudhakar, the class teacher, asks Shreyas to measure the heights of +all 24 students in his class and calculate the average height. Shreyas +informs the teacher that the average height is 150.2 cm. Sudhakar +discovers that the students were wearing uniform shoes when the +measurements were taken and the shoes add 1 cm to the height. +(i) Should the teacher get all the heights measured again without +the shoes to find the correct average height? Or is there a +simpler way? +(ii) What is the correct average height of the class? + +(a) 174.2 cm +(b) 126.2 cm +(c) 150.2 cm + +(d) 149.2 cm +(e) 151.2 cm +(f) None of the above + +(g) Insufficient information +4. The three dot plots below show the lengths, in minutes, of songs of +different albums. Which of these has a mean of 5.57 minutes? Explain +how you arrived at the answer. +0 +0.5 +1 +1.5 +2 +2.5 +3 +3.5 +4 +4.5 +5 +5.5 +6 +6.5 +7 +7.5 +A +0 +0.5 +1 +1.5 +2 +2.5 +3 +3.5 +4 +4.5 +5 +5.5 +6 +6.5 +7 +7.5 +B +0 +0.5 +1 +1.5 +2 +2.5 +3 +3.5 +4 +4.5 +5 +5.5 +6 +6.5 +7 +7.5 +C +5. Find the median of 8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, +84, 91, 92. +(i) If we include one value to the data (in the given list) without +affecting the median, what could that value be? +(ii) If we include two values to the data without affecting the +median what could the two values be? +(iii) If we remove one value from the data without affecting the +median what could the value be? +Math +Talk + +Tales by Dots and Lines +115 +6. Examine the statements below and justify if the statement is always +true, sometimes true, or never true. +(i) Removing a value less than the median will decrease the +median. +(ii) Including a value less than the mean will decrease the mean. +(iii) Including any 4 values will not affect the median. +(iv) Including 4 values less than the median will increase the +median. +7. The mean of the numbers 8, 13, 10, 4, 5, 20, y, 10 is 10.375. Find the +value of y. +8. The mean of a set of data with 15 values is 134. Find the sum of the +data. +9. Consider the data: 12, 47, 8, 73, 18, 35, 39, 8, 29, 25, p. Which of the +following number(s) could be p if the median of this data is 29? +(i) 10 +(ii) 25 + +(iii) 40 + +(iv) 100 +(v) 29 +(vi) 47 + +(vii) 30 +10. The number of times students rode their cycles in a week is shown +in the dot plot below. Four students rode their cycles twice in that +week. +(i) Find the average number of times students rode their cycles. +(ii) Find the median number of times students rode their cycles. +(iii) Which of the following statements are valid? Why? +(a) Everyone used their cycle at least once. +(b) Almost everyone used their cycle a few times. +(c) There are some students who cycled more than once on +some days. +(d) Exactly 5 students have used their cycles more than once on +some days. + +Ganita Prakash | Grade 8 | Part-II +116 +(e) The following week, if all of them cycled 1 more time than +they did the previous week, what would be the average and +median of the next week’s data? +11. A dart-throwing competition was organised in a +school. The number of throws participants took +to hit the bull’s eye (the centre circle) is given +in the table below. Describe the data using its +minimum, maximum, mean and median. +No. of trials +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +No. of students +1 +0 +0 +1 +4 +9 +12 +15 +10 +10 +5.2 Visualising and Interpreting Data +So far we have learnt how to read and make pictographs, bar graphs, +clustered-bar graphs, and dot plots. We now examine some more ways +of visualising data. +Line Graphs +Temperature +The following figure is a clustered-column graph that shows the monthly +maximum temperature in Kerala and Punjab in 2023. +Now, observe the following graph. Do both these graphs represent the +same information? + +Tales by Dots and Lines +117 +We call such a graph made up of lines a line graph. Line graphs are +generally used to visualise data across time. +How do we get the maximum temperature over a month in a state? +There could be a few weather stations across the state that regularly +track the local temperature. We can get the monthly maximum +temperature by looking at the maximum value among all the values +recorded across the state. +To identify and interpret the information presented, let us follow a +two-step process. +Step 1: Identify what is given +Notice how the graph is organised, what scale is used, and what patterns +the data shows. +• The data for each month for a city +is marked and connected by lines to +show the change over time. Kerala’s +data is shown using blue circle +marks connected by blue lines and +Punjab’s data is shown in red. +When we try to understand how the data is collected or produced, +we gain a clearer idea of its scope and can interpret it more +meaningfully. It also helps us identify any limitations, such as bias +or missing information, and decide how confidently we can draw +conclusions from the data. +The different shape markers +help in easily distinguishing +if the graph is printed in +greyscale/black-and-white, or +for people who find it difficult +to distinguish colour. + +Ganita Prakash | Grade 8 | Part-II +118 +• The horizontal line shows the months of the year. The vertical line +shows the temperature in oC. +Step 2: Infer and interpret from what is given +Analyse and interpret each of the observations you made. Share +appropriate summary/conclusion statements. +• In Punjab, the monthly maximum temperature increases from +January to June, reaching a high of 38oC. Then, it reduces to just +under 35oC in July, stays mostly flat till September, and then falls +continuously till December, reaching about 23oC. January has the +lowest among the monthly maximum temperature of about 19oC. +• Kerala’s trend is different — it stays mostly flat throughout the year. +The peak is around 33oC in April and the lowest point is around +29oC in July. Notice that the monthly maximum temperatures in +Kerala are similar both in summer and winter! +• In short, the temperature in Punjab varies more, +reaching colder and warmer temperatures +than in Kerala. +This can trigger questions in different directions, +some of which can be answered using data. Here are +some directions to think about — +• Why are the trends so distinct in these two states? +What factors determine a region’s temperature? +• You might be curious to look at the trends of a few other +states. Are there other types of trends that states exhibit? +• Which states show trends similar to Punjab’s? Is there anything +common between these states? + +Tales by Dots and Lines +119 +• What would a plot of the monthly minimum temperatures for +these states look like? +What thoughts or questions occur to you? +Space Jam: A Traffic Problem in the Future? +Take a look at the line graph below showing the annual number of +objects launched into space. +What could be the possible method used to derive this data? Discuss. +Step 1: Identify what is given +Notice how the graph is organised, what scale is used, and what patterns +the data shows. +Step 2: Infer from and interpret what is given +Analyse and interpret each of the observations you made. Once all +interpretations are made, summarising/concluding statements can be +made. +• In 2024, the worldwide count is around 2800, whereas in 2023, it +was around 2900. We can say that 2023 saw the highest number of +objects launched into space worldwide (assuming the trend before +2012 was decreasing). +• The counts of the three countries don’t add up to the worldwide +count. Therefore, we can infer that the counts of other countries +are not shown in this visualisation. +Math +Talk +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +120 +• For the USA, the increase from 2022 to 2023 is more than from 2023 +to 2024. We can say this by looking at how steep the line segments +are — the steeper the line is, the greater the increase. +Which of the following statements are valid inferences? +• From 2012 till 2024, the worldwide count of space object launches +increased every year. +• USA is a major contributor in the years 2022 – 24, launching about +3 +4th of the worldwide count. +• Nepal did not launch any object in the period 2012 – 24. +• The combined count of object launches by China and Russia in +2024 is about 400. +Identify two consecutive years where the worldwide count increased by +2 times or more. +Imagine the same data being shown as a clustered column graph. There +would be 13 clusters  — one for each year — and within each cluster, +4 columns, making a total of 52 bars! Such a graph would look crowded +and difficult to read, making it hard to interpret the trends clearly. +A line graph, on the other hand, is better suited for illustrating +changes over time. By connecting data points with line segments, +it provides a clear visual representation of trends and variations, +allowing the reader to easily track how a parameter evolves across +different years. +Catch the (Pattern in) Rain +The following graphs show monthly average rainfall data of a few cities. + +Tales by Dots and Lines +121 +Source: weather-and-climate.com +What could be the possible method to compile this data? +This data shows the monthly average rainfall in 6 cities. This means +rainfall data is collected over a few years in every city. The total rainfall +in a month, say June, across years is averaged to get the monthly average +rainfall in June in a city. +Mark these cities on a map of India. What is common to how they +are grouped in the graphs? Share your observations and inferences +about the graphs. +Kovalam, Udupi, and Mumbai are along the west coast. Rameswaram, +Chennai, and Puri are along the east coast. It appears that regions along +the west coast receive more rain. +Identify the peak months and low months of rainfall for each city. +Udupi, Mumbai, and Kovalam have peak rainfall during June – August. +Rameswaram gets most of its rain during October – December. Chennai +starts getting rain from June onwards, which peaks in November, and +continues till December. Puri, although on the east coast, gets its peak +rainfall during July – September. January – March are dry months, with +low rainfall, for all the cities. Rameswaram receives very little rain from +January – September. +Read about the south-west monsoon and north-east monsoon and which +regions come under the influence of these and when. +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +122 +Figure it Out +1. The average number of customers visiting a shop and the average +number of customers actually purchasing items over different days +of the week is shown in the table below. Visualise this data on a line +graph. +Mon +Tue +Wed +Thu +Fri +Sat +Sun +Visiting +16 +19 +10 +14 +20 +22 +35 +Purchasing +10 +8 +7 +11 +12 +16 +26 +2. The average number of days of rainfall in each month for a few cities +is shown in the table below: +Jan +Feb +Mar +Apr +May +Jun +Jul +Aug +Sep +Oct +Nov +Dec +Mangaluru +0.1 +0 +0.1 +1.8 +6.2 +24.1 +27.7 +24.5 +14 +8.8 +3.9 +0.9 +New Delhi +Port Blair +2.4 +1.3 +0.9 +3.3 +15.5 +18.7 +17.3 +18.8 +16.8 +14.1 +11.3 +5.4 +Rameswaram +2.6 +1.3 +1.9 +3.4 +2.5 +0.4 +1 +1 +1.9 +8.1 +10.4 +7.8 +(i) What could be the possible method to compile this data? +(ii) Mark the data for Mangaluru, Port Blair, and Rameswaram in +the line graph shown below. You can round off the values to the +nearest integer. +Source: weather-and-climate.com + +Tales by Dots and Lines +123 +(iii) Based on the line for New Delhi in the graph fill the data in the +table. +(iv) Which city among these receives the most number of days of +rainfall per year? Which city gets the least number of days of +rainfall per year? +(v) Looking at the table, when is the rainy season in New Delhi and +Rameswaram? +2. The following line graph shows the number of births in every month +in India over a time period: +(i) What are your observations? +(ii) What was the approximate number of births in July 2017? +(iii) What time period does the graph capture? +(iv) Compare the number of births in the month of January in the +years 2018, 2019, and 2020. +(v) Estimate the number of births in the year 2019. +Infographics +In Grade 6, we saw an example of how infographics can be used to +communicate information and insights more clearly and quickly in a +visually appealing way. Take a look at the following infographic. Are you +able to understand the information presented here? + +Ganita Prakash | Grade 8 | Part-II +124 +Map not to scale +This infographic compares the preference between rice and wheat +in different states. The colour scale at the top right indicates how to +interpret the different shades in terms of preferences. +The difference between the per capita consumption of rice and wheat +is mapped to values between – 100 and + 100. A value of + 100 doesn’t +mean that the state doesn’t consume wheat at all. It means that this +state prefers mostly rice and the difference between per capita rice and +wheat consumption is the highest. Isn’t it interesting how there is a clear +geographical split in preferences of rice vs. wheat shown by the red line? +Share your observations. Based on this infographic, answer the following: +(i) The value of Karnataka is hidden. Can you guess what it could +be? +(ii) Which are the top 5 states where rice is the most popular? +(iii) Which are the top 5 states where wheat is the most popular? +(iv) List a few states where the preference between rice and wheat +is more or less balanced. + +Tales by Dots and Lines +125 +I wonder if there is +anything common in +all the states preferring +rice or states preferring +wheat. +I want to know if the +preferences were similar +even 100 years ago. +What can a Strip Say? +Manoj has an interesting hobby. He makes note +of what he does throughout the day. He records +his activities by colouring a strip of paper with +48 boxes, marking time in 30 minute intervals from +midnight to midnight. +He has recorded five types of activities for three +different days of the week on three coloured strips, +shown to the right. +(i) Sleeping +(ii) Eating +(iii) Meeting friends, hobbies, media, time +with family +(iv) Attending classes, studying and homework +(v) Showering and getting dressed, yoga or +exercise +(vi) Travelling +Look at the three coloured strips carefully. +(i) What activity does each colour stand for? +(ii) The three strips correspond to the days +Friday – Sunday in some order. Which day +do you think each strip represents? +(iii) On one of these days, he went out with +friends to watch a long movie. When do you think this happened? +(iv) At what time does his school break for lunch? +(v) What more can the strips tell us? + +Ganita Prakash | Grade 8 | Part-II +126 +What would your strip for a weekday look like? How similar or different +is it to Manoj’s? +What would a strip of your typical day during your vacation look +like? How similar/different would it look? +What would a strip for any of the adults in your family look like? +Make a strip of a day for any adult at home. Compare your strip +with theirs. What do you find interesting? +Data Story: Sleepy-Deepy +Do you remember the sleep time pie chart from the Proportionality +chapter? +Isn’t it amusing that there are animals that sleep as little as 2 hours +per day and animals that sleep as much as 20 hours per day? How +about insects — have you seen any insect sleep/rest? Humans typically +sleep for about 7 –  9 hours a day. The sleep duration can vary greatly +among people. The amount of sleep people need depends on age, living +conditions, lifestyle (food they consume, the activities they engage in, +etc.) among other factors. You may have observed that babies sleep +longer than adults. +The line graphs below show typical sleep durations of Indians across +ages 6 to 75. The second picture is a zoomed-in version of the first picture. +Try +This +Math +Talk + +Tales by Dots and Lines +127 +Share your observations on this graph. What do you find interesting? +Unlike the earlier graphs that were made up of a few connected line +segments, this line graph looks like a smooth curve because it contains +80 data points that are placed very close to each other. Representing the +same data using a column graph would require 70 columns, making the +graph look cluttered and heavy. A line graph is a better choice here. It is +light and captures the pattern well. +From the graph we can see that the average sleep time for 6-year olds +is about 9.5 hours per day. The daily sleep time decreases as we grow +through teenage years and step into adulthood, touching about 8 hours +a day between ages 30 and 50. After 50, the daily sleep time increases, +reaching about 8.5 hours. +You might wonder — why do newborns and infants sleep for so long? +Do people in different countries sleep differently? Do animals also +exhibit such patterns in sleep durations across age? +Figure it Out +1. Mean Grids: +(i) Fill the grid with 9 distinct numbers such that the +average along each row, column, and diagonal is 10. +(ii) Can we fill the grid by changing a few numbers and +still get 10 as the average in all directions? +2. Give two examples of data that satisfy each of the following +conditions: +(i) 3 numbers whose mean is 8. +(ii) 4 numbers whose median is 15.5. +(iii) 5 numbers whose mean is 13.6. +(iv) 6 numbers whose mean = median. +(v) 6 numbers whose mean > median. +3. Fill in the blanks such that the median of the collection is 13: 5, 21, +14, _____, ______, ______. How many possibilities exist if only counting +numbers are allowed? +4. Fill in the blanks such that the mean of the collection is 6.5: 3, 11, ____, +_____, 15, 6. How many possibilities exist if only counting numbers +are allowed? +5. Check whether each of the statements below is true. Justify your +reasoning. Use algebra, if necessary, to justify. +(i) The average of two even numbers is even. +(ii) The average of any two multiples of 5 will be a multiple of 5. +(iii) The average of any 5 multiples of 5 will also be a multiple of 5. + +Ganita Prakash | Grade 8 | Part-II +128 +6. There were 2 new admissions to Sudhakar’s class just a couple of +days after the class average height was found to be 150.2 cm. +(i) Which of the following statements are correct? Why? +(a) The average height of the class will increase as there are 2 +new values. +(b) The average height of the class will remain the same. +(c) The heights of the new students have to be measured to find +out the new average height. +(d) The heights of everyone in the class has to be measured +again to calculate the new average height. +(ii) The heights of the two new joinees are 149 cm and 152 cm. +Which of the following statements about the class’ average +height are correct? Why? +(a) The average will remain the same. +(b) The average will increase. +(c) The average will decrease. +(d) The information is not sufficient to make a claim about the +average. +(iii) Which of the following statements about the new class average +height are correct? Why? +(a) The median will remain the same. +(b) The median will increase. +(c) The median will decrease. +(d) The information is not sufficient to make a claim about +median. +7. Is 17 the average of the data shown in the dot plot below? Share +the method you used to answer this question. +14 +15 +16 +17 +18 +19 +20 +21 +22 + × + × + × + × + × + × + × + × + × + × + × + × + × + × + × + × + × + × + × + × + × + × + × +23 + × + × +8. The weights of people in a group were measured every month. +The average weight for the previous month was 65.3 kg and the +median weight was 67 kg. The data for this month showed that +Math +Talk +Math +Talk + +Tales by Dots and Lines +129 +one person has lost 2 kg and two have gained 1 kg. What can we say +about the change in mean weight and median weight this month? +9. The following table shows the retail price (in ₹) of iodised salt in the +month of January in a few states over 10 years. For your calculations +and plotting you may round off values to the nearest counting +number. +Andaman and +Nicobar Islands +Assam +Gujarat +Mizoram +Uttar +Pradesh +West +Bengal +2016 +16 +6 +16.5 +20 +16.15 +9.47 +2017 +12 +12 +14.75 +20 +16.97 +11.65 +2018 +12 +12 +14.75 +22 +16.18 +11.63 +2019 +12 +12 +14.75 +22 +18.24 +11.43 +2020 +13.88 +12 +13 +20 +18.96 +11.11 +2021 +18.22 +15 +14.45 +22 +20.63 +12.79 +2022 +18.73 +14 +14.28 +25 +21.3 +16.14 +2023 +20.63 +12.02 +14.54 +27.65 +25.39 +18.43 +2024 +19.73 +13.72 +14.8 +29.03 +26.9 +21.66 +2025 +20.99 +12.35 +19.2 +29.8 +24.81 +23.99 +(i) Choose data from any 3 states you find interesting and present +it through a line graph using an appropriate scale. +(ii) What do you find interesting in this data? Share your +observations. +(iii) Compare the price variation in Gujarat and Uttar Pradesh. +(iv) In which state has the price increased the most from 2016 to +2025? +(v) What are you curious to explore further? + +Ganita Prakash | Grade 8 | Part-II +130 +10. Referring to the graph below, which of the following statements are +valid? Why? +(i) In 1983, the majority in rural areas used kerosene as a +primary lighting source while the majority in urban areas used +electricity. +(ii) The use of kerosene as a primary lighting source has decreased +over time in both rural and urban areas. +(iii) In the year 2000, 10% of the urban households used electricity +as a primary lighting source. +(iv) In 2023, there were no power cuts. +11. Answer the following questions based on the line graph. +(i) How long do children aged 10 in urban areas spend each day on +hobbies and games? +(ii) At what age is the average time spent daily on hobbies and games +by rural kids 1.5 hours? +(a) 8 years +(b) 10 years +(c) 12 years +(d) 14 years +(e) 18 years + +Tales by Dots and Lines +131 +(iii) Are the following statements correct? +(a) The average time spent daily on hobbies and games by kids +aged 15 is twice that of kids aged 10. +(b) All rural kids aged 15 spend +at least 1 hour on hobbies and +games everyday. +12. Individual project: Make your own +activity strip for different days of the +week. +(i) Do you eat and sleep at regular +times every day? Typically how +long do you spend outdoors? +(ii) Calculate the average time spent per activity. Represent this +average day using a strip. +(iii) Similarly, track the activities of any adult at home. Compare +your data with theirs. +13. Small group project: Make a group of 3 – 4 members. Do at least +one of the following: +(i) Track daily sleep time of all your family members for a week. +Daily sleep time includes night sleep, naps, and any sleep during +the day. +(a) Represent this on strips. +(b) Put together the data of all your group members. Calculate +the average and median sleep time of children, adults, +elderly. +(c) Share your findings and observations. +(ii) When do schools start and end? On a weekday, Manoj’s school +starts at 9:30 am and ends at 4:30 pm, i.e., 7 hours which include +class time and breaks. Collect information on the daily timings +of different schools for Grade 8, including class time and break +time (the schools can be anywhere in the country. You can ask +your neighbours, relatives, parents and friends to find out). +Analyse and present the data collected. +14. The following graphs show the sunrise and sunset times across the +year at 4 locations in India. Observe how the graphs are organised. +Are you able to identify which lines indicate the sunrise and which +indicate the sunset? +Hey, come play +with us — in +real life! +Try +This + +Ganita Prakash | Grade 8 | Part-II +132 +Answer the following questions based on the graphs: +(i) At which place does the sun rise the earliest in January? What is +the approximate day length at this place in January? +(ii) Which place has the longest day length over the year? +(iii) Share your observations — what do you find interesting? What +are you curious to find out? +15. We all know the typical sunrise and sunset timings. Do you +know when the moon rises and sets? Does it follow a regular +pattern like the sun? Let’s find out. The following graph shows +the moonrise and moonset time over a month: +(i) Find out on what dates amavasya (new moon) and purnima +(full moon) were in this month. +(ii) What do you notice? What do you wonder? +Try +This + +Tales by Dots and Lines +133 + +y +Last year we looked at mean as a fair-share. Here, we learnt +how the sum of the distances of the values to its left and right +are the same. + +y +We saw that when values greater than the mean are inserted, +the  mean increases. When values less than the mean are +inserted, the mean decreases. Similar phenomena can be +observed with the median. + +y +Line graphs can be used to visualise change over time. + +y +We saw that examining data can lead to new questions and +directions to probe further. +SUMMARY + +Game of Hex +Hex is a two-player strategy game played on a rhombus-shaped board +made of hexagonal cells, usually of size 11 × 11. Each player is assigned +a colour and two opposite sides of the board. Players take turns placing +a piece of their colour on any empty cell. Once placed, pieces can not +be moved or removed. The objective is to create an unbroken chain of +one’s own pieces connecting the two assigned sides. The first player to +complete such a connection wins the game. The pictures below show +two possible gameplays where blue wins in the first and red wins in the +second board. +Here is an empty board. You can use pencils to play each round and +erase the marks to play a fresh round." +class_8,13,algebra play,ncert_books/class_8/hegp2dd/hegp206.pdf,"6.1 Algebra Play +Over the last two years, we have used algebra to model different +situations. We have learned how to solve algebraic equations and find +the values of unknown letter-numbers. Let’s now have some fun with +algebra. We shall investigate tricks and puzzles, and explain why they +work using algebra. We will also see how to invent new tricks and +puzzles to entertain others. +6.2 Thinking about ‘Think of a Number’ Tricks +In Grade 7, we learned about ‘Think of a Number’ tricks, like this one. +1. Think of a number. +2. Double it. +3. Add four. +4. Divide by two. +5. Subtract the original number you thought of. +I predict you get 2. Am I right? Try it out with different starting numbers. +Do you always end up with the same value, 2? Why? +We saw that we can understand such tricks through algebra. +1. Think of a number: x +2. Double it: 2x +3. Add four: 2x + 4 +4. Divide by 2: x + 2 +5. Subtract the original number you thought of: x + 2 – x = 2 +Therefore, no matter what the starting number is, the end result will +always be 2! +ALGEBRA PLAY +6 + +Ganita Prakash | Grade 8 | Part-II +136 +How would you change this game to make the final answer 3? What +about 5? +Can you come up with more complicated steps that always lead to the +same final value? +Let us now look at a different trick of this type. +Think of a date. +26/01 +1 × 5 = 5 +5 + 6 = 11 +44 + 9 = 53 +53 × 5 = 265 +265 + 26 = 291. +11 × 4 = 44 +Multiply the month by 5. +Add 6. +Add 9. +Add the day. +Tell me your answer. +You thought of +our Republic Day, 26/01. +Multiply by 4. +Multiply by 5. +How did Shubham figure out the date chosen by Mukta? +• Let the month be M and the day be D. +• Multiply M by 5: 5M + +Algebra Play +137 +• Add 6: 5M + 6 +• Multiply by 4: 20M + 24 +• Add 9: 20M + 33 +• Multiply by 5: 100M + 165 +• Add the day: 100M + 165 + D +Mukta’s answer was 291. +• 291 = 100M + 165 + D +• 291 – 165 = 100M + D +• 126 = 100M + D +Since D is a day within a month, it is atmost 31 and requires only 2 +digits. So the last 2 digits are D and what comes before that is M. In this +case, M is 1 and D is 26, i.e., the 26th of January. +Mukta thinks of another date, follows the same steps, and reports her +answer as 1390. What date did Mukta start with this time? +• Subtracting 165 from 1390, we get 1225. +• This means that the date she thought of was 25th December. +Find the dates if the final answers are the following: +(i) 1269 +(ii) 394 +(iii) 296 +You can try this trick with your friends. Ask them to choose the starting +date as their birthday. +Can you change the steps in this trick and still find the original date? +Instead of subtracting 165 from the final answer, you might have to +subtract some other number. +Try to devise your own ‘Think of a Number’ trick. +6.3 Number Pyramids +In a number pyramid, each number is the sum of the two numbers +directly below it (see the figure). +10 +1 +23 +13 +9 +4 +Math +Talk +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +138 +Use the same rule to fill these pyramids: +6 +? +2 +? +3 +? +? +4 +3 +? +? +? +? +? +? +4 +5 +5 +0 +How do we fill this pyramid? +4 +1 +10 +? +? +? +4 +1 +10 +What will +go here? +And here? +4 +1 +10 +10 – 4 = 6 +4 – 1 = 3 +4 +1 +10 +What will go here? +4 +1 +10 +6 – 3 = 3 +6 +6 +3 +3 +4 +1 +10 +6 +3 +3 +What about filling in the numbers in this pyramid? Where do we start? +? +12 +60 +? +? +8 + +Algebra Play +139 +Let us fill the empty boxes with letter-numbers. +a +12 +60 +b +c +8 +From the rules for filling up pyramids, we get the following equations. +• a + b = 60 +• 12 + c = a +• c + 8 = b +From this, we see that, +• (12 + c) + (c + 8) = a + b = 60 +Hence, +• 20 + 2c = 60 +• 2c = 60 – 20 = 40 +• c = 20. +Once we know c, we can find a and b, and complete the pyramid. +32 +12 +60 +28 +20 +8 +Fill the following pyramids: +40 +7 +5 +9 +50 +6 +4 +22 +5 +35 +3 +7 +What is the relationship between the numbers in the bottom row and +the number at the top? +Let us start with the simplest pyramid. +a + b +b +a + +Ganita Prakash | Grade 8 | Part-II +140 +What about a pyramid with three rows? +Using letter numbers for the bottom row, we can write an expression +for the top row. +a + 2b + c +a + b +b + c +b +c +a +Figure it Out +1. Without building the entire pyramid, find the number in the topmost +row given the bottom row in each of these cases. +8 +13 +4 +11 +3 +7 +14 +25 +10 +2. Write an expression for the topmost row of a pyramid with 4 rows in +terms of the values in the bottom row. +3. Without building the entire pyramid, find the number in the topmost +row given the bottom row in each of these cases. +21 +8 +19 +13 +19 +7 +18 +6 +5 +9 +7 +11 +Recall the Virahāṅka-Fibonacci number sequence 1, 2, 3, 5, … where +each number is the sum of the two numbers before it. +4. If the first three Virahāṅka-Fibonacci numbers are written in the +bottom row of a number pyramid with three rows, fill in the rest of +the pyramid. What numbers appear in the grid? What is the number +at the top? Are they all Virahāṅka-Fibonacci numbers? +5. What can you say about the numbers in the pyramid and the number +at the top in the following cases? +(i) The first four Virahāṅka-Fibonacci numbers are written in the +bottom row of a four row pyramid. +(ii) The first 29 Virahāṅka-Fibonacci numbers are written in the +bottom row of a 29 row pyramid. +6. If the bottom row of an n row pyramid contains the first n +Virahāṅka-Fibonacci numbers, what can we say about the numbers +in the pyramid? What can we say about the number at the top? + +Algebra Play +141 +6.4 Fun with Grids +Calendar Magic +A page from a calendar is given below. Your friend picks a 2 × 2 grid from +this calendar, adds the 4 numbers in this grid and tells you the sum. +AUGUST 2025 +SUN +MON +TUE +WED +THU +FRI +SAT +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +13 +14 +15 +16 +17 +18 +19 +20 +21 +22 +23 +24 +25 +26 +27 +28 +29 +30 +31 +6 +7 +13 +14 +6 + 7 + 13 + 14 = 40 +Can we find the 4 numbers in the grid from just knowing this sum? +Let us use algebra. Consider a 2 × 2 grid. Let a represent the top left +number. What are the other numbers in terms of a? +a +a + 1 +a + 7 a + 8 +Adding all four numbers, the sum is a + (a + 1) + (a + 7) + (a + 8) = 4a + 16. +Suppose you are told that the sum is 36. Can you find the 4 numbers in +the grid? +• 4a + 16 = 36 +• 4a = 20 (subtracting 16 from both sides) +• a = 5 (dividing both sides by 4) +Now that we have found a, we know that the other three numbers are +a + 1, a + 7 and a + 8. Therefore, the grid must be the following: +5 +6 +12 +13 + +Ganita Prakash | Grade 8 | Part-II +142 +Create your own calendar trick. For instance, choose a grid of a +different size and shape. +AUGUST 2025 +SUN MON TUE +WED THU +FRI +SAT +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +13 +14 +15 +16 +17 +18 +19 +20 +21 +22 +23 +24 +25 +26 +27 +28 +29 +30 +31 +2 +3 +4 +5 +6 +7 +8 +9 +10 +12 +13 +14 +15 +16 +17 +18 +19 +20 +22 +23 +24 +25 +26 +27 +28 +29 +30 +32 +33 +34 +35 +36 +37 +38 +39 +40 +42 +43 +44 +45 +46 +47 +48 +49 +50 +Algebra Grids +In the following grid, shapes represent numbers. In each row, the last +column is the sum of the values to its left. How do we find the values of +the shapes? ++ ++ ++ ++ ++ ++ ++ += += += += += += +27 +19 +19 +9 +9 +2 × +19 +5 +9 +So, +Now, +27 +19 +In the following grids, find the values of the shapes and fill in the empty +squares: +27 +21 +18 +15 +6.5 The Largest Product +Fill the digits 2, 3, and 5 in + + × +, using each digit once. What is the +largest product possible? +Math +Talk + +Algebra Play +143 +Let us approach this problem systematically. There are six ways to place +three digits: +• We can fill the first box with 2, 3 or 5. +• For each of these choices, we have 2 ways of filling the remaining +2 digits. +• The six options are 23 × 5, 25 × 3, 32 × 5, 35 × 2, 52 × 3, 53 × 2. +How do we find the largest product among these six options? +We can group them in pairs where the multiplier is the same. +• 35 × 2 and 53 × 2 +• 25 × 3 and 52 × 3 +• 23 × 5 and 32 × 5 +In each pair, the one with the larger multiplicand generates the larger +product, so we can reduce the comparison to these three expressions. +• 53 × 2 +• 52 × 3 +• 32 × 5 +It is clear that 52 × 3 is bigger than 53 × 2, so we only need to compare +52 × 3 and 32 × 5. Let us expand these. +• 32 × 5 = (3 × 10 × 5) + (2 × 5) +• 52 × 3 = (5 × 10 × 3) + (2 × 3) +The first terms in both expressions are the same. The second term +shows that 32 × 5 is larger, and hence the largest of the six possible +products we can form with 2, 3 and 5. +In this case, we used the largest digit as the multiplier. The other two +digits were arranged in decreasing order to form the multiplicand. Will +this always be the case? Let us find out using algebra. +Suppose p, q, and r are the three digits such that p < q < r. +As before, we have six possible products, which we group by the +multiplier: +• qr × p, rq × p +• pr × q, rp × q +• pq × r, qp × r +In each pair, the multiplicand with the larger tens digit forms the +larger product. So we have three products to compare: +• rq × p +• rp × q +• qp × r + +Ganita Prakash | Grade 8 | Part-II +144 +Since q > p, we can see that rp × q is bigger than rq × p. This leaves us +with a comparison between qp × r and rp × q. If we expand these, we get +• qp × r = (10 × q × r) + (p × r) +• rp × q = (10 × r × q) + (p × q) +Once again, the first term is the same in both expressions. Since r > q +the second term of the first expression is larger, so the largest product +is qp × r. This matches our earlier observation that the largest digit +should be the multiplier and the other two digits should be arranged in +decreasing order to form the multiplicand. +Figure it Out +1. Fill the digits 1, 3, and 7 in + + × + to make the largest product +possible. +2. Fill the digits 3, 5, and 9 in + + × + to make the largest product +possible. +6.6 Decoding Divisibility Tricks +It is now Mukta’s turn to show a mathematics trick to Shubham. +Choose a 2-digit number +of different digits and +don’t reveal it! +47 +74 +74 – 47 = 27 +27 +9 = 3 +Yes, you are right. +But how did you +know? +Reverse the digits to get +another number. +Find their difference. +Divide the result by 9. +There won’t be any +remainder. + +Algebra Play +145 +If we choose other 2-digit numbers, and follow the steps, will there +always be no remainder? +Let’s try to understand how Mukta’s trick works. +Suppose the two-digit number is ab. When it is reversed, the new +number is ba. +• If b > a, then ba > ab. So, the difference is + +(10b + a) – (10a + b) + += 10b – b – 10a + a + += 9b – 9a = 9 (b – a). +The difference is divisible by 9. +Can you work out what happens if a > b? +Figure it Out +1. In the trick given above, what is the quotient when you divide by 9? +Is there a relationship between the two numbers and the quotient? +2. In the trick given above, instead of finding the difference of the two +2-digit numbers, find their sum. What will happen? For example: +• We start with 31. After reversing we get 13. Adding 31 and 13, +we get 44. +• We start with 28. After reversing we get 82. Adding 28 and 82, +we get 110. +• We start with 12. After reversing we get 21. Adding 12 and 21, +we get 33. + +Observe that all these numbers are divisible by 11. Is this always +true? Can we justify this claim using algebra? +3. Consider any 3-digit number, say abc (100a + 10b + c). Make two +other 3-digit numbers from these digits by cycling these digits +around, yielding bca and cab. Now add the three numbers. Using +algebra, justify that the sum is always divisible by 37. Will it +also always be divisible by 3? [Hint: Look at some multiples of +37.] +4. Consider any 3-digit number, say abc. Make it a 6-digit number by +repeating the digits, that is abcabc. Divide this number by 7, then +by 11, and finally by 13. What do you get? Try this with other +numbers. Figure out why it works. [Hint: Multiply 7, 11 and 13.] +5. There are 3 shrines, each with a magical pond in the front. If +anyone dips flowers into these magical ponds, the number of +flowers doubles. A person has some flowers. He dips them all in +the first pond and then places some flowers in shrine 1. Next, +he dips the remaining flowers in the second pond and places +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +146 +some flowers in shrine 2. Finally, he dips the remaining flowers in +the third pond and then places them all in shrine 3. If he placed an +equal number of flowers in each shrine, how many flowers did he +start with? How many flowers did he place in each shrine? +6. A farm has some horses and hens. The total number of heads of these +animals is 55 and the total number of legs is 150. How many horses +and how many hens are on the farm? +Can you solve this without letter-numbers? +[Hint: If all the 55 animals were hens, then how many legs +would there be? Using the difference between this number and +150, can you find the number of horses?] +7. A mother is 5 times her daughter’s age. In 6 years’ time, the mother +will be 3 times her daughter’s age. How old is the daughter now? +8. Two friends, Gauri and Naina, are cowherds. One day, they pass +each other on the road with their cows. Gauri says to Naina, “You +have twice as many cows as I do”. Naina says, “That’s true, but if I +gave you three of my cows, we would each have the same number of +cows”. How many cows do Gauri and Naina have? +9. I run a small dosa cart and my expenses are as follows: +• Rent for the dosa cart is `5000 per day. +• The cost of making one dosa (including all the ingredients and +fuel) is `10. +(i) If I can sell 100 dosas a day, what should be the selling price of +my dosa to make a profit of `2000? +(ii) If my customers are willing to pay only `50 for a dosa, how many +dosas should I aim to sell in a day to make a profit of `2000? +10. Evaluate the following sequence of fractions: + +1 +3 , (1 + 3) +(5 + 7) , (1 + 3 + 5) +(7 + 9 + 11) +What do you observe? Can you explain why this happens? +[Hint: Recall what you know about the sum of the first n odd +numbers.] +Math +Talk + +Algebra Play +147 +11. Karim and the Genie +Karim was taking a nap under a tree. He had a dream +about a magical lamp and a genie. He heard a voice +saying, “I have come to serve you, Oh master”. He woke +up and to his surprise, it was a genie! +“Do you want to make money?”, asked the genie. +Karim nodded dumbly in bewilderment. The genie +continued, “Do you see the banyan tree over there? All +you have to do is go around it once. The money in your +pocket will double”. +Karim immediately started towards the tree, only to +be stopped by the genie. “One moment!”, said the genie. +“Since I am bringing you great riches, you should share +some of your gains with me. You must give me 8 coins +each time you go around the tree.” +Thinking that was a trifling amount, Karim readily +agreed. +He went around the tree once. Just as the genie had +said, the number of coins in his pocket doubled! He gave +8 coins to the genie. He made another round. Again the +number of coins doubled. He gave 8 more coins to the +genie. He went around the tree for the third time. The +number of coins doubled again, but to his horror, he was left with only +8 coins, exactly the number of coins he owed the genie! +As Karim began to wonder how the genie tricked him, the genie let +out a loud laugh and disappeared. +(i) How many coins did Karim initially have? +(ii) For what cost per round should Karim agree to the deal, if he +wants to increase the number of coins he has? +(iii) Through its magical powers, the genie knows the number of +coins that Karim has. How should the genie set the cost per +round so that it gets all of Karim’s coins? + +y +Algebra is very useful in modeling and understanding +numerical scenarios. Because of this, it occurs in almost all +areas of mathematics, science and beyond. + +y +Algebra is an indispensable tool in justifying mathematical +statements. + +y +We applied algebra to analyse ‘Think of a Number’ tricks, +number pyramids, grids, ways of forming numbers using given +digits to maximise certain products, divisibility tricks, and +various other problems. +SUMMARY +Math +Talk" +class_8,14,area,ncert_books/class_8/hegp2dd/hegp207.pdf,"7 +7.1 Rectangle and Squares +How many different ways can you divide a square into 4 parts of equal +area? +One can actually think of infinitely many such ways! Consider a +division, such as — +and alter each part as follows. +In each part, the area is compressed along one edge and expanded +along another edge. If both the compression and expansion are of the +same magnitude, then all 4 parts still have the same area! +Try to think of different creative ways to divide a square into +4 parts of equal area. +Math +Talk +AREA + +Area +149 +You might have seen the rangoli art form, in which regions of different +shapes are beautifully coloured using rangoli powder. +Which of these rectangles requires more +rangoli powder to be coloured, if the +colouring is done evenly? +We can answer this by counting the +number of non-overlapping unit squares +(squares of sidelength 1 cm in this case) +that can be packed into each of the +rectangles. +Clearly, +the +rectangle +having +sidelengths 7 cm and 4 cm contains 7 × 4 = 28 unit squares, and the +rectangle having sidelengths 8 cm and 3 cm contains 8 × 3 = 24 unit +squares. +Thus, the rectangle of sidelengths 7 cm and 4 cm requires more +powder to be coloured. +Recall that we measure the area of a region by finding the number of +unit squares (which can also be a fraction) whose area equals that of the +given region. +We have seen that the number of unit squares contained in a rectangle +is given by the product of its length and width — +Area of a rectangle = length × width. +The areas of the rectangles as seen in the previous problem are +generally written as 28 sq. cm and 24 sq. cm, or as 28 cm2 and 24 cm2. +4 cm +7 cm +3 cm +8 cm + +Ganita Prakash | Grade 8 | Part-II +150 +What is the area of each triangle in this rectangle? +We have seen that the diagonal of a rectangle divides +it into two congruent triangles. So, the area of each +triangle is half the area of the rectangle. +In terms of unit squares, half the area fills exactly +half the number of unit squares. +So the area of each triangle is 1 +2 × 7 × 4 = 14 cm2. +Why Can’t Perimeter be a Measure of Area? +Why do we count the number of unit squares to assign measures for +area? Couldn’t we have just used the perimeter of a region, i.e., the +length of its boundary as a measure of its area? +If two regions have the same perimeter, can’t we conclude that they have +the same area? Or, if one region has a larger perimeter than another +region, can’t we conclude that it also has a larger area? +The perimeter of a region is not indicative of its area. The reason is that +regions can have the same perimeter but different areas, and vice versa. +We can even find two regions, Region 1 and Region 2, such that +Perimeter of Region 1 > Perimeter of Region 2, but +Area of Region 1 < Area of Region 2. +Find two rectangles that are examples of such regions. If needed, use a +grid paper (given at the end of the book) for this. +Also give an example of two regions of other shapes, where the +region with the larger perimeter has the smaller area! This property +should be visually clear in your example. +Figure it Out +1. Identify the missing sidelengths. +7 in +4 in +? in +3 in +2 in +14 in2 +35 in2 +28 in2 +21 in2 +(i) +4 m +? +29 m2 +? +? +11 m2 +Area = 50 m2 +(ii) +7 cm +4 cm +Math +Talk + +Area +151 +2. The figure shows a path (the shaded portion) + +laid around a rectangular park EFGH. +(i) What measurements do you need to find +the area of the path? Once you identify +the lengths to be measured, assign +possible values of your choice to these +measurements and find the area of the +path. Give a formula for the area. + +An example of a formula — Area of a rectangle = length × width. + +[Hint: There is a relation between the areas of EFGH, the path, +and ABCD.] +(ii) If the width of the path along each side is given, can you find +its area? If not, what other measurements do you need? Assign +values of your choice to these measurements and find the area of +the path. Give a formula for the area using these measurements. +[Hint: Break the path into rectangles.] +(iii) Does the area of the path change when the outer rectangle is +moved while keeping the inner rectangular park EFGH inside +it, as shown? +A +B +D +C +H +G +E +F +A +B +D +C +H +G +E +F +A +B +D +C +H +G +E +F +3. The figure shows a plot with sides 14m and 12m, and with a +crosspath. What other measurements do you need to find +the area of the crosspath? Once you identify the lengths to be +measured, assign some possible values of your choice and find +the area of the path. Give a formula for the area based on the +measurements you choose. +A +B +D +C +H +G +E +F +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +152 +4. Find the area of the spiral tube shown in the figure. The tube has the +same width throughout. +20 +20 +20 +15 +5 +10 +10 +5 +15 +1 +1 +[Hint: There are different ways of finding the area. Here is one +method.] +5 +? +5 + +What should be the length of the straight tube if it is to have the +same area as the bent tube on the left? +5. In this figure, if the sidelength of the square is +doubled, what is the increase in the areas of the +regions 1, 2 and 3? Give reasons. +6. Divide a square into 4 parts by drawing two +perpendicular lines inside the square as shown in +the figure. + +Rearrange the pieces to get a larger square, with +a hole inside. + +You can try this activity by constructing the +square using cardboard, thick chart paper, or +similar materials. +1 +2 +3 +Math +Talk + +Area +153 +Triangles +In the given figure, which triangle has a greater area: ∆XDC or ∆YDC, if +both the rectangles are identical? +A +X +D +B +C +A +Y +D +B +C +In the given figure, which triangle has a greater area: ∆XDC or ∆YBC, if +both the rectangles are identical? +A +X +D +B +C +A +Y +D +B +C +In each case, by dropping the altitudes from X and Y, it becomes clear +that each triangle has exactly half the area of the rectangle ABCD. +Find the area of ∆ XDC. +A +X +D +B +C +4 +5 +Fig. 7.1 +Y +To find the area of a triangle, what measurements do we need? +We need the sidelengths of the outer rectangle, as in Fig. 7.1. +How do we get the outer rectangle from the given triangle? +A +B +C +A +B +C +A +B +C +l +E +D +l ‖ BC + +Ganita Prakash | Grade 8 | Part-II +154 +BCDE is a rectangle (how?). Let us take its +sidelengths to be height and base. +Then, +Area ( ∆ABC) = 1 +2 × base × height +A +B +C +X +E +D +height +base +Since BXAE is a rectangle (how?), the height of the rectangle is the +same as the height of the triangle. +Thus, if the height and the base of a triangle +are known, we can find its area. +Area of a triangle = 1 +2 × base × height. +height +base +A +B +C +Will this formula hold for the kind of triangle, around which we cannot +draw a rectangle with BC as the base? +A +D +C +h +B +Here is one way to look at it. The area of ∆ABC is the difference of the +areas of ∆ADC and ∆ADB, each of which can be enclosed in a rectangle, +as in Fig. 7.1. +Area (∆ABC) = 1 +2 × h × DC – 1 +2 × h × DB += 1 +2 × h (DC – DB) += 1 +2 × h × BC +Thus, the area formula holds for all types of triangles. + +Area +155 +Some Applications of the Area Formula +Find BY. +BY can be found using the formula for the area +of a triangle. +What is Area (∆ABC)? +Area (∆ABC) = 1 +2 × AX × BC = 15 +2 sq. units. +The area of the triangle can also be written as +Area (∆ABC) = 1 +2 × BY × AC = 1 +2 × 4 × BY = 2 BY. +Thus, +2 BY = 15 +2 . +So, BY = 15 +4 = 3.75 units. +Are the 4 triangles obtained by drawing the diagonals of a rectangle +(regions 1 – 4 in the figure) of equal areas? +Clearly, the triangles 1 – 4 are not four +congruent triangles. +To compare their areas, let us consider any +two adjacent triangles, say 1 and 2. +Since the area of a triangle depends on its +height and the base, let us consider suitable +height-base pairs in each of the triangles. If OD +and OB are taken as the bases, then we see that +both triangles have the same altitude! +We also have OB = OD, since the diagonals of +a rectangle bisect each other. So, the triangles 1 and 2 have equal areas. +Arguing this way, we can show that all four triangles have equal areas. +From this problem, we can make the +following general statement. +In a triangle, the line joining a vertex to +the midpoint of its opposite side divides +the triangle into two triangles of equal +areas. +5 +3 +4 +C +A +Y +X +B +A +B +D +C +O +2 +1 +3 +4 +A +B +D +C +O +X +1 +2 +Triangles 1 and 2 have equal areas as they +have the same measures for height and base + +Ganita Prakash | Grade 8 | Part-II +156 +Triangles between Parallel Lines with a Common Base +l +C +B +Line l‖BC. Consider the different triangles that have BC as their base, +and with their third vertex lying anywhere on l. +(i) Which of these triangles has the maximum area, and which has +the minimum area? +(ii) Which of these triangles has the maximum perimeter, and +which has the minimum perimeter? +Here, we will only show how to find the triangle with the minimum +perimeter, and leave the other questions as exercises. +Intuition might suggest that the +triangle with the minimum perimeter +can be obtained by constructing the +perpendicular bisector of BC. +But how do we justify that this +triangle has the least perimeter among +all the triangles? +l +B +C +A +Firstly, note that in all these triangles, BC is a common side. Therefore, +it is enough to consider the sum of the other two sides. +Let us imagine the line l as a mirror. Then we get a reflection of all the +points and lines below it. +While studying the properties of a plane mirror, we experimentally +observed that the distance of the image behind the mirror is the same as +the distance of the object (that creates the image) in front of the mirror. +This law can be used to locate the reflections of the points lying below +the line l. + +Area +157 +What can we say about the lengths of AB and its +reflection AB ́? +Since ∆AXB ≅ ∆AXB ́, AB = AB ́. +Similarly, AC = AC ́. +Therefore, the length of the path B + A + C is the +same as the length of the path B + A + C ́. +l +B´ +C´ +B +A +C +This is true no matter where point A is +on line l. +So, finding point A that gives the shortest possible +path B +A +C is the same as finding a point A that +gives the shortest possible path B +A +C ́. +But the solution to the latter problem is clear — choose +A on the straight line BC ́, since BC ́ is the shortest +possible path from B to C ́. +Therefore, this triangle ∆ABC has the minimum +perimeter. +Analyse whether A lies on the perpendicular bisector of BC. +Figure it Out +1. Find the areas of the following triangles: +(i) +A +B +E +C +3 cm +4 cm +(iii) +N +A +T +3 cm +4 cm +(ii) +E +D +F +N +5cm +3.2 cm +l +B´ +C´ +B +C +A +Math +Talk +l +B´ +C´ +B +C +X +Y +A + +Ganita Prakash | Grade 8 | Part-II +158 +2. Find the length of the altitude BY. +3. Find the area of ∆SUB, given that it is isosceles, +SE is perpendicular to UB, and the area of ∆SEB is +24 sq. units. +In the Śulba-Sūtras, which are ancient Indian geometric texts that deal +with the construction of altars, we can find many interesting problems +on the topic of areas. When altars are built, they must have the exact +prescribed shape and area. This gives rise to problems of the kind where +one has to transform a given shape into another of the same area. The +Śulba-Sūtras give solutions to many such problems. +Such problems are also posed and solved in Euclid’s Elements. +Here are two problems of this kind. +4. [Śulba-Sūtras] Give a method to transform a rectangle into a triangle +of equal area. +5. [Śulba-Sūtras] Give a method to transform a triangle into a rectangle +of equal area. +6. ABCD, BCEF, and BFGH are identical squares. +(i) If the area of the red region is 49 sq. units, then what is the area +of the blue region? +(ii) In another version of this figure, if the total area enclosed +by the blue and red regions is 180 sq. units, then what is the +area of each square? +C +A +D +E +B +F +G +H +A +B +C +6 units +Y +4 units +8 units +X +S +U +B +E +Math +Talk + +Area +159 +7. If M and N are the midpoints of XY and XZ, what fraction of the +area of ∆XYZ is the area of ∆XMN? [Hint: Join NY] +X +M +N +Z +Y +Try +This +8. Gopal needs to carry water from the river to his water tank. He +starts from his house. What is the shortest path he can take from +his house to the river and then to the water tank? Roughly +recreate the map in your notebook and trace the shortest path. +River +Water tank +House +Area of any Polygon +How do we find the area of this quadrilateral? +What measurements do we need for this? +If we join BD, the quadrilateral ABCD gets divided +into two triangles. By finding their areas, we can find +the area of ABCD. +How do we find the area of this pentagon? +Can any polygon be divided into triangles? +It can be seen that any polygon can be divided into triangles. Thus, by +knowing how to compute the area of a triangle, we can find the area of +any polygon. +Math +Talk +A +B +D +C + +Ganita Prakash | Grade 8 | Part-II +160 +Figure it Out +1. Find the area of the quadrilateral ABCD given that AC = 22 cm, +BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is +perpendicular to AC. +A +B +D +C +N +M +22 cm +3 cm +3 cm +2. Find the area of the shaded region given that ABCD is a rectangle. +A +B +D +C +E +F +18 cm +10 cm +10 cm +8 cm +6 cm +4 cm +3. What measurements would you need to find the area of a regular +hexagon? +4. What fraction of the total area of the rectangle is the area of the +blue region? +5. Give a method to obtain a quadrilateral whose area is half that of +a given quadrilateral. +One can derive special formulae to find the areas of a parallelogram, +rhombus and trapezium. +Math +Talk +Math +Talk +Math +Talk + +Area +161 +Parallelogram +We can derive a special formula for the area of a parallelogram by +converting it into a rectangle of equal area. +Give a method to convert a parallelogram into a rectangle of equal area. +You can try this using a cut-out of a parallelogram. +Construct AX perpendicular to CD — represented in short as AX +CD +^ +. +We call this a height of the parallelogram. Cut the parallelogram into +∆AXD and trapezium ABCX. +A +B +D +C +X +A +B +D +C +X +A +X +Can ∆AXD and ABCX fit together, as shown in the figure, to get a rectangle? +One simple way to check this is to identify the triangle that can complete +ABCX to a rectangle, and then check if this triangle is congruent to ∆AXD. +A +D +X +A +X +B +C +Y +Observe that ∠X = 90°, and so ∠A = 90°, since AB‖XC. We need another +right angle to get a rectangle (what about the fourth angle?). To get +the third right angle, extend XC to the right and then construct a line +perpendicular to XC that passes through B. The ∆BYC completes ABCX to +a rectangle. Is ∆AXD congruent to it? +BY = AX (since ABYX is a rectangle) +∠BYC = ∠AXD = 90° +BC = AD (since ABCD is a parallelogram) +So, by the RHS congruency criterion, ∆BYC ≅ ∆AXD. Thus, ∆AXD will fit +exactly over the region occupied by ∆BYC, and convert the parallelogram +into a rectangle. +The process of cutting a figure into pieces and rearranging them to +get a different figure of equal area is called dissection. + +Ganita Prakash | Grade 8 | Part-II +162 +How do we find the area of a parallelogram +by dissecting it into a rectangle? +We have +Area of the parallelogram ABCD = Area of the +rectangle ABYX. +Area of the rectangle ABYX = AX × XY. +AX is the height of the parallelogram. +A +B +D +C +X +Y +height +base +Is there a relation between XY and DC? +Since DX = CY, we get DC = XY by adding the common part XC to DX +and CY. Since DC is the base of the parallelogram, we get +Area of the parallelogram = base × height. +Can the area of the parallelogram be +determined by taking another side +as the base and its corresponding +height? +A +B +D +C +Z +height +base +Can the parallelogram be cut along CZ and rearranged to form a +rectangle? +It can be seen that this is indeed possible, and so any side and its +corresponding height can be used to find the area of a parallelogram. +Figure it Out +1. Observe the parallelograms in the figure below. +(i) What can we say about the areas of all these parallelograms? +(ii) What can we say about their perimeters? Which figure appears +to have the maximum perimeter, and which has the minimum +perimeter? +(a) +(b) +(c) +(d) +(e) +(f) +(g) + +Area +163 +2. Find the areas of the following parallelograms: +4 cm +7 cm +(i) +(iii) +4.8 cm +5 cm +(iv) +4.4 cm +2 cm +3 cm +5 cm +(ii) +3. Find QN. +6 cm +7.6 cm +P +Q +S +R +M +N +12 cm +4. Consider a rectangle and a parallelogram of the same sidelengths: 5 cm +and 4 cm. Which has the greater area? [Hint: Imagine constructing +them on the same base.] +4 cm +5 cm +4 cm +5. Give a method to obtain a rectangle whose area is twice that of a +given triangle. What are the different methods that you can think of? + +Ganita Prakash | Grade 8 | Part-II +164 +6. [Śulba-Sūtras] Give a method to obtain a rectangle of the same area +as a given triangle. +7. [Śulba-Sūtras] An isosceles triangle can be converted into a rectangle +by dissection in a simpler way. Can you find out how to do it? +A +B +C +D +[Hint: Show that triangles ∆ADB and ∆ADC can be made into halves of a +rectangle. Figure out how they should be assembled to get a rectangle. +Use cut-outs if necessary.] +8. [Śulba-Sūtras] Give a method to convert a rectangle into an isosceles +triangle by dissection. +9. Which has greater area — an equilateral triangle or a square of +the same sidelength as the triangle? Which has greater area — two +identical equilateral triangles together or a square of the same +sidelength as the triangle? Give reasons. +Rhombus +Since a rhombus is a parallelogram, the area formula for a parallelogram +holds for a rhombus as well. However, the additional properties of +a rhombus give us another method to transform a rhombus into a +rectangle of the same area by dissection. This method occurs in one of +the Śulba-Sūtras. +Try working this out! +A +B +O +A +D +O +B +C +O +O +C +D +A +B +C +D +O + +Area +165 +A +B +O +A +D +O +C +D +O +C +B +O +X +Y +W +Z +Since ABCD is a rhombus, all its sides have equal length, and the diagonals +are perpendicular bisectors of each other. Therefore, ∆ABD and ∆CBD +are isosceles triangles. Each of them can be transformed into a rectangle +of equal area, and the two rectangles can then be joined to form a single +rectangle. This rectangle, say WXYZ, has the same area as the rhombus +ABCD. +What are the sidelengths of the rectangle WXYZ? +From the dissection, we can see that +XW = length of the diagonal AC, and +WZ = half the length of the other diagonal BD. +Thus, we have +Area of rhombus ABCD = Area of rectangle WXYZ + + + + + + += XW × WZ + + + + + + += AC × BD +2 + + + + + + += 1 +2 × AC × BD +Therefore, +Area of a rhombus = 1 +2 × product of diagonals. +Area of rhombus ABCD can also be determined by finding the areas of +∆ADB and ∆CDB. What formula does this give us? +Since the diagonals are perpendicular to each other, +we have +Area ( ∆ADB) = 1 +2 × AO × BD, and +Area ( ∆CDB) = 1 +2 × CO × BD, and +Area of rhombus ABCD = Area ( ∆ADB) + Area ( ∆CDB). +A +B +C +D +O + +Ganita Prakash | Grade 8 | Part-II +166 +Simplify the expression to show that we get the same formula for the +area of a rhombus in terms of its diagonals. +Trapezium +Find the areas of the following trapeziums by breaking them into figures +whose areas can be computed. +A +B +C +D +M +P +Q +S +R +T +U +W +X +Z +Y +M +N +One way of finding the area of a trapezium is by breaking it into a +rectangle and triangles. +Consider a trapezium WXYZ with WX ‖ ZY. Find its area. +Construct WM +ZY +^ +, and XN +ZY +^ + (WM and XN perpendicular to ZY). +Is WXNM a rectangle? +∠MWX = ∠NXW = 90°, since WX‖ZY and the interior angles on the +same side of a transversal (WM and XN) add up to 180°. Therefore, +WXNM is a rectangle. +We have +Area WXYZ = Area (∆WMZ) + Area WXNM + Area (∆XNY) += 1 +2 × MZ × WM + WX × WM + 1 +2 × NY × XN +Let us assign letter numbers to the lengths that we need to find the +area of the trapezium. Let MZ = x, WM = XN = h, WX = a, NY = y. +So we get +Area WXYZ = 1 +2 hx + ha + 1 +2 hy += h(1 +2 x + a + 1 +2 y) += h(x + y + 2a +2 +) += 1 +2 h(x + y + 2a). +W +X +Z +Y +M +N +Z +W +X +Y +M +N +a +h +h +x +y +b + +Area +167 +We have taken the length of one of the parallel sides as a. Let b be the +length of the other parallel side. +Can the area of the trapezium be expressed in terms of a, b and h? +To replace x + y in the area expression with a and b, observe that +b = x + y + a. Subtracting a from both sides, we get +x + y = b – a. +Using this, we get +Area WXYZ = 1 +2 h(b – a + 2a) += 1 +2 h(a + b). +Therefore, +Area of a trapezium = 1 +2 × height × sum of the parallel sides. +Will this formula hold for a trapezium that looks like this? +a +b +h +There are different ways of approaching this, which are sketched +below. Complete the arguments. +Approach 1: Rectangle and Triangles +A +D +C +a +B +F +E +b +h +Area ABCD = Area ABED + Area ∆BEC +Area ABED = Area ABEF – Area ∆AFD +Approach 2: Parallelogram and Triangle +A +D +C +a +B +F +G +b +Draw BG ‖ AD +h +Will Approach 2 work for any type of trapezium? +Math +Talk + +Ganita Prakash | Grade 8 | Part-II +168 +Finding the Area Using Two Copies of the Trapezium +There is another interesting way of finding the area of a trapezium. +Consider two copies of the given trapezium in which AB‖CD. Rotate the +second copy as shown. +What figure will we get when the two trapeziums are joined along BC? +The possibilities are either a 6-sided figure or a 4-sided figure +(quadrilateral). +Possibilities +A +D +B +C +y +x +A´ +D´ +x +y +6-sided figure +A +D +B +C +y +x +A´ +D´ +x +y +Quadrilateral +We can rule out the first possibility by looking at the sum of angles x and y. +Since AB ‖ CD, we have x + y = 180, since they are internal angles along +the same side of the transversal (BC). So ABD́ and A ́CD are straight lines. +Therefore, the resulting figure is a quadrilateral. +What type of a quadrilateral is this? +Let us look at the other two angles of the trapezium. +A +D +B +C +u +A´ +D´ +u +v +v +We have u + v = 180. Therefore, AD‖D́ A ́, since the sum of the internal +angles along the same side of the transversal A ́D is 180°. Since we already +have AD́  ‖ A ́D, the quadrilateral AD́ A ́D is actually a parallelogram. +a +b +h +b +a +a +b +h +A +D +B +C +y +x +B´ +A´ +C´ +D´ +x +y + +Area +169 +Area of the trapezium = 1 +2 × Area of the parallelogram += 1 +2 h(a + b). +Figure it Out +1. Find the area of a rhombus whose diagonals are 20 cm and 15 cm. +2. Give a method to convert a rectangle into a rhombus of equal area +using dissection. +3. Find the areas of the following figures: +4. [Śulba-Sūtras] Give a method to convert an isosceles trapezium to a +rectangle using dissection. +5. Here is one of the ways to convert trapezium ABCD into a rectangle +EFGH of equal area — +A +B +D +C +H +E +G +F +I +J +Given the trapezium ABCD, how do we find the vertices of the +rectangle EFGH? +[Hint: If ∆AHI ≅ ∆DGI and ∆BEJ ≅ ∆CFJ, then the trapezium and +rectangle have equal areas.] +Math +Talk +(ii) +24 m +36 m +14 m +(iii) +10 in +6 in +14 in +(iv) +12 ft +18 ft +8 ft +7 ft +10 ft +(i) +16 ft + +Ganita Prakash | Grade 8 | Part-II +170 +6. Using the idea of converting a trapezium into a rectangle of +equal area, and vice versa, construct a trapezium of area +144 cm2. +7. A regular hexagon is divided into a trapezium, an +equilateral triangle, and a rhombus, as shown. Find +the ratio of their areas. +8. ZYXW is a trapezium with ZY‖WX. A is the midpoint of XY. Show that +the area of the trapezium ZYXW is equal to the area of ∆ZWB. +A +B +Z +X +W +Y +Areas in Real Life +What do you think is the area of an A4 sheet? +Its sidelengths are 21 cm and 29.7 cm. Now find its area. +What do you think is the area of the tabletop that you use at school or +at home? You could perhaps try to visualise how many A4 sheets can fit +on your table. +The dimensions of furniture like tables and chairs are sometimes +measured in inches (in) and feet (ft). +1 in = 2.54 cm +1 ft = 12 in +Express the following lengths in centimeters: +(i) 5 in + +(ii) 7.4 in +Express the following lengths in inches: +(i) 5.08 cm +(ii) 11.43 cm +How many cm2 is 1 in2 ? += +1 in +2.54 cm +1 in +2.54 cm +So, 1 in2 = 2.542 cm2 = 6.4516 cm2. +Math +Talk + +Area +171 +How many cm2 is 10 in2 ? +10 in2 = 10 × 6.4516 cm2 = 64.516 cm2. +Convert 161.29 cm2 to in2. +Every 6.4516 cm2 gives an in2. Hence, +161.29 cm2 = 161.29 +6.4516 in2. +Evaluate the quotient. +What do you think is the area of your classroom? +Areas of classroom, house, etc., are generally measured in ft2 or m2. +How many in2 is 1 ft2 ? +What do you think is the area of your school? Make an estimate and +compare it with the actual data. +Larger areas of land are also measured in acres. +1 acre = 43,560 ft2. +Besides these units, different parts of India use different local units for +measuring area, such as bigha, gaj, katha, dhur, cent, ankanam, etc. +Find out the local unit of area measurement in your region. +What do you think is the area of your village/town/city? Make an estimate +and compare it with the actual data. +Larger areas are measured in km2. +How many m2 is a km2  ? +How many times is your village/town/city bigger than your school? +Find the city with the largest area in (i) India, and (ii) the world. +Find the city with the smallest area in (i) India, and (ii) the world. + +y +Area of a triangle = 1 +2 × base × height. + +y +The area of any polygon can be evaluated by breaking it into +triangles. + +y +Area of a parallelogram = base × height. + +y +Area of a rhombus = 1 +2 × product of its diagonals. + +y +Area of a trapezium = 1 +2 × height × sum of parallel sides. +SUMMARY + +LEARNING MATERIAL SHEETS + +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +13 +14 +15 +16 +17 +18 +19 +20 +21 +22 +23 +24 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +13 +14 +15 +16 +17 +18 +19 +20 +21 +22 +23 +24 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +13 +14 +15 +16 +17 +18 +19 +20 +21 +22 +23 +24 + + +Isometric Grid" +class_9,1,Number Systems,ncert_books/class_9/iemh1dd/iemh101.pdf,"NUMBER SYSTEMS +1 +CHAPTER 1 +NUMBER SYSTEMS +1.1 Introduction +In your earlier classes, you have learnt about the number line and how to represent +various types of numbers on it (see Fig. 1.1). +Fig. 1.1 : The number line +Just imagine you start from zero and go on walking along this number line in the +positive direction. As far as your eyes can see, there are numbers, numbers and +numbers! +Fig. 1.2 +Now suppose you start walking along the number line, and collecting some of the +numbers. Get a bag ready to store them! +Reprint 2025-26 + +2 +MATHEMATICS +3 +-40 +166 +22 +-75 +2 +1 +9 +0 +Z +3 +40 +16 +74 +5 +2 +601 +422 +58 +0 +-3 +-757 +-66 +-21 +-40 +31 +71 +3 +40 +16 +74 +5 +2 +601 +29 +58 +0 +W +9 +40 +16 +74 +5 +2 +601 +4652 +58 +0 +31 +1 +71 10 +N +You might begin with picking up only natural +numbers like 1, 2, 3, and so on. You know that this list +goes on for ever. (Why is this true?) So, now your +bag contains infinitely many natural numbers! Recall +that we denote this collection by the symbol N. +Now turn and walk all the way back, pick up +zero and put it into the bag. You now have the +collection of whole numbers which is denoted by +the symbol W. +Now, stretching in front of you are many, many negative integers. Put all the +negative integers into your bag. What is your new collection? Recall that it is the +collection of all integers, and it is denoted by the symbol Z. +Are there some numbers still left on the line? Of course! There are numbers like +1 3 +, +2 4 +, or even 2005 +2006 +− +. If you put all such numbers also into the bag, it will now be the +Z comes from the +German word +“zahlen”, which means +“to count”. +Q +–6721 +12 +1 +3 +–1 +9 +81 +16 +1 +4 +2005 +2006 +–12 +13 +9 +14 +–6625 +-65 +60 +19 +19 +999 +0 +–6 +7 +27 +58 +2005 +2006 +3 +–5 +16 +60 +999 +4 +–8 +–6625 +58 +0 +27 +71 +17 +981 +–12 +13 +89 +–6 +7 +2 +3 +9 +14 +– +Why Z ? +Reprint 2025-26 + +NUMBER SYSTEMS +3 +collection of rational numbers. The collection of rational numbers is denoted by Q. +‘Rational’ comes from the word ‘ratio’, and Q comes from the word ‘quotient’. +You may recall the definition of rational numbers: +A number ‘r’ is called a rational number, if it can be written in the form p +q , +where p and q are integers and q ≠ 0. (Why do we insist that q ≠ 0?) +Notice that all the numbers now in the bag can be written in the form p +q , where p +and q are integers and q ≠ 0. For example, –25 can be written as +25; +1 +− + here p = –25 +and q = 1. Therefore, the rational numbers also include the natural numbers, whole +numbers and integers. +You also know that the rational numbers do not have a unique representation in +the form p +q , where p and q are integers and q ≠ 0. For example, 1 +2 = 2 +4 = 10 +20 = 25 +50 += 47 +94 , and so on. These are equivalent rational numbers (or fractions). However, +when we say that p +q is a rational number, or when we represent p +q on the number +line, we assume that q ≠ 0 and that p and q have no common factors other than 1 +(that is, p and q are co-prime). So, on the number line, among the infinitely many +fractions equivalent to 1 +2 , we will choose 1 +2 to represent all of them. +Now, let us solve some examples about the different types of numbers, which you +have studied in earlier classes. +Example 1 : Are the following statements true or false? Give reasons for your answers. +(i) +Every whole number is a natural number. +(ii) Every integer is a rational number. +(iii) Every rational number is an integer. +Solution : (i) False, because zero is a whole number but not a natural number. +(ii) True, because every integer m can be expressed in the form 1 +m , and so it is a +rational number. +Reprint 2025-26 + +4 +MATHEMATICS +(iii) False, because 3 +5 is not an integer. +Example 2 : Find five rational numbers between 1 and 2. +We can approach this problem in at least two ways. +Solution 1 : Recall that to find a rational number between r and s, you can add r and +s and divide the sum by 2, that is +2 +r +s ++ + lies between r and s. So, 3 +2 is a number +between 1 and 2. You can proceed in this manner to find four more rational numbers +between 1 and 2. These four numbers are 5 11 13 +7. +, +, +and +4 +8 +8 +4 +Solution 2 : The other option is to find all the five rational numbers in one step. Since +we want five numbers, we write 1 and 2 as rational numbers with denominator 5 + 1, +i.e., 1 = 6 +6 and 2 = 12 +6 . Then you can check that 7 +6 , 8 +6 , 9 +6 , 10 +6 and 11 +6 are all rational +numbers between 1 and 2. So, the five numbers are 7 4 3 5 +11 +, +, +, +and +6 3 2 3 +6 . +Remark : Notice that in Example 2, you were asked to find five rational numbers +between 1 and 2. But, you must have realised that in fact there are infinitely many +rational numbers between 1 and 2. In general, there are infinitely many rational +numbers between any two given rational numbers. +Let us take a look at the number line again. Have you picked up all the numbers? +Not, yet. The fact is that there are infinitely many more numbers left on the number +line! There are gaps in between the places of the numbers you picked up, and not just +one or two but infinitely many. The amazing thing is that there are infinitely many +numbers lying between any two of these gaps too! +So we are left with the following questions: +1. What are the numbers, that are left on the number +line, called? +2. How do we recognise them? That is, how do we +distinguish them from the rationals (rational +numbers)? +These questions will be answered in the next section. +Reprint 2025-26 + +NUMBER SYSTEMS +5 +EXERCISE 1.1 +1. +Is zero a rational number? Can you write it in the form p +q , where p and q are integers +and q ≠ 0? +2. +Find six rational numbers between 3 and 4. +3. +Find five rational numbers between 3 +5 and 4 +5 . +4. +State whether the following statements are true or false. Give reasons for your answers. +(i) +Every natural number is a whole number. +(ii) Every integer is a whole number. +(iii) Every rational number is a whole number. +1.2 Irrational Numbers +We saw, in the previous section, that there may be numbers on the number line that +are not rationals. In this section, we are going to investigate these numbers. So far, all +the numbers you have come across, are of the form p +q , where p and q are integers +and q ≠ 0. So, you may ask: are there numbers which are not of this form? There are +indeed such numbers. +The Pythagoreans in Greece, followers of the famous +mathematician and philosopher Pythagoras, were the first +to discover the numbers which were not rationals, around +400 BC. These numbers are called irrational numbers +(irrationals), because they cannot be written in the form of +a ratio of integers. There are many myths surrounding the +discovery of irrational numbers by the Pythagorean, +Hippacus of Croton. In all the myths, Hippacus has an +unfortunate end, either for discovering that +2 is irrational +or for disclosing the secret about +2 to people outside the +secret Pythagorean sect! + Let us formally define these numbers. +A number ‘s’ is called irrational, if it cannot be written in the form p +q , where p +and q are integers and q ≠ 0. +Pythagoras +(569 BCE – 479 BCE) +Fig. 1.3 +Reprint 2025-26 + +6 +MATHEMATICS +2005 +2006 +3 +–5 +16 +60 +999 +4 +–8 +–6625 +58 +0 +27 +71 +17 +981 +–12 +13 +89 +–6 +7 +2 +3 +9 +14 +-65 +–66 +26 +-45 +0 +36 +19 +R +You already know that there are infinitely many rationals. It turns out that there +are infinitely many irrational numbers too. Some examples are: +2, +3, +15, π, 0.10110111011110... +Remark : Recall that when we use the symbol +, we assume that it is the +positive square root of the number. So +4 = 2, though both 2 and –2 are square +roots of 4. +Some of the irrational numbers listed above are familiar to you. For example, you +have already come across many of the square roots listed above and the number π. +The Pythagoreans proved that +2 is irrational. Later in approximately 425 BC, +Theodorus of Cyrene showed that 3, +5, +6, +7, +10, +11, +12, +13, 14, +15 +and 17 are also irrationals. Proofs of irrationality of +2 , +3 , +5 , etc., shall be +discussed in Class X. As to π, it was known to various cultures for thousands of +years, it was proved to be irrational by Lambert and Legendre only in the late 1700s. +In the next section, we will discuss why 0.10110111011110... and π are irrational. +Let us return to the questions raised at the end of +the previous section. Remember the bag of rational +numbers. If we now put all irrational numbers into +the bag, will there be any number left on the number +line? The answer is no! It turns out that the collection +of all rational numbers and irrational numbers together +make up what we call the collection of real numbers, +which is denoted by R. Therefore, a real number is either rational or irrational. So, we +can say that every real number is represented by a unique point on the number +line. Also, every point on the number line represents a unique real number. +This is why we call the number line, the real number line. +In the 1870s two German mathematicians, +Cantor and Dedekind, showed that : +Corresponding to every real number, there is a +point on the real number line, and corresponding +to every point on the number line, there exists a +unique real number. +G. Cantor (1845-1918) +Fig. 1.5 +R. Dedekind (1831-1916) +Fig. 1.4 +Reprint 2025-26 + +NUMBER SYSTEMS +7 +Let us see how we can locate some of the irrational numbers on the number line. +Example 3 : Locate +2 on the number line. +Solution : It is easy to see how the Greeks might have discovered +2 . Consider a square OABC, with each side 1 unit in length (see +Fig. 1.6). Then you can see by the Pythagoras theorem that +OB = +2 +2 +1 +1 +2 ++ += +. How do we represent +2 on the number line? +This is easy. Transfer Fig. 1.6 onto the number line making sure that the vertex O +coincides with zero (see Fig. 1.7). +Fig. 1.7 +We have just seen that OB = +2 . Using a compass with centre O and radius OB, +draw an arc intersecting the number line at the point P. Then P corresponds to +2 on +the number line. +Example 4 : Locate +3 on the number line. +Solution : Let us return to Fig. 1.7. +Fig. 1.8 +Construct BD of unit length perpendicular to OB (as in Fig. 1.8). Then using the +Pythagoras theorem, we see that OD = ( +) +2 +2 +2 +1 +3 ++ += +. Using a compass, with +centre O and radius OD, draw an arc which intersects the number line at the point Q. +Then Q corresponds to +3 . +Fig. 1.6 +Reprint 2025-26 + +8 +MATHEMATICS +In the same way, you can locate +n for any positive integer n, after +1 +n − + has been +located. +EXERCISE 1.2 +1. +State whether the following statements are true or false. Justify your answers. +(i) +Every irrational number is a real number. +(ii) Every point on the number line is of the form m , where m is a natural number. +(iii) Every real number is an irrational number. +2. +Are the square roots of all positive integers irrational? If not, give an example of the +square root of a number that is a rational number. +3. +Show how +5 can be represented on the number line. +4. +Classroom activity (Constructing the ‘square root +spiral’) : Take a large sheet of paper and construct +the ‘square root spiral’ in the following fashion. Start +with a point O and draw a line segment OP1 of unit +length. Draw a line segment P1P2 perpendicular to +OP1 of unit length (see Fig. 1.9). Now draw a line +segment P2P3 perpendicular to OP2. Then draw a line +segment P3P4 perpendicular to OP3. Continuing in +this manner, you can get the line segment Pn–1Pn by +drawing a line segment of unit length perpendicular to OPn–1. In this manner, you will +have created the points P2, P3,...., Pn,... ., and joined them to create a beautiful spiral +depicting 2, +3, +4, ... +1.3 Real Numbers and their Decimal Expansions +In this section, we are going to study rational and irrational numbers from a different +point of view. We will look at the decimal expansions of real numbers and see if we +can use the expansions to distinguish between rationals and irrationals. We will also +explain how to visualise the representation of real numbers on the number line using +their decimal expansions. Since rationals are more familiar to us, let us start with +them. Let us take three examples : +10 +7 +1 +, +, +3 +8 +7 . +Pay special attention to the remainders and see if you can find any pattern. +Fig. 1.9 : Constructing +square root spiral +Reprint 2025-26 + +NUMBER SYSTEMS +9 +Example 5 : Find the decimal expansions of 10 +3 , 7 +8 and 1 +7 . +Solution : + 3.333... + 0.875 + 0.142857... +3 10 +8 7.0 +7 +1.0 + 9 + 64 + 7 + 10 + 60 + 30 + 9 + 56 + 28 + 10 + 40 + 20 + 9 + 40 + 14 + 10 + 0 + 60 + 9 + 56 + 1 + 40 + 35 + 50 + 49 + 1 +Remainders : 1, 1, 1, 1, 1... +Remainders : 6, 4, 0 +Remainders : 3, 2, 6, 4, 5, 1, +Divisor : 3 +Divisor : 8 +3, 2, 6, 4, 5, 1,... +Divisor : 7 +What have you noticed? You should have noticed at least three things: +(i) +The remainders either become 0 after a certain stage, or start repeating themselves. +(ii) +The number of entries in the repeating string of remainders is less than the divisor +(in 10 +3 one number repeats itself and the divisor is 3, in 1 +7 there are six entries +326451 in the repeating string of remainders and 7 is the divisor). +(iii) If the remainders repeat, then we get a repeating block of digits in the quotient +(for 10 +3 , 3 repeats in the quotient and for 1 +7 , we get the repeating block 142857 +in the quotient). +Reprint 2025-26 + +10 +MATHEMATICS +Although we have noticed this pattern using only the examples above, it is true for all +rationals of the form p +q (q ≠ 0). On division of p by q, two main things happen – either +the remainder becomes zero or never becomes zero and we get a repeating string of +remainders. Let us look at each case separately. +Case (i) : The remainder becomes zero +In the example of 7 +8 , we found that the remainder becomes zero after some steps and +the decimal expansion of 7 +8 = 0.875. Other examples are 1 +2 = 0.5, 639 +250 = 2.556. In all +these cases, the decimal expansion terminates or ends after a finite number of steps. +We call the decimal expansion of such numbers terminating. +Case (ii) : The remainder never becomes zero +In the examples of 10 +3 and 1 +7 , we notice that the remainders repeat after a certain +stage forcing the decimal expansion to go on for ever. In other words, we have a +repeating block of digits in the quotient. We say that this expansion is non-terminating +recurring. For example, 10 +3 = 3.3333... and 1 +7 = 0.142857142857142857... +The usual way of showing that 3 repeats in the quotient of 10 +3 is to write it as 3.3. +Similarly, since the block of digits 142857 repeats in the quotient of 1 +7 , we write 1 +7 as +0.142857 , where the bar above the digits indicates the block of digits that repeats. +Also 3.57272... can be written as 3.572 . So, all these examples give us non-terminating +recurring (repeating) decimal expansions. +Thus, we see that the decimal expansion of rational numbers have only two choices: +either they are terminating or non-terminating recurring. +Now suppose, on the other hand, on your walk on the number line, you come across a +number like 3.142678 whose decimal expansion is terminating or a number like +1.272727... that is, 1.27 , whose decimal expansion is non-terminating recurring, can +you conclude that it is a rational number? The answer is yes! +Reprint 2025-26 + +NUMBER SYSTEMS +11 +We will not prove it but illustrate this fact with a few examples. The terminating cases +are easy. +Example 6 : Show that 3.142678 is a rational number. In other words, express 3.142678 +in the form p +q , where p and q are integers and q ≠ 0. +Solution : We have 3.142678 = 3142678 +1000000 +, and hence is a rational number. +Now, let us consider the case when the decimal expansion is non-terminating recurring. +Example 7 : Show that 0.3333... = 03. can be expressed in the form p +q , where p and +q are integers and q ≠ 0. +Solution : Since we do not know what 03. is , let us call it ‘x’ and so +x = 0.3333... +Now here is where the trick comes in. Look at +10 x = 10 × (0.333...) = 3.333... +Now, +3.3333... = 3 + x, since x = 0.3333... +Therefore, +10 x = 3 + x +Solving for x, we get +9x = 3, i.e., x = 1 +3 +Example 8 : Show that 1.272727... = 1 27 +. + can be expressed in the form p +q , where p +and q are integers and q ≠ 0. +Solution : Let x = 1.272727... Since two digits are repeating, we multiply x by 100 to +get +100 x = 127.2727... +So, +100 x = 126 + 1.272727... = 126 + x +Therefore, +100 x – x = 126, i.e., 99 x = 126 +Reprint 2025-26 + +12 +MATHEMATICS +i.e., +x = 126 +14 +99 +11 += +You can check the reverse that 14 +11 = 1 27 +. +. +Example 9 : Show that 0.2353535... = 0 235 +. + can be expressed in the form p +q , +where p and q are integers and q ≠ 0. +Solution : Let x = 0 235 +. +. Over here, note that 2 does not repeat, but the block 35 +repeats. Since two digits are repeating, we multiply x by 100 to get +100 x = 23.53535... +So, +100 x = 23.3 + 0.23535... = 23.3 + x +Therefore, +99 x = 23.3 +i.e., +99 x = 233 +10 +, which gives x = 233 +990 +You can also check the reverse that 233 +990 = 0 235 +. +. +So, every number with a non-terminating recurring decimal expansion can be expressed +in the form p +q (q ≠ 0), where p and q are integers. Let us summarise our results in the +following form : +The decimal expansion of a rational number is either terminating or non- +terminating recurring. Moreover, a number whose decimal expansion is +terminating or non-terminating recurring is rational. +So, now we know what the decimal expansion of a rational number can be. What +about the decimal expansion of irrational numbers? Because of the property above, +we can conclude that their decimal expansions are non-terminating non-recurring. +So, the property for irrational numbers, similar to the property stated above for rational +numbers, is +The decimal expansion of an irrational number is non-terminating non-recurring. +Moreover, a number whose decimal expansion is non-terminating non-recurring +is irrational. +Reprint 2025-26 + +NUMBER SYSTEMS +13 +Recall s = 0.10110111011110... from the previous section. Notice that it is non- +terminating and non-recurring. Therefore, from the property above, it is irrational. +Moreover, notice that you can generate infinitely many irrationals similar to s. +What about the famous irrationals +2 and π? Here are their decimal expansions up +to a certain stage. +2 = 1.4142135623730950488016887242096... + π = 3.14159265358979323846264338327950... +(Note that, we often take 22 +7 as an approximate value for π, but π ≠ 22 +7 .) +Over the years, mathematicians have developed various techniques to produce more +and more digits in the decimal expansions of irrational numbers. For example, you +might have learnt to find digits in the decimal expansion of 2 by the division method. +Interestingly, in the Sulbasutras (rules of chord), a mathematical treatise of the Vedic +period (800 BC - 500 BC), you find an approximation of +2 as follows: +2 = +1 +1 +1 +1 +1 +1 +1 +1 4142156 +3 +4 +3 +34 +4 +3 +. + + + + ++ ++ +× +− +× +× += + + + + + + + + +Notice that it is the same as the one given above for the first five decimal places. The +history of the hunt for digits in the decimal expansion of π is very interesting. +The Greek genius Archimedes was the first to compute +digits in the decimal expansion of π. He showed 3.140845 +< π < 3.142857. Aryabhatta (476 – 550 C.E.), the great +Indian mathematician and astronomer, found the value +of π correct to four decimal places (3.1416). Using high +speed computers and advanced algorithms, π has been +computed to over 1.24 trillion decimal places! +Now, let us see how to obtain irrational numbers. +Example 10 : Find an irrational number between 1 +7 and 2 +7 . +Solution : We saw that 1 +7 = 0142857 +. +. So, you can easily calculate 2 +0 285714 +7 +. += +. +To find an irrational number between 1 +7 and 2 +7 , we find a number which is +Archimedes (287 BCE – 212 BCE) +Fig. 1.10 +Reprint 2025-26 + +14 +MATHEMATICS +non-terminating non-recurring lying between them. Of course, you can find infinitely +many such numbers. +An example of such a number is 0.150150015000150000... +EXERCISE 1.3 +1. +Write the following in decimal form and say what kind of decimal expansion each +has : +(i) +36 +100 +(ii) +1 +11 +(iii) +1 +48 +(iv) +3 +13 +(v) +2 +11 +(vi) +329 +400 +2. +You know that 1 +7 = 0142857 +. +. Can you predict what the decimal expansions of 2 +7 , 3 +7 , +4 +7 , +5 +7 , +6 +7 are, without actually doing the long division? If so, how? +[Hint : Study the remainders while finding the value of 1 +7 carefully.] +3. +Express the following in the form p +q , where p and q are integers and q ≠ 0. +(i) +0 6. +(ii) 0 47 +. +(iii) 0 001 +. +4. +Express 0.99999 .... in the form p +q . Are you surprised by your answer? With your +teacher and classmates discuss why the answer makes sense. +5. +What can the maximum number of digits be in the repeating block of digits in the +decimal expansion of 1 +17 ? Perform the division to check your answer. +6. +Look at several examples of rational numbers in the form p +q (q ≠ 0), where p and q are +integers with no common factors other than 1 and having terminating decimal +representations (expansions). Can you guess what property q must satisfy? +7. +Write three numbers whose decimal expansions are non-terminating non-recurring. +8. +Find three different irrational numbers between the rational numbers 5 +7 and 9 +11 . +9. +Classify the following numbers as rational or irrational : +(i) +23 +(ii) +225 +(iii) 0.3796 +(iv) 7.478478... +(v) 1.101001000100001... +Reprint 2025-26 + +NUMBER SYSTEMS +15 +1.4 Operations on Real Numbers +You have learnt, in earlier classes, that rational numbers satisfy the commutative, +associative and distributive laws for addition and multiplication. Moreover, if we add, +subtract, multiply or divide (except by zero) two rational numbers, we still get a rational +number (that is, rational numbers are ‘closed’ with respect to addition, subtraction, +multiplication and division). It turns out that irrational numbers also satisfy the +commutative, associative and distributive laws for addition and multiplication. However, +the sum, difference, quotients and products of irrational numbers are not always +irrational. For example, ( +) +( +) +6 +6 ++ − +,( +) +( +) ( +) ( +) +2 +2 +3 +3 +, +− +⋅ + and +17 +17 are +rationals. +Let us look at what happens when we add and multiply a rational number with an +irrational number. For example, 3 is irrational. What about 2 +3 ++ + and 2 3 ? Since +3 has a non-terminating non-recurring decimal expansion, the same is true for +2 +3 ++ + and 2 3 . Therefore, both 2 +3 ++ + and 2 3 are also irrational numbers. +Example 11 : Check whether 7 5 , 7 +2 +21 +2 +5 +, +, ++ +π − + are irrational numbers or +not. +Solution : +5 = 2.236... , +2 = 1.4142..., π = 3.1415... +Then 7 5 = 15.652..., +7 +5 = +7 5 +7 5 +5 +5 5 += + = 3.1304... +2 + 21 = 22.4142..., π – 2 = 1.1415... +All these are non-terminating non-recurring decimals. So, all these are irrational numbers. +Now, let us see what generally happens if we add, subtract, multiply, divide, take +square roots and even nth roots of these irrational numbers, where n is any natural +number. Let us look at some examples. +Example 12 : Add 2 2 +5 3 ++ + and +2 +3 3 +– +. +Solution : ( +) ( +) +2 2 +5 3 +2 +3 3 +– ++ ++ + = ( +) +( +) +2 2 +2 +5 3 +3 3 +– ++ ++ + += (2 + 1) +2 +(5 +3) 3 +3 2 +2 3 ++ +− += ++ +Reprint 2025-26 + +16 +MATHEMATICS +Example 13 : Multiply 6 5 by 2 5 . +Solution : 6 5 × 2 5 = 6 × 2 × +5 × 5 = 12 × 5 = 60 +Example 14 : Divide 8 15 by 2 3 . +Solution : +8 3 +5 +8 15 +2 3 +4 5 +2 3 +× +÷ += += +These examples may lead you to expect the following facts, which are true: +(i) +The sum or difference of a rational number and an irrational number is irrational. +(ii) The product or quotient of a non-zero rational number with an irrational number is +irrational. +(iii) If we add, subtract, multiply or divide two irrationals, the result may be rational or +irrational. +We now turn our attention to the operation of taking square roots of real numbers. +Recall that, if a is a natural number, then +a +b += + means b2 = a and b > 0. The same +definition can be extended for positive real numbers. +Let a > 0 be a real number. Then +a = b means b2 = a and b > 0. +In Section 1.2, we saw how to represent n for any positive integer n on the number +line. We now show how to find +x for any given positive real number x geometrically. +For example, let us find it for x = 3.5, i.e., we find 3 5. geometrically. +Fig. 1.11 +Mark the distance 3.5 units from a fixed point A on a given line to obtain a point B such +that AB = 3.5 units (see Fig. 1.11). From B, mark a distance of 1 unit and mark the +new point as C. Find the mid-point of AC and mark that point as O. Draw a semicircle +with centre O and radius OC. Draw a line perpendicular to AC passing through B and +intersecting the semicircle at D. Then, BD = 3.5 . +Reprint 2025-26 + +NUMBER SYSTEMS +17 +More generally, to find +x , for any positive real +number x, we mark B so that AB = x units, and, as in +Fig. 1.12, mark C so that BC = 1 unit. Then, as we +have done for the case x = 3.5, we find BD = +x +(see Fig. 1.12). We can prove this result using the +Pythagoras Theorem. +Notice that, in Fig. 1.12, ∆ OBD is a right-angled triangle. Also, the radius of the circle +is +1 +2 +x + + units. +Therefore, OC = OD = OA = +1 +2 +x + + units. +Now, OB = +1 +1 +2 +2 +x +x +x ++ +− + + +− += +⋅ + + + + +So, by the Pythagoras Theorem, we have +BD2 = OD2 – OB2 = +2 +2 +1 +1 +4 +2 +2 +4 +x +x +x +x ++ +− + + + + +− += += + + + + + + + + +. +This shows that BD = +x . +This construction gives us a visual, and geometric way of showing that +x exists for +all real numbers x > 0. If you want to know the position of +x on the number line, +then let us treat the line BC as the number line, with B as zero, C as 1, and so on. +Draw an arc with centre B and radius BD, which intersects the number line in E +(see Fig. 1.13). Then, E represents +x . +Fig. 1.13 +Fig. 1.12 +Reprint 2025-26 + +18 +MATHEMATICS +We would like to now extend the idea of square roots to cube roots, fourth roots, +and in general nth roots, where n is a positive integer. Recall your understanding of +square roots and cube roots from earlier classes. +What is 3 8 ? Well, we know it has to be some positive number whose cube is 8, and +you must have guessed 3 8 = 2. Let us try 5 243 . Do you know some number b such +that b5 = 243? The answer is 3. Therefore, 5 243 = 3. +From these examples, can you define n a for a real number a > 0 and a positive +integer n? +Let a > 0 be a real number and n be a positive integer. Then n a = b, if bn = a and +b > 0. Note that the symbol ‘ +’ used in +3 +2, +8, n a , etc. is called the radical sign. +We now list some identities relating to square roots, which are useful in various +ways. You are already familiar with some of these from your earlier classes. The +remaining ones follow from the distributive law of multiplication over addition of real +numbers, and from the identity (x + y) (x – y) = x2 – y2, for any real numbers x and y. +Let a and b be positive real numbers. Then +(i) +ab +a b += +(ii) +a +a +b +b += +(iii) ( +) ( +) +a +b +a +b +a +b ++ +− += +− +(iv) ( +) ( +) +2 +a +b +a +b +a +b ++ +− += +− +(v) ( +) ( +) +a +b +c +d +ac +ad +bc +bd ++ ++ += ++ ++ ++ +(vi) ( +) +2 +2 +a +b +a +ab +b ++ += ++ ++ +Let us look at some particular cases of these identities. +Example 15 : Simplify the following expressions: +(i) ( +) ( +) +5 +7 +2 +5 ++ ++ +(ii) ( +) ( +) +5 +5 +5 +5 ++ +− +(iii) ( +) +2 +3 +7 ++ +(iv) ( +) ( +) +11 +7 +11 +7 +− ++ +Reprint 2025-26 + +NUMBER SYSTEMS +19 +Solution : (i) ( +) ( +) +5 +7 +2 +5 +10 +5 5 +2 7 +35 ++ ++ += ++ ++ ++ +(ii) ( +) ( +) +( +) +2 +2 +5 +5 +5 +5 +5 +5 +25 +5 +20 +– ++ +− += +− += += +(iii) ( +) +( +) +( +) +2 +2 +2 +3 +7 +3 +2 3 7 +7 +3 +2 21 +7 +10 +2 21 ++ += ++ ++ += ++ ++ += ++ +(iv) ( +) ( +) +( +) +( +) +2 +2 +11 +7 +11 +7 +11 +7 +11 +7 +4 +− ++ += +− += +− += +Remark : Note that ‘simplify’ in the example above has been used to mean that the +expression should be written as the sum of a rational and an irrational number. +We end this section by considering the following problem. Look at +1 +2 +⋅ Can you tell +where it shows up on the number line? You know that it is irrational. May be it is easier +to handle if the denominator is a rational number. Let us see, if we can ‘rationalise’ the +denominator, that is, to make the denominator into a rational number. To do so, we +need the identities involving square roots. Let us see how. +Example 16 : Rationalise the denominator of +1 +2 +⋅ +Solution : We want to write +1 +2 as an equivalent expression in which the denominator +is a rational number. We know that 2 . +2 is rational. We also know that multiplying +1 +2 by +2 +2 will give us an equivalent expression, since +2 +2 = 1. So, we put these two +facts together to get +1 +1 +2 +2 +2 +2 +2 +2 += +× += +⋅ +In this form, it is easy to locate +1 +2 on the number line. It is half way between 0 +and +2 . +Reprint 2025-26 + +20 +MATHEMATICS +Example 17 : Rationalise the denominator of +1 +2 +3 +⋅ ++ +Solution : We use the Identity (iv) given earlier. Multiply and divide +1 +2 +3 ++ + by +2 +3 +− + to get +1 +2 +3 +2 +3 +2 +3 +4 +3 +2 +3 +2 +3 +− +− +× += += +− +− ++ +− +. +Example 18 : Rationalise the denominator of +5 +3 +5 +⋅ +− +Solution : Here we use the Identity (iii) given earlier. +So, +5 +3 +5 +− + = +( +) +( +) +5 +3 +5 +5 +3 +5 +5 +3 +5 +3 +5 +2 +3 +5 +3 +5 ++ ++ +− + + +× += += ++ + + +− +− ++ + + +Example 19 : Rationalise the denominator of +1 +7 +3 2 +⋅ ++ +Solution : +1 +1 +7 +3 2 +7 +3 2 +7 +3 2 +49 +18 +31 +7 +3 2 +7 +3 2 +7 +3 2 + + +− +− +− += +× += += + + + + +− ++ ++ +− + + +So, when the denominator of an expression contains a term with a square root (or +a number under a radical sign), the process of converting it to an equivalent expression +whose denominator is a rational number is called rationalising the denominator. +EXERCISE 1.4 +1. +Classify the following numbers as rational or irrational: +(i) +2 +5 +− +(ii) ( +) +3 +23 +23 ++ +− +(iii) +2 7 +7 7 +(iv) +1 +2 +(v) 2π +Reprint 2025-26 + +NUMBER SYSTEMS +21 +2. +Simplify each of the following expressions: +(i) ( +) ( +) +3 +3 +2 +2 ++ ++ +(ii) ( +) ( +) +3 +3 +3 +3 ++ +− +(iii) ( +) +2 +5 +2 ++ +(iv) ( +) ( +) +5 +2 +5 +2 +− ++ +3. +Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter +(say d). That is, π = c +d ⋅ This seems to contradict the fact that π is irrational. How will +you resolve this contradiction? +4. +Represent +9 3. on the number line. +5. +Rationalise the denominators of the following: +(i) +1 +7 +(ii) +1 +7 +6 +− +(iii) +1 +5 +2 ++ +(iv) +1 +7 +2 +− +1.5 Laws of Exponents for Real Numbers +Do you remember how to simplify the following? +(i) 172 . 175 = +(ii) (52)7 = +(iii) +10 +7 +23 +23 = +(iv) 73 . 93 = +Did you get these answers? They are as follows: +(i) 172 . 175 = 177 +(ii) (52)7 = 514 +(iii) +10 +3 +7 +23 +23 +23 += +(iv) 73 . 93 = 633 +To get these answers, you would have used the following laws of exponents, +which you have learnt in your earlier classes. (Here a, n and m are natural numbers. +Remember, a is called the base and m and n are the exponents.) +(i) +am . an = am + n +(ii) (am)n = amn +(iii) +m +m +n +n +a +a +, m +n +a +− += +> +(iv) ambm = (ab)m +Reprint 2025-26 + +22 +MATHEMATICS +What is (a)0? Yes, it is 1! So you have learnt that (a)0 = 1. So, using (iii), we can +get 1 +. +n +n +a +a +− += + We can now extend the laws to negative exponents too. +So, for example : +(i) +2 +–5 +–3 +3 +1 +17 +17 +17 +17 +⋅ += += +(ii) +2 +–7 +–14 +(5 ) +5 += +(iii) +–10 +–17 +7 +23 +23 +23 += +(iv) +–3 +–3 +–3 +(7) +(9) +(63) +⋅ += +Suppose we want to do the following computations: +(i) +2 +1 +3 +3 +2 +2 +⋅ +(ii) +4 +1 +5 +3 + + + + + + +(iii) +1 +5 +1 +3 +7 +7 +(iv) +1 +1 +5 +5 +13 +17 +⋅ +How would we go about it? It turns out that we can extend the laws of exponents +that we have studied earlier, even when the base is a positive real number and the +exponents are rational numbers. (Later you will study that it can further to be extended +when the exponents are real numbers.) But before we state these laws, and to even +make sense of these laws, we need to first understand what, for example +3 +2 +4 is. So, +we have some work to do! +We define n a for a real number a > 0 as follows: +Let a > 0 be a real number and n a positive integer. Then n a = b, if bn = a and +b > 0. +In the language of exponents, we define n a = +1 +n +a . So, in particular, +1 +3 +3 +2 +2 += +. +There are now two ways to look at +3 +2 +4 . +3 +2 +4 = +3 +1 +3 +2 +4 +2 +8 + += += + + + + +3 +2 +4 = ( +) +( +) +1 +1 +3 +2 +2 +4 +64 +8 += += +Reprint 2025-26 + +NUMBER SYSTEMS +23 +Therefore, we have the following definition: +Let a > 0 be a real number. Let m and n be integers such that m and n have no +common factors other than 1, and n > 0. Then, +m +n +a = ( +) +m +n +m +n a +a += +We now have the following extended laws of exponents: +Let a > 0 be a real number and p and q be rational numbers. Then, we have +(i) ap . aq = ap+q +(ii) (ap)q = apq +(iii) +p +p +q +q +a +a +a +− += +(iv) apbp = (ab)p +You can now use these laws to answer the questions asked earlier. +Example 20 : Simplify (i) +2 +1 +3 +3 +2 +2 +⋅ +(ii) +4 +1 +5 +3 + + + + + + + (iii) +1 +5 +1 +3 +7 +7 +(iv) +1 +1 +5 +5 +13 +17 +⋅ +Solution : +(i) +2 +1 +2 +1 +3 +1 +3 +3 +3 +3 +3 +2 +2 +2 +2 +2 +2 + + ++ + + + + +⋅ += += += += +(ii) +4 +1 +4 +5 +5 +3 +3 + += + + + + +(iii) +1 +1 +1 +3 +5 +2 +5 +5 +3 +15 +15 +1 +3 +7 +7 +7 +7 +7 + + +− +− +− + + + + += += += +(iv) +1 +1 +1 +1 +5 +5 +5 +5 +13 +17 +(13 +17) +221 +⋅ += +× += +EXERCISE 1.5 +1. +Find : +(i) +1 +2 +64 +(ii) +1 +5 +32 +(iii) +1 +3 +125 +2. +Find : +(i) +3 +2 +9 +(ii) +2 +5 +32 +(iii) +3 +4 +16 +(iv) +1 +3 +125 +− +3. +Simplify : +(i) +2 +1 +3 +5 +2 +2 +⋅ +(ii) +7 +3 +1 +3 + + + + + + +(iii) +1 +2 +1 +4 +11 +11 +(iv) +1 +1 +2 +2 +7 +8 +⋅ +Reprint 2025-26 + +24 +MATHEMATICS +1.6 Summary +In this chapter, you have studied the following points: +1. +A number r is called a rational number, if it can be written in the form p +q , where p and q are +integers and q ≠ 0. +2. +A number s is called a irrational number, if it cannot be written in the form p +q , where p and +q are integers and q ≠ 0. +3. +The decimal expansion of a rational number is either terminating or non-terminating recurring. +Moreover, a number whose decimal expansion is terminating or non-terminating recurring +is rational. +4. +The decimal expansion of an irrational number is non-terminating non-recurring. Moreover, +a number whose decimal expansion is non-terminating non-recurring is irrational. +5. +All the rational and irrational numbers make up the collection of real numbers. +6. +If r is rational and s is irrational, then r + s and r – s are irrational numbers, and rs and r +s are +irrational numbers, r ≠ 0. +7. +For positive real numbers a and b, the following identities hold: +(i) +ab +a b += +(ii) +a +a +b +b += +(iii) ( +) ( +) +a +b +a +b +a +b ++ +− += +− +(iv) ( +) ( +) +2 +a +b +a +b +a +b ++ +− += +− +(v) ( +) +2 +2 +a +b +a +ab +b ++ += ++ ++ +8. +To rationalise the denominator of +1 +, +a +b ++ + we multiply this by +, +a +b +a +b +− +− + where a and b are +integers. +9. +Let a > 0 be a real number and p and q be rational numbers. Then +(i) +ap . aq = ap + q +(ii) (ap)q = apq +(iii) +p +p +q +q +a +a +a +− += +(iv) apbp = (ab)p +Reprint 2025-26" +class_9,2,Polynomials,ncert_books/class_9/iemh1dd/iemh102.pdf,"POLYNOMIALS +25 +CHAPTER 2 +POLYNOMIALS +2.1 Introduction +You have studied algebraic expressions, their addition, subtraction, multiplication and +division in earlier classes. You also have studied how to factorise some algebraic +expressions. You may recall the algebraic identities : +(x + y)2 = x2 + 2xy + y2 +(x – y)2 = x2 – 2xy + y2 +and +x2 – y2 = (x + y) (x – y) +and their use in factorisation. In this chapter, we shall start our study with a particular +type of algebraic expression, called polynomial, and the terminology related to it. We +shall also study the Remainder Theorem and Factor Theorem and their use in the +factorisation of polynomials. In addition to the above, we shall study some more algebraic +identities and their use in factorisation and in evaluating some given expressions. +2.2 Polynomials in One Variable +Let us begin by recalling that a variable is denoted by a symbol that can take any real +value. We use the letters x, y, z, etc. to denote variables. Notice that 2x, 3x, – x, – 1 +2 x +are algebraic expressions. All these expressions are of the form (a constant) × x. Now +suppose we want to write an expression which is (a constant) × (a variable) and we do +not know what the constant is. In such cases, we write the constant as a, b, c, etc. So +the expression will be ax, say. +However, there is a difference between a letter denoting a constant and a letter +denoting a variable. The values of the constants remain the same throughout a particular +situation, that is, the values of the constants do not change in a given problem, but the +value of a variable can keep changing. +Reprint 2025-26 + +26 +MATHEMATICS +Now, consider a square of side 3 units (see Fig. 2.1). +What is its perimeter? You know that the perimeter of a square +is the sum of the lengths of its four sides. Here, each side is +3 units. So, its perimeter is 4 × 3, i.e., 12 units. What will be the +perimeter if each side of the square is 10 units? The perimeter +is 4 × 10, i.e., 40 units. In case the length of each side is x +units (see Fig. 2.2), the perimeter is given by 4x units. So, as +the length of the side varies, the perimeter varies. +Can you find the area of the square PQRS? It is +x × x = x2 square units. x2 is an algebraic expression. You are +also familiar with other algebraic expressions like +2x, x2 + 2x, x3 – x2 + 4x + 7. Note that, all the algebraic +expressions we have considered so far have only whole +numbers as the exponents of the variable. Expressions of this +form are called polynomials in one variable. In the examples +above, the variable is x. For instance, x3 – x2 + 4x + 7 is a +polynomial in x. Similarly, 3y2 + 5y is a polynomial in the +variable y and t2 + 4 is a polynomial in the variable t. +In the polynomial x2 + 2x, the expressions x2 and 2x are called the terms of the +polynomial. Similarly, the polynomial 3y2 + 5y + 7 has three terms, namely, 3y2, 5y and +7. Can you write the terms of the polynomial –x3 + 4x2 + 7x – 2 ? This polynomial has +4 terms, namely, –x3, 4x2, 7x and –2. +Each term of a polynomial has a coefficient. So, in –x3 + 4x2 + 7x – 2, the +coefficient of x3 is –1, the coefficient of x2 is 4, the coefficient of x is 7 and –2 is the +coefficient of x0 (Remember, x0 = 1). Do you know the coefficient of x in x2 – x + 7? +It is –1. +2 is also a polynomial. In fact, 2, –5, 7, etc. are examples of constant polynomials. +The constant polynomial 0 is called the zero polynomial. This plays a very important +role in the collection of all polynomials, as you will see in the higher classes. +Now, consider algebraic expressions such as x + +2 +3 +1 , +3 and +. ++ ++ +x +y +y +x + Do you +know that you can write x + 1 +x = x + x–1? Here, the exponent of the second term, i.e., +x–1 is –1, which is not a whole number. So, this algebraic expression is not a polynomial. +Again, +3 +x + + can be written as +1 +2 +3 +x ++ +. Here the exponent of x is 1 +2 , which is +not a whole number. So, is +3 +x + + a polynomial? No, it is not. What about +3 y + y2? It is also not a polynomial (Why?). +Fig. 2.1 +Fig. 2.2 +3 +3 +3 +3 +x +x +x +x +S +R +P +Q +Reprint 2025-26 + +POLYNOMIALS +27 +If the variable in a polynomial is x, we may denote the polynomial by p(x), or q(x), +or r(x), etc. So, for example, we may write : +p(x) = 2x2 + 5x – 3 +q(x) = x3 –1 +r(y) = y3 + y + 1 +s(u) = 2 – u – u2 + 6u5 +A polynomial can have any (finite) number of terms. For instance, x150 + x149 + ... ++ x2 + x + 1 is a polynomial with 151 terms. +Consider the polynomials 2x, 2, 5x3, –5x2, y and u4. Do you see that each of these +polynomials has only one term? Polynomials having only one term are called monomials +(‘mono’ means ‘one’). +Now observe each of the following polynomials: +p(x) = x + 1, +q(x) = x2 – x, +r(y) = y9 + 1, +t(u) = u15 – u2 +How many terms are there in each of these? Each of these polynomials has only +two terms. Polynomials having only two terms are called binomials (‘bi’ means ‘two’). +Similarly, polynomials having only three terms are called trinomials +(‘tri’ means ‘three’). Some examples of trinomials are +p(x) = x + x2 + π, +q(x) = +2 + x – x2, +r(u) = u + u2 – 2, +t(y) = y4 + y + 5. +Now, look at the polynomial p(x) = 3x7 – 4x6 + x + 9. What is the term with the +highest power of x ? It is 3x7. The exponent of x in this term is 7. Similarly, in the +polynomial q(y) = 5y6 – 4y2 – 6, the term with the highest power of y is 5y6 and the +exponent of y in this term is 6. We call the highest power of the variable in a polynomial +as the degree of the polynomial. So, the degree of the polynomial 3x7 – 4x6 + x + 9 +is 7 and the degree of the polynomial 5y6 – 4y2 – 6 is 6. The degree of a non-zero +constant polynomial is zero. +Example 1 : Find the degree of each of the polynomials given below: +(i) x5 – x4 + 3 +(ii) 2 – y2 – y3 + 2y8 +(iii) 2 +Solution : (i) The highest power of the variable is 5. So, the degree of the polynomial +is 5. +(ii) The highest power of the variable is 8. So, the degree of the polynomial is 8. +(iii) The only term here is 2 which can be written as 2x0. So the exponent of x is 0. +Therefore, the degree of the polynomial is 0. +Reprint 2025-26 + +28 +MATHEMATICS +Now observe the polynomials p(x) = 4x + 5, q(y) = 2y, r(t) = t + +2 and +s(u) = 3 – u. Do you see anything common among all of them? The degree of each of +these polynomials is one. A polynomial of degree one is called a linear polynomial. +Some more linear polynomials in one variable are 2x – 1, 2 y + 1, 2 – u. Now, try and +find a linear polynomial in x with 3 terms? You would not be able to find it because a +linear polynomial in x can have at most two terms. So, any linear polynomial in x will +be of the form ax + b, where a and b are constants and a ≠ 0 (why?). Similarly, +ay + b is a linear polynomial in y. +Now consider the polynomials : +2x2 + 5, 5x2 + 3x + π, x2 and x2 + 2 +5 x +Do you agree that they are all of degree two? A polynomial of degree two is called +a quadratic polynomial. Some examples of a quadratic polynomial are 5 – y2, +4y + 5y2 and 6 – y – y2. Can you write a quadratic polynomial in one variable with four +different terms? You will find that a quadratic polynomial in one variable will have at +most 3 terms. If you list a few more quadratic polynomials, you will find that any +quadratic polynomial in x is of the form ax2 + bx + c, where a ≠ 0 and a, b, c are +constants. Similarly, quadratic polynomial in y will be of the form ay2 + by + c, provided +a ≠ 0 and a, b, c are constants. +We call a polynomial of degree three a cubic polynomial. Some examples of a +cubic polynomial in x are 4x3, 2x3 + 1, 5x3 + x2, 6x3 – x, 6 – x3, 2x3 + 4x2 + 6x + 7. How +many terms do you think a cubic polynomial in one variable can have? It can have at +most 4 terms. These may be written in the form ax3 + bx2 + cx + d, where a ≠ 0 and +a, b, c and d are constants. +Now, that you have seen what a polynomial of degree 1, degree 2, or degree 3 +looks like, can you write down a polynomial in one variable of degree n for any natural +number n? A polynomial in one variable x of degree n is an expression of the form + anxn + an–1xn–1 + . . . + a1x + a0 +where a0, a1, a2, . . ., an are constants and an ≠ 0. +In particular, if a0 = a1 = a2 = a3 = . . . = an = 0 (all the constants are zero), we get +the zero polynomial, which is denoted by 0. What is the degree of the zero polynomial? +The degree of the zero polynomial is not defined. +So far we have dealt with polynomials in one variable only. We can also have +polynomials in more than one variable. For example, x2 + y2 + xyz (where variables +are x, y and z) is a polynomial in three variables. Similarly p2 + q10 + r (where the +variables are p, q and r), u3 + v2 (where the variables are u and v) are polynomials in +three and two variables, respectively. You will be studying such polynomials in detail +later. +Reprint 2025-26 + +POLYNOMIALS +29 +EXERCISE 2.1 +1. +Which of the following expressions are polynomials in one variable and which are +not? State reasons for your answer. +(i) +4x2 – 3x + 7 +(ii) y2 + 2 +(iii) 3 +2 +t +t ++ +(iv) y + 2 +y +(v) x10 + y3 + t50 +2. +Write the coefficients of x2 in each of the following: +(i) +2 + x2 + x +(ii) 2 – x2 + x3 +(iii) +2 +2 x +x +π ++ +(iv) +2 +1 +x − +3. +Give one example each of a binomial of degree 35, and of a monomial of degree 100. +4. +Write the degree of each of the following polynomials: +(i) +5x3 + 4x2 + 7x +(ii) 4 – y2 +(iii) 5t – 7 +(iv) 3 +5. +Classify the following as linear, quadratic and cubic polynomials: +(i) +x2 + x +(ii) x – x3 +(iii) y + y2 + 4 +(iv) 1 + x +(v) 3t +(vi) r2 +(vii) 7x3 +2.3 Zeroes of a Polynomial +Consider the polynomial +p(x) = 5x3 – 2x2 + 3x – 2. +If we replace x by 1 everywhere in p(x), we get +p(1) = 5 × (1)3 – 2 × (1)2 + 3 × (1) – 2 += 5 – 2 + 3 –2 += 4 +So, we say that the value of p(x) at x = 1 is 4. +Similarly, +p(0) = 5(0)3 – 2(0)2 + 3(0) –2 += –2 +Can you find p(–1)? +Example 2 : Find the value of each of the following polynomials at the indicated value +of variables: +(i) +p(x) = 5x2 – 3x + 7 at x = 1. +(ii) q(y) = 3y3 – 4y + +11 at y = 2. +(iii) p(t) = 4t4 + 5t3 – t2 + 6 at t = a. +Reprint 2025-26 + +30 +MATHEMATICS +Solution : (i) p(x) = 5x2 – 3x + 7 +The value of the polynomial p(x) at x = 1 is given by +p(1) = 5(1)2 – 3(1) + 7 += 5 – 3 + 7 = 9 +(ii) q(y) = 3y3 – 4y + 11 +The value of the polynomial q(y) at y = 2 is given by +q(2) = 3(2)3 – 4(2) + 11 = 24 – 8 + 11 = 16 + 11 +(iii) p(t) = 4t4 + 5t3 – t2 + 6 +The value of the polynomial p(t) at t = a is given by + p(a) = 4a4 + 5a3 – a2 + 6 +Now, consider the polynomial p(x) = x – 1. +What is p(1)? Note that : p(1) = 1 – 1 = 0. +As p(1) = 0, we say that 1 is a zero of the polynomial p(x). +Similarly, you can check that 2 is a zero of q(x), where q(x) = x – 2. +In general, we say that a zero of a polynomial p(x) is a number c such that p(c) = 0. +You must have observed that the zero of the polynomial x – 1 is obtained by +equating it to 0, i.e., x – 1 = 0, which gives x = 1. We say p(x) = 0 is a polynomial +equation and 1 is the root of the polynomial equation p(x) = 0. So we say 1 is the zero +of the polynomial x – 1, or a root of the polynomial equation x – 1 = 0. +Now, consider the constant polynomial 5. Can you tell what its zero is? It has no +zero because replacing x by any number in 5x0 still gives us 5. In fact, a non-zero +constant polynomial has no zero. What about the zeroes of the zero polynomial? By +convention, every real number is a zero of the zero polynomial. +Example 3 : Check whether –2 and 2 are zeroes of the polynomial x + 2. +Solution : Let p(x) = x + 2. +Then p(2) = 2 + 2 = 4, p(–2) = –2 + 2 = 0 +Therefore, –2 is a zero of the polynomial x + 2, but 2 is not. +Example 4 : Find a zero of the polynomial p(x) = 2x + 1. +Solution : Finding a zero of p(x), is the same as solving the equation +p(x) = 0 +Reprint 2025-26 + +POLYNOMIALS +31 +Now, +2x + 1 = 0 gives us x = +1 +– 2 +So, +1 +– 2 is a zero of the polynomial 2x + 1. +Now, if p(x) = ax + b, a ≠ 0, is a linear polynomial, how can we find a zero of +p(x)? Example 4 may have given you some idea. Finding a zero of the polynomial p(x), +amounts to solving the polynomial equation p(x) = 0. +Now, p(x) = 0 means +ax + b = 0, a ≠ 0 +So, +ax = –b +i.e., +x = – b +a +. +So, x = +b +a +− + is the only zero of p(x), i.e., a linear polynomial has one and only one zero. +Now we can say that 1 is the zero of x – 1, and –2 is the zero of x + 2. +Example 5 : Verify whether 2 and 0 are zeroes of the polynomial x2 – 2x. +Solution : Let +p(x) = x2 – 2x +Then +p(2) = 22 – 4 = 4 – 4 = 0 +and +p(0) = 0 – 0 = 0 +Hence, 2 and 0 are both zeroes of the polynomial x2 – 2x. +Let us now list our observations: +(i) A zero of a polynomial need not be 0. +(ii) 0 may be a zero of a polynomial. +(iii) Every linear polynomial has one and only one zero. +(iv) A polynomial can have more than one zero. +EXERCISE 2.2 +1. +Find the value of the polynomial 5x – 4x2 + 3 at +(i) +x = 0 +(ii) x = –1 +(iii) x = 2 +2. +Find p(0), p(1) and p(2) for each of the following polynomials: +(i) +p(y) = y2 – y + 1 +(ii) p(t) = 2 + t + 2t2 – t3 +(iii) p(x) = x3 +(iv) p(x) = (x – 1) (x + 1) +Reprint 2025-26 + +32 +MATHEMATICS +3. +Verify whether the following are zeroes of the polynomial, indicated against them. +(i) +p(x) = 3x + 1, x = +1 +– 3 +(ii) p(x) = 5x – π, x = 4 +5 +(iii) p(x) = x2 – 1, x = 1, –1 +(iv) p(x) = (x + 1) (x – 2), x = – 1, 2 +(v) p(x) = x2, x = 0 +(vi) p(x) = lx + m, x = – m +l +(vii) p(x) = 3x2 – 1, x = +1 +2 +, +3 +3 +− +(viii) p(x) = 2x + 1, x = 1 +2 +4. +Find the zero of the polynomial in each of the following cases: +(i) +p(x) = x + 5 +(ii) p(x) = x – 5 +(iii) p(x) = 2x + 5 +(iv) p(x) = 3x – 2 +(v) p(x) = 3x +(vi) p(x) = ax, a ≠ 0 +(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers. +2.4 Factorisation of Polynomials +Let us now look at the situation of Example 10 above more closely. It tells us that since +the remainder, +1 +2 +q + +− + + + + = 0, (2t + 1) is a factor of q(t), i.e., q(t) = (2t + 1) g(t) +for some polynomial g(t). This is a particular case of the following theorem. +Factor Theorem : If p(x) is a polynomial of degree n > 1 and a is any real number, +then (i) x – a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x – a is a factor of p(x). +Proof: By the Remainder Theorem, p(x)=(x – a) q(x) + p(a). +(i) If p(a) = 0, then p(x) = (x – a) q(x), which shows that x – a is a factor of p(x). +(ii) Since x – a is a factor of p(x), p(x) = (x – a) g(x) for same polynomial g(x). +In this case, p(a) = (a – a) g(a) = 0. +Example 6 : Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6 and of 2x + 4. +Solution : The zero of x + 2 is –2. Let p(x) = x3 + 3x2 + 5x + 6 and s(x) = 2x + 4 +Then, +p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6 += –8 + 12 – 10 + 6 += 0 +Reprint 2025-26 + +POLYNOMIALS +33 +So, by the Factor Theorem, x + 2 is a factor of x3 + 3x2 + 5x + 6. +Again, +s(–2) = 2(–2) + 4 = 0 +So, x + 2 is a factor of 2x + 4. In fact, you can check this without applying the Factor +Theorem, since 2x + 4 = 2(x + 2). +Example 7 : Find the value of k, if x – 1 is a factor of 4x3 + 3x2 – 4x + k. +Solution : As x – 1 is a factor of p(x) = 4x3 + 3x2 – 4x + k, p(1) = 0 +Now, +p(1) = 4(1)3 + 3(1)2 – 4(1) + k +So, +4 + 3 – 4 + k = 0 +i.e., +k = –3 +We will now use the Factor Theorem to factorise some polynomials of degree 2 and 3. +You are already familiar with the factorisation of a quadratic polynomial like +x2 + lx + m. You had factorised it by splitting the middle term lx as ax + bx so that +ab = m. Then x2 + lx + m = (x + a) (x + b). We shall now try to factorise quadratic +polynomials of the type ax2 + bx + c, where a ≠ 0 and a, b, c are constants. +Factorisation of the polynomial ax2 + bx + c by splitting the middle term is as +follows: +Let its factors be (px + q) and (rx + s). Then +ax2 + bx + c = (px + q) (rx + s) = pr x2 + (ps + qr) x + qs +Comparing the coefficients of x2, we get a = pr. +Similarly, comparing the coefficients of x, we get b = ps + qr. +And, on comparing the constant terms, we get c = qs. +This shows us that b is the sum of two numbers ps and qr, whose product is +(ps)(qr) = (pr)(qs) = ac. +Therefore, to factorise ax2 + bx + c, we have to write b as the sum of two +numbers whose product is ac. This will be clear from Example 13. +Example 8 : Factorise 6x2 + 17x + 5 by splitting the middle term, and by using the +Factor Theorem. +Solution 1 : (By splitting method) : If we can find two numbers p and q such that +p + q = 17 +and +pq = 6 × 5 = 30, then we can get the factors. +So, let us look for the pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5 +and 6. Of these pairs, 2 and 15 will give us p + q = 17. +2 +3x +x = 3x = first term of quotient +Reprint 2025-26 + +34 +MATHEMATICS +So, 6x2 + 17x + 5 = 6x2 + (2 + 15)x + 5 += 6x2 + 2x + 15x + 5 += 2x(3x + 1) + 5(3x + 1) += (3x + 1) (2x + 5) +Solution 2 : (Using the Factor Theorem) +6x2 + 17x + 5 = +2 +17 +5 +6 +6 +6 +x +x + + ++ ++ + + + + + = 6 p(x), say. If a and b are the zeroes of p(x), then +6x2 + 17x + 5 = 6(x – a) (x – b). So, ab = 5. +6 Let us look at some possibilities for a and +b. They could be +1 +1 +5 +5 +, +, +, +, +1 +2 +3 +3 +2 +± +± +± +± +± . Now, +1 +1 +17 1 +5 +2 +4 +6 +2 +6 +p + + + += ++ ++ + + + + + + + + + ≠ 0. But +1 +3 +p − + + + + + + + = 0. So, +1 +3 +x + + ++ + + + + is a factor of p(x). Similarly, by trial, you can find that +5 +2 +x + + ++ + + + + + is a factor of p(x). +Therefore, +6x2 + 17x + 5 = 6 +1 +5 +3 +2 +x +x + + + ++ ++ + + + + + + += +3 +1 +2 +5 +6 +3 +2 +x +x ++ ++ +�� + + + + + + + + += (3x + 1) (2x + 5) +For the example above, the use of the splitting method appears more efficient. However, +let us consider another example. +Example 9 : Factorise y2 – 5y + 6 by using the Factor Theorem. +Solution : Let p(y) = y2 – 5y + 6. Now, if p(y) = (y – a) (y – b), you know that the +constant term will be ab. So, ab = 6. So, to look for the factors of p(y), we look at the +factors of 6. +The factors of 6 are 1, 2 and 3. +Now, +p(2) = 22 – (5 × 2) + 6 = 0 +So, +y – 2 is a factor of p(y). +Reprint 2025-26 + +POLYNOMIALS +35 +Also, +p(3) = 32 – (5 × 3) + 6 = 0 +So, +y – 3 is also a factor of y2 – 5y + 6. +Therefore, +y2 – 5y + 6 = (y – 2)(y – 3) +Note that y2 – 5y + 6 can also be factorised by splitting the middle term –5y. +Now, let us consider factorising cubic polynomials. Here, the splitting method will not +be appropriate to start with. We need to find at least one factor first, as you will see in +the following example. +Example 10 : Factorise x3 – 23x2 + 142x – 120. +Solution : Let +p(x) = x3 – 23x2 + 142x – 120 +We shall now look for all the factors of –120. Some of these are ±1, ±2, ±3, +±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60. +By trial, we find that p(1) = 0. So x – 1 is a factor of p(x). +Now we see that x3 – 23x2 + 142x – 120 = x3 – x2 – 22x2 + 22x + 120x – 120 + = x2(x –1) – 22x(x – 1) + 120(x – 1) +(Why?) + = (x – 1) (x2 – 22x + 120) +[Taking (x – 1) common] +We could have also got this by dividing p(x) by x – 1. +Now x2 – 22x + 120 can be factorised either by splitting the middle term or by using +the Factor theorem. By splitting the middle term, we have: +x2 – 22x + 120 = x2 – 12x – 10x + 120 += x(x – 12) – 10(x – 12) += (x – 12) (x – 10) +So, +x3 – 23x2 – 142x – 120 = (x – 1)(x – 10)(x – 12) +EXERCISE 2.3 +1. +Determine which of the following polynomials has (x + 1) a factor : +(i) +x3 + x2 + x + 1 +(ii) x4 + x3 + x2 + x + 1 +(iii) x4 + 3x3 + 3x2 + x + 1 +(iv) x3 – x2 – ( +) +2 +2 +2 +x ++ ++ +2. +Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the +following cases: +(i) +p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1 +Reprint 2025-26 + +36 +MATHEMATICS +(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2 +(iii) p(x) = x3 – 4x2 + x + 6, g(x) = x – 3 +3. +Find the value of k, if x – 1 is a factor of p(x) in each of the following cases: +(i) +p(x) = x2 + x + k +(ii) p(x) = 2x2 + kx + +2 +(iii) p(x) = kx2 – +2 x + 1 +(iv) p(x) = kx2 – 3x + k +4. +Factorise : +(i) +12x2 – 7x + 1 +(ii) 2x2 + 7x + 3 +(iii) 6x2 + 5x – 6 +(iv) 3x2 – x – 4 +5. +Factorise : +(i) +x3 – 2x2 – x + 2 +(ii) x3 – 3x2 – 9x – 5 +(iii) x3 + 13x2 + 32x + 20 +(iv) 2y3 + y2 – 2y – 1 +2.5 Algebraic Identities +From your earlier classes, you may recall that an algebraic identity is an algebraic +equation that is true for all values of the variables occurring in it. You have studied the +following algebraic identities in earlier classes: +Identity I +: (x + y)2 = x2 + 2xy + y2 +Identity II +: (x – y)2 = x2 – 2xy + y2 +Identity III : x2 – y2 = (x + y) (x – y) +Identity IV : (x + a) (x + b) = x2 + (a + b)x + ab +You must have also used some of these algebraic identities to factorise the algebraic +expressions. You can also see their utility in computations. +Example 11 : Find the following products using appropriate identities: +(i) (x + 3) (x + 3) +(ii) (x – 3) (x + 5) +Solution : (i) Here we can use Identity I : (x + y)2 = x2 + 2xy + y2. Putting y = 3 in it, +we get +(x + 3) (x + 3) = (x + 3)2 = x2 + 2(x)(3) + (3)2 += x2 + 6x + 9 +(ii) Using Identity IV above, i.e., (x + a) (x + b) = x2 + (a + b)x + ab, we have +(x – 3) (x + 5) = x2 + (–3 + 5)x + (–3)(5) += x2 + 2x – 15 +Reprint 2025-26 + +POLYNOMIALS +37 +Example 12 : Evaluate 105 × 106 without multiplying directly. +Solution : +105 × 106 = (100 + 5) × (100 + 6) += (100)2 + (5 + 6) (100) + (5 × 6), using Identity IV += 10000 + 1100 + 30 += 11130 +You have seen some uses of the identities listed above in finding the product of some +given expressions. These identities are useful in factorisation of algebraic expressions +also, as you can see in the following examples. +Example 13 : Factorise: +(i) 49a2 + 70ab + 25b2 +(ii) +2 +2 +25 +4 +9 +y +x − +Solution : (i) Here you can see that +49a2 = (7a)2, 25b2 = (5b)2, 70ab = 2(7a) (5b) +Comparing the given expression with x2 + 2xy + y2, we observe that x = 7a and y = 5b. +Using Identity I, we get +49a2 + 70ab + 25b2 = (7a + 5b)2 = (7a + 5b) (7a + 5b) +(ii) We have +2 +2 +2 +2 +25 +5 +– +– +4 +9 +2 +3 +y +y +x +x + + + + +=  + + + + + + + +Now comparing it with Identity III, we get +2 +2 +25 +– +4 +9 +y +x + = +2 +2 +5 +– +2 +3 +y +x + + + + + + + + + + + + += +5 +5 +2 +3 +2 +3 +y +y +x +x + + + ++ +− + + + + + + +So far, all our identities involved products of binomials. Let us now extend the Identity +I to a trinomial x + y + z. We shall compute (x + y + z)2 by using Identity I. +Let x + y = t. Then, +(x + y + z)2 = (t + z)2 += t2 + 2tz + t2 +(Using Identity I) += (x + y)2 + 2(x + y)z + z2 +(Substituting the value of t) +Reprint 2025-26 + +38 +MATHEMATICS += x2 + 2xy + y2 + 2xz + 2yz + z2 +(Using Identity I) += x2 + y2 + z2 + 2xy + 2yz + 2zx +(Rearranging the terms) +So, we get the following identity: +Identity V : (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx +Remark : We call the right hand side expression the expanded form of the left hand +side expression. Note that the expansion of (x + y + z)2 consists of three square terms +and three product terms. +Example 14 : Write (3a + 4b + 5c)2 in expanded form. +Solution : Comparing the given expression with (x + y + z)2, we find that +x = 3a, y = 4b and z = 5c. +Therefore, using Identity V, we have +(3a + 4b + 5c)2 = (3a)2 + (4b)2 + (5c)2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a) += 9a2 + 16b2 + 25c2 + 24ab + 40bc + 30ac +Example 15 : Expand (4a – 2b – 3c)2. +Solution : Using Identity V, we have +(4a – 2b – 3c)2 = [4a + (–2b) + (–3c)]2 += (4a)2 + (–2b)2 + (–3c)2 + 2(4a)(–2b) + 2(–2b)(–3c) + 2(–3c)(4a) += 16a2 + 4b2 + 9c2 – 16ab + 12bc – 24ac +Example 16 : Factorise 4x2 + y2 + z2 – 4xy – 2yz + 4xz. +Solution : We have 4x2 + y2 + z2 – 4xy – 2yz + 4xz = (2x)2 + (–y)2 + (z)2 + 2(2x)(–y) ++ 2(–y)(z) + 2(2x)(z) += [2x + (–y) + z]2 +(Using Identity V) += (2x – y + z)2 = (2x – y + z)(2x – y + z) +So far, we have dealt with identities involving second degree terms. Now let us +extend Identity I to compute (x + y)3. We have: +(x + y)3 = (x + y) (x + y)2 += (x + y)(x2 + 2xy + y2) += x(x2 + 2xy + y2) + y(x2 + 2xy + y2) += x3 + 2x2y + xy2 + x2y + 2xy2 + y3 += x3 + 3x2y + 3xy2 + y3 += x3 + y3 + 3xy(x + y) +Reprint 2025-26 + +POLYNOMIALS +39 +So, we get the following identity: +Identity VI : +(x + y)3 = x3 + y3 + 3xy (x + y) +Also, by replacing y by –y in the Identity VI, we get +Identity VII : +(x – y)3 = x3 – y3 – 3xy(x – y) += x3 – 3x2y + 3xy2 – y3 +Example 17 : Write the following cubes in the expanded form: +(i) (3a + 4b)3 +(ii) (5p – 3q)3 +Solution : (i) Comparing the given expression with (x + y)3, we find that +x = 3a and y = 4b. +So, using Identity VI, we have: +(3a + 4b)3 = (3a)3 + (4b)3 + 3(3a)(4b)(3a + 4b) += 27a3 + 64b3 + 108a2b + 144ab2 +(ii) Comparing the given expression with (x – y)3, we find that +x = 5p, y = 3q. +So, using Identity VII, we have: +(5p – 3q)3 = (5p)3 – (3q)3 – 3(5p)(3q)(5p – 3q) += 125p3 – 27q3 – 225p2q + 135pq2 +Example 18 : Evaluate each of the following using suitable identities: +(i) (104)3 +(ii) (999)3 +Solution : (i) We have +(104)3 = (100 + 4)3 += (100)3 + (4)3 + 3(100)(4)(100 + 4) +(Using Identity VI) += 1000000 + 64 + 124800 += 1124864 +(ii) We have +(999)3 = (1000 – 1)3 += (1000)3 – (1)3 – 3(1000)(1)(1000 – 1) +(Using Identity VII) += 1000000000 – 1 – 2997000 += 997002999 +Reprint 2025-26 + +40 +MATHEMATICS +Example 19 : Factorise 8x3 + 27y3 + 36x2y + 54xy2 +Solution : The given expression can be written as +(2x)3 + (3y)3 + 3(4x2)(3y) + 3(2x)(9y2) += (2x)3 + (3y)3 + 3(2x)2(3y) + 3(2x)(3y)2 += (2x + 3y)3 +(Using Identity VI) += (2x + 3y)(2x + 3y)(2x + 3y) +Now consider (x + y + z)(x2 + y2 + z2 – xy – yz – zx) +On expanding, we get the product as +x(x2 + y2 + z2 – xy – yz – zx) + y(x2 + y2 + z2 – xy – yz – zx) ++ z(x2 + y2 + z2 – xy – yz – zx) = x3 + xy2 + xz2 – x2y – xyz – zx2 + x2y ++ y3 + yz2 – xy2 – y2z – xyz + x2z + y2z + z3 – xyz – yz2 – xz2 += x3 + y3 + z3 – 3xyz +(On simplification) +So, we obtain the following identity: +Identity VIII : x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) +Example 20 : Factorise : 8x3 + y3 + 27z3 – 18xyz +Solution : Here, we have +8x3 + y3 + 27z3 – 18xyz += (2x)3 + y3 + (3z)3 – 3(2x)(y)(3z) += (2x + y + 3z)[(2x)2 + y2 + (3z)2 – (2x)(y) – (y)(3z) – (2x)(3z)] += (2x + y + 3z) (4x2 + y2 + 9z2 – 2xy – 3yz – 6xz) +EXERCISE 2.4 +1. +Use suitable identities to find the following products: +(i) +(x + 4) (x + 10) +(ii) (x + 8) (x – 10) +(iii) (3x + 4) (3x – 5) +(iv) (y2 + 3 +2 ) (y2 – 3 +2 ) +(v) (3 – 2x) (3 + 2x) +2. +Evaluate the following products without multiplying directly: +(i) +103 × 107 +(ii) 95 × 96 +(iii) 104 × 96 +3. +Factorise the following using appropriate identities: +(i) +9x2 + 6xy + y2 +(ii) 4y2 – 4y + 1 +(iii) x2 – +2 +100 +y +Reprint 2025-26 + +POLYNOMIALS +41 +4. +Expand each of the following, using suitable identities: +(i) +(x + 2y + 4z)2 +(ii) (2x – y + z)2 +(iii) (–2x + 3y + 2z)2 +(iv) (3a – 7b – c)2 +(v) (–2x + 5y – 3z)2 +(vi) +2 +1 +1 +1 +4 +2 +a +b + + +− ++ + + + + +5. +Factorise: +(i) +4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz +(ii) 2x2 + y2 + 8z2 – 2 2 xy + 4 2 yz – 8xz +6. +Write the following cubes in expanded form: +(i) +(2x + 1)3 +(ii) (2a – 3b)3 +(iii) +3 +3 +1 +2 x + + ++ + + + + +(iv) +3 +2 +3 +x +y + + +− + + + + +7. +Evaluate the following using suitable identities: +(i) +(99)3 +(ii) (102)3 +(iii) (998)3 +8. +Factorise each of the following: +(i) +8a3 + b3 + 12a2b + 6ab2 +(ii) 8a3 – b3 – 12a2b + 6ab2 +(iii) 27 – 125a3 – 135a + 225a2 +(iv) 64a3 – 27b3 – 144a2b + 108ab2 +(v) 27p3 – 1 +216 – +2 +9 +1 +2 +4 +p +p ++ +9. +Verify : (i) x3 + y3 = (x + y) (x2 – xy + y2) +(ii) x3 – y3 = (x – y) (x2 + xy + y2) +10. Factorise each of the following: + (i) 27y3 + 125z3 +(ii) 64m3 – 343n3 +[Hint : See Question 9.] +11. Factorise : 27x3 + y3 + z3 – 9xyz +12. Verify that x3 + y3 + z3 – 3xyz = +2 +2 +2 +1 ( +) ( +) +( +) +( +) +2 x +y +z +x +y +y +z +z +x + + ++ ++ +− ++ +− ++ +− + + +13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz. +14. Without actually calculating the cubes, find the value of each of the following: +(i) +(–12)3 + (7)3 + (5)3 +(ii) (28)3 + (–15)3 + (–13)3 +15. Give possible expressions for the length and breadth of each of the following +rectangles, in which their areas are given: +Area : 25a2 – 35a + 12 + Area : 35y2 + 13y –12 + (i) +(ii) +Reprint 2025-26 + +42 +MATHEMATICS +16. What are the possible expressions for the dimensions of the cuboids whose volumes +are given below? + Volume : 3x2 – 12x + Volume : 12ky2 + 8ky – 20k + (i) +(ii) +2.6 Summary +In this chapter, you have studied the following points: +1. +A polynomial p(x) in one variable x is an algebraic expression in x of the form +p(x) = anxn + an–1xn – 1 + . . . + a2x2 + a1x + a0, +where a0, a1, a2, . . ., an are constants and an ≠ 0. +a0, a1, a2, . . ., an are respectively the coefficients of x0, x, x2, . . ., xn, and n is called the degree +of the polynomial. Each of anxn, an–1 xn–1, ..., a0, with an ≠ 0, is called a term of the polynomial +p(x). +2. +A polynomial of one term is called a monomial. +3. +A polynomial of two terms is called a binomial. +4. +A polynomial of three terms is called a trinomial. +5. +A polynomial of degree one is called a linear polynomial. +6. +A polynomial of degree two is called a quadratic polynomial. +7. +A polynomial of degree three is called a cubic polynomial. +8. +A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0. In this case, a is also called a +root of the equation p(x) = 0. +9. +Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial +has no zero, and every real number is a zero of the zero polynomial. +10. Factor Theorem : x – a is a factor of the polynomial p(x), if p(a) = 0. Also, if x – a is a factor +of p(x), then p(a) = 0. +11. +(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx +12. (x + y)3 = x3 + y3 + 3xy(x + y) +13. (x – y)3 = x3 – y3 – 3xy(x – y) +14. x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) +Reprint 2025-26" +class_9,3,Coordinate Geometry,ncert_books/class_9/iemh1dd/iemh103.pdf,"CHAPTER 3 +COORDINATE GEOMETRY +What’s the good of Mercator’s North Poles and Equators, Tropics, Zones and +Meridian Lines?’ So the Bellman would cry; and crew would reply ‘ They are +merely conventional signs!’ +LEWIS CARROLL, The Hunting of the Snark +3.1 Introduction +You have already studied how to locate a point on a number line. You also know how +to describe the position of a point on the line. There are many other situations, in which +to find a point we are required to describe its position with reference to more than one +line. For example, consider the following situations: +I. In Fig. 3.1, there is a main road running +in the East-West direction and streets with +numbering from West to East. Also, on each +street, house numbers are marked. To look for +a friend’s house here, is it enough to know only +one reference point? For instance, if we only +know that she lives on Street 2, will we be able +to find her house easily? Not as easily as when +we know two pieces of information about it, +namely, the number of the street on which it is +situated, and the house number. If we want to +reach the house which is situated in the 2nd +street and has the number 5, first of all we +would identify the 2nd street and then the house +numbered 5 on it. In Fig. 3.1, H shows the +location of the house. Similarly, P shows the +location of the house corresponding to Street +number 7 and House number 4. +Fig. 3.1 +Reprint 2025-26 + +44 +MATHEMATICS +II. Suppose you put a dot on a sheet of paper [Fig.3.2 (a)]. If we ask you to tell us +the position of the dot on the paper, how will you do this? Perhaps you will try in some +such manner: “The dot is in the upper half of the paper”, or “It is near the left edge of +the paper”, or “It is very near the left hand upper corner of the sheet”. Do any of +these statements fix the position of the dot precisely? No! But, if you say “ The dot is +nearly 5 cm away from the left edge of the paper”, it helps to give some idea but still +does not fix the position of the dot. A little thought might enable you to say that the dot +is also at a distance of 9 cm above the bottom line. We now know exactly where the dot is! +Fig. 3.2 +For this purpose, we fixed the position of the dot by specifying its distances from two +fixed lines, the left edge of the paper and the bottom line of the paper [Fig.3.2 (b)]. In +other words, we need two independent informations for finding the position of the dot. +Now, perform the following classroom activity known as ‘Seating Plan’. +Activity 1 (Seating Plan) : Draw a plan of the seating in your classroom, pushing all +the desks together. Represent each desk by a square. In each square, write the name +of the student occupying the desk, which the square represents. Position of each +student in the classroom is described precisely by using two independent informations: +(i) the column in which she or he sits, +(ii) the row in which she or he sits. +If you are sitting on the desk lying in the 5th column and 3rd row (represented by +the shaded square in Fig. 3.3), your position could be written as (5, 3), first writing the +column number, and then the row number. Is this the same as (3, 5)? Write down the +names and positions of other students in your class. For example, if Sonia is sitting in +the 4th column and 1st row, write S(4,1). The teacher’s desk is not part of your seating +plan. We are treating the teacher just as an observer. +Reprint 2025-26 + +COORDINATE GEOMETRY +45 +Fig. 3.3 +In the discussion above, you observe that position of any object lying in a plane +can be represented with the help of two perpendicular lines. In case of ‘dot’, we +require distance of the dot from bottom line as well as from left edge of the paper. In +case of seating plan, we require the number of the column and that of the row. This +simple idea has far reaching consequences, and has given rise to a very important +branch of Mathematics known as Coordinate Geometry. In this chapter, we aim to +introduce some basic concepts of coordinate geometry. You will study more about +these in your higher classes. This study was initially developed by the French philosopher +and mathematician René Déscartes. +René Déscartes, the great French mathematician of the +seventeenth century, liked to lie in bed and think! One +day, when resting in bed, he solved the problem of +describing the position of a point in a plane. His method +was a development of the older idea of latitude and +longitude. In honour of Déscartes, the system used for +describing the position of a point in a plane is also +known as the Cartesian system. +EXERCISE 3.1 +1. +How will you describe the position of a table lamp on your study table to another +person? +2. +(Street Plan) : A city has two main roads which cross each other at the centre of the +city. These two roads are along the North-South direction and East-West direction. +René Déscartes (1596 -1650) +Fig. 3.4 +Reprint 2025-26 + +46 +MATHEMATICS +All the other streets of the city run parallel to these roads and are 200 m apart. There +are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your +notebook. Represent the roads/streets by single lines. +There are many cross- streets in your model. A particular cross-street is made by +two streets, one running in the North - South direction and another in the East - West +direction. Each cross street is referred to in the following manner : If the 2nd street +running in the North - South direction and 5th in the East - West direction meet at some +crossing, then we will call this cross-street (2, 5). Using this convention, find: +(i) +how many cross - streets can be referred to as (4, 3). +(ii) how many cross - streets can be referred to as (3, 4). +3.2 Cartesian System +You have studied the number line in the chapter on ‘Number System’. On the number +line, distances from a fixed point are marked in equal units positively in one direction +and negatively in the other. The point from which the distances are marked is called +the origin. We use the number line to represent the numbers by marking points on a +line at equal distances. If one unit distance represents the number ‘1’, then 3 units +distance represents the number ‘3’, ‘0’ being at the origin. The point in the positive +direction at a distance r from the origin represents the number r. The point in the +negative direction at a distance r from the origin represents the number −r. Locations +of different numbers on the number line are shown in Fig. 3.5. +Fig. 3.5 +Descartes invented the idea of placing two such lines perpendicular to each other +on a plane, and locating points on the plane by referring them to these lines. The +perpendicular lines may be in any direction such as in Fig.3.6. But, when we choose +Fig. 3.6 +Reprint 2025-26 + +COORDINATE GEOMETRY +47 +these two lines to locate a point in a plane in this chapter, one line +will be horizontal and the other will be vertical, as in Fig. 3.6(c). +These lines are actually obtained as follows : Take two number +lines, calling them X′X and Y′Y. Place X′X horizontal [as in Fig. 3.7(a)] +and write the numbers on it just as written on the number line. We do +the same thing with Y′Y except that Y′Y is vertical, not horizontal +[Fig. 3.7(b)]. +Fig. 3.7 +Combine both the lines in such +a way that the two lines cross each +other at their zeroes, or origins +(Fig. 3.8). The horizontal line X′X +is called the x - axis and the vertical +line YY′ is called the y - axis. The +point where X′X and Y′Y cross is +called the origin, and is denoted +by O. Since the positive numbers +lie on the directions OX and OY, +OX and OY are called the positive +directions of the x - axis and the +y - axis, respectively. Similarly, OX′ +and OY′ are called the negative +directions of the x - axis and the +y - axis, respectively. +Fig. 3.8 +Reprint 2025-26 + +48 +MATHEMATICS +You observe that the axes (plural of the word +‘axis’) divide the plane into four parts. These four +parts are called the quadrants (one fourth part), +numbered I, II, III and IV anticlockwise from OX +(see Fig.3.9). So, the plane consists of the axes and +these quadrants. We call the plane, the Cartesian +plane, or the coordinate plane, or the xy-plane. +The axes are called the coordinate axes. +Now, let us see why this system is so basic to mathematics, and how it is useful. +Consider the following diagram where the axes are drawn on graph paper. Let us see +the distances of the points P and Q from the axes. For this, we draw perpendiculars +PM on the x - axis and PN on the y - axis. Similarly, we draw perpendiculars QR and +QS as shown in Fig. 3.10. +Fig.3.10 +You find that +(i) The perpendicular distance of the point P from the y - axis measured along the +positive direction of the x - axis is PN = OM = 4 units. +(ii) The perpendicular distance of the point P from the x - axis measured along the +positive direction of the y - axis is PM = ON = 3 units. +Fig. 3.9 +Reprint 2025-26 + +COORDINATE GEOMETRY +49 +(iii) The perpendicular distance of the point Q from the y - axis measured along +the negative direction of the x - axis is OR = SQ = 6 units. +(iv) The perpendicular distance of the point Q from the x - axis measured along +the negative direction of the y - axis is OS = RQ = 2 units. +Now, using these distances, how can we describe the points so that there is no +confusion? +We write the coordinates of a point, using the following conventions: +(i) The x - coordinate of a point is its perpendicular distance from the y - axis +measured along the x -axis (positive along the positive direction of the x - axis +and negative along the negative direction of the x - axis). For the point P, it is ++ 4 and for Q, it is – 6. The x - coordinate is also called the abscissa. +(ii) The y - coordinate of a point is its perpendicular distance from the x - axis +measured along the y - axis (positive along the positive direction of the y - axis +and negative along the negative direction of the y - axis). For the point P, it is ++ 3 and for Q, it is –2. The y - coordinate is also called the ordinate. +(iii) In stating the coordinates of a point in the coordinate plane, the x - coordinate +comes first, and then the y - coordinate. We place the coordinates in brackets. +Hence, the coordinates of P are (4, 3) and the coordinates of Q are (– 6, – 2). +Note that the coordinates describe a point in the plane uniquely. (3, 4) is not the +same as (4, 3). +Example 1 : See Fig. 3.11 and complete the following statements: +(i) +The abscissa and the ordinate of the point B are _ _ _ and _ _ _, respectively. +Hence, the coordinates of B are (_ _, _ _). +(ii) +The x-coordinate and the y-coordinate of the point M are _ _ _ and _ _ _, +respectively. Hence, the coordinates of M are (_ _, _ _). +(iii) The x-coordinate and the y-coordinate of the point L are _ _ _ and _ _ _, +respectively. Hence, the coordinates of L are (_ _, _ _). +(iv) The x-coordinate and the y-coordinate of the point S are _ _ _ and _ _ _, +respectively. Hence, the coordinates of S are (_ _, _ _). +Reprint 2025-26 + +50 +MATHEMATICS +Fig. 3.11 +Solution: (i) Since the distance of the point B from the y - axis is 4 units, the +x - coordinate or abscissa of the point B is 4. The distance of the point B from the +x - axis is 3 units; therefore, the y - coordinate, i.e., the ordinate, of the point B is 3. +Hence, the coordinates of the point B are (4, 3). +As in (i) above : +(ii) +The x - coordinate and the y - coordinate of the point M are –3 and 4, respectively. +Hence, the coordinates of the point M are (–3, 4). +(iii) The x - coordinate and the y - coordinate of the point L are –5 and – 4, respectively. +Hence, the coordinates of the point L are (–5, – 4). +(iv) The x - coordinate and the y- coordinate of the point S are 3 and – 4, respectively. +Hence, the coordinates of the point S are (3, – 4). +Reprint 2025-26 + +COORDINATE GEOMETRY +51 +Example 2 : Write the coordinates of the +points marked on the axes in Fig. 3.12. +Solution : You can see that : +(i) The point A is at a distance of + 4 units +from the y - axis and at a distance zero +from the x - axis. Therefore, the +x - coordinate of A is 4 and the +y - coordinate is 0. Hence, the +coordinates of A are (4, 0). +(ii) The coordinates of B are (0, 3). Why? +(iii) The coordinates of C are (– 5, 0). +Why? +(iv) The coordinates of D are (0, – 4). Why? +(v) The coordinates of E are +2, 0 +3 + + + + + +. Why? +Since every point on the x - axis has no distance (zero distance) from the x - axis, +therefore, the y - coordinate of every point lying on the x - axis is always zero. Thus, the +coordinates of any point on the x - axis are of the form (x, 0), where x is the distance of +the point from the y - axis. Similarly, the coordinates of any point on the y - axis are of +the form (0, y), where y is the distance of the point from the x - axis. Why? +What are the coordinates of the origin O? It has zero distance from both the +axes so that its abscissa and ordinate are both zero. Therefore, the coordinates of +the origin are (0, 0). +In the examples above, you may have observed the following relationship between +the signs of the coordinates of a point and the quadrant of a point in which it lies. +(i) +If a point is in the 1st quadrant, then the point will be in the form (+, +), since the +1st quadrant is enclosed by the positive x - axis and the positive y - axis. +(ii) +If a point is in the 2nd quadrant, then the point will be in the form (–, +), since the +2nd quadrant is enclosed by the negative x - axis and the positive y - axis. +(iii) If a point is in the 3rd quadrant, then the point will be in the form (–, –), since the +3rd quadrant is enclosed by the negative x - axis and the negative y - axis. +(iv) If a point is in the 4th quadrant, then the point will be in the form (+, –), since the +4th quadrant is enclosed by the positive x - axis and the negative y - axis +(see Fig. 3.13). +Fig. 3.12 +Reprint 2025-26 + +52 +MATHEMATICS +Fig. 3.13 +Remark : The system we have discussed above for describing a point in a plane is +only a convention, which is accepted all over the world. The system could also have +been, for example, the ordinate first, and the abscissa second. However, the whole +world sticks to the system we have described to avoid any confusion. +EXERCISE 3.2 +1. +Write the answer of each of the following questions: +(i) +What is the name of horizontal and the vertical lines drawn to determine the +position of any point in the Cartesian plane? +(ii) What is the name of each part of the plane formed by these two lines? +(iii) Write the name of the point where these two lines intersect. +2. +See Fig.3.14, and write the following: +(i) +The coordinates of B. +(ii) The coordinates of C. +(iii) The point identified by the coordinates (–3, –5). +Reprint 2025-26 + +COORDINATE GEOMETRY +53 +(iv) The point identified by the coordinates (2, – 4). +(v) The abscissa of the point D. +(vi) The ordinate of the point H. +(vii) The coordinates of the point L. +(viii) The coordinates of the point M. +Fig. 3.14 +3.3 Summary +In this chapter, you have studied the following points : +1. +To locate the position of an object or a point in a plane, we require two perpendicular +lines. One of them is horizontal, and the other is vertical. +2. +The plane is called the Cartesian, or coordinate plane and the lines are called the coordinate +axes. +3. +The horizontal line is called the x -axis, and the vertical line is called the y - axis. +Reprint 2025-26 + +54 +MATHEMATICS +4. +The coordinate axes divide the plane into four parts called quadrants. +5. +The point of intersection of the axes is called the origin. +6. +The distance of a point from the y - axis is called its x-coordinate, or abscissa, and the +distance of the point from the x-axis is called its y-coordinate, or ordinate. +7. +If the abscissa of a point is x and the ordinate is y, then (x, y) are called the coordinates of +the point. +8. +The coordinates of a point on the x-axis are of the form (x, 0) and that of the point on the +y-axis are (0, y). +9. +The coordinates of the origin are (0, 0). +10. The coordinates of a point are of the form (+ , +) in the first quadrant, (–, +) in the second +quadrant, (–, –) in the third quadrant and (+, –) in the fourth quadrant, where + denotes a +positive real number and – denotes a negative real number. +11. +If x ≠ y, then (x, y) ≠ (y, x), and (x, y) = (y, x), if x = y. +Reprint 2025-26" +class_9,4,Linear Equations in Two Variables,ncert_books/class_9/iemh1dd/iemh104.pdf,"LINEAR EQUATIONS IN TWO VARIABLES +55 +CHAPTER 4 +LINEAR EQUATIONS IN TWO VARIABLES +The principal use of the Analytic Art is to bring Mathematical Problems to +Equations and to exhibit those Equations in the most simple terms that can be. +—Edmund Halley +4.1 Introduction +In earlier classes, you have studied linear equations in one variable. Can you write +down a linear equation in one variable? You may say that x + 1 = 0, x + +2 = 0 and +2 y + 3 = 0 are examples of linear equations in one variable. You also know that +such equations have a unique (i.e., one and only one) solution. You may also remember +how to represent the solution on a number line. In this chapter, the knowledge of linear +equations in one variable shall be recalled and extended to that of two variables. You +will be considering questions like: Does a linear equation in two variables have a +solution? If yes, is it unique? What does the solution look like on the Cartesian plane? +You shall also use the concepts you studied in Chapter 3 to answer these questions. +4.2 Linear Equations +Let us first recall what you have studied so far. Consider the following equation: +2x + 5 = 0 +Its solution, i.e., the root of the equation, is +5 +2 +− +. This can be represented on the +number line as shown below: +Fig. 4.1 +Reprint 2025-26 + +56 +MATHEMATICS +While solving an equation, you must always keep the following points in mind: +The solution of a linear equation is not affected when: +(i) +the same number is added to (or subtracted from) both the sides of the equation. +(ii) you multiply or divide both the sides of the equation by the same non-zero +number. +Let us now consider the following situation: +In a One-day International Cricket match between India and Sri Lanka played in +Nagpur, two Indian batsmen together scored 176 runs. Express this information in the +form of an equation. +Here, you can see that the score of neither of them is known, i.e., there are two +unknown quantities. Let us use x and y to denote them. So, the number of runs scored +by one of the batsmen is x, and the number of runs scored by the other is y. We know +that +x + y = 176, +which is the required equation. +This is an example of a linear equation in two variables. It is customary to denote +the variables in such equations by x and y, but other letters may also be used. Some +examples of linear equations in two variables are: +1.2s + 3t = 5, p + 4q = 7, πu + 5v = 9 and 3 = +2 x – 7y. +Note that you can put these equations in the form 1.2s + 3t – 5 = 0, +p + 4q – 7 = 0, πu + 5v – 9 = 0 and +2 x – 7y – 3 = 0, respectively. +So, any equation which can be put in the form ax + by + c = 0, where a, b and c +are real numbers, and a and b are not both zero, is called a linear equation in two +variables. This means that you can think of many many such equations. +Example 1 : Write each of the following equations in the form ax + by + c = 0 and +indicate the values of a, b and c in each case: +(i) 2x + 3y = 4.37 +(ii) x – 4 = +3 y +(iii) 4 = 5x – 3y +(iv) 2x = y +Solution : (i) 2x + 3y = 4.37 can be written as 2x + 3y – 4.37 = 0. Here a = 2, b = 3 +and c = – 4.37. +(ii) +The equation x – 4 = +3 y can be written as x – +3 y – 4 = 0. Here a = 1, +b = – +3 and c = – 4. +(iii) The equation 4 = 5x – 3y can be written as 5x – 3y – 4 = 0. Here a = 5, b = –3 +and c = – 4. Do you agree that it can also be written as –5x + 3y + 4 = 0 ? In this +case a = –5, b = 3 and c = 4. +Reprint 2025-26 + +LINEAR EQUATIONS IN TWO VARIABLES +57 +(iv) The equation 2x = y can be written as 2x – y + 0 = 0. Here a = 2, b = –1 and +c = 0. +Equations of the type ax + b = 0 are also examples of linear equations in two variables +because they can be expressed as +ax + 0.y + b = 0 +For example, 4 – 3x = 0 can be written as –3x + 0.y + 4 = 0. +Example 2 : Write each of the following as an equation in two variables: +(i) x = –5 +(ii) y = 2 +(iii) 2x = 3 +(iv) 5y = 2 +Solution : (i) x = –5 can be written as 1.x + 0.y = –5, or 1.x + 0.y + 5 = 0. +(ii) y = 2 can be written as 0.x + 1.y = 2, or 0.x + 1.y – 2 = 0. +(iii) 2x = 3 can be written as 2x + 0.y – 3 = 0. +(iv) 5y = 2 can be written as 0.x + 5y – 2 = 0. +EXERCISE 4.1 +1. +The cost of a notebook is twice the cost of a pen. Write a linear equation in two +variables to represent this statement. +(Take the cost of a notebook to be ` x and that of a pen to be ` y). +2. +Express the following linear equations in the form ax + by + c = 0 and indicate the +values of a, b and c in each case: +(i) +2x + 3y = 9.35 +(ii) x – 5 +y – 10 = 0 +(iii) –2x + 3y = 6 +(iv) x = 3y +(v) 2x = –5y +(vi) 3x + 2 = 0 +(vii) y – 2 = 0 +(viii) 5 = 2x +4.3 Solution of a Linear Equation +You have seen that every linear equation in one variable has a unique solution. What +can you say about the solution of a linear equation involving two variables? As there +are two variables in the equation, a solution means a pair of values, one for x and one +for y which satisfy the given equation. Let us consider the equation 2x + 3y = 12. +Here, x = 3 and y = 2 is a solution because when you substitute x = 3 and y = 2 in the +equation above, you find that +2x + 3y = (2 × 3) + (3 × 2) = 12 +This solution is written as an ordered pair (3, 2), first writing the value for x and +then the value for y. Similarly, (0, 4) is also a solution for the equation above. +Reprint 2025-26 + +58 +MATHEMATICS +On the other hand, (1, 4) is not a solution of 2x + 3y = 12, because on putting +x = 1 and y = 4 we get 2x + 3y = 14, which is not 12. Note that (0, 4) is a solution but +not (4, 0). +You have seen at least two solutions for 2x + 3y = 12, i.e., (3, 2) and (0, 4). Can +you find any other solution? Do you agree that (6, 0) is another solution? Verify the +same. In fact, we can get many many solutions in the following way. Pick a value of +your choice for x (say x = 2) in 2x + 3y = 12. Then the equation reduces to 4 + 3y = 12, +which is a linear equation in one variable. On solving this, you get y = 8 +3. So +8 +2, 3 + + + + + + + is +another solution of 2x + 3y = 12. Similarly, choosing x = – 5, you find that the equation +becomes –10 + 3y = 12. This gives y = 22 +3 . So, +22 +5, 3 + + +− + + + + + is another solution of +2x + 3y = 12. So there is no end to different solutions of a linear equation in two +variables. That is, a linear equation in two variables has infinitely many solutions. +Example 3 : Find four different solutions of the equation x + 2y = 6. +Solution : By inspection, x = 2, y = 2 is a solution because for x = 2, y = 2 +x + 2y = 2 + 4 = 6 +Now, let us choose x = 0. With this value of x, the given equation reduces to 2y = 6 +which has the unique solution y = 3. So x = 0, y = 3 is also a solution of x + 2y = 6. +Similarly, taking y = 0, the given equation reduces to x = 6. So, x = 6, y = 0 is a solution +of x + 2y = 6 as well. Finally, let us take y = 1. The given equation now reduces to +x + 2 = 6, whose solution is given by x = 4. Therefore, (4, 1) is also a solution of the +given equation. So four of the infinitely many solutions of the given equation are: +(2, 2), (0, 3), (6, 0) and (4, 1). +Remark : Note that an easy way of getting a solution is to take x = 0 and get the +corresponding value of y. Similarly, we can put y = 0 and obtain the corresponding +value of x. +Example 4 : Find two solutions for each of the following equations: +(i) +4x + 3y = 12 +(ii) 2x + 5y = 0 +(iii) 3y + 4 = 0 +Solution : (i) Taking x = 0, we get 3y = 12, i.e., y = 4. So, (0, 4) is a solution of the +given equation. Similarly, by taking y = 0, we get x = 3. Thus, (3, 0) is also a solution. +Reprint 2025-26 + +LINEAR EQUATIONS IN TWO VARIABLES +59 +(ii) Taking x = 0, we get 5y = 0, i.e., y = 0. So (0, 0) is a solution of the given equation. +Now, if you take y = 0, you again get (0, 0) as a solution, which is the same as the +earlier one. To get another solution, take x = 1, say. Then you can check that the +corresponding value of y is +2. +5 +− + So +2 +1, +5 + + +− + + + + + is another solution of 2x + 5y = 0. +(iii) Writing the equation 3y + 4 = 0 as 0.x + 3y + 4 = 0, you will find that y = +4 +– 3 for +any value of x. Thus, two solutions can be given as +4 +4 +0, – +and 1, – +3 +3 + + + + + + + + + + + + +. +EXERCISE 4.2 +1. +Which one of the following options is true, and why? +y = 3x + 5 has +(i) a unique solution, +(ii) only two solutions, +(iii) infinitely many solutions +2. +Write four solutions for each of the following equations: +(i) +2x + y = 7 +(ii) πx + y = 9 +(iii) x = 4y +3. +Check which of the following are solutions of the equation x – 2y = 4 and which are +not: +(i) +(0, 2) +(ii) (2, 0) +(iii) (4, 0) +(iv) ( +) +2 , 4 2 +(v) (1, 1) +4. +Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k. +4.4 Summary +In this chapter, you have studied the following points: +1. +An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and +b are not both zero, is called a linear equation in two variables. +2. +A linear equation in two variables has infinitely many solutions. +3. +Every point on the graph of a linear equation in two variables is a solution of the linear +equation. Moreover, every solution of the linear equation is a point on the graph of the +linear equation. +Reprint 2025-26" +class_9,5,Introduction to Euclid's Geometry,ncert_books/class_9/iemh1dd/iemh105.pdf,"60 +MATHEMATICS +CHAPTER 5 +INTRODUCTION TO EUCLID’S GEOMETRY +5.1 Introduction +The word ‘geometry’ comes form the Greek words ‘geo’, meaning the ‘earth’, +and ‘metrein’, meaning ‘to measure’. Geometry appears to have originated from +the need for measuring land. This branch of mathematics was studied in various +forms in every ancient civilisation, be it in Egypt, Babylonia, China, India, Greece, +the Incas, etc. The people of these civilisations faced several practical problems +which required the development of geometry in various ways. +For example, whenever the river Nile +overflowed, it wiped out the boundaries between +the adjoining fields of different land owners. After +such flooding, these boundaries had to be +redrawn. For this purpose, the Egyptians +developed a number of geometric techniques and +rules for calculating simple areas and also for +doing simple constructions. The knowledge of +geometry was also used by them for computing +volumes of granaries, and for constructing canals +and pyramids. They also knew the correct formula +to find the volume of a truncated pyramid (see +Fig. 5.1).You know that a pyramid is a solid figure, +the base of which is a triangle, or square, or some +other polygon, and its side faces are triangles +converging to a point at the top. +Fig. 5.1 : A Truncated Pyramid +Reprint 2025-26 + +INTRODUCTION TO EUCLID’S GEOMETRY +61 +In the Indian subcontinent, the excavations at Harappa and Mohenjo-Daro, etc. +show that the Indus Valley Civilisation (about 3000 BCE) made extensive use of +geometry. It was a highly organised society. The cities were highly developed and +very well planned. For example, the roads were parallel to each other and there was +an underground drainage system. The houses had many rooms of different types. This +shows that the town dwellers were skilled in mensuration and practical arithmetic. +The bricks used for constructions were kiln fired and the ratio length : breadth : thickness, +of the bricks was found to be 4 : 2 : 1. +In ancient India, the Sulbasutras (800 BCE to 500 BCE) were the manuals of +geometrical constructions. The geometry of the Vedic period originated with the +construction of altars (or vedis) and fireplaces for performing Vedic rites. The location +of the sacred fires had to be in accordance to the clearly laid down instructions about +their shapes and areas, if they were to be effective instruments. Square and circular +altars were used for household rituals, while altars whose shapes were combinations +of rectangles, triangles and trapeziums were required for public worship. The sriyantra +(given in the Atharvaveda) consists of nine interwoven isosceles triangles. These +triangles are arranged in such a way that they produce 43 subsidiary triangles. Though +accurate geometric methods were used for the constructions of altars, the principles +behind them were not discussed. +These examples show that geometry was being developed and applied everywhere +in the world. But this was happening in an unsystematic manner. What is interesting +about these developments of geometry in the ancient world is that they were passed +on from one generation to the next, either orally or through palm leaf messages, or by +other ways. Also, we find that in some civilisations like Babylonia, geometry remained +a very practical oriented discipline, as was the case in India and Rome. The geometry +developed by Egyptians mainly consisted of the statements of results. There were no +general rules of the procedure. In fact, Babylonians and Egyptians used geometry +mostly for practical purposes and did very little to develop it as a systematic science. +But in civilisations like Greece, the emphasis was on the reasoning behind why certain +constructions work. The Greeks were interested in establishing the truth of the +statements they discovered using deductive reasoning (see Appendix 1). +A Greek mathematician, Thales is credited with giving the +first known proof. This proof was of the statement that a circle +is bisected (i.e., cut into two equal parts) by its diameter. One of +Thales’ most famous pupils was Pythagoras (572 BCE), whom +you have heard about. Pythagoras and his group discovered many +geometric properties and developed the theory of geometry to a +great extent. This process continued till 300 BCE. At that time +Euclid, a teacher of mathematics at Alexandria in Egypt, collected +all the known work and arranged it in his famous treatise, +Thales +(640 BCE – 546 BCE) +Fig. 5.2 +Reprint 2025-26 + +62 +MATHEMATICS +called ‘Elements’. He divided the ‘Elements’ into thirteen +chapters, each called a book. These books influenced +the whole world’s understanding of geometry for +generations to come. +In this chapter, we shall discuss Euclid’s approach +to geometry and shall try to link it with the present day +geometry. +5.2 Euclid’s Definitions, Axioms and Postulates +The Greek mathematicians of Euclid’s time thought of geometry as an abstract model +of the world in which they lived. The notions of point, line, plane (or surface) and so on +were derived from what was seen around them. From studies of the space and solids +in the space around them, an abstract geometrical notion of a solid object was developed. +A solid has shape, size, position, and can be moved from one place to another. Its +boundaries are called surfaces. They separate one part of the space from another, +and are said to have no thickness. The boundaries of the surfaces are curves or +straight lines. These lines end in points. +Consider the three steps from solids to points (solids-surfaces-lines-points). In +each step we lose one extension, also called a dimension. So, a solid has three +dimensions, a surface has two, a line has one and a point has none. Euclid summarised +these statements as definitions. He began his exposition by listing 23 definitions in +Book 1 of the ‘Elements’. A few of them are given below : +1. +A point is that which has no part. +2. +A line is breadthless length. +3. +The ends of a line are points. +4. +A straight line is a line which lies evenly with the points on itself. +5. +A surface is that which has length and breadth only. +6. +The edges of a surface are lines. +7. +A plane surface is a surface which lies evenly with the straight lines on itself. +If you carefully study these definitions, you find that some of the terms like part, +breadth, length, evenly, etc. need to be further explained clearly. For example, consider +his definition of a point. In this definition, ‘a part’ needs to be defined. Suppose if you +define ‘a part’ to be that which occupies ‘area’, again ‘an area’ needs to be defined. +So, to define one thing, you need to define many other things, and you may get a long +chain of definitions without an end. For such reasons, mathematicians agree to leave +Euclid (325 BCE – 265 BCE) +Fig. 5.3 +Reprint 2025-26 + +INTRODUCTION TO EUCLID’S GEOMETRY +63 +some geometric terms undefined. However, we do have a intuitive feeling for the +geometric concept of a point than what the ‘definition’ above gives us. So, we represent +a point as a dot, even though a dot has some dimension. +A similar problem arises in Definition 2 above, since it refers to breadth and length, +neither of which has been defined. Because of this, a few terms are kept undefined +while developing any course of study. So, in geometry, we take a point, a line and a +plane (in Euclid‘s words a plane surface) as undefined terms. The only thing is +that we can represent them intuitively, or explain them with the help of ‘physical +models’. +Starting with his definitions, Euclid assumed certain properties, which were not to +be proved. These assumptions are actually ‘obvious universal truths’. He divided them +into two types: axioms and postulates. He used the term ‘postulate’ for the assumptions +that were specific to geometry. Common notions (often called axioms), on the other +hand, were assumptions used throughout mathematics and not specifically linked to +geometry. For details about axioms and postulates, refer to Appendix 1. Some of +Euclid’s axioms, not in his order, are given below : +(1) Things which are equal to the same thing are equal to one another. +(2) If equals are added to equals, the wholes are equal. +(3) If equals are subtracted from equals, the remainders are equal. +(4) Things which coincide with one another are equal to one another. +(5) The whole is greater than the part. +(6) Things which are double of the same things are equal to one another. +(7) Things which are halves of the same things are equal to one another. +These ‘common notions’ refer to magnitudes of some kind. The first common +notion could be applied to plane figures. For example, if an area of a triangle equals the +area of a rectangle and the area of the rectangle equals that of a square, then the area +of the triangle also equals the area of the square. +Magnitudes of the same kind can be compared and added, but magnitudes of +different kinds cannot be compared. For example, a line cannot be compared to a +rectangle, nor can an angle be compared to a pentagon. +The 4th axiom given above seems to say that if two things are identical (that is, +they are the same), then they are equal. In other words, everything equals itself. It is +the justification of the principle of superposition. Axiom (5) gives us the definition of +‘greater than’. For example, if a quantity B is a part of another quantity A, then A can +be written as the sum of B and some third quantity C. Symbolically, A > B means that +there is some C such that A = B + C. +Reprint 2025-26 + +64 +MATHEMATICS +Now let us discuss Euclid’s five postulates. They are : +Postulate 1 : A straight line may be drawn from any one point to any other point. +Note that this postulate tells us that at least one straight line passes through two +distinct points, but it does not say that there cannot be more than one such line. However, +in his work, Euclid has frequently assumed, without mentioning, that there is a unique +line joining two distinct points. We state this result in the form of an axiom as follows: +Axiom 5.1 : Given two distinct points, there is a unique line that passes through +them. +How many lines passing through P also pass through Q (see Fig. 5.4)? Only one, +that is, the line PQ. How many lines passing through Q also pass through P? Only one, +that is, the line PQ. Thus, the statement above is self-evident, and so is taken as an +axiom. +Fig. 5.4 +Postulate 2 : A terminated line can be produced indefinitely. +Note that what we call a line segment now-a-days is what Euclid called a terminated +line. So, according to the present day terms, the second postulate says that a line +segment can be extended on either side to form a line (see Fig. 5.5). +Fig. 5.5 +Postulate 3 : A circle can be drawn with any centre and any radius. +Postulate 4 : All right angles are equal to one another. +Postulate 5 : If a straight line falling on two straight lines makes the interior +angles on the same side of it taken together less than two right angles, then the +two straight lines, if produced indefinitely, meet on that side on which the sum of +angles is less than two right angles. +Reprint 2025-26 + +INTRODUCTION TO EUCLID’S GEOMETRY +65 +For example, the line PQ in Fig. 5.6 falls on lines +AB and CD such that the sum of the interior angles 1 +and 2 is less than 180° on the left side of PQ. +Therefore, the lines AB and CD will eventually +intersect on the left side of PQ. +A brief look at the five postulates brings to your notice that Postulate 5 is far more +complex than any other postulate. On the other hand, Postulates 1 through 4 are so +simple and obvious that these are taken as ‘self-evident truths’. However, it is not +possible to prove them. So, these statements are accepted without any proof +(see Appendix 1). Because of its complexity, the fifth postulate will be given more +attention in the next section. +Now-a-days, ‘postulates’ and ‘axioms’ are terms that are used interchangeably +and in the same sense. ‘Postulate’ is actually a verb. When we say “let us postulate”, +we mean, “let us make some statement based on the observed phenomenon in the +Universe”. Its truth/validity is checked afterwards. If it is true, then it is accepted as a +‘Postulate’. +A system of axioms is called consistent (see Appendix 1), if it is impossible to +deduce from these axioms a statement that contradicts any axiom or previously proved +statement. So, when any system of axioms is given, it needs to be ensured that the +system is consistent. +After Euclid stated his postulates and axioms, he used them to prove other results. +Then using these results, he proved some more results by applying deductive reasoning. +The statements that were proved are called propositions or theorems. Euclid +deduced 465 propositions in a logical chain using his axioms, postulates, definitions and +theorems proved earlier in the chain. In the next few chapters on geometry, you will +be using these axioms to prove some theorems. +Now, let us see in the following examples how Euclid used his axioms and postulates +for proving some of the results: +Example 1 : If A, B and C are three points on a line, and B lies between A and C +(see Fig. 5.7), then prove that AB + BC = AC. +Fig. 5.7 +Fig. 5.6 +Reprint 2025-26 + +66 +MATHEMATICS +Solution : In the figure given above, AC coincides with AB + BC. +Also, Euclid’s Axiom (4) says that things which coincide with one another are equal to +one another. So, it can be deduced that +AB + BC = AC +Note that in this solution, it has been assumed that there is a unique line passing +through two points. +Example 2 : Prove that an equilateral triangle can be constructed on any given line +segment. +Solution : In the statement above, a line segment of any length is given, say AB +[see Fig. 5.8(i)]. +Fig. 5.8 +Here, you need to do some construction. Using Euclid’s Postulate 3, you can draw a +circle with point A as the centre and AB as the radius [see Fig. 5.8(ii)]. Similarly, draw +another circle with point B as the centre and BA as the radius. The two circles meet at +a point, say C. Now, draw the line segments AC and BC to form ∆ ABC +[see Fig. 5.8 (iii)]. +So, you have to prove that this triangle is equilateral, i.e., AB = AC = BC. +Now, +AB = AC, since they are the radii of the same circle +(1) +Similarly, AB = BC +(Radii of the same circle) +(2) +From these two facts, and Euclid’s axiom that things which are equal to the same thing +are equal to one another, you can conclude that AB = BC = AC. +So, ∆ ABC is an equilateral triangle. +Note that here Euclid has assumed, without mentioning anywhere, that the two circles +drawn with centres A and B will meet each other at a point. +Now we prove a theorem, which is frequently used in different results: +Reprint 2025-26 + +INTRODUCTION TO EUCLID’S GEOMETRY +67 +Theorem 5.1 : Two distinct lines cannot have more than one point in common. +Proof : Here we are given two lines l and m. We need to prove that they have only one +point in common. +For the time being, let us suppose that the two lines intersect in two distinct points, +say P and Q. So, you have two lines passing through two distinct points P and Q. But +this assumption clashes with the axiom that only one line can pass through two distinct +points. So, the assumption that we started with, that two lines can pass through two +distinct points is wrong. +From this, what can we conclude? We are forced to conclude that two distinct +lines cannot have more than one point in common. +EXERCISE 5.1 +1. +Which of the following statements are true and which are false? Give reasons for your +answers. +(i) +Only one line can pass through a single point. +(ii) There are an infinite number of lines which pass through two distinct points. +(iii) A terminated line can be produced indefinitely on both the sides. +(iv) If two circles are equal, then their radii are equal. +(v) In Fig. 5.9, if AB = PQ and PQ = XY, then AB = XY. +Fig. 5.9 +2. +Give a definition for each of the following terms. Are there other terms that need to be +defined first? What are they, and how might you define them? +(i) +parallel lines +(ii) perpendicular lines +(iii) line segment +(iv) radius of a circle +(v) square +3. +Consider two ‘postulates’ given below: +(i) +Given any two distinct points A and B, there exists a third point C which is in +between A and B. +(ii) There exist at least three points that are not on the same line. +Do these postulates contain any undefined terms? Are these postulates consistent? +Do they follow from Euclid’s postulates? Explain. +Reprint 2025-26 + +68 +MATHEMATICS +4. +If a point C lies between two points A and B such that AC = BC, then prove that +AC = 1 +2 AB. Explain by drawing the figure. +5. +In Question 4, point C is called a mid-point of line segment AB. Prove that every line +segment has one and only one mid-point. +6. +In Fig. 5.10, if AC = BD, then prove that AB = CD. +Fig. 5.10 +7. +Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that +the question is not about the fifth postulate.) +5.3 Summary +In this chapter, you have studied the following points: +1. +Though Euclid defined a point, a line, and a plane, the definitions are not accepted by +mathematicians. Therefore, these terms are now taken as undefined. +2. +Axioms or postulates are the assumptions which are obvious universal truths. They are not +proved. +3. +Theorems are statements which are proved, using definitions, axioms, previously proved +statements and deductive reasoning. +4. +Some of Euclid’s axioms were : +(1) Things which are equal to the same thing are equal to one another. +(2) If equals are added to equals, the wholes are equal. +(3) If equals are subtracted from equals, the remainders are equal. +(4) Things which coincide with one another are equal to one another. +(5) The whole is greater than the part. +(6) Things which are double of the same things are equal to one another. +(7) Things which are halves of the same things are equal to one another. +5. +Euclid’s postulates were : +Postulate 1 : A straight line may be drawn from any one point to any other point. +Postulate 2 : A terminated line can be produced indefinitely. +Postulate 3 : A circle can be drawn with any centre and any radius. +Postulate 4 : All right angles are equal to one another. +Reprint 2025-26" +class_9,6,Lines and Angles,ncert_books/class_9/iemh1dd/iemh106.pdf,"CHAPTER 6 +LINES AND ANGLES +6.1 Introduction +In Chapter 5, you have studied that a minimum of two points are required to draw a +line. You have also studied some axioms and, with the help of these axioms, you +proved some other statements. In this chapter, you will study the properties of the +angles formed when two lines intersect each other, and also the properties of the +angles formed when a line intersects two or more parallel lines at distinct points. +Further you will use these properties to prove some statements using deductive reasoning +(see Appendix 1). You have already verified these statements through some activities +in the earlier classes. +In your daily life, you see different types of angles formed between the edges of +plane surfaces. For making a similar kind of model using the plane surfaces, you need +to have a thorough knowledge of angles. For instance, suppose you want to make a +model of a hut to keep in the school exhibition using bamboo sticks. Imagine how you +would make it? You would keep some of the sticks parallel to each other, and some +sticks would be kept slanted. Whenever an architect has to draw a plan for a multistoried +building, she has to draw intersecting lines and parallel lines at different angles. Without +the knowledge of the properties of these lines and angles, do you think she can draw +the layout of the building? +In science, you study the properties of light by drawing the ray diagrams. +For example, to study the refraction property of light when it enters from one medium +to the other medium, you use the properties of intersecting lines and parallel lines. +When two or more forces act on a body, you draw the diagram in which forces are +represented by directed line segments to study the net effect of the forces on the +body. At that time, you need to know the relation between the angles when the rays +(or line segments) are parallel to or intersect each other. To find the height of a tower +or to find the distance of a ship from the light house, one needs to know the angle +Reprint 2025-26 + +70 +MATHEMATICS +formed between the horizontal and the line of sight. Plenty of other examples can be +given where lines and angles are used. In the subsequent chapters of geometry, you +will be using these properties of lines and angles to deduce more and more useful +properties. +Let us first revise the terms and definitions related to lines and angles learnt in +earlier classes. +6.2 Basic Terms and Definitions +Recall that a part (or portion) of a line with two end points is called a line-segment +and a part of a line with one end point is called a ray. Note that the line segment AB is +denoted by AB , and its length is denoted by AB. The ray AB is denoted by AB + , and +a line is denoted by AB + . However, we will not use these symbols, and will denote +the line segment AB, ray AB, length AB and line AB by the same symbol, AB. The +meaning will be clear from the context. Sometimes small letters l, m, n, etc. will be +used to denote lines. +If three or more points lie on the same line, they are called collinear points; +otherwise they are called non-collinear points. +Recall that an angle is formed when two rays originate from the same end point. +The rays making an angle are called the arms of the angle and the end point is called +the vertex of the angle. You have studied different types of angles, such as acute +angle, right angle, obtuse angle, straight angle and reflex angle in earlier classes +(see Fig. 6.1). +(i) acute angle : 0° < x < 90° +(ii) right angle : y = 90° +(iii) obtuse angle : 90° < z < 180° +(iv) straight angle : s = 180° +(v) reflex angle : 180° < t < 360° +Fig. 6.1 : Types of Angles +Reprint 2025-26 + +LINES AND ANGLES +71 +An acute angle measures between 0° and 90°, whereas a right angle is exactly +equal to 90°. An angle greater than 90° but less than 180° is called an obtuse angle. +Also, recall that a straight angle is equal to 180°. An angle which is greater than 180° +but less than 360° is called a reflex angle. Further, two angles whose sum is 90° are +called complementary angles, and two angles whose sum is 180° are called +supplementary angles. +You have also studied about adjacent angles +in the earlier classes (see Fig. 6.2). Two angles +are adjacent, if they have a common vertex, a +common arm and their non-common arms are +on different sides of the common arm. In +Fig. 6.2, ∠ ABD and ∠ DBC are adjacent +angles. Ray BD is their common arm and point +B is their common vertex. Ray BA and ray BC +are non common arms. Moreover, when two +angles are adjacent, then their sum is always +equal to the angle formed by the two non- +common arms. So, we can write +∠ ABC = ∠ ABD + ∠ DBC. +Note that ∠ ABC and ∠ ABD are not +adjacent angles. Why? Because their non- +common arms BD and BC lie on the same side +of the common arm BA. +If the non-common arms BA and BC in +Fig. 6.2, form a line then it will look like Fig. 6.3. +In this case, ∠ ABD and ∠ DBC are called +linear pair of angles. +You may also recall the vertically opposite +angles formed when two lines, say AB and CD, +intersect each other, say at the point O +(see Fig. 6.4). There are two pairs of vertically +opposite angles. +One pair is ∠AOD and ∠BOC. Can you +find the other pair? +Fig. 6.2 : Adjacent angles +Fig. 6.3 : Linear pair of angles +Fig. 6.4 : Vertically opposite +angles +Reprint 2025-26 + +72 +MATHEMATICS +6.3 Intersecting Lines and Non-intersecting Lines +Draw two different lines PQ and RS on a paper. You will see that you can draw them +in two different ways as shown in Fig. 6.5 (i) and Fig. 6.5 (ii). +(i) Intersecting lines +(ii) Non-intersecting (parallel) lines +Fig. 6.5 : Different ways of drawing two lines +Recall the notion of a line, that it extends indefinitely in both directions. Lines PQ +and RS in Fig. 6.5 (i) are intersecting lines and in Fig. 6.5 (ii) are parallel lines. Note +that the lengths of the common perpendiculars at different points on these parallel +lines is the same. This equal length is called the distance between two parallel lines. +6.4 Pairs of Angles +In Section 6.2, you have learnt the definitions of +some of the pairs of angles such as +complementary angles, supplementary angles, +adjacent angles, linear pair of angles, etc. Can +you think of some relations between these +angles? Now, let us find out the relation between +the angles formed when a ray stands on a line. +Draw a figure in which a ray stands on a line as +shown in Fig. 6.6. Name the line as AB and the +ray as OC. What are the angles formed at the +point O? They are ∠ AOC, ∠ BOC and ∠ AOB. +Can we write ∠ AOC + ∠ BOC = ∠ AOB? +(1) +Yes! (Why? Refer to adjacent angles in Section 6.2) +What is the measure of ∠ AOB? It is 180°. +(Why?) +(2) +From (1) and (2), can you say that ∠ AOC + ∠ BOC = 180°? +Yes! (Why?) +From the above discussion, we can state the following Axiom: +Fig. 6.6 : Linear pair of angles +Reprint 2025-26 + +LINES AND ANGLES +73 +Axiom 6.1 : If a ray stands on a line, then the sum of two adjacent angles so +formed is 180°. +Recall that when the sum of two adjacent angles is 180°, then they are called a +linear pair of angles. +In Axiom 6.1, it is given that ‘a ray stands on a line’. From this ‘given’, we have +concluded that ‘the sum of two adjacent angles so formed is 180°’. Can we write +Axiom 6.1 the other way? That is, take the ‘conclusion’ of Axiom 6.1 as ‘given’ and +the ‘given’ as the ‘conclusion’. So it becomes: +(A) If the sum of two adjacent angles is 180°, then a ray stands on a line (that is, +the non-common arms form a line). +Now you see that the Axiom 6.1 and statement (A) are in a sense the reverse of +each others. We call each as converse of the other. We do not know whether the +statement (A) is true or not. Let us check. Draw adjacent angles of different measures +as shown in Fig. 6.7. Keep the ruler along one of the non-common arms in each case. +Does the other non-common arm also lie along the ruler? +Fig. 6.7 : Adjacent angles with different measures +Reprint 2025-26 + +74 +MATHEMATICS +You will find that only in Fig. 6.7 (iii), both the non-common arms lie along the +ruler, that is, points A, O and B lie on the same line and ray OC stands on it. Also see +that ∠ AOC + ∠ COB = 125° + 55° = 180°. From this, you may conclude that statement +(A) is true. So, you can state in the form of an axiom as follows: +Axiom 6.2 : If the sum of two adjacent angles is 180°, then the non-common arms +of the angles form a line. +For obvious reasons, the two axioms above together is called the Linear Pair +Axiom. +Let us now examine the case when two lines intersect each other. +Recall, from earlier classes, that when two lines intersect, the vertically opposite +angles are equal. Let us prove this result now. See Appendix 1 for the ingredients of a +proof, and keep those in mind while studying the proof given below. +Theorem 6.1 : If two lines intersect each other, then the vertically opposite +angles are equal. +Proof : In the statement above, it is given +that ‘two lines intersect each other’. So, let +AB and CD be two lines intersecting at O as +shown in Fig. 6.8. They lead to two pairs of +vertically opposite angles, namely, +(i) ∠ AOC and ∠ BOD (ii) ∠ AOD and +∠ BOC. +We need to prove that ∠ AOC = ∠ BOD +and ∠ AOD = ∠ BOC. +Now, ray OA stands on line CD. +Therefore, ∠ AOC + ∠ AOD = 180° +(Linear pair axiom) (1) +Can we write ∠ AOD + ∠ BOD = 180°? Yes! (Why?) +(2) +From (1) and (2), we can write + ∠ AOC + ∠ AOD = ∠ AOD + ∠ BOD +This implies that ∠ AOC = ∠ BOD +(Refer Section 5.2, Axiom 3) +Similarly, it can be proved that ∠AOD = ∠BOC +Now, let us do some examples based on Linear Pair Axiom and Theorem 6.1. +Fig. 6.8 : Vertically opposite angles +Reprint 2025-26 + +LINES AND ANGLES +75 +Fig. 6.9 +Example 1 : In Fig. 6.9, lines PQ and RS +intersect each other at point O. If +∠ POR : ∠ ROQ = 5 : 7, find all the angles. +Solution : +∠ POR +∠ ROQ = 180° +(Linear pair of angles) +But +∠ POR : ∠ ROQ = 5 : 7 +(Given) +Therefore, +∠ POR = +5 +12 × 180° = 75° +Similarly, +∠ ROQ = +7 +12 × 180° = 105° +Now, +∠ POS = ∠ROQ = 105° +(Vertically opposite angles) +and +∠ SOQ = ∠POR = 75° +(Vertically opposite angles) +Example 2 : In Fig. 6.10, ray OS stands on a line POQ. Ray OR and ray OT are +angle bisectors of ∠ POS and ∠ SOQ, respectively. If ∠ POS = x, find ∠ ROT. +Solution : Ray OS stands on the line POQ. +Therefore, +∠ POS + ∠ SOQ = 180° +But, +∠ POS = x +Therefore, +x + ∠ SOQ = 180° +So, +∠ SOQ = 180° – x +Now, ray OR bisects ∠ POS, therefore, +∠ ROS = 1 +2 × ∠ POS += 1 +2 × x = 2 +x +Similarly, +∠ SOT = 1 +2 × ∠ SOQ += 1 +2 × (180° – x) += 90 +2 +x +° − +Fig. 6.10 +Reprint 2025-26 + +76 +MATHEMATICS +Now, +∠ ROT = ∠ ROS + ∠ SOT += +90 – +2 +2 +x +x ++ +° += 90° +Example 3 : In Fig. 6.11, OP, OQ, OR and OS are +four rays. Prove that ∠ POQ + ∠ QOR + ∠ SOR + +∠ POS = 360°. +Solution : In Fig. 6.11, you need to produce any of +the rays OP, OQ, OR or OS backwards to a point. +Let us produce ray OQ backwards to a point T so +that TOQ is a line (see Fig. 6.12). +Now, ray OP stands on line TOQ. +Therefore, +∠ TOP + ∠ POQ = 180° +(1) +(Linear pair axiom) +Similarly, ray OS stands on line TOQ. +Therefore, +∠ TOS + ∠ SOQ = 180° +(2) +But +∠ SOQ = ∠ SOR + ∠ QOR +So, (2) becomes +∠ TOS + ∠ SOR + ∠ QOR = 180° +(3) +Now, adding (1) and (3), you get +∠ TOP + ∠ POQ + ∠ TOS + ∠ SOR + ∠ QOR = 360° +(4) +But +∠ TOP + ∠ TOS = ∠ POS +Therefore, (4) becomes +∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = 360° +EXERCISE 6.1 +1. +In Fig. 6.13, lines AB and CD intersect at O. If +∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°, find +∠ BOE and reflex ∠ COE. +Fig. 6.11 +Fig. 6.12 +Fig. 6.13 +Reprint 2025-26 + +LINES AND ANGLES +77 +2. +In Fig. 6.14, lines XY and MN intersect at O. If +∠ POY = 90° and a : b = 2 : 3, find c. +3. +In Fig. 6.15, ∠ PQR = ∠ PRQ, then prove that +∠ PQS = ∠ PRT. +4. +In Fig. 6.16, if x + y = w + z, then prove that AOB +is a line. +5. +In Fig. 6.17, POQ is a line. Ray OR is perpendicular +to line PQ. OS is another ray lying between rays +OP and OR. Prove that +∠ ROS = 1 +2 (∠ QOS – ∠ POS). +6. +It is given that ∠ XYZ = 64° and XY is produced +to point P. Draw a figure from the given +information. If ray YQ bisects ∠ ZYP, find ∠ XYQ +and reflex ∠ QYP. +Fig. 6.14 +Fig. 6.15 +Fig. 6.16 +Fig. 6.17 +Reprint 2025-26 + +78 +MATHEMATICS +6.5 Lines Parallel to the Same Line +If two lines are parallel to the same line, will they be parallel to each other? Let us +check it. See Fig. 6.18 in which line m || line l and line n || line l. +Let us draw a line t transversal for the lines, l, m and n. It is given that +line m || line l and line n || line l. +Therefore, ∠ 1 = ∠ 2 and ∠ 1 = ∠ 3 +(Corresponding angles axiom) +So, +∠ 2 = ∠ 3 (Why?) +But ∠ 2 and ∠ 3 are corresponding angles and they +are equal. +Therefore, you can say that +Line m || Line n +(Converse of corresponding angles axiom) +This result can be stated in the form of the following theorem: +Theorem 6.6 : Lines which are parallel to the same line are parallel to each +other. +Note : The property above can be extended to more than two lines also. +Now, let us solve some examples related to parallel lines. +Example 4 : In Fig. 6.19, if PQ || RS, ∠ MXQ = 135° and ∠ MYR = 40°, find ∠ XMY. +Fig. 6.19 +Fig. 6.20 +Solution : Here, we need to draw a line AB parallel to line PQ, through point M as +shown in Fig. 6.20. Now, AB || PQ and PQ || RS. +Fig. 6.18 +Reprint 2025-26 + +LINES AND ANGLES +79 +Therefore, +AB || RS +(Why?) +Now, +∠ QXM + ∠ XMB = 180° + (AB || PQ, Interior angles on the same side of the transversal XM) +But +∠ QXM = 135° +So, +135° + ∠ XMB = 180° +Therefore, +∠ XMB = 45° + (1) +Now, +∠ BMY = ∠ MYR + (AB || RS, Alternate angles) +Therefore, +∠ BMY = 40° + (2) +Adding (1) and (2), you get +∠ XMB + ∠ BMY = 45° + 40° +That is, +∠ XMY = 85° +Example 5 : If a transversal intersects two lines such that the bisectors of a pair of +corresponding angles are parallel, then prove that the two lines are parallel. +Solution : In Fig. 6.21, a transversal AD intersects two lines PQ and RS at points B +and C respectively. Ray BE is the bisector of ∠ ABQ and ray CG is the bisector of +∠ BCS; and BE || CG. +We are to prove that PQ || RS. +It is given that ray BE is the bisector of ∠ ABQ. +Therefore, +∠ ABE = 1 +2 ∠ ABQ + (1) +Similarly, ray CG is the bisector of ∠ BCS. +Therefore, +∠ BCG = 1 +2 ∠ BCS + (2) +But BE || CG and AD is the transversal. +Therefore, +∠ ABE = ∠ BCG + +(Corresponding angles axiom) + (3) +Substituting (1) and (2) in (3), you get +1 +2 ∠ ABQ = 1 +2 ∠ BCS +That is, +∠ ABQ = ∠ BCS +Fig. 6.21 +Reprint 2025-26 + +80 +MATHEMATICS +But, they are the corresponding angles formed by transversal AD with PQ and RS; +and are equal. +Therefore, +PQ || RS +(Converse of corresponding angles axiom) +Example 6 : In Fig. 6.22, AB || CD and CD || EF. Also EA ⊥ AB. If ∠ BEF = 55°, find +the values of x, y and z. +Solution : +y + 55° = 180° +(Interior angles on the same side of the +transversal ED) +Therefore, +y = 180º – 55º = 125º +Again +x = y + (AB || CD, Corresponding angles axiom) +Therefore +x = 125º +Now, since AB || CD and CD || EF, therefore, AB || EF. +So, +∠ EAB + ∠ FEA = 180° +(Interior angles on the same +side of the transversal EA) +Therefore, +90° + z + 55° = 180° +Which gives +z = 35° +EXERCISE 6.2 +1. +In Fig. 6.23, if AB || CD, CD || EF and y : z = 3 : 7, +find x. +Fig. 6.23 +Fig. 6.22 +Reprint 2025-26 + +LINES AND ANGLES +81 +2. +In Fig. 6.24, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE. +Fig. 6.24 +3. +In Fig. 6.25, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS. +[Hint : Draw a line parallel to ST through point R.] +Fig. 6.25 +4. +In Fig. 6.26, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y. +Fig. 6.26 +5. +In Fig. 6.27, PQ and RS are two mirrors placed parallel to each other. An incident ray +AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes +the mirror RS at C and again reflects back along CD. Prove that +AB || CD. +Fig. 6.27 +Reprint 2025-26 + +82 +MATHEMATICS +6.6 Summary +In this chapter, you have studied the following points: +1. +If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and vice- +versa. This property is called as the Linear pair axiom. +2. +If two lines intersect each other, then the vertically opposite angles are equal. +3. +Lines which are parallel to a given line are parallel to each other. +Reprint 2025-26" +class_9,7,Triangles,ncert_books/class_9/iemh1dd/iemh107.pdf,"CHAPTER 7 +TRIANGLES +7.1 Introduction +You have studied about triangles and their various properties in your earlier classes. +You know that a closed figure formed by three intersecting lines is called a triangle. +(‘Tri’ means ‘three’). A triangle has three sides, three angles and three vertices. For +example, in triangle ABC, denoted as ∆ ABC (see Fig. 7.1); AB, BC, CA are the three +sides, ∠ A, ∠ B, ∠ C are the three angles and A, B, C are three vertices. +In Chapter 6, you have also studied some properties +of triangles. In this chapter, you will study in details +about the congruence of triangles, rules of congruence, +some more properties of triangles and inequalities in +a triangle. You have already verified most of these +properties in earlier classes. We will now prove some +of them. +7.2 Congruence of Triangles +You must have observed that two copies of your photographs of the same size are +identical. Similarly, two bangles of the same size, two ATM cards issued by the same +bank are identical. You may recall that on placing a one rupee coin on another minted +in the same year, they cover each other completely. +Do you remember what such figures are called? Indeed they are called congruent +figures (‘congruent’ means equal in all respects or figures whose shapes and sizes +are both the same). +Now, draw two circles of the same radius and place one on the other. What do +you observe? They cover each other completely and we call them as congruent circles. +Fig. 7.1 +Reprint 2025-26 + +84 +MATHEMATICS +Repeat this activity by placing one +square on the other with sides of the same +measure (see Fig. 7.2) or by placing two +equilateral triangles of equal sides on each +other. You will observe that the squares are +congruent to each other and so are the +equilateral triangles. +You may wonder why we are studying congruence. You all must have seen the ice +tray in your refrigerator. Observe that the moulds for making ice are all congruent. +The cast used for moulding in the tray also has congruent depressions (may be all are +rectangular or all circular or all triangular). So, whenever identical objects have to be +produced, the concept of congruence is used in making the cast. +Sometimes, you may find it difficult to replace the refill in your pen by a new one +and this is so when the new refill is not of the same size as the one you want to +remove. Obviously, if the two refills are identical or congruent, the new refill fits. +So, you can find numerous examples where congruence of objects is applied in +daily life situations. +Can you think of some more examples of congruent figures? +Now, which of the following figures are not congruent to the square in +Fig 7.3 (i) : +Fig. 7.3 +The large squares in Fig. 7.3 (ii) and (iii) are obviously not congruent to the one in +Fig 7.3 (i), but the square in Fig 7.3 (iv) is congruent to the one given in Fig 7.3 (i). +Let us now discuss the congruence of two triangles. +You already know that two triangles are congruent if the sides and angles of one +triangle are equal to the corresponding sides and angles of the other triangle. +Fig. 7.2 +Reprint 2025-26 + +TRIANGLES +85 +Now, which of the triangles given below are congruent to triangle ABC in +Fig. 7.4 (i)? +Fig. 7.4 +Cut out each of these triangles from Fig. 7.4 (ii) to (v) and turn them around and +try to cover ∆ ABC. Observe that triangles in Fig. 7.4 (ii), (iii) and (iv) are congruent +to ∆ ABC while ∆ TSU of Fig 7.4 (v) is not congruent to ∆ ABC. +If ∆ PQR is congruent to ∆ ABC, we write ∆ PQR ≅ ∆ ABC. +Notice that when ∆ PQR ≅ ∆ ABC, then sides of ∆ PQR fall on corresponding +equal sides of ∆ ABC and so is the case for the angles. +That is, PQ covers AB, QR covers BC and RP covers CA; ∠ P covers ∠ A, +∠ Q covers ∠ B and ∠ R covers ∠ C. Also, there is a one-one correspondence +between the vertices. That is, P corresponds to A, Q to B, R to C and so on which is +written as +P ↔ A, Q ↔ B, R ↔ C +Note that under this correspondence, ∆ PQR ≅ ∆ ABC; but it will not be correct to +write ∆QRP ≅ ∆ ABC. +Similarly, for Fig. 7.4 (iii), +Reprint 2025-26 + +86 +MATHEMATICS +FD ↔ AB, DE ↔ BC and EF ↔ CA +and +F ↔ A, D ↔ B and E ↔ C +So, ∆ FDE ≅ ∆ ABC but writing ∆ DEF ≅ ∆ ABC is not correct. +Give the correspondence between the triangle in Fig. 7.4 (iv) and ∆ ABC. +So, it is necessary to write the correspondence of vertices correctly for writing of +congruence of triangles in symbolic form. +Note that in congruent triangles corresponding parts are equal and we write +in short ‘CPCT’ for corresponding parts of congruent triangles. +7.3 Criteria for Congruence of Triangles +In earlier classes, you have learnt four criteria for congruence of triangles. Let us +recall them. +Draw two triangles with one side 3 cm. Are these triangles congruent? Observe +that they are not congruent (see Fig. 7.5). +Fig. 7.5 +Now, draw two triangles with one side 4 cm and one angle 50° (see Fig. 7.6). Are +they congruent? +Fig. 7.6 +Reprint 2025-26 + +TRIANGLES +87 +See that these two triangles are not congruent. +Repeat this activity with some more pairs of triangles. +So, equality of one pair of sides or one pair of sides and one pair of angles is not +sufficient to give us congruent triangles. +What would happen if the other pair of arms (sides) of the equal angles are also +equal? +In Fig 7.7, BC = QR, ∠ B = ∠ Q and also, AB = PQ. Now, what can you say +about congruence of ∆ABC and ∆PQR? +Recall from your earlier classes that, in this case, the two triangles are congruent. +Verify this for ∆ ABC and ∆ PQR in Fig. 7.7. +Repeat this activity with other pairs of triangles. Do you observe that the equality +of two sides and the included angle is enough for the congruence of triangles? Yes, it +is enough. +Fig. 7.7 +This is the first criterion for congruence of triangles. +Axiom 7.1 (SAS congruence rule) : Two triangles are congruent if two sides +and the included angle of one triangle are equal to the two sides and the included +angle of the other triangle. +This result cannot be proved with the help of previously known results and so it is +accepted true as an axiom (see Appendix 1). +Let us now take some examples. +Example 1 : In Fig. 7.8, OA = OB and OD = OC. Show that +(i) ∆ AOD ≅ ∆ BOC and +(ii) AD || BC. +Solution : (i) You may observe that in ∆ AOD and ∆ BOC, +OA = OB +(Given) +OD = OC +Fig. 7.8 + + + +Reprint 2025-26 + +88 +MATHEMATICS +Also, since ∠ AOD and ∠ BOC form a pair of vertically opposite angles, we have +∠ AOD = ∠ BOC. +So, +∆ AOD ≅∆ BOC +(by the SAS congruence rule) +(ii) In congruent triangles AOD and BOC, the other corresponding parts are also +equal. +So, +∠ OAD = ∠ OBC and these form a pair of alternate angles for line segments +AD and BC. +Therefore, +AD || BC. +Example 2 : AB is a line segment and line l is its perpendicular bisector. If a point P +lies on l, show that P is equidistant from A and B. +Solution : Line l ⊥ AB and passes through C which +is the mid-point of AB (see Fig. 7.9). You have to +show that PA = PB. Consider ∆ PCA and ∆ PCB. +We have +AC = BC +(C is the mid-point of AB) +∠ PCA = ∠ PCB = 90° + (Given) +PC = PC + (Common) +So, +∆ PCA ≅ ∆ PCB (SAS rule) +and so, PA = PB, as they are corresponding sides of +congruent triangles. +Now, let us construct two triangles, whose sides are 4 cm and 5 cm and one of the +angles is 50° and this angle is not included in between the equal sides (see Fig. 7.10). +Are the two triangles congruent? +Fig. 7.10 +Fig. 7.9 +Reprint 2025-26 + +TRIANGLES +89 +Notice that the two triangles are not congruent. +Repeat this activity with more pairs of triangles. You will observe that for triangles +to be congruent, it is very important that the equal angles are included between the +pairs of equal sides. +So, SAS congruence rule holds but not ASS or SSA rule. +Next, try to construct the two triangles in which two angles are 60° and 45° and +the side included between these angles is 4 cm (see Fig. 7.11). +Fig. 7.11 +Cut out these triangles and place one triangle on the other. What do you observe? +See that one triangle covers the other completely; that is, the two triangles are congruent. +Repeat this activity with more pairs of triangles. You will observe that equality of two +angles and the included side is sufficient for congruence of triangles. +This result is the Angle-Side-Angle criterion for congruence and is written as +ASA criterion. You have verified this criterion in earlier classes, but let us state and +prove this result. +Since this result can be proved, it is called a theorem and to prove it, we use the +SAS axiom for congruence. +Theorem 7.1 (ASA congruence rule) : Two triangles are congruent if two angles +and the included side of one triangle are equal to two angles and the included +side of other triangle. +Proof : We are given two triangles ABC and DEF in which: +∠ B = ∠ E, ∠ C = ∠ F +and +BC = EF +We need to prove that +∆ ABC ≅∆ DEF +For proving the congruence of the two triangles see that three cases arise. +Reprint 2025-26 + +90 +MATHEMATICS +Case (i) : Let AB = DE (see Fig. 7.12). +Now what do you observe? You may observe that +AB = DE +(Assumed) +∠ B = ∠ E +(Given) +BC = EF +(Given) +So, +∆ ABC ≅∆ DEF +(By SAS rule) +Fig. 7.12 +Case (ii) : Let if possible AB > DE. So, we can take a point P on AB such that +PB = DE. Now consider ∆ PBC and ∆ DEF (see Fig. 7.13). +Fig. 7.13 +Observe that in ∆ PBC and ∆ DEF, +PB = DE +(By construction) +∠ B = ∠ E +(Given) +BC = EF +(Given) +So, we can conclude that: +∆ PBC ≅ ∆ DEF, by the SAS axiom for congruence. +Reprint 2025-26 + +TRIANGLES +91 +Since the triangles are congruent, their corresponding parts will be equal. +So, +∠ PCB = ∠ DFE +But, we are given that +∠ ACB = ∠ DFE +So, +∠ ACB = ∠ PCB +Is this possible? +This is possible only if P coincides with A. +or, +BA = ED +So, +∆ ABC ≅∆ DEF +(by SAS axiom) +Case (iii) : If AB < DE, we can choose a point M on DE such that ME = AB and +repeating the arguments as given in Case (ii), we can conclude that AB = DE and so, +∆ ABC ≅ ∆ DEF. +Suppose, now in two triangles two pairs of angles and one pair of corresponding +sides are equal but the side is not included between the corresponding equal pairs of +angles. Are the triangles still congruent? You will observe that they are congruent. +Can you reason out why? +You know that the sum of the three angles of a triangle is 180°. So if two pairs of +angles are equal, the third pair is also equal (180° – sum of equal angles). +So, two triangles are congruent if any two pairs of angles and one pair of +corresponding sides are equal. We may call it as the AAS Congruence Rule. +Now let us perform the following activity : +Draw triangles with angles 40°, 50° and 90°. How many such triangles can you +draw? +In fact, you can draw as many triangles as you want with different lengths of +sides (see Fig. 7.14). +Fig. 7.14 +Reprint 2025-26 + +92 +MATHEMATICS +Observe that the triangles may or may not be congruent to each other. +So, equality of three angles is not sufficient for congruence of triangles. Therefore, +for congruence of triangles out of three equal parts, one has to be a side. +Let us now take some more examples. +Example 3 : Line-segment AB is parallel to another line-segment CD. O is the +mid-point of AD (see Fig. 7.15). Show that (i) ∆AOB ≅ ∆DOC (ii) O is also the +mid-point of BC. +Solution : (i) Consider ∆ AOB and ∆ DOC. +∠ ABO = ∠ DCO +(Alternate angles as AB || CD + and BC is the transversal) +∠ AOB = ∠ DOC +(Vertically opposite angles) +OA = OD +(Given) +Therefore, +∆AOB ≅∆DOC +(AAS rule) +(ii) +OB = OC +(CPCT) +So, O is the mid-point of BC. +EXERCISE 7.1 +1. +In quadrilateral ACBD, +AC = AD and AB bisects ∠A +(see Fig. 7.16). Show that ∆ ABC ≅ ∆ ABD. +What can you say about BC and BD? +Fig. 7.15 +Fig. 7.16 +Reprint 2025-26 + +TRIANGLES +93 +2. +ABCD is a quadrilateral in which AD = BC and +∠ DAB = ∠ CBA (see Fig. 7.17). Prove that +(i) +∆ ABD ≅ ∆ BAC +(ii) BD = AC +(iii) ∠ ABD = ∠ BAC. +3. +AD and BC are equal perpendiculars to a line +segment AB (see Fig. 7.18). Show that CD bisects +AB. +4. +l and m are two parallel lines intersected by +another pair of parallel lines p and q +(see Fig. 7.19). Show that ∆ ABC ≅ ∆ CDA. +5. +Line l is the bisector of an angle ∠ A and B is any +point on l. BP and BQ are perpendiculars from B +to the arms of ∠ A (see Fig. 7.20). Show that: +(i) +∆ APB ≅ ∆ AQB +(ii) BP = BQ or B is equidistant from the arms +of ∠ A. +Fig. 7.17 +Fig. 7.18 +Fig. 7.19 +Fig. 7.20 +Reprint 2025-26 + +94 +MATHEMATICS +6. +In Fig. 7.21, AC = AE, AB = AD and +∠ BAD = ∠ EAC. Show that BC = DE. +7. +AB is a line segment and P is its mid-point. D and +E are points on the same side of AB such that +∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB +(see Fig. 7.22). Show that +(i) +∆ DAP ≅ ∆ EBP +(ii) AD = BE +8. +In right triangle ABC, right angled at C, M is +the mid-point of hypotenuse AB. C is joined +to M and produced to a point D such that +DM = CM. Point D is joined to point B +(see Fig. 7.23). Show that: +(i) ∆ AMC ≅ ∆ BMD +(ii) ∠ DBC is a right angle. +(iii) ∆ DBC ≅ ∆ ACB +(iv) CM = 1 +2 AB +7.4 Some Properties of a Triangle +In the above section you have studied two criteria for congruence of triangles. Let us +now apply these results to study some properties related to a triangle whose two sides +are equal. +Fig. 7.22 +Fig. 7.23 +Fig. 7.21 +Reprint 2025-26 + +TRIANGLES +95 +Perform the activity given below: +Construct a triangle in which two sides are +equal, say each equal to 3.5 cm and the third side +equal to 5 cm (see Fig. 7.24). You have done such +constructions in earlier classes. +Do you remember what is such a triangle +called? +A triangle in which two sides are equal is called +an isosceles triangle. So, ∆ ABC of Fig. 7.24 is +an isosceles triangle with AB = AC. +Now, measure ∠ B and ∠ C. What do you observe? +Repeat this activity with other isosceles triangles with different sides. +You may observe that in each such triangle, the angles opposite to the equal sides +are equal. +This is a very important result and is indeed true for any isosceles triangle. It can +be proved as shown below. +Theorem 7.2 : Angles opposite to equal sides of an isosceles triangle are equal. +This result can be proved in many ways. One of +the proofs is given here. +Proof : We are given an isosceles triangle ABC +in which AB = AC. We need to prove that +∠ B = ∠ C. +Let us draw the bisector of ∠ A and let D be +the point of intersection of this bisector of +∠ A and BC (see Fig. 7.25). +In ∆ BAD and ∆ CAD, +AB = AC +(Given) +∠ BAD = ∠ CAD +(By construction) +AD = AD +(Common) +So, +∆ BAD ≅∆ CAD +(By SAS rule) +So, ∠ ABD = ∠ ACD, since they are corresponding angles of congruent triangles. +So, +∠ B = ∠ C +Fig. 7.24 +Fig. 7.25 +Reprint 2025-26 + +96 +MATHEMATICS +Is the converse also true? That is: +If two angles of any triangle are equal, can we conclude that the sides opposite to +them are also equal? +Perform the following activity. +Construct a triangle ABC with BC of any length and ∠ B = ∠ C = 50°. Draw the +bisector of ∠ A and let it intersect BC at D (see Fig. 7.26). +Cut out the triangle from the sheet of paper and fold it along AD so that vertex C +falls on vertex B. +What can you say about sides AC and AB? +Observe that AC covers AB completely +So, +AC = AB +Repeat this activity with some more triangles. +Each time you will observe that the sides opposite +to equal angles are equal. So we have the +following: +Theorem 7.3 : The sides opposite to equal angles of a triangle are equal. +This is the converse of Theorem 7.2. +You can prove this theorem by ASA congruence rule. +Let us take some examples to apply these results. +Example 4 : In ∆ ABC, the bisector AD of ∠ A is perpendicular to side BC +(see Fig. 7.27). Show that AB = AC and ∆ ABC is isosceles. +Solution : In ∆ABD and ∆ACD, +∠ BAD = ∠ CAD +(Given) +AD = AD +(Common) +∠ ADB = ∠ ADC = 90° +(Given) +So, +∆ ABD ≅∆ ACD +(ASA rule) +So, +AB = AC +(CPCT) +or, ∆ ABC is an isosceles triangle. +Fig. 7.26 +Fig. 7.27 +Reprint 2025-26 + +TRIANGLES +97 +Example 5 : E and F are respectively the mid-points +of equal sides AB and AC of ∆ ABC (see Fig. 7.28). +Show that BF = CE. +Solution : In ∆ ABF and ∆ ACE, +AB = AC +(Given) +∠ A = ∠ A +(Common) +AF = AE +(Halves of equal sides) +So, +∆ ABF ≅∆ ACE +(SAS rule) +Therefore, +BF = CE +(CPCT) +Example 6 : In an isosceles triangle ABC with AB = AC, D and E are points on BC +such that BE = CD (see Fig. 7.29). Show that AD = AE. +Solution : In ∆ ABD and ∆ ACE, +AB = AC +(Given) +(1) +∠ B = ∠ C +(Angles opposite to equal sides) (2) +Also, +BE = CD +So, +BE – DE = CD – DE +That is, +BD = CE +(3) +So, +∆ ABD ≅∆ ACE +(Using (1), (2), (3) and SAS rule). +This gives +AD = AE +(CPCT) +EXERCISE 7.2 +1. +In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect +each other at O. Join A to O. Show that : +(i) +OB = OC +(ii) AO bisects ∠ A +2. +In ∆ ABC, AD is the perpendicular bisector of BC +(see Fig. 7.30). Show that ∆ ABC is an isosceles +triangle in which AB = AC. +Fig. 7.28 +Fig. 7.29 +Fig. 7.30 +Reprint 2025-26 + +98 +MATHEMATICS +3. +ABC is an isosceles triangle in which altitudes +BE and CF are drawn to equal sides AC and AB +respectively (see Fig. 7.31). Show that these +altitudes are equal. +4. +ABC is a triangle in which altitudes BE and CF to +sides AC and AB are equal (see Fig. 7.32). Show +that +(i) +∆ ABE ≅ ∆ ACF +(ii) AB = AC, i.e., ABC is an isosceles triangle. +5. +ABC and DBC are two isosceles triangles on the +same base BC (see Fig. 7.33). Show that +∠ ABD = ∠ ACD. +6. +∆ABC is an isosceles triangle in which AB = AC. +Side BA is produced to D such that AD = AB +(see Fig. 7.34). Show that ∠ BCD is a right angle. +7. +ABC is a right angled triangle in which ∠ A = 90° +and AB = AC. Find ∠ B and ∠ C. +8. +Show that the angles of an equilateral triangle +are 60° each. +Fig. 7.32 +Fig. 7.33 +Fig. 7.34 +Fig. 7.31 +Reprint 2025-26 + +TRIANGLES +99 +7.5 Some More Criteria for Congruence of Triangles +You have seen earlier in this chapter that equality of three angles of one triangle to +three angles of the other is not sufficient for the congruence of the two triangles. You +may wonder whether equality of three sides of one triangle to three sides of another +triangle is enough for congruence of the two triangles. You have already verified in +earlier classes that this is indeed true. +To be sure, construct two triangles with sides 4 cm, 3.5 cm and 4.5 cm +(see Fig. 7.35). Cut them out and place them on each other. What do you observe? +They cover each other completely, if the equal sides are placed on each other. So, the +triangles are congruent. +Fig. 7.35 +Repeat this activity with some more triangles. We arrive at another rule for +congruence. +Theorem 7.4 (SSS congruence rule) : If three sides of one triangle are equal to +the three sides of another triangle, then the two triangles are congruent. +This theorem can be proved using a suitable construction. +You have already seen that in the SAS congruence rule, the pair of equal angles +has to be the included angle between the pairs of corresponding pair of equal sides and +if this is not so, the two triangles may not be congruent. +Perform this activity: +Construct two right angled triangles with hypotenuse equal to 5 cm and one side +equal to 4 cm each (see Fig. 7.36). +Reprint 2025-26 + +100 +MATHEMATICS +Fig. 7.36 +Cut them out and place one triangle over the other with equal side placed on each +other. Turn the triangles, if necessary. What do you observe? +The two triangles cover each other completely and so they are congruent. Repeat +this activity with other pairs of right triangles. What do you observe? +You will find that two right triangles are congruent if one pair of sides and the +hypotenuse are equal. You have verified this in earlier classes. +Note that, the right angle is not the included angle in this case. +So, you arrive at the following congruence rule: +Theorem 7.5 (RHS congruence rule) : If in two right triangles the hypotenuse +and one side of one triangle are equal to the hypotenuse and one side of the +other triangle, then the two triangles are congruent. +Note that RHS stands for Right angle - Hypotenuse - Side. +Let us now take some examples. +Example 7 : AB is a line-segment. P and Q are +points on opposite sides of AB such that each of them +is equidistant from the points A and B (see Fig. 7.37). +Show that the line PQ is the perpendicular bisector +of AB. +Solution : You are given that PA = PB and +QA = QB and you are to show that PQ ⊥ AB and +PQ bisects AB. Let PQ intersect AB at C. +Can you think of two congruent triangles in this figure? +Let us take ∆ PAQ and ∆ PBQ. +In these triangles, +Fig. 7.37 +Reprint 2025-26 + +TRIANGLES +101 +AP = BP + (Given) +AQ = BQ + (Given) +PQ = PQ + (Common) +So, +∆ PAQ ≅∆ PBQ +(SSS rule) +Therefore, +∠ APQ = ∠ BPQ +(CPCT). +Now let us consider ∆PAC and ∆PBC. +You have : +AP = BP +(Given) +∠ APC = ∠ BPC (∠ APQ = ∠ BPQ proved above) +PC = PC +(Common) +So, +∆ PAC ≅∆ PBC +(SAS rule) +Therefore, +AC = BC +(CPCT) (1) +and +∠ ACP = ∠ BCP +(CPCT) +Also, +∠ ACP + ∠ BCP = 180° +(Linear pair) +So, +2∠ ACP = 180° +or, +∠ ACP = 90° +(2) +From (1) and (2), you can easily conclude that PQ is the perpendicular bisector of AB. +[Note that, without showing the congruence of ∆PAQ and ∆PBQ, you cannot show +that ∆PAC ≅ ∆PBC even though AP = BP +(Given) +PC = PC + (Common) +and +∠ PAC = ∠ PBC (Angles opposite to equal sides in +∆APB) +It is because these results give us SSA rule which is not always valid or true for +congruence of triangles. Also the angle is not included between the equal pairs of +sides.] +Let us take some more examples. +Example 8 : P is a point equidistant from two lines l and m intersecting at point A +(see Fig. 7.38). Show that the line AP bisects the angle between them. +Solution : You are given that lines l and m intersect each other at A. Let PB ⊥ l, +PC ⊥ m. It is given that PB = PC. +You are to show that ∠ PAB = ∠ PAC. +Reprint 2025-26 + +102 +MATHEMATICS +Let us consider ∆ PAB and ∆ PAC. In these two +triangles, +PB = PC +(Given) +∠ PBA = ∠ PCA = 90° +(Given) +PA = PA +(Common) +So, +∆ PAB ≅∆ PAC +(RHS rule) +So, +∠ PAB = ∠ PAC +(CPCT) +Note that this result is the converse of the result proved in Q.5 of Exercise 7.1. +EXERCISE 7.3 +1. +∆ ABC and ∆ DBC are two isosceles triangles on +the same base BC and vertices A and D are on the +same side of BC (see Fig. 7.39). If AD is extended +to intersect BC at P, show that +(i) +∆ ABD ≅ ∆ ACD +(ii) ∆ ABP ≅ ∆ ACP +(iii) AP bisects ∠ A as well as ∠ D. +(iv) AP is the perpendicular bisector of BC. +2. +AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that +(i) +AD bisects BC +(ii) AD bisects ∠ A. +3. +Two sides AB and BC and median AM +of one triangle ABC are respectively +equal to sides PQ and QR and median +PN of ∆ PQR (see Fig. 7.40). Show that: +(i) +∆ ABM ≅ ∆ PQN +(ii) ∆ ABC ≅ ∆ PQR +4. +BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence +rule, prove that the triangle ABC is isosceles. +5. +ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that +∠ B = ∠ C. +Fig. 7.38 +Fig. 7.39 +Fig. 7.40 +Reprint 2025-26 + +TRIANGLES +103 +7.6 Summary +In this chapter, you have studied the following points : +1. +Two figures are congruent, if they are of the same shape and of the same size. +2. +Two circles of the same radii are congruent. +3. +Two squares of the same sides are congruent. +4. +If two triangles ABC and PQR are congruent under the correspondence A ↔ P, +B ↔ Q and C ↔ R, then symbolically, it is expressed as ∆ ABC ≅ ∆ PQR. +5. +If two sides and the included angle of one triangle are equal to two sides and the included +angle of the other triangle, then the two triangles are congruent (SAS Congruence Rule). +6. +If two angles and the included side of one triangle are equal to two angles and the +included side of the other triangle, then the two triangles are congruent (ASA Congruence +Rule). +7. +If two angles and one side of one triangle are equal to two angles and the corresponding +side of the other triangle, then the two triangles are congruent (AAS Congruence Rule). +8. +Angles opposite to equal sides of a triangle are equal. +9. +Sides opposite to equal angles of a triangle are equal. +10. Each angle of an equilateral triangle is of 60°. +11. +If three sides of one triangle are equal to three sides of the other triangle, then the two +triangles are congruent (SSS Congruence Rule). +12. If in two right triangles, hypotenuse and one side of a triangle are equal to the hypotenuse +and one side of other triangle, then the two triangles are congruent (RHS Congruence +Rule). +Reprint 2025-26" +class_9,8,Quadrilaterals,ncert_books/class_9/iemh1dd/iemh108.pdf,"104 +MATHEMATICS +CHAPTER 8 +QUADRILATERALS +8.1 Properties of a Parallelogram +You have already studied quadrilaterals and their types in Class VIII. A quadrilateral +has four sides, four angles and four vertices. A parallelogram is a quadrilateral in +which both pairs of opposite sides are parallel. +Let us perform an activity. +Cut out a parallelogram from a sheet of paper +and cut it along a diagonal (see Fig. 8.1). You obtain +two triangles. What can you say about these +triangles? +Place one triangle over the other. Turn one around, +if necessary. What do you observe? +Observe that the two triangles are congruent to +each other. +Repeat this activity with some more parallelograms. Each time you will observe +that each diagonal divides the parallelogram into two congruent triangles. +Let us now prove this result. +Theorem 8.1 : A diagonal of a parallelogram divides it into two congruent +triangles. +Proof : Let ABCD be a parallelogram and AC be a diagonal (see Fig. 8.2). Observe +that the diagonal AC divides parallelogram ABCD into two triangles, namely, ∆ ABC +and ∆ CDA. We need to prove that these triangles are congruent. +Fig. 8.1 +Reprint 2025-26 + +QUADRILATERALS +105 +In ∆ ABC and ∆ CDA, note that BC || AD and AC is a transversal. +So, +∠ BCA = ∠ DAC (Pair of alternate angles) +Also, +AB || DC and AC is a transversal. +So, +∠ BAC = ∠ DCA (Pair of alternate angles) +and +AC = CA +(Common) +So, +∆ ABC ≅ ∆ CDA +(ASA rule) +or, +diagonal AC divides parallelogram ABCD into two congruent +triangles ABC and CDA. +Now, measure the opposite sides of parallelogram ABCD. What do you observe? +You will find that AB = DC and AD = BC. +This is another property of a parallelogram stated below: +Theorem 8.2 : In a parallelogram, opposite sides are equal. +You have already proved that a diagonal divides the parallelogram into two congruent +triangles; so what can you say about the corresponding parts say, the corresponding +sides? They are equal. +So, +AB = DC +and +AD = BC +Now what is the converse of this result? You already know that whatever is given +in a theorem, the same is to be proved in the converse and whatever is proved in the +theorem it is given in the converse. Thus, Theorem 8.2 can be stated as given below : +If a quadrilateral is a parallelogram, then each pair of its opposite sides is equal. So +its converse is : +Theorem 8.3 : If each pair of opposite sides of a quadrilateral is equal, then it +is a parallelogram. +Can you reason out why? +Let sides AB and CD of the quadrilateral ABCD +be equal and also AD = BC (see Fig. 8.3). Draw +diagonal AC. +Clearly, +∆ ABC ≅∆ CDA +(Why?) +So, +∠ BAC = ∠ DCA +and +∠ BCA = ∠ DAC +(Why?) +Can you now say that ABCD is a parallelogram? Why? +Fig. 8.2 +Fig. 8.3 +Reprint 2025-26 + +106 +MATHEMATICS +You have just seen that in a parallelogram each pair of opposite sides is equal and +conversely if each pair of opposite sides of a quadrilateral is equal, then it is a +parallelogram. Can we conclude the same result for the pairs of opposite angles? +Draw a parallelogram and measure its angles. What do you observe? +Each pair of opposite angles is equal. +Repeat this with some more parallelograms. We arrive at yet another result as +given below. +Theorem 8.4 : In a parallelogram, opposite angles are equal. +Now, is the converse of this result also true? Yes. Using the angle sum property of +a quadrilateral and the results of parallel lines intersected by a transversal, we can see +that the converse is also true. So, we have the following theorem : +Theorem 8.5 : If in a quadrilateral, each pair of opposite angles is equal, then +it is a parallelogram. +There is yet another property of a parallelogram. Let us study the same. Draw a +parallelogram ABCD and draw both its diagonals intersecting at the point O +(see Fig. 8.4). +Measure the lengths of OA, OB, OC and OD. +What do you observe? You will observe that +OA = OC +and +OB = OD. +or, +O is the mid-point of both the diagonals. +Repeat this activity with some more parallelograms. +Each time you will find that O is the mid-point of +both the diagonals. +So, we have the following theorem : +Theorem 8.6 : The diagonals of a parallelogram bisect each other. +Now, what would happen, if in a quadrilateral the diagonals bisect each other? +Will it be aparallelogram? Indeed this is true. +This result is the converse of the result of Theorem 8.6. It is given below: +Theorem 8.7 : If the diagonals of a quadrilateral bisect each other, then it is a +parallelogram. +You can reason out this result as follows: +Fig. 8.4 +Reprint 2025-26 + +QUADRILATERALS +107 +Note that in Fig. 8.5, it is given that OA = OC +and OB = OD. +So, + ∆ AOB ≅ ∆ COD +(Why?) +Therefore, ∠ ABO = ∠ CDO +(Why?) +From this, we get AB || CD +Similarly, + BC || AD +Therefore ABCD is a parallelogram. +Let us now take some examples. +Example 1 : Show that each angle of a rectangle is a right angle. +Solution : Let us recall what a rectangle is. +A rectangle is a parallelogram in which one angle is a right angle. + Let ABCD be a rectangle in which ∠ A = 90°. +We have to show that ∠ B = ∠ C = ∠ D = 90° +We have, AD || BC and AB is a transversal +(see Fig. 8.6). +So, +∠ A + ∠ B = 180° +(Interior angles on the same + side of the transversal) +But, +∠ A = 90° +So, +∠ B = 180° – ∠ A = 180° – 90° = 90° +Now, +∠ C = ∠ A and ∠ D = ∠ B +(Opposite angles of the parallellogram) +So, +∠ C = 90° and ∠ D = 90°. +Therefore, each of the angles of a rectangle is a right angle. +Example 2 : Show that the diagonals of a rhombus are perpendicular to each other. +Solution : Consider the rhombus ABCD (see Fig. 8.7). +You know that AB = BC = CD = DA (Why?) +Now, in ∆ AOD and ∆ COD, +OA = OC (Diagonals of a parallelogram +bisect each other) +OD = OD +(Common) +AD = CD +Fig. 8.5 +Fig. 8.6 +Fig. 8.7 +Reprint 2025-26 + +108 +MATHEMATICS +Therefore, ∆ AOD ≅ ∆ COD +(SSS congruence rule) +This gives, ∠ AOD = ∠ COD +(CPCT) +But, +∠ AOD + ∠ COD = 180° (Linear pair) +So, +2∠ AOD = 180° +or, +∠ AOD = 90° +So, the diagonals of a rhombus are perpendicular to each other. +Example 3 : ABC is an isosceles triangle in which AB = AC. AD bisects exterior +angle PAC and CD || AB (see Fig. 8.8). Show that +(i) ∠ DAC = ∠ BCA and +(ii) ABCD is a parallelogram. +Solution : (i) ∆ ABC is isosceles in which AB = AC (Given) +So, +∠ ABC = ∠ ACB +(Angles opposite to equal sides) +Also, +∠ PAC = ∠ ABC + ∠ ACB +(Exterior angle of a triangle) +or, +∠ PAC = 2∠ ACB +(1) +Now, AD bisects ∠ PAC. +So, +∠ PAC = 2∠ DAC +(2) +Therefore, +2∠ DAC = 2∠ ACB +[From (1) and (2)] +or, +∠ DAC = ∠ ACB +(ii) Now, these equal angles form a pair of alternate angles when line segments BC +and AD are intersected by a transversal AC. +So, +BC || AD +Also, +BA || CD +(Given) +Now, both pairs of opposite sides of quadrilateral ABCD are parallel. +So, ABCD is a parallelogram. +Example 4 : Two parallel lines l and m are intersected by a transversal p +(see Fig. 8.9). Show that the quadrilateral formed by the bisectors of interior angles is +a rectangle. +Fig. 8.8 +Reprint 2025-26 + +QUADRILATERALS +109 +Solution : It is given that PS || QR and transversal p intersects them at points A and +C respectively. +The bisectors of ∠ PAC and ∠ ACQ intersect at B and bisectors of ∠ ACR and +∠ SAC intersect at D. +We are to show that quadrilateral ABCD is a +rectangle. +Now, +∠ PAC = ∠ ACR +(Alternate angles as l || m and p is a transversal) +So, +1 +2 ∠ PAC = 1 +2 ∠ ACR +i.e., +∠ BAC = ∠ ACD +These form a pair of alternate angles for lines AB +and DC with AC as transversal and they are equal also. +So, +AB || DC +Similarly, +BC || AD +(Considering ∠ ACB and ∠ CAD) +Therefore, quadrilateral ABCD is a parallelogram. +Also, +∠ PAC + ∠ CAS = 180° +(Linear pair) +So, +1 +2 ∠ PAC + 1 +2 ∠ CAS = 1 +2 × 180° = 90° +or, +∠ BAC + ∠ CAD = 90° +or, +∠ BAD = 90° +So, ABCD is a parallelogram in which one angle is 90°. +Therefore, ABCD is a rectangle. +Example 5 : Show that the bisectors of angles of a parallelogram form a rectangle. +Solution : Let P, Q, R and S be the points of +intersection of the bisectors of ∠ A and ∠ B, ∠ B +and ∠ C, ∠ C and ∠ D, and ∠ D and ∠ A respectively +of parallelogram ABCD (see Fig. 8.10). +In ∆ ASD, what do you observe? +Since DS bisects ∠ D and AS bisects ∠ A, therefore, +Fig. 8.9 +Fig. 8.10 +Reprint 2025-26 + +110 +MATHEMATICS +∠ DAS + ∠ ADS = 1 +2 ∠ A + 1 +2 ∠ D += 1 +2 (∠ A + ∠ D) += 1 +2 × 180° +(∠ A and ∠ D are interior angles +on the same side of the transversal) += 90° +Also, ∠ DAS + ∠ ADS + ∠ DSA = 180° +(Angle sum property of a triangle) +or, +90° + ∠ DSA = 180° +or, +∠ DSA = 90° +So, +∠ PSR = 90° +(Being vertically opposite to ∠ DSA) +Similarly, it can be shown that ∠ APB = 90° or ∠ SPQ = 90° (as it was shown for +∠ DSA). Similarly, ∠ PQR = 90° and ∠ SRQ = 90°. +So, PQRS is a quadrilateral in which all angles are right angles. +Can we conclude that it is a rectangle? Let us examine. We have shown that +∠ PSR = ∠ PQR = 90° and ∠ SPQ = ∠ SRQ = 90°. So both pairs of opposite angles +are equal. +Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is 90° and +so, PQRS is a rectangle. +EXERCISE 8.1 +1. +If the diagonals of a parallelogram are equal, then show that it is a rectangle. +2. +Show that the diagonals of a square are equal and bisect each other at right angles. +3. +Diagonal AC of a parallelogram ABCD bisects +∠ A (see Fig. 8.11). Show that +(i) +it bisects ∠ C also, +(ii) ABCD is a rhombus. +4. +ABCD is a rectangle in which diagonal AC bisects +∠ A as well as ∠ C. Show that: (i) ABCD is a square +(ii) diagonal BD bisects ∠ B as well as ∠ D. +Fig. 8.11 +Reprint 2025-26 + +QUADRILATERALS +111 +5. +In parallelogram ABCD, two points P and Q are +taken on diagonal BD such that DP = BQ +(see Fig. 8.12). Show that: +(i) +∆ APD ≅ ∆ CQB +(ii) AP = CQ +(iii) ∆ AQB ≅ ∆ CPD +(iv) AQ = CP +(v) APCQ is a parallelogram +6. +ABCD is a parallelogram and AP and CQ are +perpendiculars from vertices A and C on diagonal +BD (see Fig. 8.13). Show that +(i) +∆ APB ≅ ∆ CQD +(ii) AP = CQ +7. +ABCD is a trapezium in which AB || CD and +AD = BC (see Fig. 8.14). Show that +(i) +∠ A = ∠ B +(ii) ∠ C = ∠ D +(iii) ∆ ABC ≅ ∆ BAD +(iv) diagonal AC = diagonal BD +[Hint : Extend AB and draw a line through C +parallel to DA intersecting AB produced at E.] +8.2 The Mid-point Theorem +You have studied many properties of a triangle as well as a quadrilateral. Now let us +study yet another result which is related to the mid-point of sides of a triangle. Perform +the following activity. +Draw a triangle and mark the mid-points E and F of two sides of the triangle. Join +the points E and F (see Fig. 8.15). +Measure EF and BC. Measure ∠ AEF and ∠ ABC. +What do you observe? You will find that : +EF = 1 +2 BC and ∠ AEF = ∠ ABC +so, +EF || BC +Repeat this activity with some more triangles. +So, you arrive at the following theorem: +Fig. 8.12 +Fig. 8.13 +Fig. 8.14 +Fig. 8.15 +Reprint 2025-26 + +112 +MATHEMATICS +Theorem 8.8 : The line segment joining the mid-points of two sides of a triangle +is parallel to the third side. +You can prove this theorem using the following +clue: +Observe Fig 8.16 in which E and F are mid-points +of AB and AC respectively and CD || BA. +∆ AEF ≅∆ CDF +(ASA Rule) +So, +EF = DF and BE = AE = DC +(Why?) +Therefore, BCDE is a parallelogram. +(Why?) +This gives EF || BC. +In this case, also note that EF = 1 +2 ED = 1 +2 BC. +Can you state the converse of Theorem 8.8? Is the converse true? +You will see that converse of the above theorem is also true which is stated as +below: +Theorem 8.9 : The line drawn through the mid-point of one side of a triangle, +parallel to another side bisects the third side. +In Fig 8.17, observe that E is the mid-point of +AB, line l is passsing through E and is parallel to BC +and CM || BA. +Prove that AF = CF by using the congruence of +∆ AEF and ∆ CDF. +Example 6 : In ∆ ABC, D, E and F are respectively +the mid-points of sides AB, BC and CA +(see Fig. 8.18). Show that ∆ ABC is divided into four +congruent triangles by joining D, E and F. +Solution : As D and E are mid-points of sides AB +and BC of the triangle ABC, by Theorem 8.8, +DE || AC +Similarly, +DF || BC and EF || AB +Fig. 8.16 +Fig. 8.17 +Fig. 8.18 +Reprint 2025-26 + +QUADRILATERALS +113 +Therefore ADEF, BDFE and DFCE are all parallelograms. +Now DE is a diagonal of the parallelogram BDFE, +therefore, +∆ BDE ≅∆ FED +Similarly +∆ DAF ≅∆ FED +and +∆ EFC ≅∆ FED +So, all the four triangles are congruent. +Example 7 : l, m and n are three parallel lines +intersected by transversals p and q such that l, m +and n cut off equal intercepts AB and BC on p +(see Fig. 8.19). Show that l, m and n cut off equal +intercepts DE and EF on q also. +Solution : We are given that AB = BC and have +to prove that DE = EF. +Let us join A to F intersecting m at G.. +The trapezium ACFD is divided into two triangles; +namely ∆ ACF and ∆ AFD. +In ∆ ACF, it is given that B is the mid-point of AC (AB = BC) +and +BG || CF +(since m || n). +So, G is the mid-point of AF +(by using Theorem 8.9) +Now, in ∆ AFD, we can apply the same argument as G is the mid-point of AF, +GE || AD and so by Theorem 8.9, E is the mid-point of DF, +i.e., +DE = EF. +In other words, l, m and n cut off equal intercepts on q also. +EXERCISE 8.2 +1. +ABCD is a quadrilateral in which P, Q, R and S are +mid-points of the sides AB, BC, CD and DA +(see Fig 8.20). AC is a diagonal. Show that : +(i) +SR || AC and SR = 1 +2 AC +(ii) PQ = SR +(iii) PQRS is a parallelogram. +Fig. 8.19 +Fig. 8.20 +Reprint 2025-26 + +114 +MATHEMATICS +2. +ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and +DA respectively. Show that the quadrilateral PQRS is a rectangle. +3. +ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA +respectively. Show that the quadrilateral PQRS is a rhombus. +4. +ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. +A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.21). Show that +F is the mid-point of BC. +Fig. 8.21 +5. +In a parallelogram ABCD, E and F are the +mid-points of sides AB and CD respectively +(see Fig. 8.22). Show that the line segments AF +and EC trisect the diagonal BD. +6. +ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB +and parallel to BC intersects AC at D. Show that +(i) +D is the mid-point of AC +(ii) MD ⊥ AC +(iii) CM = MA = 1 +2 AB +Fig. 8.22 +Reprint 2025-26 + +QUADRILATERALS +115 +8.3 Summary +In this chapter, you have studied the following points : +1. +A diagonal of a parallelogram divides it into two congruent triangles. +2. +In a parallelogram, +(i) +opposite sides are equal +(ii) opposite angles are equal +(iii) diagonals bisect each other +3. +Diagonals of a rectangle bisect each other and are equal and vice-versa. +4. +Diagonals of a rhombus bisect each other at right angles and vice-versa. +5. +Diagonals of a square bisect each other at right angles and are equal, and vice-versa. +6. +The line-segment joining the mid-points of any two sides of a triangle is parallel to the +third side and is half of it. +7. +A line through the mid-point of a side of a triangle parallel to another side bisects the third +side. +Reprint 2025-26" +class_9,9,Circles,ncert_books/class_9/iemh1dd/iemh109.pdf,"116 +MATHEMATICS +CHAPTER 9 +CIRCLES +9.1 Angle Subtended by a Chord at a Point +You have already studied about circles and its parts in Class VI. +Take a line segment PQ and a point R not on the line containing PQ. Join PR and QR +(see Fig. 9.1). Then ∠ PRQ is called the angle subtended by the line segment PQ at +the point R. What are angles POQ, PRQ and PSQ called in Fig. 9.2? ∠ POQ is the +angle subtended by the chord PQ at the centre O, ∠ PRQ and ∠ PSQ are respectively +the angles subtended by PQ at points R and S on the major and minor arcs PQ. +Fig. 9.1 + Fig. 9.2 +Let us examine the relationship between the size of the chord and the angle +subtended by it at the centre. You may see by drawing different chords of a circle +and angles subtended by them at the centre that the longer is the chord, the bigger +will be the angle subtended by it at the centre. What will happen if you take two +equal chords of a circle? Will the angles subtended at the centre be the same +or not? +Reprint 2025-26 + +CIRCLES +117 +Draw two or more equal chords of a circle and +measure the angles subtended by them at the centre +(see Fig.9.3). You will find that the angles subtended +by them at the centre are equal. Let us give a proof +of this fact. +Theorem 9.1 : Equal chords of a circle subtend +equal angles at the centre. +Proof : You are given two equal chords AB and CD +of a circle with centre O (see Fig.9.4). You want to +prove that ∠ AOB = ∠ COD. +In triangles AOB and COD, +OA = OC +(Radii of a circle) +OB = OD +(Radii of a circle) +AB = CD +(Given) +Therefore, +∆ AOB ≅∆ COD +(SSS rule) +This gives +∠ AOB = ∠ COD + (Corresponding parts of congruent triangles) +Remark : For convenience, the abbreviation CPCT will be used in place of +‘Corresponding parts of congruent triangles’, because we use this very frequently as +you will see. +Now if two chords of a circle subtend equal angles at the centre, what can you +say about the chords? Are they equal or not? Let us examine this by the following +activity: +Take a tracing paper and trace a circle on it. Cut +it along the circle to get a disc. At its centre O, draw +an angle AOB where A, B are points on the circle. +Make another angle POQ at the centre equal to +∠AOB. Cut the disc along AB and PQ +(see Fig. 9.5). You will get two segments ACB and +PRQ of the circle. If you put one on the other, what +do you observe? They cover each other, i.e., they +are congruent. So AB = PQ. +Fig. 9.3 +Fig. 9.4 +Fig. 9.5 +Reprint 2025-26 + +118 +MATHEMATICS +Though you have seen it for this particular case, try it out for other equal angles +too. The chords will all turn out to be equal because of the following theorem: +Theorem 9.2 : If the angles subtended by the chords of a circle at the centre are +equal, then the chords are equal. +The above theorem is the converse of the Theorem 9.1. Note that in Fig. 9.4, if +you take ∠ AOB = ∠ COD, then + ∆ AOB ≅ ∆ COD (Why?) +Can you now see that AB = CD? +EXERCISE 9.1 +1. +Recall that two circles are congruent if they have the same radii. Prove that equal +chords of congruent circles subtend equal angles at their centres. +2. +Prove that if chords of congruent circles subtend equal angles at their centres, then +the chords are equal. +9.2 Perpendicular from the Centre to a Chord +Activity : Draw a circle on a tracing paper. Let O +be its centre. Draw a chord AB. Fold the paper along +a line through O so that a portion of the chord falls on +the other. Let the crease cut AB at the point M. Then, +∠ OMA = ∠ OMB = 90° or OM is perpendicular to +AB. Does the point B coincide with A (see Fig.9.6)? +Yes it will. So MA = MB. +Give a proof yourself by joining OA and OB and proving the right triangles OMA +and OMB to be congruent. This example is a particular instance of the following +result: +Theorem 9.3 : The perpendicular from the centre of a circle to a chord bisects +the chord. +What is the converse of this theorem? To write this, first let us be clear what is +assumed in Theorem 9.3 and what is proved. Given that the perpendicular from the +centre of a circle to a chord is drawn and to prove that it bisects the chord. Thus in the +converse, what the hypothesis is ‘if a line from the centre bisects a chord of a +circle’ and what is to be proved is ‘the line is perpendicular to the chord’. So the +converse is: +Fig. 9.6 +Reprint 2025-26 + +CIRCLES +119 +Theorem 9.4 : The line drawn through the centre of a circle to bisect a chord is +perpendicular to the chord. +Is this true? Try it for few cases and see. You will +see that it is true for these cases. See if it is true, in +general, by doing the following exercise. We will write +the stages and you give the reasons. +Let AB be a chord of a circle with centre O and +O is joined to the mid-point M of AB. You have to +prove that OM ⊥ AB. Join OA and OB +(see Fig. 9.7). In triangles OAM and OBM, +OA = OB +(Why ?) +AM = BM +(Why ?) +OM = OM +(Common) +Therefore, ∆OAM ≅∆OBM +(How ?) +This gives ∠OMA = ∠OMB = 90° (Why ?) +9.3 Equal Chords and their Distances from the Centre +Let AB be a line and P be a point. Since there are +infinite numbers of points on a line, if you join these +points to P, you will get infinitely many line segments +PL1, PL2, PM, PL3, PL4, etc. Which of these is the +distance of AB from P? You may think a while and +get the answer. Out of these line segments, the +perpendicular from P to AB, namely PM in Fig. 9.8, +will be the least. In Mathematics, we define this least +length PM to be the distance of AB from P. So you +may say that: +The length of the perpendicular from a point to a line is the distance of the +line from the point. +Note that if the point lies on the line, the distance of the line from the point is zero. +A circle can have infinitely many chords. You may observe by drawing chords of +a circle that longer chord is nearer to the centre than the smaller chord. You may +observe it by drawing several chords of a circle of different lengths and measuring +their distances from the centre. What is the distance of the diameter, which is the +Fig. 9.7 +Fig. 9.8 +Reprint 2025-26 + +120 +MATHEMATICS +longest chord from the centre? Since the centre lies on it, the distance is zero. Do you +think that there is some relationship between the length of chords and their distances +from the centre? Let us see if this is so. +Fig. 9.9 +Activity : Draw a circle of any radius on a tracing paper. Draw two equal chords +AB and CD of it and also the perpendiculars OM and ON on them from the centre +O. Fold the figure so that D falls on B and C falls on A [see Fig.9.9 (i)]. You may +observe that O lies on the crease and N falls on M. Therefore, OM = ON. Repeat +the activity by drawing congruent circles with centres O and O′ and taking equal +chords AB and CD one on each. Draw perpendiculars OM and O′N on them [see +Fig. 9.9(ii)]. Cut one circular disc and put it on the other so that AB coincides with +CD. Then you will find that O coincides with O′ and M coincides with N. In this +way you verified the following: +Theorem 9.5 : Equal chords of a circle (or of congruent circles) are equidistant +from the centre (or centres). +Next, it will be seen whether the converse of this theorem is true or not. For +this, draw a circle with centre O. From the centre O, draw two line segments OL +and OM of equal length and lying inside the circle [see Fig. 9.10(i)]. Then draw +chords PQ and RS of the circle perpendicular to OL and OM respectively [see Fig +9.10(ii)]. Measure the lengths of PQ and RS. Are these different? No, both are +equal. Repeat the activity for more equal line segments and drawing the chords +perpendicular to them. This verifies the converse of the Theorem 9.5 which is stated +as follows: +Reprint 2025-26 + +CIRCLES +121 +Theorem 9.6 : Chords equidistant from the centre of a circle are equal in length. +We now take an example to illustrate the use of the above results: +Example 1 : If two intersecting chords of a circle make equal angles with the diameter +passing through their point of intersection, prove that the chords are equal. +Solution : Given that AB and CD are two chords of +a circle, with centre O intersecting at a point E. PQ +is a diameter through E, such that ∠ AEQ = ∠ DEQ +(see Fig.9.11). You have to prove that AB = CD. +Draw perpendiculars OL and OM on chords AB and +CD, respectively. Now +∠ LOE = 180° – 90° – ∠ LEO = 90° – ∠ LEO +(Angle sum property of a triangle) += 90° – ∠ AEQ = 90° – ∠ DEQ += 90° – ∠ MEO = ∠ MOE +In triangles OLE and OME, +∠ LEO = ∠ MEO +(Why ?) +∠ LOE = ∠ MOE +(Proved above) +EO = EO +(Common) +Therefore, +∆ OLE ≅∆ OME +(Why ?) +This gives +OL = OM +(CPCT) +So, +AB = CD +(Why ?) +Fig. 9.10 +Fig. 9.11 +Reprint 2025-26 + +122 +MATHEMATICS +EXERCISE 9.2 +1. +Two circles of radii 5 cm and 3 cm intersect at two points and the distance between +their centres is 4 cm. Find the length of the common chord. +2. +If two equal chords of a circle intersect within the circle, prove that the segments of +one chord are equal to corresponding segments of the other chord. +3. +If two equal chords of a circle intersect within the circle, prove that the line +joining the point of intersection to the centre makes equal angles with the chords. +4. +If a line intersects two concentric circles (circles +with the same centre) with centre O at A, B, C and +D, prove that AB = CD (see Fig. 9.12). +5. +Three girls Reshma, Salma and Mandip are +playing a game by standing on a circle of radius +5m drawn in a park. Reshma throws a ball to +Salma, Salma to Mandip, Mandip to Reshma. If +the distance between Reshma and Salma and +between Salma and Mandip is 6m each, what is +the distance between Reshma and Mandip? +6. + A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and +David are sitting at equal distance on its boundary each having a toy telephone in +his hands to talk each other. Find the length of the string of each phone. +9.4 Angle Subtended by an Arc of a Circle +You have seen that the end points of a chord other than diameter of a circle cuts it into +two arcs – one major and other minor. If you take two equal chords, what can you say +about the size of arcs? Is one arc made by first chord equal to the corresponding arc +made by another chord? In fact, they are more than just equal in length. They are +congruent in the sense that if one arc is put on the other, without bending or twisting, +one superimposes the other completely. +You can verify this fact by cutting the arc, +corresponding to the chord CD from the circle along +CD and put it on the corresponding arc made by equal +chord AB. You will find that the arc CD superimpose +the arc AB completely (see Fig. 9.13). This shows +that equal chords make congruent arcs and +conversely congruent arcs make equal chords of a +circle. You can state it as follows: +If two chords of a circle are equal, then their corresponding arcs are congruent +and conversely, if two arcs are congruent, then their corresponding chords are equal. +Fig. 9.12 +Fig. 9.13 +Reprint 2025-26 + +CIRCLES +123 +Also the angle subtended by an arc at the centre +is defined to be angle subtended by the corresponding +chord at the centre in the sense that the minor arc +subtends the angle and the major arc subtends the +reflex angle. Therefore, in Fig 9.14, the angle +subtended by the minor arc PQ at O is ∠POQ and +the angle subtended by the major arc PQ at O is +reflex angle POQ. + In view of the property above and Theorem 9.1, +the following result is true: +Congruent arcs (or equal arcs) of a circle subtend equal angles at the centre. +Therefore, the angle subtended by a chord of a circle at its centre is equal to the +angle subtended by the corresponding (minor) arc at the centre. The following theorem +gives the relationship between the angles subtended by an arc at the centre and at a +point on the circle. +Theorem 9.7 : The angle subtended by an arc at the centre is double the angle +subtended by it at any point on the remaining part of the circle. +Proof : Given an arc PQ of a circle subtending angles POQ at the centre O and +PAQ at a point A on the remaining part of the circle. We need to prove that +∠ POQ = 2 ∠ PAQ. +Fig. 9.15 +Consider the three different cases as given in Fig. 9.15. In (i), arc PQ is minor; in (ii), +arc PQ is a semicircle and in (iii), arc PQ is major. +Let us begin by joining AO and extending it to a point B. +In all the cases, +∠ BOQ = ∠ OAQ + ∠ AQO +because an exterior angle of a triangle is equal to the sum of the two interior opposite +angles. +Fig. 9.14 +Reprint 2025-26 + +124 +MATHEMATICS +Also in ∆ OAQ, +OA = OQ +(Radii of a circle) +Therefore, +∠ OAQ = ∠ OQA +(Theorem 7.5) +This gives +∠ BOQ = 2 ∠ OAQ +(1) +Similarly, +∠ BOP = 2 ∠ OAP +(2) +From (1) and (2), +∠ BOP + ∠ BOQ = 2(∠ OAP + ∠ OAQ) +This is the same as +∠ POQ = 2 ∠ PAQ +(3) +For the case (iii), where PQ is the major arc, (3) is replaced by +reflex angle POQ = 2 ∠ PAQ +Remark : Suppose we join points P and Q and +form a chord PQ in the above figures. Then +∠ PAQ is also called the angle formed in the +segment PAQP. +In Theorem 9.7, A can be any point on the +remaining part of the circle. So if you take any +other point C on the remaining part of the circle +(see Fig. 9.16), you have +∠ POQ = 2 ∠ PCQ = 2 ∠ PAQ +Therefore, +∠ PCQ = ∠ PAQ. +This proves the following: +Theorem 9.8 : Angles in the same segment of a circle are equal. +Again let us discuss the case (ii) of Theorem 10.8 separately. Here ∠PAQ is an angle +in the segment, which is a semicircle. Also, ∠ PAQ = 1 +2 ∠ POQ = 1 +2 × 180° = 90°. +If you take any other point C on the semicircle, again you get that +∠ PCQ = 90° +Therefore, you find another property of the circle as: +Angle in a semicircle is a right angle. +The converse of Theorem 9.8 is also true. It can be stated as: +Theorem 9.9 : If a line segment joining two points subtends equal angles at two +other points lying on the same side of the line containing the line segment, the +four points lie on a circle (i.e. they are concyclic). +Fig. 9.16 +Reprint 2025-26 + +CIRCLES +125 +You can see the truth of this result as follows: +In Fig. 9.17, AB is a line segment, which subtends equal angles at two points C and D. +That is +∠ ACB = ∠ ADB +To show that the points A, B, C and D lie on a circle +let us draw a circle through the points A, C and B. +Suppose it does not pass through the point D. Then it +will intersect AD (or extended AD) at a point, say E +(or E′). +If points A, C, E and B lie on a circle, +∠ ACB = ∠ AEB +(Why?) +But it is given that ∠ ACB = ∠ ADB. +Therefore, +∠ AEB = ∠ ADB. +This is not possible unless E coincides with D. (Why?) +Similarly, E′ should also coincide with D. +9.5 Cyclic Quadrilaterals +A quadrilateral ABCD is called cyclic if all the four vertices +of it lie on a circle (see Fig 9.18). You will find a peculiar +property in such quadrilaterals. Draw several cyclic +quadrilaterals of different sides and name each of these +as ABCD. (This can be done by drawing several circles +of different radii and taking four points on each of them.) +Measure the opposite angles and write your observations +in the following table. + S.No. of Quadrilateral +∠ A +∠ B +∠ C +∠ D +∠ A +∠ C +∠ B +∠ D +1. +2. +3. +4. +5. +6. +What do you infer from the table? +Fig. 9.17 +Fig. 9.18 +Reprint 2025-26 + +126 +MATHEMATICS +You find that ∠A + ∠C = 180° and ∠B + ∠D = 180°, neglecting the error in +measurements. This verifies the following: +Theorem 9.10 : The sum of either pair of opposite angles of a cyclic quadrilateral +is 180º. +In fact, the converse of this theorem, which is stated below is also true. +Theorem 9.11 : If the sum of a pair of opposite angles of a quadrilateral is 180º, +the quadrilateral is cyclic. +You can see the truth of this theorem by following a method similar to the method +adopted for Theorem 9.9. +Example 2 : In Fig. 9.19, AB is a diameter of the circle, CD is a chord equal to the +radius of the circle. AC and BD when extended intersect at a point E. Prove that +∠ AEB = 60°. +Solution : Join OC, OD and BC. +Triangle ODC is equilateral +(Why?) +Therefore, +∠ COD = 60° +Now, +∠ CBD = 1 +2 ∠ COD +(Theorem 9.7) +This gives +∠ CBD = 30° +Again, +∠ ACB = 90° +(Why ?) +So, +∠ BCE = 180° – ∠ ACB = 90° +Which gives +∠ CEB = 90° – 30° = 60°, i.e., ∠ AEB = 60° +Example 3 : In Fig 9.20, ABCD is a cyclic +quadrilateral in which AC and BD are its diagonals. +If ∠ DBC = 55° and ∠ BAC = 45°, find ∠ BCD. +Solution : +∠ CAD = ∠ DBC = 55° +(Angles in the same segment) +Therefore, +∠ DAB = ∠ CAD + ∠ BAC += 55° + 45° = 100° +But ∠ DAB + ∠ BCD = 180° +(Opposite angles of a cyclic quadrilateral) +So, +∠ BCD = 180° – 100° = 80° +Fig. 9.19 +Fig. 9.20 +Reprint 2025-26 + +CIRCLES +127 +Example 4 : Two circles intersect at two points A +and B. AD and AC are diameters to the two circles +(see Fig. 9.21). Prove that B lies on the line segment +DC. +Solution : Join AB. +∠ ABD = 90° (Angle in a semicircle) +∠ ABC = 90° (Angle in a semicircle) +So, +∠ ABD + ∠ ABC = 90° + 90° = 180° +Therefore, DBC is a line. That is B lies on the line segment DC. +Example 5 : Prove that the quadrilateral formed (if possible) by the internal angle +bisectors of any quadrilateral is cyclic. +Solution : In Fig. 9.22, ABCD is a quadrilateral in +which the angle bisectors AH, BF, CF and DH of +internal angles A, B, C and D respectively form a +quadrilateral EFGH. +Now, ∠ FEH = ∠ AEB = 180° – ∠ EAB – ∠ EBA (Why ?) + = 180° – 1 +2 (∠ A + ∠ B) +and ∠ FGH = ∠ CGD = 180° – ∠ GCD – ∠ GDC (Why ?) + = 180° – 1 +2 (∠ C + ∠ D) +Therefore, ∠ FEH + ∠ FGH = 180° – 1 +2 (∠ A + ∠ B) + 180° – 1 +2 (∠ C + ∠ D) + = 360° – 1 +2 (∠ A+ ∠ B +∠ C +∠ D) = 360° – 1 +2 × 360° + = 360° – 180° = 180° +Therefore, by Theorem 9.11, the quadrilateral EFGH is cyclic. +EXERCISE 9.3 +1. +In Fig. 9.23, A,B and C are three points on a circle +with centre O such that ∠ BOC = 30° and +∠ AOB = 60°. If D is a point on the circle other +than the arc ABC, find ∠ADC. +Fig. 9.22 +Fig. 9.23 +Fig. 9.21 +Reprint 2025-26 + +128 +MATHEMATICS +2. +A chord of a circle is equal to the radius of the +circle. Find the angle subtended by the chord at +a point on the minor arc and also at a point on the +major arc. +3. +In Fig. 9.24, ∠ PQR = 100°, where P, Q and R are +points on a circle with centre O. Find ∠ OPR. +4. +In Fig. 9.25, ∠ ABC = 69°, ∠ ACB = 31°, find +∠ BDC. +5. +In Fig. 9.26, A, B, C and D are four points on a +circle. AC and BD intersect at a point E such +that ∠ BEC = 130° and ∠ ECD = 20°. Find +∠ BAC. +6. +ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°, +∠ BAC is 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD. +7. +If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of +the quadrilateral, prove that it is a rectangle. +8. +If the non-parallel sides of a trapezium are equal, prove that it is cyclic. +Fig. 9.24 +Fig. 9.25 +Fig. 9.26 +Reprint 2025-26 + +CIRCLES +129 +9. +Two circles intersect at two points B and C. +Through B, two line segments ABD and PBQ +are drawn to intersect the circles at A, D and P, +Q respectively (see Fig. 9.27). Prove that +∠ ACP = ∠ QCD. +10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of +intersection of these circles lie on the third side. +11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that +∠ CAD = ∠ CBD. +12. Prove that a cyclic parallelogram is a rectangle. +9.6 Summary +In this chapter, you have studied the following points: +1. +A circle is the collection of all points in a plane, which are equidistant from a fixed point in +the plane. +2. +Equal chords of a circle (or of congruent circles) subtend equal angles at the centre. +3. +If the angles subtended by two chords of a circle (or of congruent circles) at the centre +(corresponding centres) are equal, the chords are equal. +4. +The perpendicular from the centre of a circle to a chord bisects the chord. +5. +The line drawn through the centre of a circle to bisect a chord is perpendicular to +the chord. +6. +Equal chords of a circle (or of congruent circles) are equidistant from the centre (or +corresponding centres). +7. +Chords equidistant from the centre (or corresponding centres) of a circle (or of congruent +circles) are equal. +8. +If two arcs of a circle are congruent, then their corresponding chords are equal and +conversely if two chords of a circle are equal, then their corresponding arcs (minor, major) +are congruent. +9. +Congruent arcs of a circle subtend equal angles at the centre. +10. The angle subtended by an arc at the centre is double the angle subtended by it at any +point on the remaining part of the circle. +11. +Angles in the same segment of a circle are equal. +Fig. 9.27 +Reprint 2025-26 + +130 +MATHEMATICS +12. Angle in a semicircle is a right angle. +13. If a line segment joining two points subtends equal angles at two other points lying on +the same side of the line containing the line segment, the four points lie on a circle. +14. The sum of either pair of opposite angles of a cyclic quadrilateral is 1800. +15. If sum of a pair of opposite angles of a quadrilateral is 1800, the quadrilateral is cyclic. +Reprint 2025-26" +class_9,10,Heron's Formula,ncert_books/class_9/iemh1dd/iemh110.pdf,"CHAPTER 10 +HERON’S FORMULA +10.1 Area of a Triangle — by Heron’s Formula +We know that the area of triangle when its height is given, is 1 +2 × base × height. Now +suppose that we know the lengths of the sides of a scalene triangle and not the height. +Can you still find its area? For instance, you have a triangular park whose sides are 40 +m, 32 m, and 24 m. How will you calculate its area? Definitely if you want to apply the +formula, you will have to calculate its height. But we do not have a clue to calculate +the height. Try doing so. If you are not able to get it, then go to the next section. +Heron was born in about 10AD possibly in Alexandria in +Egypt. He worked in applied mathematics. His works on +mathematical and physical subjects are so numerous and +varied that he is considered to be an encyclopedic writer +in these fields. His geometrical works deal largely with +problems on mensuration written in three books. Book I +deals with the area of squares, rectangles, triangles, +trapezoids (trapezia), various other specialised +quadrilaterals, the regular polygons, circles, surfaces of +cylinders, cones, spheres etc. In this book, Heron has +derived the famous formula for the area of a triangle in +terms of its three sides. +The formula given by Heron about the area of a triangle, is also known as Hero’s +formula. It is stated as: +Area of a triangle = +( +) ( +) ( +) +s s +a +s +b +s +c +− +− +− +(I) +Heron (10 C.E. – 75 C.E.) +Fig. 10.1 +Reprint 2025-26 + +132 +MATHEMATICS +where a, b and c are the sides of the triangle, and s = semi-perimeter, i.e., half the +perimeter of the triangle = +2 +a +b +c ++ ++ +, +This formula is helpful where it is not possible to find the height of the triangle +easily. Let us apply it to calculate the area of the triangular park ABC, mentioned +above (see Fig. 10.2). +Let us take a = 40 m, b = 24 m, c = 32 m, +so that we have s = 40 +24 +32 +2 ++ ++ + m = 48 m. +s – a = (48 – 40) m = 8 m, +s – b = (48 – 24) m = 24 m, +s – c = (48 – 32) m = 16 m. +Therefore, area of the park ABC += +( +) ( +) ( +) +− +− +− +s s +a +s +b +s +c += +2 +2 +48 +8 +24 +16 m +384m +× +× +× += +We see that 322 + 242 = 1024 + 576 = 1600 = 402. Therefore, the sides of the park +make a right triangle. The largest side, i.e., BC which is 40 m will be the hypotenuse +and the angle between the sides AB and AC will be 90°. +We can check that the area of the park is 1 +2 × 32 × 24 m2 = 384 m2. +We find that the area we have got is the same as we found by using Heron’s +formula. +Now using Heron’s formula, you verify this fact by finding the areas of other +triangles discussed earlier viz., +(i) equilateral triangle with side 10 cm. +(ii) isosceles triangle with unequal side as 8 cm and each equal side as 5 cm. +You will see that +For (i), we have s = 10 +10 +10 +2 ++ ++ + cm = 15 cm. +Fig. 10.2 +Reprint 2025-26 + +HERON’S FORMULA +133 +Area of triangle = 15(15 +10) (15 +10) (15 +10) +− +− +− + cm2 + = +2 +2 +15 +5 +5 +5 cm +25 3 cm +× +× +× += +For (ii), we have s = 8 +5 +5 cm +9 cm +2 ++ ++ += +Area of triangle = +9(9 +8) (9 +5) (9 +5) +− +− +− +cm2 = +2 +2 +9 +1 +4 +4 cm +12 cm . +× × +× += +Let us now solve some more examples: +Example 1 : Find the area of a triangle, two sides of which are 8 cm and 11 cm and +the perimeter is 32 cm (see Fig. 10.3). +Solution : Here we have perimeter of the triangle = 32 cm, a = 8 cm and b = 11 cm. +Third side c = 32 cm – (8 + 11) cm = 13 cm +So, +2s = 32, i.e., s = 16 cm, +s – a = (16 – 8) cm = 8 cm, +s – b = (16 – 11) cm = 5 cm, +s – c = (16 – 13) cm = 3 cm. +Therefore, area of the triangle = +( +) ( +) ( +) +s s +a +s +b +s +c +− +− +− += +2 +2 +16 +8 +5 +3cm +8 30 cm +× +× +× += +Example 2 : A triangular park ABC has sides 120m, 80m and 50m (see Fig. 10.4). A +gardener Dhania has to put a fence all around it and also plant grass inside. How +much area does she need to plant? Find the cost of fencing it with barbed wire at the +rate of `20 per metre leaving a space 3m wide for a gate on one side. +Solution : For finding area of the park, we have +2s = 50 m + 80 m + 120 m = 250 m. +i.e., +s = 125 m +Now, +s – a = (125 – 120) m = 5 m, +s – b = (125 – 80) m = 45 m, +s – c = (125 – 50) m = 75 m. +Fig. 10.3 +Fig. 10.4 +Reprint 2025-26 + +134 +MATHEMATICS +Therefore, area of the park = +( +) ( +) ( +) +s s +a +s +b +s +c +− +− +− += +125 +5 +45 +75 +× +× +× +m2 += +2 +375 15 m +Also, perimeter of the park = AB + BC + CA = 250 m +Therefore, length of the wire needed for fencing = 250 m – 3 m (to be left for gate) += 247 m +And so the cost of fencing = `20 × 247 = `4940 +Example 3 : The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeter +is 300 m. Find its area. +Solution : Suppose that the sides, in metres, are 3x, 5x and 7x (see Fig. 10.5). +Then, we know that 3x + 5x + 7x = 300 (perimeter of the triangle) +Therefore, 15x = 300, which gives x = 20. +So the sides of the triangle are 3 × 20 m, 5 × 20 m and 7 × 20 m +i.e., +60 m, 100 m and 140 m. +Can you now find the area [Using Heron’s formula]? +We have s = 60 +100 +140 +2 ++ ++ + m = 150 m, +and area will be +150(150 +60) (150 +100) (150 +140) +− +− +− +m2 += +150 +90 +50 +10 +× +× +× +m2 += +2 +1500 3 m +EXERCISE 10.1 +1. +A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with +side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is +180 cm, what will be the area of the signal board? +Fig. 10.5 +Reprint 2025-26 + +HERON’S FORMULA +135 +2. +The triangular side walls of a flyover have been used for advertisements. The sides of +the walls are 122 m, 22 m and 120 m (see Fig. 10.6). The advertisements yield an +earning of ` 5000 per m2 per year. A company hired one of its walls for 3 months. How +much rent did it pay? +Fig. 10.6 +3. +There is a slide in a park. One of its side walls has been painted in some colour with a +message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 10.7 ). If the sides of the +wall are 15 m, 11 m and 6 m, find the area painted in colour. +Fig. 10.7 +4. +Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is +42cm. +5. +Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area. +6. +An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find +the area of the triangle. +Reprint 2025-26 + +136 +MATHEMATICS +10.2 Summary +In this chapter, you have studied the following points : +1. +Area of a triangle with its sides as a, b and c is calculated by using Heron’s formula, +stated as +Area of triangle = +( +) ( +) ( +) +− +− +− +s s +a +s +b +s +c +where +s = +2 ++ ++ +a +b +c +Reprint 2025-26" +class_9,11,Surface Areas and Volumes,ncert_books/class_9/iemh1dd/iemh111.pdf,"SURFACE AREAS AND VOLUMES +137 +CHAPTER 11 +SURFACE AREAS AND VOLUMES +11.1 Surface Area of a Right Circular Cone +We have already studied the surface areas of cube, cuboid and cylinder. We will now +study the surface area of cone. +So far, we have been generating solids by stacking up congruent figures. Incidentally, +such figures are called prisms. Now let us look at another kind of solid which is not a +prism (These kinds of solids are called pyramids.). Let us see how we can generate +them. +Activity : Cut out a right-angled triangle ABC right angled at B. Paste a long thick +string along one of the perpendicular sides say AB of the triangle [see Fig. 11.1(a)]. +Hold the string with your hands on either sides of the triangle and rotate the triangle +about the string a number of times. What happens? Do you recognize the shape that +the triangle is forming as it rotates around the string [see Fig. 11.1(b)]? Does it remind +you of the time you had eaten an ice-cream heaped into a container of that shape [see +Fig. 11.1 (c) and (d)]? +Fig. 11.1 +Reprint 2025-26 + +138 +MATHEMATICS +This is called a right circular cone. In Fig. 11.1(c) +of the right circular cone, the point A is called the +vertex, AB is called the height, BC is called the radius +and AC is called the slant height of the cone. Here B +will be the centre of circular base of the cone. The +height, radius and slant height of the cone are usually +denoted by h, r and l respectively. Once again, let us +see what kind of cone we can not call a right circular +cone. Here, you are (see Fig. 11.2)! What you see in +these figures are not right circular cones; because in +(a), the line joining its vertex to the centre of its base +is not at right angle to the base, and in (b) the base is +not circular. +As in the case of cylinder, since we will be studying only about right circular cones, +remember that by ‘cone’ in this chapter, we shall mean a ‘right circular cone.’ +Activity : (i) Cut out a neatly made paper cone that does not have any overlapped +paper, straight along its side, and opening it out, to see the shape of paper that forms +the surface of the cone. (The line along which you cut the cone is the slant height of +the cone which is represented by l). It looks like a part of a round cake. +(ii) +If you now bring the sides marked A and B at the tips together, you can see that +the curved portion of Fig. 11.3 (c) will form the circular base of the cone. +Fig. 11.3 +(iii) If the paper like the one in Fig. 11.3 (c) is now cut into hundreds of little pieces, +along the lines drawn from the point O, each cut portion is almost a small triangle, +whose height is the slant height l of the cone. +(iv) Now the area of each triangle = +1 +2 × base of each triangle × l. +So, area of the entire piece of paper +Fig. 11.2 +Reprint 2025-26 + +SURFACE AREAS AND VOLUMES +139 += sum of the areas of all the triangles += +1 +2 +3 +1 +1 +1 +2 +2 +2 +bl +b l +b l ++ ++ ++ ⋯ = +( +) +1 +2 +3 +1 +2l b +b +b ++ ++ ++ ⋯ += 1 +2 × l × length of entire curved boundary of Fig. 11.3(c) +(as b1 + b2 + b3 + . . . makes up the curved portion of the figure) +But the curved portion of the figure makes up the perimeter of the base of the cone +and the circumference of the base of the cone = 2πr, where r is the base radius of the +cone. +So, Curved Surface Area of a Cone = 1 +2 × l × 2πr = πrl +where r is its base radius and l its slant height. +Note that l2 = r2 + h2 (as can be seen from Fig. 11.4), by +applying Pythagoras Theorem. Here h is the height of the +cone. +Therefore, l = +2 +2 +r +h ++ +Now if the base of the cone is to be closed, then a circular piece of paper of radius r +is also required whose area is πr2. +So, Total Surface Area of a Cone = πrl + πr2 = πr(l + r) +Example 1 : Find the curved surface area of a right circular cone whose slant height +is 10 cm and base radius is 7 cm. +Solution : Curved surface area = πrl += 22 +7 × 7 × 10 cm2 += 220 cm2 +Example 2 : The height of a cone is 16 cm and its base radius is 12 cm. Find the +curved surface area and the total surface area of the cone (Use π = 3.14). +Solution : Here, h = 16 cm and r = 12 cm. +So, from l2 = h2 + r2, we have +l = +2 +2 +16 +12 ++ + cm = 20 cm +Fig. 11.4 +Reprint 2025-26 + +140 +MATHEMATICS +So, curved surface area = πrl += 3.14 × 12 × 20 cm2 += 753.6 cm2 +Further, total surface area = πrl + πr2 += (753.6 + 3.14 × 12 × 12) cm2 += (753.6 + 452.16) cm2 += 1205.76 cm2 +Example 3 : A corn cob (see Fig. 11.5), shaped somewhat +like a cone, has the radius of its broadest end as 2.1 cm and +length (height) as 20 cm. If each 1 cm2 of the surface of the +cob carries an average of four grains, find how many grains +you would find on the entire cob. +Solution : Since the grains of corn are found only on the curved surface of the corn +cob, we would need to know the curved surface area of the corn cob to find the total +number of grains on it. In this question, we are given the height of the cone, so we +need to find its slant height. +Here, +l = +2 +2 +r +h ++ + = +2 +2 +(2.1) +20 ++ + cm += +404.41 cm = 20.11 cm +Therefore, the curved surface area of the corn cob = πrl += 22 +7 × 2.1 × 20.11 cm2 = 132.726 cm2 = 132.73 cm2 (approx.) +Number of grains of corn on 1 cm2 of the surface of the corn cob = 4 +Therefore, number of grains on the entire curved surface of the cob += 132.73 × 4 = 530.92 = 531 (approx.) +So, there would be approximately 531 grains of corn on the cob. +EXERCISE 11.1 +Assume π = 22 +7 , unless stated otherwise. +1. +Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved +surface area. +2. +Find the total surface area of a cone, if its slant height is 21 m and diameter of its base +is 24 m. +Fig. 11.5 +Reprint 2025-26 + +SURFACE AREAS AND VOLUMES +141 +3. +Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find +(i) radius of the base and (ii) total surface area of the cone. +4. +A conical tent is 10 m high and the radius of its base is 24 m. Find +(i) slant height of the tent. +(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ` 70. +5. +What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m +and base radius 6 m? Assume that the extra length of material that will be required for +stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14). +6. +The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. +Find the cost of white-washing its curved surface at the rate of ` 210 per 100 m2. +7. +A joker’s cap is in the form of a right circular cone of base radius 7 cm and height +24 cm. Find the area of the sheet required to make 10 such caps. +8. +A bus stop is barricaded from the remaining part of the road, by using 50 hollow +cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height +1 m. If the outer side of each of the cones is to be painted and the cost of painting is +` 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take +1.04 = 1.02) +11.2 Surface Area of a Sphere +What is a sphere? Is it the same as a circle? Can you draw a circle on a paper? Yes, +you can, because a circle is a plane closed figure whose every point lies at a constant +distance (called radius) from a fixed point, which is called the centre of the circle. +Now if you paste a string along a diameter of a circular disc and rotate it as you had +rotated the triangle in the previous section, you see a new solid (see Fig 11.6). What +does it resemble? A ball? Yes. It is called a sphere. +Fig. 11.6 +Reprint 2025-26 + +142 +MATHEMATICS +Can you guess what happens to the centre of the circle, when it forms a sphere on +rotation? Of course, it becomes the centre of the sphere. So, a sphere is a three +dimensional figure (solid figure), which is made up of all points in the space, +which lie at a constant distance called the radius, from a fixed point called the +centre of the sphere. +Note : A sphere is like the surface of a ball. The word solid sphere is used for the +solid whose surface is a sphere. +Activity : Have you ever played with a top or have you at least watched someone +play with one? You must be aware of how a string is wound around it. Now, let us take +a rubber ball and drive a nail into it. Taking support of the nail, let us wind a string +around the ball. When you have reached the ‘fullest’ part of the ball, use pins to keep +the string in place, and continue to wind the string around the remaining part of the ball, +till you have completely covered the ball [see Fig. 11.7(a)]. Mark the starting and +finishing points on the string, and slowly unwind the string from the surface of the ball. +Now, ask your teacher to help you in measuring the diameter of the ball, from which +you easily get its radius. Then on a sheet of paper, draw four circles with radius equal +to the radius of the ball. Start filling the circles one by one, with the string you had +wound around the ball [see Fig. 11.7(b)]. +Fig. 11.7 +What have you achieved in all this? +The string, which had completely covered the surface area of the sphere, has been +used to completely fill the regions of four circles, all of the same radius as of the sphere. +So, what does that mean? This suggests that the surface area of a sphere of radius r += 4 times the area of a circle of radius r = 4 × (π r2) +So, +Surface Area of a Sphere = 4 π r2 +Reprint 2025-26 + +SURFACE AREAS AND VOLUMES +143 +where r is the radius of the sphere. +How many faces do you see in the surface of a sphere? There is only one, which is +curved. +Now, let us take a solid sphere, and slice it exactly ‘through +the middle’ with a plane that passes through its centre. What +happens to the sphere? +Yes, it gets divided into two equal parts (see Fig. 11.8)! What +will each half be called? It is called a hemisphere. (Because +‘hemi’ also means ‘half’) +And what about the surface of a hemisphere? How many faces does it have? +Two! There is a curved face and a flat face (base). +The curved surface area of a hemisphere is half the surface area of the sphere, which +is 1 +2 of 4πr2. +Therefore, +Curved Surface Area of a Hemisphere = 2πr2 +where r is the radius of the sphere of which the hemisphere is a part. +Now taking the two faces of a hemisphere, its surface area 2πr2 + πr2 +So, Total Surface Area of a Hemisphere = 3πr2 +Example 4 : Find the surface area of a sphere of radius 7 cm. +Solution : The surface area of a sphere of radius 7 cm would be +4πr2 = 4 × 22 +7 × 7 × 7 cm2 = 616 cm2 +Example 5 : Find (i) the curved surface area and (ii) the total surface area of a +hemisphere of radius 21 cm. +Solution : The curved surface area of a hemisphere of radius 21 cm would be += 2πr2 = 2 × 22 +7 × 21 × 21 cm2 = 2772 cm2 +Fig. 11.8 +Reprint 2025-26 + +144 +MATHEMATICS +(ii) the total surface area of the hemisphere would be +3πr2 = 3 × 22 +7 × 21 × 21 cm2 = 4158 cm2 +Example 6 : The hollow sphere, in which the circus motorcyclist performs his stunts, +has a diameter of 7 m. Find the area available to the motorcyclist for riding. +Solution : Diameter of the sphere = 7 m. Therefore, radius is 3.5 m. So, the riding +space available for the motorcyclist is the surface area of the ‘sphere’ which is +given by +4πr2 = 4 × 22 +7 × 3.5 × 3.5 m2 += 154 m2 +Example 7 : A hemispherical dome of a building needs to be painted +(see Fig. 11.9). If the circumference of the base of the dome is 17.6 m, find the cost of +painting it, given the cost of painting is ` 5 per 100 cm2. +Solution : Since only the rounded surface of the dome is to be painted, we would need +to find the curved surface area of the hemisphere to know the extent of painting that +needs to be done. Now, circumference of the dome = 17.6 m. Therefore, 17.6 = 2πr. +So, the radius of the dome = 17.6 × +7 +2 +22 +× + m = 2.8 m +The curved surface area of the dome = 2πr2 += 2 × 22 +7 × 2.8 × 2.8 m2 += 49.28 m2 +Now, cost of painting 100 cm2 is ` 5. +So, cost of painting 1 m2 = ` 500 +Therefore, cost of painting the whole dome += ` 500 × 49.28 += ` 24640 +EXERCISE 11.2 +Assume π = 22 +7 , unless stated otherwise. +1. +Find the surface area of a sphere of radius: +(i) +10.5 cm +(ii) 5.6 cm +(iii) 14 cm +Fig. 11.9 +Reprint 2025-26 + +SURFACE AREAS AND VOLUMES +145 +2. +Find the surface area of a sphere of diameter: +(i) +14 cm +(ii) 21 cm +(iii) 3.5 m +3. +Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14) +4. +The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped +into it. Find the ratio of surface areas of the balloon in the two cases. +5. +A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of +tin-plating it on the inside at the rate of ` 16 per 100 cm2. +6. +Find the radius of a sphere whose surface area is 154 cm2. +7. +The diameter of the moon is approximately one fourth of the diameter of the earth. +Find the ratio of their surface areas. +8. +A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is +5 cm. Find the outer curved surface area of the bowl. +9. +A right circular cylinder just encloses a sphere of +radius r (see Fig. 11.10). Find +(i) +surface area of the sphere, +(ii) curved surface area of the cylinder, +(iii) ratio of the areas obtained in (i) and (ii). +11.3 Volume of a Right Circular Cone +In earlier classes we have studied the volumes of +cube, cuboid and cylinder +In Fig 11.11, can you see that there is a right circular +cylinder and a right circular cone of the same base +radius and the same height? +Activity : Try to make a hollow cylinder and a hollow cone like this with the same +base radius and the same height (see Fig. 11.11). Then, we can try out an experiment +that will help us, to see practically what the volume of a right circular cone would be! +Fig. 11.12 +Fig. 11.10 +Fig. 11.11 +Reprint 2025-26 + +146 +MATHEMATICS +So, let us start like this. +Fill the cone up to the brim with sand once, and empty it into the cylinder. We find +that it fills up only a part of the cylinder [see Fig. 11.12(a)]. +When we fill up the cone again to the brim, and empty it into the cylinder, we see +that the cylinder is still not full [see Fig. 11.12(b)]. +When the cone is filled up for the third time, and emptied into the cylinder, it can be +seen that the cylinder is also full to the brim [see Fig. 11.12(c)]. +With this, we can safely come to the conclusion that three times the volume of a +cone, makes up the volume of a cylinder, which has the same base radius and the +same height as the cone, which means that the volume of the cone is one-third the +volume of the cylinder. +So, +Volume of a Cone = 1 +3 πr2h +where r is the base radius and h is the height of the cone. +Example 8 : The height and the slant height of a cone are 21 cm and 28 cm respectively. +Find the volume of the cone. +Solution : From l2 = r2 + h2, we have +r = +2 +2 +2 +2 +28 +21 cm +7 7 cm +l +h +− += +− += +So, volume of the cone = 1 +3 πr2h = 1 +3 × 22 +7 7 +7 7 +21 +7 × +× +× + cm3 + = 7546 cm3 +Example 9 : Monica has a piece of canvas whose area is 551 m2. She uses it to have +a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins +and the wastage incurred while cutting, amounts to approximately 1 m2, find the volume +of the tent that can be made with it. +Solution : Since the area of the canvas = 551 m2 and area of the canvas lost in +wastage is 1 m2, therefore the area of canvas available for making the tent is +(551 – 1) m2 = 550 m2. +Now, the surface area of the tent = 550 m2 and the required base radius of the conical +tent = 7 m +Note that a tent has only a curved surface (the floor of a tent is not covered by +canvas!!). +Reprint 2025-26 + +SURFACE AREAS AND VOLUMES +147 +Therefore, curved surface area of tent = 550 m2. +That is, +πrl = 550 +or, +22 +7 × 7 × l = 550 +or, +l = 3 550 +22 m = 25 m +Now, +l2 = r2 + h2 +Therefore, +h = +2 +2 +l +r +− + = +2 +2 +25 +7 m +625 +49 m +576 m +− += +− += += 24 m +So, the volume of the conical tent = +2 +3 +1 +1 +22 +7 +7 +24 m +3 +3 +7 +r h +π += +× +× +× +× += 1232 m3. +EXERCISE 11.3 +Assume π = 22 +7 , unless stated otherwise. +1. +Find the volume of the right circular cone with +(i) +radius 6 cm, height 7 cm +(ii) radius 3.5 cm, height 12 cm +2. +Find the capacity in litres of a conical vessel with +(i) +radius 7 cm, slant height 25 cm +(ii) height 12 cm, slant height 13 cm +3. +The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. +(Use π = 3.14) +4. +If the volume of a right circular cone of height 9 cm is 48 π cm3, find the diameter of its +base. +5. +A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? +6. +The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, +find +(i) +height of the cone +(ii) slant height of the cone +(iii) curved surface area of the cone +7. +A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. +Find the volume of the solid so obtained. +8. +If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find +the volume of the solid so obtained. Find also the ratio of the volumes of the two +solids obtained in Questions 7 and 8. +9. +A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. +Find its volume. The heap is to be covered by canvas to protect it from rain. Find the +area of the canvas required. +Reprint 2025-26 + +148 +MATHEMATICS +11.4 Volume of a Sphere +Now, let us see how to go about measuring the volume of a sphere. First, take two or +three spheres of different radii, and a container big enough to be able to put each of +the spheres into it, one at a time. Also, take a large trough in which you can place the +container. Then, fill the container up to the brim with water [see Fig. 11.13(a)]. +Now, carefully place one of the spheres in the container. Some of the water from +the container will over flow into the trough in which it is kept [see Fig. 11.13(b)]. +Carefully pour out the water from the trough into a measuring cylinder (i.e., a graduated +cylindrical jar) and measure the water over flowed [see Fig. 11.13(c)]. Suppose the +radius of the immersed sphere is r (you can find the radius by measuring the diameter +of the sphere). Then evaluate 4 +3 πr3. Do you find this value almost equal to the +measure of the volume over flowed? +Fig. 11.13 +Once again repeat the procedure done just now, with a different size of sphere. +Find the radius R of this sphere and then calculate the value of +3 +4 R . +3 π + Once again this +value is nearly equal to the measure of the volume of the water displaced (over flowed) +by the sphere. What does this tell us? We know that the volume of the sphere is the +same as the measure of the volume of the water displaced by it. By doing this experiment +repeatedly with spheres of varying radii, we are getting the same result, namely, the +volume of a sphere is equal to +4 +3 π times the cube of its radius. This gives us the idea +that +Volume of a Sphere = +3 +4 +3 +r +π +where r is the radius of the sphere. +Later, in higher classes it can be proved also. But at this stage, we will just take it +as true. +Reprint 2025-26 + +SURFACE AREAS AND VOLUMES +149 +Since a hemisphere is half of a sphere, can you guess what the volume of a +hemisphere will be? Yes, it is +3 +1 +4 +of +2 +3 +r +π + = +2 +3 πr3. +So, +Volume of a Hemisphere = +3 +2 +3 +r +π +where r is the radius of the hemisphere. +Let us take some examples to illustrate the use of these formulae. +Example 10 : Find the volume of a sphere of radius 11.2 cm. +Solution : Required volume = 4 +3 πr3 += 4 +22 +11.2 +11.2 +11.2 +3 +7 +× +× +× +× + cm3 = 5887.32 cm3 +Example 11 : A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the +metal is 7.8 g per cm3, find the mass of the shot-putt. +Solution : Since the shot-putt is a solid sphere made of metal and its mass is equal to +the product of its volume and density, we need to find the volume of the sphere. +Now, volume of the sphere = +3 +4 +3 +r +π += +3 +4 +22 +4.9 +4.9 +4.9 cm +3 +7 +× +× +× +× += 493 cm3 (nearly) +Further, mass of 1 cm3 of metal is 7.8 g. +Therefore, mass of the shot-putt = 7.8 × 493 g += 3845.44 g = 3.85 kg (nearly) +Example 12 : A hemispherical bowl has a radius of 3.5 cm. What would be the +volume of water it would contain? +Solution : The volume of water the bowl can contain += +3 +2 +3 +r +π += 2 +22 +3.5 +3.5 +3.5 +3 +7 +× +× +× +× + cm3 = 89.8 cm3 +Reprint 2025-26 + +150 +MATHEMATICS +EXERCISE 11.4 +Assume π = +22 +7 , unless stated otherwise. +1. +Find the volume of a sphere whose radius is +(i) +7 cm +(ii) 0.63 m +2. +Find the amount of water displaced by a solid spherical ball of diameter +(i) +28 cm +(ii) 0.21 m +3. +The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of +the metal is 8.9 g per cm3? +4. +The diameter of the moon is approximately one-fourth of the diameter of the earth. +What fraction of the volume of the earth is the volume of the moon? +5. +How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold? +6. +A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, +then find the volume of the iron used to make the tank. +7. +Find the volume of a sphere whose surface area is 154 cm2. +8. +A dome of a building is in the form of a hemisphere. From inside, it was white-washed +at the cost of ` 4989.60. If the cost of white-washing is ` 20 per square metre, find the +(i) +inside surface area of the dome, +(ii) volume of the air inside the dome. +9. +Twenty seven solid iron spheres, each of radius r and surface area S are melted to +form a sphere with surface area S′. Find the +(i) +radius r′ of the new sphere, +(ii) ratio of S and S′. +10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much +medicine (in mm3) is needed to fill this capsule? +11.5 Summary +In this chapter, you have studied the following points: +1. +Curved surface area of a cone = πrl +2. +Total surface area of a right circular cone = πrl + πr2, i.e., π π π π πr (l + r) +3. +Surface area of a sphere of radius r = 4 π r2 +4. +Curved surface area of a hemisphere = 2πr2 +5. +Total surface area of a hemisphere = 3πr2 +6. +Volume of a cone = 1 +3 πr2h +7. +Volume of a sphere of radius r = +3 +4 +3 +r +π +8. +Volume of a hemisphere = +3 +2 +3 +r +π +[Here, letters l, b, h, a, r, etc. have been used in their usual meaning, depending on the +context.] +Reprint 2025-26" +class_9,12,Statistics,ncert_books/class_9/iemh1dd/iemh112.pdf,"STATISTICS +151 +CHAPTER 12 +STATISTICS +12.1 Graphical Representation of Data +The representation of data by tables has already been discussed. Now let us turn our +attention to another representation of data, i.e., the graphical representation. It is well +said that one picture is better than a thousand words. Usually comparisons among the +individual items are best shown by means of graphs. The representation then becomes +easier to understand than the actual data. We shall study the following graphical +representations in this section. +(A) Bar graphs +(B) Histograms of uniform width, and of varying widths +(C) Frequency polygons +(A) Bar Graphs +In earlier classes, you have already studied and constructed bar graphs. Here we +shall discuss them through a more formal approach. Recall that a bar graph is a +pictorial representation of data in which usually bars of uniform width are drawn with +equal spacing between them on one axis (say, the x-axis), depicting the variable. The +values of the variable are shown on the other axis (say, the y-axis) and the heights of +the bars depend on the values of the variable. +Example 1 : In a particular section of Class IX, 40 students were asked about the +months of their birth and the following graph was prepared for the data so obtained: +Reprint 2025-26 + +152 +MATHEMATICS +Fig. 12.1 +Observe the bar graph given above and answer the following questions: +(i) +How many students were born in the month of November? +(ii) +In which month were the maximum number of students born? +Solution : Note that the variable here is the ‘month of birth’, and the value of the +variable is the ‘Number of students born’. +(i) 4 students were born in the month of November. +(ii) The Maximum number of students were born in the month of August. +Let us now recall how a bar graph is constructed by considering the following example. +Example 2 : A family with a monthly income of ` 20,000 had planned the following +expenditures per month under various heads: +Table 12.1 +Heads +Expenditure +(in thousand rupees) +Grocery +4 +Rent +5 +Education of children +5 +Medicine +2 +Fuel +2 +Entertainment +1 +Miscellaneous +1 +Draw a bar graph for the data above. +Reprint 2025-26 + +STATISTICS +153 +Solution : We draw the bar graph of this data in the following steps. Note that the unit +in the second column is thousand rupees. So, ‘4’ against ‘grocery’ means `4000. +1. +We represent the Heads (variable) on the horizontal axis choosing any scale, +since the width of the bar is not important. But for clarity, we take equal widths +for all bars and maintain equal gaps in between. Let one Head be represented by +one unit. +2. +We represent the expenditure (value) on the vertical axis. Since the maximum +expenditure is `5000, we can choose the scale as 1 unit = `1000. +3. +To represent our first Head, i.e., grocery, we draw a rectangular bar with width +1 unit and height 4 units. +4. +Similarly, other Heads are represented leaving a gap of 1 unit in between two +consecutive bars. +The bar graph is drawn in Fig. 12.2. +Fig. 12.2 +Here, you can easily visualise the relative characteristics of the data at a glance, e.g., +the expenditure on education is more than double that of medical expenses. Therefore, +in some ways it serves as a better representation of data than the tabular form. +Activity 1 : Continuing with the same four groups of Activity 1, represent the data by +suitable bar graphs. +Let us now see how a frequency distribution table for continuous class intervals +can be represented graphically. +Reprint 2025-26 + +154 +MATHEMATICS +(B) Histogram +This is a form of representation like the bar graph, but it is used for continuous class +intervals. For instance, consider the frequency distribution Table 12.2, representing +the weights of 36 students of a class: +Table 12.2 +Weights (in kg) +Number of students +30.5 - 35.5 +9 +35.5 - 40.5 +6 +40.5 - 45.5 +15 +45.5 - 50.5 +3 +50.5 - 55.5 +1 +55.5 - 60.5 +2 +Total +36 +Let us represent the data given above graphically as follows: +(i) +We represent the weights on the horizontal axis on a suitable scale. We can choose +the scale as 1 cm = 5 kg. Also, since the first class interval is starting from 30.5 +and not zero, we show it on the graph by marking a kink or a break on the axis. +(ii) We represent the number of students (frequency) on the vertical axis on a suitable +scale. Since the maximum frequency is 15, we need to choose the scale to +accomodate this maximum frequency. +(iii) We now draw rectangles (or rectangular bars) of width equal to the class-size +and lengths according to the frequencies of the corresponding class intervals. For +example, the rectangle for the class interval 30.5 - 35.5 will be of width 1 cm and +length 4.5 cm. +(iv) In this way, we obtain the graph as shown in Fig. 12.3: +Reprint 2025-26 + +STATISTICS +155 +Fig. 12.3 +Observe that since there are no gaps in between consecutive rectangles, the resultant +graph appears like a solid figure. This is called a histogram, which is a graphical +representation of a grouped frequency distribution with continuous classes. Also, unlike +a bar graph, the width of the bar plays a significant role in its construction. +Here, in fact, areas of the rectangles erected are proportional to the corresponding +frequencies. However, since the widths of the rectangles are all equal, the lengths of +the rectangles are proportional to the frequencies. That is why, we draw the lengths +according to (iii) above. +Now, consider a situation different from the one above. +Example 3 : A teacher wanted to analyse the performance of two sections of students +in a mathematics test of 100 marks. Looking at their performances, she found that a +few students got under 20 marks and a few got 70 marks or above. So she decided to +group them into intervals of varying sizes as follows: 0 - 20, 20 - 30, . . ., 60 - 70, +70 - 100. Then she formed the following table: +Reprint 2025-26 + +156 +MATHEMATICS +Table 12.3 +Marks +Number of students + 0 - 20 +7 +20 - 30 +10 +30 - 40 +10 +40 - 50 +20 +50 - 60 +20 +60 - 70 +15 + 70 - above +8 +Total +90 +A histogram for this table was prepared by a student as shown in Fig. 12.4. +Fig. 12.4 +Carefully examine this graphical representation. Do you think that it correctly represents +the data? No, the graph is giving us a misleading picture. As we have mentioned +earlier, the areas of the rectangles are proportional to the frequencies in a histogram. +Earlier this problem did not arise, because the widths of all the rectangles were equal. +But here, since the widths of the rectangles are varying, the histogram above does not +Reprint 2025-26 + +STATISTICS +157 +give a correct picture. For example, it shows a greater frequency in the interval +70 - 100, than in 60 - 70, which is not the case. +So, we need to make certain modifications in the lengths of the rectangles so that +the areas are again proportional to the frequencies. +The steps to be followed are as given below: +1. +Select a class interval with the minimum class size. In the example above, the +minimum class-size is 10. +2. +The lengths of the rectangles are then modified to be proportionate to the +class-size 10. +For instance, when the class-size is 20, the length of the rectangle is 7. So when +the class-size is 10, the length of the rectangle will be 7 +20 × 10 = 3.5. +Similarly, proceeding in this manner, we get the following table: +Table 12.4 +Marks +Frequency +Width of +Length of the rectangle +the class +0 - 20 +7 +20 +7 +20 × 10 = 3.5 +20 - 30 +10 +10 +10 +10 × 10 = 10 +30 - 40 +10 +10 +10 +10 × 10 = 10 +40 - 50 +20 +10 +20 +10 × 10 = 20 +50 - 60 +20 +10 +20 +10 × 10 = 20 +60 - 70 +15 +10 +15 +10 × 10 = 15 +70 - 100 +8 +30 +8 +30 × 10 = 2.67 +Reprint 2025-26 + +158 +MATHEMATICS +Since we have calculated these lengths for an interval of 10 marks in each case, +we may call these lengths as “proportion of students per 10 marks interval”. +So, the correct histogram with varying width is given in Fig. 12.5. +Fig. 12.5 +(C) Frequency Polygon +There is yet another visual way of representing quantitative data and its frequencies. +This is a polygon. To see what we mean, consider the histogram represented by +Fig. 12.3. Let us join the mid-points of the upper sides of the adjacent rectangles of +this histogram by means of line segments. Let us call these mid-points B, C, D, E, F +and G. When joined by line segments, we obtain the figure BCDEFG (see Fig. 12.6). +To complete the polygon, we assume that there is a class interval with frequency zero +Reprint 2025-26 + +STATISTICS +159 +before 30.5 - 35.5, and one after 55.5 - 60.5, and their mid-points are A and H, +respectively. ABCDEFGH is the frequency polygon corresponding to the data shown +in Fig. 12.3. We have shown this in Fig. 12.6. +Fig. 12.6 +Although, there exists no class preceding the lowest class and no class succeeding +the highest class, addition of the two class intervals with zero frequency enables us to +make the area of the frequency polygon the same as the area of the histogram. Why +is this so? (Hint : Use the properties of congruent triangles.) +Now, the question arises: how do we complete the polygon when there is no class +preceding the first class? Let us consider such a situation. +Reprint 2025-26 + +160 +MATHEMATICS +Example 4 : Consider the marks, out of 100, obtained by 51 students of a class in a +test, given in Table 12.5. +Table 12.5 +Marks +Number of students +0 - 10 +5 +10 - 20 +10 +20 - 30 +4 +30 - 40 +6 +40 - 50 +7 +50 - 60 +3 +60 - 70 +2 +70 - 80 +2 +80 - 90 +3 +90 - 100 +9 +Total +51 +Draw a frequency polygon corresponding to this frequency distribution table. +Solution : Let us first draw a histogram for this data and mark the mid-points of the +tops of the rectangles as B, C, D, E, F, G, H, I, J, K, respectively. Here, the first class is +0-10. So, to find the class preceeding 0-10, we extend the horizontal axis in the negative +direction and find the mid-point of the imaginary class-interval (–10) - 0. The first end +point, i.e., B is joined to this mid-point with zero frequency on the negative direction of +the horizontal axis. The point where this line segment meets the vertical axis is marked +as A. Let L be the mid-point of the class succeeding the last class of the given data. +Then OABCDEFGHIJKL is the frequency polygon, which is shown in Fig. 12.7. +Fig. 12.7 +Reprint 2025-26 + +STATISTICS +161 +Frequency polygons can also be drawn independently without drawing +histograms. For this, we require the mid-points of the class-intervals used in the data. +These mid-points of the class-intervals are called class-marks. +To find the class-mark of a class interval, we find the sum of the upper limit and +lower limit of a class and divide it by 2. Thus, +Class-mark = Upper limit + Lower limit +2 +Let us consider an example. +Example 5 : In a city, the weekly observations made in a study on the cost of living +index are given in the following table: +Table 12.6 +Cost of living index +Number of weeks +140 - 150 +5 +150 - 160 +10 +160 - 170 +20 +170 - 180 +9 +180 - 190 +6 +190 - 200 +2 +Total +52 +Draw a frequency polygon for the data above (without constructing a histogram). +Solution : Since we want to draw a frequency polygon without a histogram, let us find +the class-marks of the classes given above, that is of 140 - 150, 150 - 160,.... +For 140 - 150, the upper limit = 150, and the lower limit = 140 +So, the class-mark = 150 + 140 +2 + = 290 +2 = 145. +Continuing in the same manner, we find the class-marks of the other classes as well. +So, the new table obtained is as shown in the following table: +Reprint 2025-26 + +162 +MATHEMATICS +Table 12.7 +Classes +Class-marks +Frequency +140 - 150 +145 +5 +150 - 160 +155 +10 +160 - 170 +165 +20 +170 - 180 +175 +9 +180 - 190 +185 +6 +190 - 200 +195 +2 +Total +52 +We can now draw a frequency polygon by plotting the class-marks along the horizontal +axis, the frequencies along the vertical-axis, and then plotting and joining the points +B(145, 5), C(155, 10), D(165, 20), E(175, 9), F(185, 6) and G(195, 2) by line segments. +We should not forget to plot the point corresponding to the class-mark of the class +130 - 140 (just before the lowest class 140 - 150) with zero frequency, that is, +A(135, 0), and the point H (205, 0) occurs immediately after G(195, 2). So, the resultant +frequency polygon will be ABCDEFGH (see Fig. 12.8). +Fig. 12.8 +Reprint 2025-26 + +STATISTICS +163 +Frequency polygons are used when the data is continuous and very large. It is +very useful for comparing two different sets of data of the same nature, for example, +comparing the performance of two different sections of the same class. +EXERCISE 12.1 +1. +A survey conducted by an organisation for the cause of illness and death among +the women between the ages 15 - 44 (in years) worldwide, found the following +figures (in %): +S.No. +Causes +Female fatality rate (%) +1. +Reproductive health conditions +31.8 +2. +Neuropsychiatric conditions +25.4 +3. +Injuries +12.4 +4. +Cardiovascular conditions +4.3 +5. +Respiratory conditions +4.1 +6. +Other causes +22.0 +(i) +Represent the information given above graphically. +(ii) Which condition is the major cause of women’s ill health and death worldwide? +(iii) Try to find out, with the help of your teacher, any two factors which play a major +role in the cause in (ii) above being the major cause. +2. +The following data on the number of girls (to the nearest ten) per thousand boys in +different sections of Indian society is given below. +Section +Number of girls per thousand boys +Scheduled Caste (SC) +940 +Scheduled Tribe (ST) +970 +Non SC/ST +920 +Backward districts +950 +Non-backward districts +920 +Rural +930 +Urban +910 +Reprint 2025-26 + +164 +MATHEMATICS +(i) Represent the information above by a bar graph. +(ii) In the classroom discuss what conclusions can be arrived at from the graph. +3. +Given below are the seats won by different political parties in the polling outcome of +a state assembly elections: +Political Party +A +B +C +D +E +F +Seats Won +75 +55 +37 +29 +10 +37 +(i) Draw a bar graph to represent the polling results. +(ii) Which political party won the maximum number of seats? +4. +The length of 40 leaves of a plant are measured correct to one millimetre, and the +obtained data is represented in the following table: +Length (in mm) +Number of leaves +118 - 126 +3 +127 - 135 +5 +136 - 144 +9 +145 - 153 +12 +154 - 162 +5 +163 - 171 +4 +172 - 180 +2 +(i) Draw a histogram to represent the given data. [Hint: First make the class intervals +continuous] +(ii) Is there any other suitable graphical representation for the same data? +(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? +Why? +5. +The following table gives the life times of 400 neon lamps: +Life time (in hours) +Number of lamps +300 - 400 +14 +400 - 500 +56 +500 - 600 +60 +600 - 700 +86 +700 - 800 +74 +800 - 900 +62 +900 - 1000 +48 +Reprint 2025-26 + +STATISTICS +165 +(i) Represent the given information with the help of a histogram. +(ii) How many lamps have a life time of more than 700 hours? +6. +The following table gives the distribution of students of two sections according to +the marks obtained by them: + Section A + Section B +Marks +Frequency +Marks +Frequency + 0 - 10 +3 + 0 - 10 +5 +10 - 20 +9 +10 - 20 +19 +20 - 30 +17 +20 - 30 +15 +30 - 40 +12 +30 - 40 +10 +40 - 50 +9 +40 - 50 +1 +Represent the marks of the students of both the sections on the same graph by two +frequency polygons. From the two polygons compare the performance of the two +sections. +7. +The runs scored by two teams A and B on the first 60 balls in a cricket match are given +below: +Number of balls +Team A +Team B +1 - 6 +2 +5 +7 - 12 +1 +6 +13 - 18 +8 +2 +19 - 24 +9 +10 +25 - 30 +4 +5 +31 - 36 +5 +6 +37 - 42 +6 +3 +43 - 48 +10 +4 +49 - 54 +6 +8 +55 - 60 +2 +10 +Represent the data of both the teams on the same graph by frequency polygons. +[Hint : First make the class intervals continuous.] +Reprint 2025-26 + +166 +MATHEMATICS +8. +A random survey of the number of children of various age groups playing in a park +was found as follows: +Age (in years) +Number of children +1 - 2 +5 +2 - 3 +3 +3 - 5 +6 +5 - 7 +12 +7 - 10 +9 +10 - 15 +10 +15 - 17 +4 +Draw a histogram to represent the data above. +9. +100 surnames were randomly picked up from a local telephone directory and a frequency +distribution of the number of letters in the English alphabet in the surnames was found +as follows: +Number of letters +Number of surnames +1 - 4 +6 +4 - 6 +30 +6 - 8 +44 +8 - 12 +16 +12 - 20 +4 +(i) Draw a histogram to depict the given information. +(ii) Write the class interval in which the maximum number of surnames lie. +12.2 Summary +In this chapter, you have studied the following points: +1. How data can be presented graphically in the form of bar graphs, histograms and frequency +polygons. +Reprint 2025-26" +class_10,1,Real Numbers,ncert_books/class_10/jemh1dd/jemh101.pdf,"REAL NUMBERS +1 +1 +1.1 Introduction +In Class IX, you began your exploration of the world of real numbers and encountered +irrational numbers. We continue our discussion on real numbers in this chapter. We +begin with very important properties of positive integers in Sections 1.2, namely the +Euclid’s division algorithm and the Fundamental Theorem of Arithmetic. +Euclid’s division algorithm, as the name suggests, has to do with divisibility of +integers. Stated simply, it says any positive integer a can be divided by another positive +integer b in such a way that it leaves a remainder r that is smaller than b. Many of you +probably recognise this as the usual long division process. Although this result is quite +easy to state and understand, it has many applications related to the divisibility properties +of integers. We touch upon a few of them, and use it mainly to compute the HCF of +two positive integers. +The Fundamental Theorem of Arithmetic, on the other hand, has to do something +with multiplication of positive integers. You already know that every composite number +can be expressed as a product of primes in a unique way—this important fact is the +Fundamental Theorem of Arithmetic. Again, while it is a result that is easy to state and +understand, it has some very deep and significant applications in the field of mathematics. +We use the Fundamental Theorem of Arithmetic for two main applications. First, we +use it to prove the irrationality of many of the numbers you studied in Class IX, such as +2, +3 and 5 . Second, we apply this theorem to explore when exactly the decimal +expansion of a rational number, say +( +0) +p q +q + +, is terminating and when it is non- +terminating repeating. We do so by looking at the prime factorisation of the denominator +q of p +q . You will see that the prime factorisation of q will completely reveal the nature +of the decimal expansion of p +q . +So let us begin our exploration. +REAL NUMBERS +Reprint 2025-26 + +2 +MATHEMATICS +1.2 The Fundamental Theorem of Arithmetic +In your earlier classes, you have seen that any natural number can be written as a +product of its prime factors. For instance, 2 = 2, 4 = 2 × 2, 253 = 11 × 23, and so on. +Now, let us try and look at natural numbers from the other direction. That is, can any +natural number be obtained by multiplying prime numbers? Let us see. +Take any collection of prime numbers, say 2, 3, 7, 11 and 23. If we multiply +some or all of these numbers, allowing them to repeat as many times as we wish, +we can produce a large collection of positive integers (In fact, infinitely many). +Let us list a few : +7 × 11 × 23 = 1771 +3 × 7 × 11 × 23 = 5313 +2 × 3 × 7 × 11 × 23 = 10626 +23 × 3 × 73 = 8232 +22 × 3 × 7 × 11 × 23 = 21252 +and so on. +Now, let us suppose your collection of primes includes all the possible primes. +What is your guess about the size of this collection? Does it contain only a finite +number of integers, or infinitely many? Infact, there are infinitely many primes. So, +if we combine all these primes in all possible ways, we will get an infinite +collection of numbers, all +the primes and all possible +products of primes. The +question is – can we +produce all the composite +numbers this way? What +do you think? Do you +think that there may be a +composite number which +is not the product of +powers +of +primes? +Before we answer this, +let us factorise positive +integers, that is, do the +opposite of what we have +done so far. +We are going to use +the factor tree with which +you are all familiar. Let us +take some large number, +say, 32760, and factorise +it as shown. +Reprint 2025-26 + +REAL NUMBERS +3 +Carl Friedrich Gauss +(1777 – 1855) +An equivalent version of Theorem 1.2 was probably +first recorded as Proposition 14 of Book IX in Euclid’s +Elements, before it came to be known as the +Fundamental Theorem of Arithmetic. However, the +first correct proof was given by Carl Friedrich Gauss +in his Disquisitiones Arithmeticae. +Carl Friedrich Gauss is often referred to as the ‘Prince +of Mathematicians’ and is considered one of the three +greatest mathematicians of all time, along with +Archimedes and Newton. He has made fundamental +contributions to both mathematics and science. +So we have factorised 32760 as 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 as a product of +primes, i.e., 32760 = 23 × 32 × 5 × 7 × 13 as a product of powers of primes. Let us try +another number, say, 123456789. This can be written as 32 × 3803 × 3607. Of course, +you have to check that 3803 and 3607 are primes! (Try it out for several other natural +numbers yourself.) This leads us to a conjecture that every composite number can be +written as the product of powers of primes. In fact, this statement is true, and is called +the Fundamental Theorem of Arithmetic because of its basic crucial importance +to the study of integers. Let us now formally state this theorem. +Theorem 1.1 (Fundamental Theorem of Arithmetic) : Every composite +number can be expressed (factorised) as a product of primes, and this factorisation +is unique, apart from the order in which the prime factors occur. +The Fundamental Theorem of Arithmetic says that every composite number can +be factorised as a product of primes. Actually it says more. It says that given any +composite number it can be factorised as a product of prime numbers in a ‘unique’ +way, except for the order in which the primes occur. That is, given any composite +number there is one and only one way to write it as a product of primes, as long as we +are not particular about the order in which the primes occur. So, for example, we +regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other possible order in which +these primes are written. This fact is also stated in the following form: +The prime factorisation of a natural number is unique, except for the order +of its factors. +Reprint 2025-26 + +4 +MATHEMATICS +In general, given a composite number x, we factorise it as x = p1p2 ... pn, where +p1, p2,..., pn are primes and written in ascending order, i.e., p1  p2 + . . .  pn. If we combine the same primes, we will get powers of primes. For example, +32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13 +Once we have decided that the order will be ascending, then the way the number +is factorised, is unique. +The Fundamental Theorem of Arithmetic has many applications, both within +mathematics and in other fields. Let us look at some examples. +Example 1 : Consider the numbers 4n, where n is a natural number. Check whether +there is any value of n for which 4n ends with the digit zero. +Solution : If the number 4n, for any n, were to end with the digit zero, then it would be +divisible by 5. That is, the prime factorisation of 4n would contain the prime 5. This is +not possible because 4n = (2)2n; so the only prime in the factorisation of 4n is 2. So, the +uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no +other primes in the factorisation of 4n. So, there is no natural number n for which 4n +ends with the digit zero. +You have already learnt how to find the HCF and LCM of two positive integers +using the Fundamental Theorem of Arithmetic in earlier classes, without realising it! +This method is also called the prime factorisation method. Let us recall this method +through an example. +Example 2 : Find the LCM and HCF of 6 and 20 by the prime factorisation method. +Solution : We have : +6 = 21 × 31 and 20 = 2 × 2 × 5 = 22 × 51. +You can find HCF(6, 20) = 2 and LCM(6, 20) = 2 × 2 × 3 × 5 = 60, as done in your +earlier classes. +Note that HCF(6, 20) = 21 = Product of the smallest power of each common +prime factor in the numbers. +LCM (6, 20) = 22 × 31 × 51 = Product of the greatest power of each prime factor, +involved in the numbers. +From the example above, you might have noticed that HCF(6, 20) × LCM(6, 20) += 6 × 20. In fact, we can verify that for any two positive integers a and b, +HCF (a, b) × LCM (a, b) = a × b. We can use this result to find the LCM of two +positive integers, if we have already found the HCF of the two positive integers. +Example 3: Find the HCF of 96 and 404 by the prime factorisation method. Hence, +find their LCM. +Reprint 2025-26 + +REAL NUMBERS +5 +Solution : The prime factorisation of 96 and 404 gives : +96 = 25 × 3, 404 = 22 × 101 +Therefore, the HCF of these two integers is 22 = 4. +Also, +LCM (96, 404) = +96 +404 +96 +404 +9696 +HCF(96, 404) +4 + + + + +Example 4 : Find the HCF and LCM of 6, 72 and 120, using the prime factorisation +method. +Solution : We have : +6 = 2 × 3, 72 = 23 × 32, 120 = 23 × 3 × 5 +Here, 21 and 31 are the smallest powers of the common factors 2 and 3, respectively. +So, +HCF (6, 72, 120) = 21 × 31 = 2 × 3 = 6 +23, 32 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively +involved in the three numbers. +So, +LCM (6, 72, 120) = 23 × 32 × 51 = 360 +Remark : Notice, 6 × 72 × 120  HCF (6, 72, 120) × LCM (6, 72, 120). So, the +product of three numbers is not equal to the product of their HCF and LCM. +EXERCISE 1.1 +1. Express each number as a product of its prime factors: +(i) 140 +(ii) 156 +(iii) 3825 +(iv) 5005 +(v) 7429 +2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = +product of the two numbers. +(i) 26 and 91 +(ii) 510 and 92 +(iii) 336 and 54 +3. Find the LCM and HCF of the following integers by applying the prime factorisation +method. +(i) 12, 15 and 21 +(ii) 17, 23 and 29 +(iii) 8, 9 and 25 +4. Given that HCF (306, 657) = 9, find LCM (306, 657). +5. Check whether 6n can end with the digit 0 for any natural number n. +6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. +7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round +of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the +Reprint 2025-26 + +6 +MATHEMATICS +* Not from the examination point of view. +same point and at the same time, and go in the same direction. After how many minutes +will they meet again at the starting point? +1.3 Revisiting Irrational Numbers +In Class IX, you were introduced to irrational numbers and many of their properties. +You studied about their existence and how the rationals and the irrationals together +made up the real numbers. You even studied how to locate irrationals on the number +line. However, we did not prove that they were irrationals. In this section, we will +prove that +2 , +3 , +5 and, in general, +p is irrational, where p is a prime. One of +the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic. +Recall, a number ‘s’ is called irrational if it cannot be written in the form +, +p +q +where p and q are integers and q ¹ 0. Some examples of irrational numbers, with +which you are already familiar, are : +2 , +2, +3, +15, +, +0.10110111011110 . . . +3 + +, etc. +Before we prove that +2 is irrational, we need the following theorem, whose +proof is based on the Fundamental Theorem of Arithmetic. +Theorem 1.2 : Let p be a prime number. If p divides a2, then p divides a, where +a is a positive integer. +*Proof : Let the prime factorisation of a be as follows : +a = p1p2 . . . pn, where p1,p2, . . ., pn are primes, not necessarily distinct. +Therefore, a2 = (p1p2 . . . pn)( p1p2 . . . pn) = p2 +1p2 +2 . . . p2 +n. +Now, we are given that p divides a2. Therefore, from the Fundamental Theorem of +Arithmetic, it follows that p is one of the prime factors of a2. However, using the +uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only +prime factors of a2 are p1, p2, . . ., pn. So p is one of p1, p2, . . ., pn. +Now, since a = p1 p2 . . . pn, p divides a. +We are now ready to give a proof that +2 is irrational. +The proof is based on a technique called ‘proof by contradiction’. (This technique is +discussed in some detail in Appendix 1). +Theorem 1.3 : +2 is irrational. +Proof : Let us assume, to the contrary, that +2 is rational. +Reprint 2025-26 + +REAL NUMBERS +7 +So, we can find integers r and s (¹ 0) such that +2 = r +s . +Suppose r and s have a common factor other than 1. Then, we divide by the common +factor to get +, +2 +a +b + + where a and b are coprime. +So, b +2 = a. +Squaring on both sides and rearranging, we get 2b2 = a2. Therefore, 2 divides a2. +Now, by Theorem 1.2, it follows that 2 divides a. +So, we can write a = 2c for some integer c. +Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2. +This means that 2 divides b2, and so 2 divides b (again using Theorem 1.2 with p = 2). +Therefore, a and b have at least 2 as a common factor. +But this contradicts the fact that a and b have no common factors other than 1. +This contradiction has arisen because of our incorrect assumption that 2 is rational. +So, we conclude that +2 is irrational. +Example 5 : Prove that +3 is irrational. +Solution : Let us assume, to the contrary, that +3 is rational. +That is, we can find integers a and b (¹ 0) such that +3 = a +b  +Suppose a and b have a common factor other than 1, then we can divide by the +common factor, and assume that a and b are coprime. +So, +3 +b +a + + +Squaring on both sides, and rearranging, we get 3b2 = a2. +Therefore, a2 is divisible by 3, and by Theorem 1.2, it follows that a is also divisible +by 3. +So, we can write a = 3c for some integer c. +Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2. +This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.2 +with p = 3). +Reprint 2025-26 + +8 +MATHEMATICS +Therefore, a and b have at least 3 as a common factor. +But this contradicts the fact that a and b are coprime. +This contradiction has arisen because of our incorrect assumption that 3 is rational. +So, we conclude that 3 is irrational. +In Class IX, we mentioned that : +the sum or difference of a rational and an irrational number is irrational and +the product and quotient of a non-zero rational and irrational number is +irrational. +We prove some particular cases here. +Example 6 : Show that 5 – +3 is irrational. +Solution : Let us assume, to the contrary, that 5 – +3 is rational. +That is, we can find coprime a and b (b  0) such that 5 +3 +a +b + + + +Therefore, 5 +3 +a +b + + + +Rearranging this equation, we get +5 +3 +5 – a +b +a +b +b + + + + +Since a and b are integers, we get 5 – a +b + is rational, and so 3 is rational. +But this contradicts the fact that +3 is irrational. +This contradiction has arisen because of our incorrect assumption that 5 – +3 is +rational. +So, we conclude that 5 +3 + + is irrational. +Example 7 : Show that 3 2 is irrational. +Solution : Let us assume, to the contrary, that 3 2 is rational. +That is, we can find coprime a and b (b  0) such that 3 2 +a +b + + +Rearranging, we get +2 +3 +a +b + + +Since 3, a and b are integers, 3 +a +b is rational, and so +2 is rational. +Reprint 2025-26 + +REAL NUMBERS +9 +But this contradicts the fact that +2 is irrational. +So, we conclude that 3 2 is irrational. +EXERCISE 1.2 +1. Prove that +5 is irrational. +2. Prove that 3 +2 5 + + is irrational. +3. Prove that the following are irrationals : +(i) +1 +2 +(ii) 7 5 +(iii) 6 +2 + +1.4 Summary +In this chapter, you have studied the following points: +1. The Fundamental Theorem of Arithmetic : +Every composite number can be expressed (factorised) as a product of primes, and this +factorisation is unique, apart from the order in which the prime factors occur. +2. If p is a prime and p divides a2, then p divides a, where a is a positive integer. +3. To prove that +2, +3 are irrationals. +A NOTE TO THE READER +You have seen that : +HCF ( p, q, r) × LCM (p, q, r)  p × q × r, where p, q, r are positive integers +(see Example 8). However, the following results hold good for three numbers +p, q and r : +LCM (p, q, r) = +HCF( , , ) +HCF( , +) HCF( , ) HCF( , ) +p q r +p q r +p q +q r +p r + + + +HCF (p, q, r) = +LCM( , , ) +LCM( , ) LCM( , +) LCM( , ) +p q r +p q r +p q +q r +p r + + + +Reprint 2025-26" +class_10,2,Polynomials,ncert_books/class_10/jemh1dd/jemh102.pdf,"10 +MATHEMATICS +2 +2.1 Introduction +In Class IX, you have studied polynomials in one variable and their degrees. Recall +that if p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of +the polynomial p(x). For example, 4x + 2 is a polynomial in the variable x of +degree 1, 2y2 – 3y + 4 is a polynomial in the variable y of degree 2, 5x3 – 4x2 + x – +2 +is a polynomial in the variable x of degree 3 and 7u6 – +4 +2 +3 +4 +8 +2u +u +u + + + + is a polynomial +in the variable u of degree 6. Expressions like +1 +1 +x  +, +2 +x  +, +2 +1 +2 +3 +x +x + + + etc., are +not polynomials. +A polynomial of degree 1 is called a linear polynomial. For example, 2x – 3, +3 +5, +x  + +2 +y +, +2 +11 +x  +, 3z + 4, 2 +1 +3u , etc., are all linear polynomials. Polynomials +such as 2x + 5 – x2, x3 + 1, etc., are not linear polynomials. +A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’ +has been derived from the word ‘quadrate’, which means ‘square’. +2 +2, +2 +3 +5 +x +x + + +y2 – 2, +2 +2 +3 , +x +x + + + +2 +2 +2 +2 +1 +2 +5, +5 +, 4 +3 +3 +7 +u +u +v +v +z + + + + + are some examples of +quadratic polynomials (whose coefficients are real numbers). More generally, any +quadratic polynomial in x is of the form ax2 + bx + c, where a, b, c are real numbers +and a  0. A polynomial of degree 3 is called a cubic polynomial. Some examples of +POLYNOMIALS +Reprint 2025-26 + +POLYNOMIALS +11 +a cubic polynomial are 2 – x3, x3, +3 +2 +, +x + 3 – x2 + x3, 3x3 – 2x2 + x – 1. In fact, the most +general form of a cubic polynomial is +ax3 + bx2 + cx + d, +where, a, b, c, d are real numbers and a  0. +Now consider the polynomial p(x) = x2 – 3x – 4. Then, putting x = 2 in the +polynomial, we get p(2) = 22 – 3 × 2 – 4 = – 6. The value ‘– 6’, obtained by replacing +x by 2 in x2 – 3x – 4, is the value of x2 – 3x – 4 at x = 2. Similarly, p(0) is the value of +p(x) at x = 0, which is – 4. +If p(x) is a polynomial in x, and if k is any real number, then the value obtained by +replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k). +What is the value of p(x) = x2 –3x – 4 at x = –1? We have : +p(–1) = (–1)2 –{3 × (–1)} – 4 = 0 +Also, note that +p(4) = 42 – (3  4) – 4 = 0. +As p(–1) = 0 and p(4) = 0, –1 and 4 are called the zeroes of the quadratic +polynomial x2 – 3x – 4. More generally, a real number k is said to be a zero of a +polynomial p(x), if p(k) = 0. +You have already studied in Class IX, how to find the zeroes of a linear +polynomial. For example, if k is a zero of p(x) = 2x + 3, then p(k) = 0 gives us +2k + 3 = 0, i.e., k = +3 +2 + + +In general, if k is a zero of p(x) = ax + b, then p(k) = ak + b = 0, i.e., +b +k +a + + + +So, the zero of the linear polynomial ax + b is +(Constant term) +Coefficient of +b +a +x + + + +. +Thus, the zero of a linear polynomial is related to its coefficients. Does this +happen in the case of other polynomials too? For example, are the zeroes of a quadratic +polynomial also related to its coefficients? +In this chapter, we will try to answer these questions. We will also study the +division algorithm for polynomials. +2.2 Geometrical Meaning of the Zeroes of a Polynomial +You know that a real number k is a zero of the polynomial p(x) if p(k) = 0. But why +are the zeroes of a polynomial so important? To answer this, first we will see the +geometrical representations of linear and quadratic polynomials and the geometrical +meaning of their zeroes. +Reprint 2025-26 + +12 +MATHEMATICS +Consider first a linear polynomial ax + b, a  0. You have studied in Class IX that the +graph of y = ax + b is a straight line. For example, the graph of y = 2x + 3 is a straight +line passing through the points (– 2, –1) and (2, 7). +x +–2 +2 +y = 2x + 3 +–1 +7 +From Fig. 2.1, you can see +that the graph of y = 2x + 3 +intersects the x-axis mid-way +between x = –1 and x = – 2, +that is, at the point +3, 0 +2 + + + + + + + +. +You also know that the zero of +2x + 3 is +3 +2 + +. Thus, the zero of +the polynomial 2x + 3 is the +x-coordinate of the point where the +graph of y = 2x + 3 intersects the +x-axis. +In general, for a linear polynomial ax + b, a  0, the graph of y = ax + b is a +straight line which intersects the x-axis at exactly one point, namely, +, 0 +b +a + + + + + + + +. +Therefore, the linear polynomial ax + b, a  0, has exactly one zero, namely, the +x-coordinate of the point where the graph of y = ax + b intersects the x-axis. +Now, let us look for the geometrical meaning of a zero of a quadratic polynomial. +Consider the quadratic polynomial x2 – 3x – 4. Let us see what the graph* of +y = x2 – 3x – 4 looks like. Let us list a few values of y = x2 – 3x – 4 corresponding to +a few values for x as given in Table 2.1. +* Plotting of graphs of quadratic or cubic polynomials is not meant to be done by the students, +nor is to be evaluated. +Fig. 2.1 +Reprint 2025-26 + +POLYNOMIALS +13 +Table 2.1 +x +– 2 +–1 +0 +1 +2 +3 +4 +5 +y = x2 – 3x – 4 +6 +0 +– 4 +– 6 +– 6 +– 4 +0 +6 +If we locate the points listed +above on a graph paper and draw +the graph, it will actually look like +the one given in Fig. 2.2. +In fact, for any quadratic +polynomial ax2 + bx + c, a  0, the +graph of the corresponding +equation y = ax2 + bx + c has one +of the two shapes either open +upwards like + or open +downwards like + depending on +whether a > 0 or a < 0. (These +curves are called parabolas.) +You can see from Table 2.1 +that –1 and 4 are zeroes of the +quadratic polynomial. Also +note from Fig. 2.2 that –1 and 4 +are the x-coordinates of the points +where the graph of y = x2 – 3x – 4 +intersects the x-axis. Thus, the +zeroes of the quadratic polynomial +x2 – 3x – 4 are x-coordinates of +the points where the graph of +y = x2 – 3x – 4 intersects the +x-axis. +This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic +polynomial ax2 + bx + c, a  0, are precisely the x-coordinates of the points where the +parabola representing y = ax2 + bx + c intersects the x-axis. +From our observation earlier about the shape of the graph of y = ax2 + bx + c, the +following three cases can happen: +Fig. 2.2 +Reprint 2025-26 + +14 +MATHEMATICS +Case (i) : Here, the graph cuts x-axis at two distinct points A and A. +The x-coordinates of A and A are the two zeroes of the quadratic polynomial +ax2 + bx + c in this case (see Fig. 2.3). +Fig. 2.3 +Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e., at two coincident +points. So, the two points A and A of Case (i) coincide here to become one point A +(see Fig. 2.4). +Fig. 2.4 +The x-coordinate of A is the only zero for the quadratic polynomial ax2 + bx + c +in this case. +Reprint 2025-26 + +POLYNOMIALS +15 +Case (iii) : Here, the graph is either completely above the x-axis or completely below +the x-axis. So, it does not cut the x-axis at any point (see Fig. 2.5). +Fig. 2.5 +So, the quadratic polynomial ax2 + bx + c has no zero in this case. +So, you can see geometrically that a quadratic polynomial can have either two +distinct zeroes or two equal zeroes (i.e., one zero), or no zero. This also means that a +polynomial of degree 2 has atmost two zeroes. +Now, what do you expect the geometrical meaning of the zeroes of a cubic +polynomial to be? Let us find out. Consider the cubic polynomial x3 – 4x. To see what +the graph of y = x3 – 4x looks like, let us list a few values of y corresponding to a few +values for x as shown in Table 2.2. +Table 2.2 +x +–2 +–1 +0 +1 +2 +y = x3 – 4x +0 +3 +0 +–3 +0 +Locating the points of the table on a graph paper and drawing the graph, we see +that the graph of y = x3 – 4x actually looks like the one given in Fig. 2.6. +Reprint 2025-26 + +16 +MATHEMATICS +We see from the table above +that – 2, 0 and 2 are zeroes of the +cubic polynomial x3 – 4x. Observe +that – 2, 0 and 2 are, in fact, the +x-coordinates of the only points +where the graph of y = x3 – 4x +intersects the x-axis. Since the curve +meets the x-axis in only these 3 +points, their x-coordinates are the +only zeroes of the polynomial. +Let us take a few more +examples. Consider the cubic +polynomials x3 and x3 – x2. We draw +the graphs of y = x3 and y = x3 – x2 +in Fig. 2.7 and Fig. 2.8 respectively. +Fig. 2.7 +Fig. 2.8 +Fig. 2.6 +Reprint 2025-26 + +POLYNOMIALS +17 +Note that 0 is the only zero of the polynomial x3. Also, from Fig. 2.7, you can see +that 0 is the x-coordinate of the only point where the graph of y = x3 intersects the +x-axis. Similarly, since x3 – x2 = x2 (x – 1), 0 and 1 are the only zeroes of the polynomial +x3 – x2. Also, from Fig. 2.8, these values are the x-coordinates of the only points +where the graph of y = x3 – x2 intersects the x-axis. +From the examples above, we see that there are at most 3 zeroes for any cubic +polynomial. In other words, any polynomial of degree 3 can have at most three zeroes. +Remark : In general, given a polynomial p(x) of degree n, the graph of y = p(x) +intersects the x-axis at atmost n points. Therefore, a polynomial p(x) of degree n has +at most n zeroes. +Example 1 : Look at the graphs in Fig. 2.9 given below. Each is the graph of y = p(x), +where p(x) is a polynomial. For each of the graphs, find the number of zeroes of p(x). +Fig. 2.9 +Solution : +(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only. +(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points. +(iii) The number of zeroes is 3. (Why?) +Reprint 2025-26 + +18 +MATHEMATICS +(iv) The number of zeroes is 1. (Why?) +(v) The number of zeroes is 1. (Why?) +(vi) The number of zeroes is 4. (Why?) +EXERCISE 2.1 +1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the +number of zeroes of p(x), in each case. +Fig. 2.10 +2.3 Relationship between Zeroes and Coefficients of a Polynomial +You have already seen that zero of a linear polynomial ax + b is +b +a + +. We will now try +to answer the question raised in Section 2.1 regarding the relationship between zeroes +and coefficients of a quadratic polynomial. For this, let us take a quadratic polynomial, +say p(x) = 2x2 – 8x + 6. In Class IX, you have learnt how to factorise quadratic +polynomials by splitting the middle term. So, here we need to split the middle term +‘– 8x’ as a sum of two terms, whose product is 6 × 2x2 = 12x2. So, we write +2x2 – 8x + 6 = 2x2 – 6x – 2x + 6 = 2x(x – 3) – 2(x – 3) += (2x – 2)(x – 3) = 2(x – 1)(x – 3) +Reprint 2025-26 + +POLYNOMIALS +19 +So, the value of p(x) = 2x2 – 8x + 6 is zero when x – 1 = 0 or x – 3 = 0, i.e., when +x = 1 or x = 3. So, the zeroes of 2x2 – 8x + 6 are 1 and 3. Observe that : +Sum of its zeroes += +2 +( 8) +(Coefficient of +) +1 +3 +4 +2 +Coefficient of +x +x + + + + + + +Product of its zeroes = +2 +6 +Constant term +1 +3 +3 +2 +Coefficient of x + + + + +Let us take one more quadratic polynomial, say, p(x) = 3x2 + 5x – 2. By the +method of splitting the middle term, +3x2 + 5x – 2 = 3x2 + 6x – x – 2 = 3x(x + 2) –1(x + 2) += (3x – 1)(x + 2) +Hence, the value of 3x2 + 5x – 2 is zero when either 3x – 1 = 0 or x + 2 = 0, i.e., +when x = 1 +3 or x = –2. So, the zeroes of 3x2 + 5x – 2 are 1 +3 and – 2. Observe that : +Sum of its zeroes += +2 +1 +5 +(Coefficient of +) +( 2) +3 +3 +Coefficient of +x +x + + + + + +Product of its zeroes = +2 +1 +2 +Constant term +( 2) +3 +3 +Coefficient of x + + + + +In general, if * and * are the zeroes of the quadratic polynomial p(x) = ax2 + bx + c, +a  0, then you know that x –  and x –  are the factors of p(x). Therefore, +ax2 + bx + c = k(x – ) (x – ), where k is a constant += k[x2 – ( + )x + ] += kx2 – k( + )x + k  +Comparing the coefficients of x2, x and constant terms on both the sides, we get +a = k, b = – k( + ) and c = k +This gives + +  = –b +a , +  + + + + = c +a +* ,are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively. We will use later one +more letter ‘’ pronounced as ‘gamma’. +Reprint 2025-26 + +20 +MATHEMATICS +i.e., +sum of zeroes =  +  = +2 +(Coefficient of +) +Coefficient of +b +x +a +x + + + +, +product of zeroes =  = +2 +Constant term +Coefficient of +c +a +x + +. +Let us consider some examples. +Example 2 : Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the +relationship between the zeroes and the coefficients. +Solution : We have +x2 + 7x + 10 = (x + 2)(x + 5) +So, the value of x2 + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0, i.e., when x = – 2 or +x = –5. Therefore, the zeroes of x2 + 7x + 10 are – 2 and – 5. Now, +sum of zeroes = +2 +(7) +–(Coefficient of +) , +–2 +(–5) +– (7) +1 +Coefficient of +x +x + + + + + +product of zeroes = +2 +10 +Constant term +( 2) +( 5) +10 +1 +Coefficient of x + + + + + + +Example 3 : Find the zeroes of the polynomial x2 – 3 and verify the relationship +between the zeroes and the coefficients. +Solution : Recall the identity a2 – b2 = (a – b)(a + b). Using it, we can write: +x2 – 3 =  + + +3 +3 +x +x + + +So, the value of x2 – 3 is zero when x = +3 or x = – +3 +Therefore, the zeroes of x2 – 3 are +3 and +3 + + +Now, +sum of zeroes = +2 +(Coefficient of +) , +3 +3 +0 +Coefficient of +x +x + + + + +product of zeroes =  + + +2 +3 +Constant term +3 +3 +– 3 +1 +Coefficient of x + + + + + + +Reprint 2025-26 + +POLYNOMIALS +21 +Example 4 : Find a quadratic polynomial, the sum and product of whose zeroes are +– 3 and 2, respectively. +Solution : Let the quadratic polynomial be ax2 + bx + c, and its zeroes be  and . +We have + +  = – 3 = b +a +, +and + = 2 = c +a +. +If a = 1, then b = 3 and c = 2. +So, one quadratic polynomial which fits the given conditions is x2 + 3x + 2. +You can check that any other quadratic polynomial that fits these conditions will +be of the form k(x2 + 3x + 2), where k is real. +Let us now look at cubic polynomials. Do you think a similar relation holds +between the zeroes of a cubic polynomial and its coefficients? +Let us consider p(x) = 2x3 – 5x2 – 14x + 8. +You can check that p(x) = 0 for x = 4, – 2, 1 +2  Since p(x) can have atmost three +zeroes, these are the zeores of 2x3 – 5x2 – 14x + 8. Now, +sum of the zeroes = +2 +3 +1 +5 +( 5) +(Coefficient of +) +4 +( 2) +2 +2 +2 +Coefficient of +x +x + + + + + + + +, +product of the zeroes = +3 +1 +8 +– Constant term +4 +( 2) +4 +2 +2 +Coefficient of x + + + + + + +. +However, there is one more relationship here. Consider the sum of the products +of the zeroes taken two at a time. We have + + +1 +1 +4 +( 2) +( 2) +4 +2 +2 + + + + + + + + + + + + + + + + + + += +14 +– 8 +1 +2 +7 +2 + + + + + = +3 +Coefficient of +Coefficient of +x +x . +In general, it can be proved that if , ,  are the zeroes of the cubic polynomial +ax3 + bx2 + cx + d, then +Reprint 2025-26 + +22 +MATHEMATICS + +  +  = –b +a , + + + + + +  + + + + +  + + + + = c +a , + + + + + = –d +a . +Let us consider an example. +Example 5* : Verify that 3, –1, +1 +3 + + are the zeroes of the cubic polynomial +p(x) = 3x3 – 5x2 – 11x – 3, and then verify the relationship between the zeroes and the +coefficients. +Solution : Comparing the given polynomial with ax3 + bx2 + cx + d, we get +a = 3, b = – 5, c = –11, d = – 3. Further +p(3) = 3 × 33 – (5 × 32) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0, +p(–1) = 3 × (–1)3 – 5 × (–1)2 – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0, +3 +2 +1 +1 +1 +1 +3 +5 +11 +3 +3 +3 +3 +3 +p + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +, + = +1 +5 +11 +2 +2 +– +3 +– +0 +9 +9 +3 +3 +3 + + + + + + +Therefore, 3, –1 and +1 +3 + + are the zeroes of 3x3 – 5x2 – 11x – 3. +So, we take  = 3,  = –1 and  = +1 +3 + +Now, +1 +1 +5 +( 5) +3 +( 1) +2 +3 +3 +3 +3 +b +a + + + + + + + + + + + + + + + + +, +1 +1 +1 +11 +3 +( 1) +( 1) +3 +3 +1 +3 +3 +3 +3 +c +a + + + + + + + + + + + + + + + + + + + + + + + + +, +1 +( 3) +3 +( 1) +1 +3 +3 +d +a + + + + + + + + + + + + + + +. +* Not from the examination point of view. +Reprint 2025-26 + +POLYNOMIALS +23 +EXERCISE 2.2 +1. Find the zeroes of the following quadratic polynomials and verify the relationship between +the zeroes and the coefficients. +(i) x2 – 2x – 8 +(ii) 4s2 – 4s + 1 +(iii) 6x2 – 3 – 7x +(iv) 4u2 + 8u +(v) t2 – 15 +(vi) 3x2 – x – 4 +2. Find a quadratic polynomial each with the given numbers as the sum and product of its +zeroes respectively. +(i) +1 , +1 +4  +(ii) +1 +2 , 3 +(iii) 0, +5 +(iv) 1, 1 +(v) +1 +1 +, +4 +4 + +(vi) 4, 1 +2.4 Summary +In this chapter, you have studied the following points: +1. Polynomials of degrees 1, 2 and 3 are called linear, quadratic and cubic polynomials +respectively. +2. A quadratic polynomial in x with real coefficients is of the form ax2 + bx + c, where a, b, c +are real numbers with a  0. +3. The zeroes of a polynomial p(x) are precisely the x-coordinates of the points, where the +graph of y = p(x) intersects the x-axis. +4. A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can have +at most 3 zeroes. +5. If  and  are the zeroes of the quadratic polynomial ax2 + bx + c, then +b +a + +, +c +a + +. +6. If , ,  are the zeroes of the cubic polynomial ax3 + bx2 + cx + d, then +b +a + + +, +c +a + +, +and +d +a + + +. +Reprint 2025-26" +class_10,3,Pair of Linear Equations in Two Variables,ncert_books/class_10/jemh1dd/jemh103.pdf,"24 +MATHEMATICS +3 +3.1 Introduction +You must have come across situations like the one given below : +Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel +and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if +the ring covers any object completely, you get it). The number of times she played +Hoopla is half the number of rides she had on the Giant Wheel. If each ride costs +` 3, and a game of Hoopla costs ` 4, how would you find out the number of rides she +had and how many times she played Hoopla, provided she spent ` 20. +May be you will try it by considering different cases. If she has one ride, is it +possible? Is it possible to have two rides? And so on. Or you may use the knowledge +of Class IX, to represent such situations as linear equations in two variables. +PAIR OF LINEAR EQUATIONS +IN TWO VARIABLES +Reprint 2025-26 + +PAIR OF LINEAR EQUATIONS IN TWO VARIABLES +25 +Let us try this approach. +Denote the number of rides that Akhila had by x, and the number of times she +played Hoopla by y. Now the situation can be represented by the two equations: +y = 1 +2 x +(1) +3x + 4y = 20 +(2) +Can we find the solutions of this pair of equations? There are several ways of +finding these, which we will study in this chapter. +3.2 Graphical Method of Solution of a Pair of Linear Equations +A pair of linear equations which has no solution, is called an inconsistent pair of linear +equations. A pair of linear equations in two variables, which has a solution, is called a +consistent pair of linear equations. A pair of linear equations which are equivalent has +infinitely many distinct common solutions. Such a pair is called a dependent pair of +linear equations in two variables. Note that a dependent pair of linear equations is +always consistent. +We can now summarise the behaviour of lines representing a pair of linear equations +in two variables and the existence of solutions as follows: +(i) the lines may intersect in a single point. In this case, the pair of equations +has a unique solution (consistent pair of equations). +(ii) the lines may be parallel. In this case, the equations have no solution +(inconsistent pair of equations). +(iii) the lines may be coincident. In this case, the equations have infinitely many +solutions [dependent (consistent) pair of equations]. +Consider the following three pairs of equations. +(i) x – 2y = 0 and 3x + 4y – 20 = 0 +(The lines intersect) +(ii) 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0 +(The lines coincide) +(iii) x + 2y – 4 = 0 and 2x + 4y – 12 = 0 +(The lines are parallel) +Let us now write down, and compare, the values of +1 +1 +1 +2 +2 +2 +, +and +a +b +c +c +a +b + in all the +three examples. Here, a1, b1, c1 and a2, b2, c2 denote the coefficents of equations +given in the general form in Section 3.2. +Reprint 2025-26 + +26 +MATHEMATICS +Table 3.1 +Sl +Pair of lines +1 +2 +a +a +1 +2 +b +b +1 +2 +c +c +Compare the +Graphical +Algebraic +No. +ratios +representation interpretation +1. +x – 2y = 0 +1 +3 +2 +4 +− +0 +20 +− +1 +1 +2 +2 +a +b +a +b +≠ +Intersecting +Exactly one +3x + 4y – 20 = 0 +lines +solution +(unique) +2. +2x + 3y – 9 = 0 +2 +4 +3 +6 +9 +18 +− +− +1 +1 +1 +2 +2 +2 +a +b +c +a +b +c += += +Coincident +Infinitely +4x + 6y – 18 = 0 +lines +many solutions +3. +x + 2y – 4 = 0 +1 +2 +2 +4 +4 +12 +− +− +1 +1 +1 +2 +2 +2 +a +b +c +a +b +c += +≠ +Parallel lines +No solution +2x + 4y – 12 = 0 +From the table above, you can observe that if the lines represented by the equation + a1x + b1y + c1 = 0 +and +a2x + b2y + c2 = 0 +are +(i) intersecting, then +1 +1 +2 +2 +a +b +a +b +≠ +⋅ +(ii) coincident, then +1 +1 +1 +2 +2 +2 +a +b +c +a +b +c += += +⋅ +(iii) parallel, then +1 +1 +1 +2 +2 +2 +a +b +c +a +b +c += +≠ +⋅ +In fact, the converse is also true for any pair of lines. You can verify them by +considering some more examples by yourself. +Let us now consider some more examples to illustrate it. +Example 1 : Check graphically whether the pair of equations +x + 3y = 6 +(1) +and +2x – 3y = 12 +(2) +is consistent. If so, solve them graphically. +Solution : Let us draw the graphs of the Equations (1) and (2). For this, we find two +solutions of each of the equations, which are given in Table 3.2 +Reprint 2025-26 + +PAIR OF LINEAR EQUATIONS IN TWO VARIABLES +27 +Fig. 3.1 +Table 3.2 +x +0 +6 +x +0 +3 +y = 6 +3 +x +− +2 +0 +y = 2 +12 +3 +x − +– 4 +–2 +Plot the points A(0, 2), B(6, 0), +P(0, – 4) and Q(3, – 2) on graph +paper, and join the points to form the +lines AB and PQ as shown in +Fig. 3.1. +We observe that there is a point +B (6, 0) common to both the lines +AB and PQ. So, the solution of the +pair of linear equations is x = 6 and +y = 0, i.e., the given pair of equations +is consistent. +Example 2 : Graphically, find whether the following pair of equations has no solution, +unique solution or infinitely many solutions: +5x – 8y + 1 = 0 +(1) +3x – 24 +5 y + 3 +5 = 0 +(2) +Solution : Multiplying Equation (2) by 5 , +3 we get +5x – 8y + 1 = 0 +But, this is the same as Equation (1). Hence the lines represented by Equations (1) +and (2) are coincident. Therefore, Equations (1) and (2) have infinitely many solutions. +Plot few points on the graph and verify it yourself. +Example 3 : Champa went to a ‘Sale’ to purchase some pants and skirts. When her +friends asked her how many of each she had bought, she answered, “The number of +skirts is two less than twice the number of pants purchased. Also, the number of skirts +is four less than four times the number of pants purchased”. Help her friends to find +how many pants and skirts Champa bought. +Reprint 2025-26 + +28 +MATHEMATICS +Solution : Let us denote the number of pants by x and the number of skirts by y. Then +the equations formed are : +y = 2x – 2 +(1) + and +y = 4x – 4 +(2) +Let us draw the graphs of +Equations (1) and (2) by finding two +solutions for each of the equations. +They are given in Table 3.3. +Table 3.3 +x +2 +0 +y = 2x – 2 +2 +– 2 +x +0 +1 +y = 4x – 4 +– 4 +0 +Plot the points and draw the lines passing through them to represent the equations, +as shown in Fig. 3.2. +The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution +of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did +not buy any skirt. +Verify the answer by checking whether it satisfies the conditions of the given +problem. +EXERCISE 3.1 +1. Form the pair of linear equations in the following problems, and find their solutions +graphically. +(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 +more than the number of boys, find the number of boys and girls who took part in +the quiz. +Fig. 3.2 +Reprint 2025-26 + +PAIR OF LINEAR EQUATIONS IN TWO VARIABLES +29 +(ii) 5 pencils and 7 pens together cost ` 50, whereas 7 pencils and 5 pens together +cost ` 46. Find the cost of one pencil and that of one pen. +2. On comparing the ratios +1 +1 +1 +2 +2 +2 +, +and +a +b +c +a +b +c , find out whether the lines representing the +following pairs of linear equations intersect at a point, are parallel or coincident: +(i) 5x – 4y + 8 = 0 +(ii) 9x + 3y + 12 = 0 +7x + 6y – 9 = 0 +18x + 6y + 24 = 0 +(iii) 6x – 3y + 10 = 0 +2x – y + 9 = 0 +3. On comparing the ratios +1 +1 +2 +2 +, +a +b +a +b + and +1 +2 +c +c , find out whether the following pair of linear +equations are consistent, or inconsistent. +(i) 3x + 2y = 5 ; +2x – 3y = 7 +(ii) 2x – 3y = 8 ; +4x – 6y = 9 +(iii) +3 +5 +7 +2 +3 +x +y + + +; 9x – 10y = 14 +(iv) 5x – 3y = 11 ; – 10x + 6y = –22 +(v) +4 +2 +8 +3 x +y + + + ; 2x + 3y = 12 +4. Which of the following pairs of linear equations are consistent/inconsistent? If +consistent, obtain the solution graphically: +(i) x + y = 5, +2x + 2y = 10 +(ii) x – y = 8, +3x – 3y = 16 +(iii) 2x + y – 6 = 0, +4x – 2y – 4 = 0 +(iv) 2x – 2y – 2 = 0, +4x – 4y – 5 = 0 +5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is +36 m. Find the dimensions of the garden. +6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables +such that the geometrical representation of the pair so formed is: +(i) intersecting lines +(ii) parallel lines +(iii) coincident lines +7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the +coordinates of the vertices of the triangle formed by these lines and the x-axis, and +shade the triangular region. +Reprint 2025-26 + +30 +MATHEMATICS +3.3 Algebraic Methods of Solving a Pair of Linear Equations +In the previous section, we discussed how to solve a pair of linear equations graphically. +The graphical method is not convenient in cases when the point representing the +solution of the linear equations has non-integral coordinates like ( +) +3, 2 7 , +(–1.75, 3.3), +4 +1 +, +13 19 + + + + + +, etc. There is every possibility of making mistakes while reading +such coordinates. Is there any alternative method of finding the solution? There are +several algebraic methods, which we shall now discuss. +3.3.1 Substitution Method : We shall explain the method of substitution by taking +some examples. +Example 4 : Solve the following pair of equations by substitution method: +7x – 15y = 2 +(1) +x + 2y = 3 +(2) +Solution : +Step 1 : We pick either of the equations and write one variable in terms of the other. +Let us consider the Equation (2) : +x + 2y = 3 +and write it as +x = 3 – 2y +(3) +Step 2 : Substitute the value of x in Equation (1). We get +7(3 – 2y) – 15y = 2 +i.e., +21 – 14y – 15y = 2 +i.e., +– 29y = –19 +Therefore, +y = 19 +29 +Step 3 : Substituting this value of y in Equation (3), we get +x = 3 – +19 +2 29 + + + + + + + = 49 +29 +Therefore, the solution is x = 49 +29 , y = 19 +29 . +Reprint 2025-26 + +PAIR OF LINEAR EQUATIONS IN TWO VARIABLES +31 +Verification : Substituting x = 49 +29 and y = 19 +29 , you can verify that both the Equations +(1) and (2) are satisfied. +To understand the substitution method more clearly, let us consider it stepwise: +Step 1 : Find the value of one variable, say y in terms of the other variable, i.e., x from +either equation, whichever is convenient. +Step 2 : Substitute this value of y in the other equation, and reduce it to an equation in +one variable, i.e., in terms of x, which can be solved. Sometimes, as in Examples 9 and +10 below, you can get statements with no variable. If this statement is true, you can +conclude that the pair of linear equations has infinitely many solutions. If the statement +is false, then the pair of linear equations is inconsistent. +Step 3 : Substitute the value of x (or y) obtained in Step 2 in the equation used in +Step 1 to obtain the value of the other variable. +Remark : We have substituted the value of one variable by expressing it in terms of +the other variable to solve the pair of linear equations. That is why the method is +known as the substitution method. +Example 5 : Solve the following question—Aftab tells his daughter, “Seven years +ago, I was seven times as old as you were then. Also, three years from now, I shall be +three times as old as you will be.” (Isn’t this interesting?) Represent this situation +algebraically and graphically by the method of substitution. +Solution : Let s and t be the ages (in years) of Aftab and his daughter, respectively. +Then, the pair of linear equations that represent the situation is +s – 7 = 7 (t – 7), i.e., s – 7t + 42 = 0 +(1) +and +s + 3 = 3 (t + 3), i.e., s – 3t = 6 +(2) +Using Equation (2), we get s = 3t + 6. +Putting this value of s in Equation (1), we get +(3t + 6) – 7t + 42 = 0, +i.e., +4t = 48, which gives t = 12. +Putting this value of t in Equation (2), we get +s = 3 (12) + 6 = 42 +Reprint 2025-26 + +32 +MATHEMATICS +So, Aftab and his daughter are 42 and 12 years old, respectively. +Verify this answer by checking if it satisfies the conditions of the given problems. +Example 6 : In a shop the cost of 2 pencils and 3 erasers is `9 and the cost of 4 +pencils and 6 erasers is `18. Find the cost of each pencil and each eraser. +Solution : The pair of linear equations formed were: +2x + 3y = 9 +(1) +4x + 6y = 18 +(2) +We first express the value of x in terms of y from the equation 2x + 3y = 9, to get +x = 9 +3 +2 +y +− +(3) +Now we substitute this value of x in Equation (2), to get +4(9 +3 ) +2 +y +− + + 6y = 18 +i.e., +18 – 6y + 6y = 18 +i.e., +18 = 18 +This statement is true for all values of y. However, we do not get a specific +value of y as a solution. Therefore, we cannot obtain a specific value of x. This +situation has arisen because both the given equations are the same. Therefore, +Equations (1) and (2) have infinitely many solutions. We cannot find a unique +cost of a pencil and an eraser, because there are many common solutions, to the +given situation. +Example 7 : Two rails are represented by the equations +x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the rails cross each other? +Solution : The pair of linear equations formed were: +x + 2y – 4 = 0 +(1) +2x + 4y – 12 = 0 +(2) +We express x in terms of y from Equation (1) to get +x = 4 – 2y +Now, we substitute this value of x in Equation (2) to get +2(4 – 2y) + 4y – 12 = 0 +Reprint 2025-26 + +PAIR OF LINEAR EQUATIONS IN TWO VARIABLES +33 +i.e., +8 – 12 = 0 +i.e., +– 4 = 0 +which is a false statement. +Therefore, the equations do not have a common solution. So, the two rails will not +cross each other. +EXERCISE 3.2 +1. Solve the following pair of linear equations by the substitution method. +(i) x + y = 14 +(ii) s – t = 3 +x – y = 4 +6 +3 +2 +s +t + + +(iii) 3x – y = 3 +(iv) 0.2x + 0.3y = 1.3 +9x – 3y = 9 +0.4x + 0.5y = 2.3 +(v) +2 +3 +0 +x +y + + +(vi) 3 +5 +2 +2 +3 +x +y + + +3 +8 +0 +x +y + + +13 +3 +2 +6 +x +y + + +2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which +y = mx + 3. +3. Form the pair of linear equations for the following problems and find their solution by +substitution method. +(i) The difference between two numbers is 26 and one number is three times the other. +Find them. +(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find +them. +(iii) The coach of a cricket team buys 7 bats and 6 balls for ` 3800. Later, she buys 3 +bats and 5 balls for ` 1750. Find the cost of each bat and each ball. +(iv) The taxi charges in a city consist of a fixed charge together with the charge for the +distance covered. For a distance of 10 km, the charge paid is ` 105 and for a +journey of 15 km, the charge paid is ` 155. What are the fixed charges and the +charge per km? How much does a person have to pay for travelling a distance of +25 km? +(v) A fraction becomes 9 +11 +, if 2 is added to both the numerator and the denominator. +If, 3 is added to both the numerator and the denominator it becomes 5 +6 +. Find the +fraction. +Reprint 2025-26 + +34 +MATHEMATICS +(vi) Five years hence, the age of Jacob will be three times that of his son. Five years +ago, Jacob’s age was seven times that of his son. What are their present ages? +3.3.2 Elimination Method +Now let us consider another method of eliminating (i.e., removing) one variable. This +is sometimes more convenient than the substitution method. Let us see how this method +works. +Example 8 : The ratio of incomes of two persons is 9 : 7 and the ratio of their +expenditures is 4 : 3. If each of them manages to save ` 2000 per month, find their +monthly incomes. +Solution : Let us denote the incomes of the two person by ` 9x and ` 7x and their +expenditures by ` 4y and ` 3y respectively. Then the equations formed in the situation +is given by : +9x – 4y = 2000 +(1) +and +7x – 3y = 2000 +(2) +Step 1 : Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of +y equal. Then we get the equations: +27x – 12y = 6000 +(3) +28x – 12y = 8000 +(4) +Step 2 : Subtract Equation (3) from Equation (4) to eliminate y, because the coefficients +of y are the same. So, we get +(28x – 27x) – (12y – 12y) = 8000 – 6000 +i.e., +x = 2000 +Step 3 : Substituting this value of x in (1), we get +9(2000) – 4y = 2000 +i.e., +y = 4000 +So, the solution of the equations is x = 2000, y = 4000. Therefore, the monthly incomes +of the persons are ` 18,000 and ` 14,000, respectively. +Verification : 18000 : 14000 = 9 : 7. Also, the ratio of their expenditures = +18000 – 2000 : 14000 – 2000 = 16000 : 12000 = 4 : 3 +Remarks : +1. The method used in solving the example above is called the elimination method, +because we eliminate one variable first, to get a linear equation in one variable. +Reprint 2025-26 + +PAIR OF LINEAR EQUATIONS IN TWO VARIABLES +35 +In the example above, we eliminated y. We could also have eliminated x. Try +doing it that way. +2. You could also have used the substitution, or graphical method, to solve this +problem. Try doing so, and see which method is more convenient. +Let us now note down these steps in the elimination method : +Step 1 : First multiply both the equations by some suitable non-zero constants to make +the coefficients of one variable (either x or y) numerically equal. +Step 2 : Then add or subtract one equation from the other so that one variable gets +eliminated. If you get an equation in one variable, go to Step 3. +If in Step 2, we obtain a true statement involving no variable, then the original +pair of equations has infinitely many solutions. +If in Step 2, we obtain a false statement involving no variable, then the original +pair of equations has no solution, i.e., it is inconsistent. +Step 3 : Solve the equation in one variable (x or y) so obtained to get its value. +Step 4 : Substitute this value of x (or y) in either of the original equations to get the +value of the other variable. +Now to illustrate it, we shall solve few more examples. +Example 9 : Use elimination method to find all possible solutions of the following pair +of linear equations : +2x + 3y = 8 +(1) +4x + 6y = 7 +(2) +Solution : +Step 1 : Multiply Equation (1) by 2 and Equation (2) by 1 to make the +coefficients of x equal. Then we get the equations as : +4x + 6y = 16 +(3) +4x + 6y = 7 +(4) +Step 2 : Subtracting Equation (4) from Equation (3), +(4x – 4x) + (6y – 6y) = 16 – 7 +i.e., +0 = 9, which is a false statement. +Therefore, the pair of equations has no solution. +Example 10 : The sum of a two-digit number and the number obtained by reversing +the digits is 66. If the digits of the number differ by 2, find the number. How many such +numbers are there? +Reprint 2025-26 + +36 +MATHEMATICS +Solution : Let the ten’s and the unit’s digits in the first number be x and y, respectively. +So, the first number may be written as 10 x + y in the expanded form (for example, +56 = 10(5) + 6). +When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s +digit. This number, in the expanded notation is 10y + x (for example, when 56 is +reversed, we get 65 = 10(6) + 5). +According to the given condition. +(10x + y) + (10y + x) = 66 +i.e., +11(x + y) = 66 +i.e., +x + y = 6 +(1) +We are also given that the digits differ by 2, therefore, +either +x – y = 2 +(2) +or +y – x = 2 +(3) +If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2. +In this case, we get the number 42. +If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4. + In this case, we get the number 24. +Thus, there are two such numbers 42 and 24. +Verification : Here 42 + 24 = 66 and 4 – 2 = 2. Also 24 + 42 = 66 and 4 – 2 = 2. +EXERCISE 3.3 +1. Solve the following pair of linear equations by the elimination method and the substitution +method : +(i) x + y = 5 and 2x – 3y = 4 +(ii) 3x + 4y = 10 and 2x – 2y = 2 +(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 +(iv) +2 +1 and +3 +2 +3 +3 +x +y +y +x + + + + +2. Form the pair of linear equations in the following problems, and find their solutions +(if they exist) by the elimination method : +(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces +to 1. It becomes 1 +2 if we only add 1 to the denominator. What is the fraction? +(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as +old as Sonu. How old are Nuri and Sonu? +(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is +twice the number obtained by reversing the order of the digits. Find the number. +Reprint 2025-26 + +PAIR OF LINEAR EQUATIONS IN TWO VARIABLES +37 +(iv) Meena went to a bank to withdraw ` 2000. She asked the cashier to give her +` 50 and ` 100 notes only. Meena got 25 notes in all. Find how many notes of +` 50 and ` 100 she received. +(v) A lending library has a fixed charge for the first three days and an additional charge +for each day thereafter. Saritha paid ` 27 for a book kept for seven days, while Susy +paid ` 21 for the book she kept for five days. Find the fixed charge and the charge +for each extra day. +3.4 Summary +In this chapter, you have studied the following points: +1. A pair of linear equations in two variables can be represented, and solved, by the: +(i) graphical method +(ii) algebraic method +2. Graphical Method : +The graph of a pair of linear equations in two variables is represented by two lines. +(i) If the lines intersect at a point, then that point gives the unique solution of the two +equations. In this case, the pair of equations is consistent. +(ii) If the lines coincide, then there are infinitely many solutions — each point on +the line being a solution. In this case, the pair of equations is dependent +(consistent). +(iii) If the lines are parallel, then the pair of equations has no solution. In this case, the +pair of equations is inconsistent. +3. Algebraic Methods : We have discussed the following methods for finding the solution(s) +of a pair of linear equations : +(i) Substitution Method +(ii) Elimination Method +4. If a pair of linear equations is given by a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then the +following situations can arise : +(i) +1 +1 +2 +1 +a +b +a +b + + : In this case, the pair of linear equations is consistent. +(ii) +1 +1 +1 +2 +2 +2 +a +b +c +a +b +c + + + : In this case, the pair of linear equations is inconsistent. +(iii) +1 +1 +1 +2 +2 +2 +a +b +c +a +b +c + + + : In this case, the pair of linear equations is dependent and consistent. +5. There are several situations which can be mathematically represented by two equations +that are not linear to start with. But we alter them so that they are reduced to a pair of +linear equations. +Reprint 2025-26" +class_10,4,Quadratic Equations,ncert_books/class_10/jemh1dd/jemh104.pdf,"38 +MATHEMATICS +4 +4.1 Introduction +In Chapter 2, you have studied different types of polynomials. One type was the +quadratic polynomial of the form ax2 + bx + c, a  0. When we equate this polynomial +to zero, we get a quadratic equation. Quadratic equations come up when we deal with +many real-life situations. For instance, suppose a +charity trust decides to build a prayer hall having +a carpet area of 300 square metres with its length +one metre more than twice its breadth. What +should be the length and breadth of the hall? +Suppose the breadth of the hall is x metres. Then, +its length should be (2x + 1) metres. We can depict +this information pictorially as shown in Fig. 4.1. +Now, +area of the hall = (2x + 1). x m2 = (2x2 + x) m2 +So, +2x2 + x = 300 +(Given) +Therefore, +2x2 + x – 300 = 0 +So, the breadth of the hall should satisfy the equation 2x2 + x – 300 = 0 which is a +quadratic equation. +Many people believe that Babylonians were the first to solve quadratic equations. +For instance, they knew how to find two positive numbers with a given positive sum +and a given positive product, and this problem is equivalent to solving a quadratic +equation of the form x2 – px + q = 0. Greek mathematician Euclid developed a +geometrical approach for finding out lengths which, in our present day terminology, +are solutions of quadratic equations. Solving of quadratic equations, in general form, is +often credited to ancient Indian mathematicians. In fact, Brahmagupta (C.E.598–665) +gave an explicit formula to solve a quadratic equation of the form ax2 + bx = c. Later, +QUADRATIC EQUATIONS +Fig. 4.1 +Reprint 2025-26 + +QUADRATIC EQUATIONS +39 +Sridharacharya (C.E. 1025) derived a formula, now known as the quadratic formula, +(as quoted by Bhaskara II) for solving a quadratic equation by the method of completing +the square. An Arab mathematician Al-Khwarizmi (about C.E. 800) also studied +quadratic equations of different types. Abraham bar Hiyya Ha-Nasi, in his book +‘Liber embadorum’ published in Europe in C.E. 1145 gave complete solutions of +different quadratic equations. +In this chapter, you will study quadratic equations, and various ways of finding +their roots. You will also see some applications of quadratic equations in daily life +situations. +4.2 Quadratic Equations +A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0, where +a, b, c are real numbers, a  0. For example, 2x2 + x – 300 = 0 is a quadratic equation. +Similarly, 2x2 – 3x + 1 = 0, 4x – 3x2 + 2 = 0 and 1 – x2 + 300 = 0 are also quadratic +equations. +In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree +2, is a quadratic equation. But when we write the terms of p(x) in descending order of +their degrees, then we get the standard form of the equation. That is, ax2 + bx + c = 0, +a  0 is called the standard form of a quadratic equation. +Quadratic equations arise in several situations in the world around us and in +different fields of mathematics. Let us consider a few examples. +Example 1 : Represent the following situations mathematically: +(i) +John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and +the product of the number of marbles they now have is 124. We would like to find +out how many marbles they had to start with. +(ii) +A cottage industry produces a certain number of toys in a day. The cost of +production of each toy (in rupees) was found to be 55 minus the number of toys +produced in a day. On a particular day, the total cost of production was +` 750. We would like to find out the number of toys produced on that day. +Solution : +(i) +Let the number of marbles John had be x. +Then the number of marbles Jivanti had = 45 – x (Why?). +The number of marbles left with John, when he lost 5 marbles = x – 5 +The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5 += 40 – x +Reprint 2025-26 + +40 +MATHEMATICS +Therefore, their product = (x – 5) (40 – x) += 40x – x2 – 200 + 5x += – x2 + 45x – 200 +So, +– x2 + 45x – 200 = 124 +(Given that product = 124) +i.e., +– x2 + 45x – 324 = 0 +i.e., +x2 – 45x + 324 = 0 +Therefore, the number of marbles John had, satisfies the quadratic equation +x2 – 45x + 324 = 0 +which is the required representation of the problem mathematically. +(ii) Let the number of toys produced on that day be x. +Therefore, the cost of production (in rupees) of each toy that day = 55 – x +So, the total cost of production (in rupees) that day = x (55 – x) +Therefore, +x (55 – x) = 750 +i.e., +55x – x2 = 750 +i.e., +– x2 + 55x – 750 = 0 +i.e., +x2 – 55x + 750 = 0 +Therefore, the number of toys produced that day satisfies the quadratic equation +x2 – 55x + 750 = 0 +which is the required representation of the problem mathematically. +Example 2 : Check whether the following are quadratic equations: +(i) (x – 2)2 + 1 = 2x – 3 +(ii) x(x + 1) + 8 = (x + 2) (x – 2) +(iii) x (2x + 3) = x2 + 1 +(iv) (x + 2)3 = x3 – 4 +Solution : +(i) LHS = (x – 2)2 + 1 = x2 – 4x + 4 + 1 = x2 – 4x + 5 +Therefore, (x – 2)2 + 1 = 2x – 3 can be rewritten as +x2 – 4x + 5 = 2x – 3 +i.e., +x2 – 6x + 8 = 0 +It is of the form ax2 + bx + c = 0. +Therefore, the given equation is a quadratic equation. +Reprint 2025-26 + +QUADRATIC EQUATIONS +41 +(ii) +Since x(x + 1) + 8 = x2 + x + 8 and (x + 2)(x – 2) = x2 – 4 +Therefore, +x2 + x + 8 = x2 – 4 +i.e., +x + 12 = 0 +It is not of the form ax2 + bx + c = 0. +Therefore, the given equation is not a quadratic equation. +(iii) Here, +LHS = x (2x + 3) = 2x2 + 3x +So, +x (2x + 3) = x2 + 1 can be rewritten as +2x2 + 3x = x2 + 1 +Therefore, we get +x2 + 3x – 1 = 0 +It is of the form ax2 + bx + c = 0. +So, the given equation is a quadratic equation. +(iv) Here, +LHS = (x + 2)3 = x3 + 6x2 + 12x + 8 +Therefore, +(x + 2)3 = x3 – 4 can be rewritten as +x3 + 6x2 + 12x + 8 = x3 – 4 +i.e., +6x2 + 12x + 12 = 0 +or, +x2 + 2x + 2 = 0 +It is of the form ax2 + bx + c = 0. +So, the given equation is a quadratic equation. +Remark : Be careful! In (ii) above, the given equation appears to be a quadratic +equation, but it is not a quadratic equation. +In (iv) above, the given equation appears to be a cubic equation (an equation of +degree 3) and not a quadratic equation. But it turns out to be a quadratic equation. As +you can see, often we need to simplify the given equation before deciding whether it +is quadratic or not. +EXERCISE 4.1 +1. Check whether the following are quadratic equations : +(i) (x + 1)2 = 2(x – 3) +(ii) x2 – 2x = (–2) (3 – x) +(iii) (x – 2)(x + 1) = (x – 1)(x + 3) +(iv) (x – 3)(2x +1) = x(x + 5) +(v) (2x – 1)(x – 3) = (x + 5)(x – 1) +(vi) x2 + 3x + 1 = (x – 2)2 +(vii) (x + 2)3 = 2x (x2 – 1) +(viii) x3 – 4x2 – x + 1 = (x – 2)3 +2. Represent the following situations in the form of quadratic equations : +(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one +more than twice its breadth. We need to find the length and breadth of the plot. +Reprint 2025-26 + +42 +MATHEMATICS +(ii) The product of two consecutive positive integers is 306. We need to find the +integers. +(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) +3 years from now will be 360. We would like to find Rohan’s present age. +(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been +8 km/h less, then it would have taken 3 hours more to cover the same distance. We +need to find the speed of the train. +4.3 Solution of a Quadratic Equation by Factorisation +Consider the quadratic equation 2x2 – 3x + 1 = 0. If we replace x by 1 on the +LHS of this equation, we get (2 × 12) – (3 × 1) + 1 = 0 = RHS of the equation. +We say that 1 is a root of the quadratic equation 2x2 – 3x + 1 = 0. This also means that +1 is a zero of the quadratic polynomial 2x2 – 3x + 1. +In general, a real number  is called a root of the quadratic equation +ax2 + bx + c = 0, a  0 if a 2 + b + c = 0. We also say that x =  is a solution of +the quadratic equation, or that  satisfies the quadratic equation. Note that the +zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic +equation ax2 + bx + c = 0 are the same. +You have observed, in Chapter 2, that a quadratic polynomial can have at most +two zeroes. So, any quadratic equation can have atmost two roots. +You have learnt in Class IX, how to factorise quadratic polynomials by splitting +their middle terms. We shall use this knowledge for finding the roots of a quadratic +equation. Let us see how. +Example 3 : Find the roots of the equation 2x2 – 5x + 3 = 0, by factorisation. +Solution : Let us first split the middle term – 5x as –2x –3x [because (–2x) × (–3x) = +6x2 = (2x2) × 3]. +So, +2x2 – 5x + 3 = 2x2 – 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1) +Now, 2x2 – 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0. +So, the values of x for which 2x2 – 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0, +i.e., either 2x – 3 = 0 or x – 1 = 0. +Now, 2x – 3 = 0 gives +3 +2 +x  + and x – 1 = 0 gives x = 1. +So, +3 +2 +x  + and x = 1 are the solutions of the equation. +In other words, 1 and 3 +2 are the roots of the equation 2x2 – 5x + 3 = 0. +Verify that these are the roots of the given equation. +Reprint 2025-26 + +QUADRATIC EQUATIONS +43 +Note that we have found the roots of 2x2 – 5x + 3 = 0 by factorising +2x2 – 5x + 3 into two linear factors and equating each factor to zero. +Example 4 : Find the roots of the quadratic equation 6x2 – x – 2 = 0. +Solution : We have +6x2 – x – 2 = 6x2 + 3x – 4x – 2 += 3x (2x + 1) – 2 (2x + 1) += (3x – 2)(2x + 1) +The roots of 6x2 – x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0 +Therefore, 3x – 2 = 0 or 2x + 1 = 0, +i.e., +x = 2 +3 +or +x = +1 +2 + +Therefore, the roots of 6x2 – x – 2 = 0 are 2 +1. +and – +3 +2 +We verify the roots, by checking that 2 +1 +and +3 +2 + + satisfy 6x2 – x – 2 = 0. +Example 5 : Find the roots of the quadratic equation +2 +3 +2 6 +2 +0 +x +x + + + +. +Solution : +2 +3 +2 6 +2 +x +x + + + = +2 +3 +6 +6 +2 +x +x +x + + + += + + + + +3 +3 +2 +2 +3 +2 +x +x +x + + + +=  + + +3 +2 +3 +2 +x +x + + +So, the roots of the equation are the values of x for which + + + +3 +2 +3 +2 +0 +x +x + + + +Now, +3 +2 +0 +x  + + for +2 +3 +x  +. +So, this root is repeated twice, one for each repeated factor +3 +2 +x  +. +Therefore, the roots of +2 +3 +2 6 +2 +0 +x +x + + + + are +2 +3 +, +2 +3 +. +Reprint 2025-26 + +44 +MATHEMATICS +Example 6 : Find the dimensions of the prayer hall discussed in Section 4.1. +Solution : In Section 4.1, we found that if the breadth of the hall is x m, then x +satisfies the equation 2x2 + x – 300 = 0. Applying the factorisation method, we write +this equation as +2x2 – 24x + 25x – 300 = 0 +2x (x – 12) + 25 (x – 12) = 0 +i.e., +(x – 12)(2x + 25) = 0 +So, the roots of the given equation are x = 12 or x = – 12.5. Since x is the breadth +of the hall, it cannot be negative. +Thus, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m. +EXERCISE 4.2 +1. Find the roots of the following quadratic equations by factorisation: +(i) x2 – 3x – 10 = 0 +(ii) 2x2 + x – 6 = 0 +(iii) +2 +2 +7 +5 2 +0 +x +x + + + +(iv) 2x2 – x + +1 +8 = 0 +(v) 100x2 – 20x + 1 = 0 +2. Solve the problems given in Example 1. +3. Find two numbers whose sum is 27 and product is 182. +4. Find two consecutive positive integers, sum of whose squares is 365. +5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find +the other two sides. +6. A cottage industry produces a certain number of pottery articles in a day. It was observed +on a particular day that the cost of production of each article (in rupees) was 3 more than +twice the number of articles produced on that day. If the total cost of production on that +day was ` 90, find the number of articles produced and the cost of each article. +4.4 Nature of Roots +The equation ax2 + bx + c = 0 are given by +x = +2 +– +4 +2 +b +b +ac +a + + +If b2 – 4ac > 0, we get two distinct real roots +2 +4 +2 +2 +b +ac +b +a +a + + + + and +2 +4 +– +2 +2 +b +ac +b +a +a + + +. +Reprint 2025-26 + +QUADRATIC EQUATIONS +45 +If b2 – 4ac = 0, then x = +0 +2 +b +a + + +, i.e., +or – +2 +2 +b +b +x +a +a + + +So, the roots of the equation ax2 + bx + c = 0 are both 2 +b +a + +Therefore, we say that the quadratic equation ax2 + bx + c = 0 has two equal +real roots in this case. +If b2 – 4ac < 0, then there is no real number whose square is b2 – 4ac. Therefore, +there are no real roots for the given quadratic equation in this case. +Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has +real roots or not, b2 – 4ac is called the discriminant of this quadratic equation. +So, a quadratic equation ax2 + bx + c = 0 has +(i) two distinct real roots, if b2 – 4ac > 0, +(ii) two equal real roots, if b2 – 4ac = 0, +(iii) no real roots, if b2 – 4ac < 0. +Let us consider some examples. +Example 7: Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0, and +hence find the nature of its roots. +Solution : The given equation is of the form ax2 + bx + c = 0, where a = 2, b = – 4 and +c = 3. Therefore, the discriminant +b2 – 4ac = (– 4)2 – (4 × 2 × 3) = 16 – 24 = – 8 < 0 +So, the given equation has no real roots. +Example 8 : A pole has to be erected at a point on the boundary of a circular park of +diameter 13 metres in such a way that the differences of its distances from two +diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to +do so? If yes, at what distances from the two gates should the pole be erected? +Solution : Let us first draw the diagram +(see Fig. 4.2). +Let P be the required location of the +pole. Let the distance of the pole from the +gate B be x m, i.e., BP = x m. Now the +difference of the distances of the pole from +the two gates = AP – BP (or, BP – AP) = +7 m. Therefore, AP = (x + 7) m. +Fig. 4.2 +Reprint 2025-26 + +46 +MATHEMATICS +Now, AB = 13m, and since AB is a diameter, +APB = 90° +(Why?) +Therefore, +AP2 + PB2 = AB2 +(By Pythagoras theorem) +i.e., +(x + 7)2 + x2 = 132 +i.e., +x2 + 14x + 49 + x2 = 169 +i.e., +2x2 + 14x – 120 = 0 +So, the distance ‘x’ of the pole from gate B satisfies the equation +x2 + 7x – 60 = 0 +So, it would be possible to place the pole if this equation has real roots. To see if this +is so or not, let us consider its discriminant. The discriminant is +b2 – 4ac = 72 – 4 × 1 × (– 60) = 289 > 0. +So, the given quadratic equation has two real roots, and it is possible to erect the +pole on the boundary of the park. +Solving the quadratic equation x2 + 7x – 60 = 0, by the quadratic formula, we get +x = +7 +289 +2 + + = 7 +17 +2 + +Therefore, x = 5 or – 12. +Since x is the distance between the pole and the gate B, it must be positive. +Therefore, x = – 12 will have to be ignored. So, x = 5. +Thus, the pole has to be erected on the boundary of the park at a distance of 5m +from the gate B and 12m from the gate A. +Example 9 : Find the discriminant of the equation 3x2 – 2x + 1 +3 = 0 and hence find the +nature of its roots. Find them, if they are real. +Solution : Here a = 3, b = – 2 and +1 +3 +c  +. +Therefore, discriminant b2 – 4ac = (– 2)2 – 4 × 3 × 1 +3 = 4 – 4 = 0. +Hence, the given quadratic equation has two equal real roots. +The roots are +2 +2 +1 +1 +, +, +, +, +. +i.e., +, i.e., +2 +2 +6 +6 +3 +3 +b +b +a +a + + +Reprint 2025-26 + +QUADRATIC EQUATIONS +47 +EXERCISE 4.3 +1. Find the nature of the roots of the following quadratic equations. If the real roots exist, +find them: +(i) 2x2 – 3x + 5 = 0 +(ii) 3x2 – 4 +3 x + 4 = 0 +(iii) 2x2 – 6x + 3 = 0 +2. Find the values of k for each of the following quadratic equations, so that they have two +equal roots. +(i) 2x2 + kx + 3 = 0 +(ii) kx (x – 2) + 6 = 0 +3. Is it possible to design a rectangular mango grove whose length is twice its breadth, +and the area is 800 m2? If so, find its length and breadth. +4. Is the following situation possible? If so, determine their present ages. +The sum of the ages of two friends is 20 years. Four years ago, the product of their ages +in years was 48. +5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find +its length and breadth. +4.5 Summary +In this chapter, you have studied the following points: +1. A quadratic equation in the variable x is of the form ax2 + bx + c = 0, where a, b, c are real +numbers and a  0. +2. A real number  is said to be a root of the quadratic equation ax2 + bx + c = 0, if +a2 + b + c = 0. The zeroes of the quadratic polynomial ax2 + bx + c and the roots of the +quadratic equation ax2 + bx + c = 0 are the same. +3. If we can factorise ax2 + bx + c, a  0, into a product of two linear factors, then the roots +of the quadratic equation ax2 + bx + c = 0 can be found by equating each factor to zero. +4. Quadratic formula: The roots of a quadratic equation ax2 + bx + c = 0 are given by +2 +4 +, +2 +b +b +ac +a + + + provided b2 – 4ac  0. +5. A quadratic equation ax2 + bx + c = 0 has +(i) two distinct real roots, if b2 – 4ac > 0, +(ii) two equal roots (i.e., coincident roots), if b2 – 4ac = 0, and +(iii) no real roots, if b2 – 4ac < 0. +Reprint 2025-26 + +48 +MATHEMATICS +NOTE +Reprint 2025-26" +class_10,5,Arithmetic Progressions,ncert_books/class_10/jemh1dd/jemh105.pdf,"ARITHMETIC PROGRESSIONS +49 +5 +5.1 Introduction +You must have observed that in nature, many things follow a certain pattern, such as +the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the +spirals on a pineapple and on a pine cone, etc. +We now look for some patterns which occur in our day-to-day life. Some such +examples are : +(i) Reena applied for a job and got selected. She +has been offered a job with a starting monthly +salary of ` 8000, with an annual increment of +` 500 in her salary. Her salary (in `) for the 1st, +2nd, 3rd, . . . years will be, respectively +8000, +8500, +9000, . . . . +(ii) The lengths of the rungs of a ladder decrease +uniformly by 2 cm from bottom to top +(see Fig. 5.1). The bottom rung is 45 cm in +length. The lengths (in cm) of the 1st, 2nd, +3rd, . . ., 8th rung from the bottom to the top +are, respectively +45, 43, 41, 39, 37, 35, 33, 31 +(iii) In a savings scheme, the amount becomes 5 +4 times of itself after every 3 years. +The maturity amount (in `) of an investment of ` 8000 after 3, 6, 9 and 12 years +will be, respectively : +10000, +12500, +15625, 19531.25 +ARITHMETIC PROGRESSIONS +Fig. 5.1 +Reprint 2025-26 + +50 +MATHEMATICS +(iv) The number of unit squares in squares with side 1, 2, 3, . . . units (see Fig. 5.2) +are, respectively +12, 22, 32, . . . . +Fig. 5.2 +(v) Shakila puts ` 100 into her daughter’s money box when she was one year old +and increased the amount by ` 50 every year. The amounts of money (in `) in the +box on the 1st, 2nd, 3rd, 4th, . . . birthday were +100, +150, +200, +250, . . ., respectively. +(vi) A pair of rabbits are too young to produce in their first month. In the second, and +every subsequent month, they produce a new pair. Each new pair of rabbits +produce a new pair in their second month and in every subsequent month (see +Fig. 5.3). Assuming no rabbit dies, the number of pairs of rabbits at the start of +the 1st, 2nd, 3rd, . . ., 6th month, respectively are : +1, 1, 2, 3, 5, 8 +Fig. 5.3 +Reprint 2025-26 + +ARITHMETIC PROGRESSIONS +51 +In the examples above, we observe some patterns. In some, we find that the +succeeding terms are obtained by adding a fixed number, in other by multiplying +with a fixed number, in another we find that they are squares of consecutive +numbers, and so on. +In this chapter, we shall discuss one of these patterns in which succeeding terms +are obtained by adding a fixed number to the preceding terms. We shall also see how +to find their nth terms and the sum of n consecutive terms, and use this knowledge in +solving some daily life problems. +5.2 Arithmetic Progressions +Consider the following lists of numbers : +(i) 1, 2, 3, 4, . . . +(ii) 100, 70, 40, 10, . . . +(iii) –3, –2, –1, 0, . . . +(iv) 3, 3, 3, 3, . . . +(v) –1.0, –1.5, –2.0, –2.5, . . . +Each of the numbers in the list is called a term. +Given a term, can you write the next term in each of the lists above? If so, how +will you write it? Perhaps by following a pattern or rule. Let us observe and write the +rule. +In (i), each term is 1 more than the term preceding it. +In (ii), each term is 30 less than the term preceding it. +In (iii), each term is obtained by adding 1 to the term preceding it. +In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding +(or subtracting) 0 to the term preceding it. +In (v), each term is obtained by adding – 0.5 to (i.e., subtracting 0.5 from) the +term preceding it. +In all the lists above, we see that successive terms are obtained by adding a fixed +number to the preceding terms. Such list of numbers is said to form an Arithmetic +Progression ( AP ). +So, an arithmetic progression is a list of numbers in which each term is +obtained by adding a fixed number to the preceding term except the first +term. +This fixed number is called the common difference of the AP. Remember that +it can be positive, negative or zero. +Reprint 2025-26 + +52 +MATHEMATICS +Let us denote the first term of an AP by a1, second term by a2, . . ., nth term by +an and the common difference by d. Then the AP becomes a1, a2, a3, . . ., an. +So, +a2 – a1 = a3 – a2 = . . . = an – an – 1 = d. +Some more examples of AP are: +(a) The heights ( in cm ) of some students of a school standing in a queue in the +morning assembly are 147 , 148, 149, . . ., 157. +(b) The minimum temperatures ( in degree celsius ) recorded for a week in the +month of January in a city, arranged in ascending order are +– 3.1, – 3.0, – 2.9, – 2.8, – 2.7, – 2.6, – 2.5 +(c) The balance money ( in ` ) after paying 5 % of the total loan of ` 1000 every +month is 950, 900, 850, 800, . . ., 50. +(d) The cash prizes ( in ` ) given by a school to the toppers of Classes I to XII are, +respectively, 200, 250, 300, 350, . . ., 750. +(e) The total savings (in `) after every month for 10 months when ` 50 are saved +each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500. +It is left as an exercise for you to explain why each of the lists above is an AP. +You can see that +a, a + d, a + 2d, a + 3d, . . . +represents an arithmetic progression where a is the first term and d the common +difference. This is called the general form of an AP. +Note that in examples (a) to (e) above, there are only a finite number of terms. +Such an AP is called a finite AP. Also note that each of these Arithmetic Progressions +(APs) has a last term. The APs in examples (i) to (v) in this section, are not finite APs +and so they are called infinite Arithmetic Progressions. Such APs do not have a +last term. +Now, to know about an AP, what is the minimum information that you need? Is it +enough to know the first term? Or, is it enough to know only the common difference? +You will find that you will need to know both – the first term a and the common +difference d. +For instance if the first term a is 6 and the common difference d is 3, then +the AP is + 6, 9,12, 15, . . . +and if a is 6 and d is – 3, then the AP is +6, 3, 0, –3, . . . +Reprint 2025-26 + +ARITHMETIC PROGRESSIONS +53 +Similarly, when +a = – 7, +d = – 2, +the AP is – 7, – 9, – 11, – 13, . . . +a = 1.0, +d = 0.1, +the AP is 1.0, 1.1, 1.2, 1.3, . . . +a = 0, +d = 1 1 +2 , +the AP is 0, 1 1 +2 , 3, 4 1 +2 , 6, . . . +a = 2, +d = 0, +the AP is 2, 2, 2, 2, . . . +So, if you know what a and d are, you can list the AP. What about the other way +round? That is, if you are given a list of numbers can you say that it is an AP and then +find a and d? Since a is the first term, it can easily be written. We know that in an AP, +every succeeding term is obtained by adding d to the preceding term. So, d found by +subtracting any term from its succeeding term, i.e., the term which immediately follows +it should be same for an AP. +For example, for the list of numbers : +6, 9, 12, 15, . . . , +We have +a2 – a1 = 9 – 6 = 3, +a3 – a2 = 12 – 9 = 3, +a4 – a3 = 15 – 12 = 3 +Here the difference of any two consecutive terms in each case is 3. So, the +given list is an AP whose first term a is 6 and common difference d is 3. +For the list of numbers : 6, 3, 0, – 3, . . ., +a2 – a1 = 3 – 6 = – 3 +a3 – a2 = 0 – 3 = – 3 +a4 – a3 = –3 – 0 = –3 +Similarly this is also an AP whose first term is 6 and the common difference +is –3. +In general, for an AP a1, a2, . . ., an, we have +d = ak + 1 – ak +where ak + 1 and ak are the ( k + 1)th and the kth terms respectively. +To obtain d in a given AP, we need not find all of a2 – a1, a3 – a2, a4 – a3, . . . . +It is enough to find only one of them. +Consider the list of numbers 1, 1, 2, 3, 5, . . . . By looking at it, you can tell that the +difference between any two consecutive terms is not the same. So, this is not an AP. +Reprint 2025-26 + +54 +MATHEMATICS +Note that to find d in the AP : 6, 3, 0, – 3, . . ., we have subtracted 6 from 3 +and not 3 from 6, i.e., we should subtract the kth term from the (k + 1) th term +even if the (k + 1) th term is smaller. +Let us make the concept more clear through some examples. +Example 1 : For the AP : 3 +2 +, 1 +2 +, – 1 +2 +, – 3 +2 + , . . ., write the first term a and the +common difference d. +Solution : Here, +a = 3 +2 , d = 1 +2 – 3 +2 = – 1. +Remember that we can find d using any two consecutive terms, once we know that +the numbers are in AP. +Example 2 : Which of the following list of numbers form an AP? If they form an AP, +write the next two terms : +(i) 4, 10, 16, 22, . . . +(ii) 1, – 1, – 3, – 5, . . . +(iii) – 2, 2, – 2, 2, – 2, . . . +(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . . +Solution : +(i) We have a2 – a1 = 10 – 4 = 6 +a3 – a2 = 16 – 10 = 6 +a4 – a3 = 22 – 16 = 6 +i.e., +ak + 1 – ak is the same every time. +So, the given list of numbers forms an AP with the common difference d = 6. +The next two terms are: 22 + 6 = 28 and 28 + 6 = 34. +(ii) +a2 – a1 = – 1 – 1 = – 2 +a3 – a2 = – 3 – ( –1 ) = – 3 + 1 = – 2 + a4 – a3 = – 5 – ( –3 ) = – 5 + 3 = – 2 +i.e., ak + 1 – ak is the same every time. +So, the given list of numbers forms an AP with the common difference d = – 2. +The next two terms are: +– 5 + (– 2 ) = – 7 +and +– 7 + (– 2 ) = – 9 +(iii) a2 – a1 = 2 – (– 2) = 2 + 2 = 4 +a3 – a2 = – 2 – 2 = – 4 +As a2 – a1  a3 – a2 , the given list of numbers does not form an AP. +Reprint 2025-26 + +ARITHMETIC PROGRESSIONS +55 +(iv) a2 – a1 = 1 – 1 = 0 +a3 – a2 = 1 – 1 = 0 +a4 – a3 = 2 – 1 = 1 +Here, a2 – a1 = a3 – a2  a4 – a3. +So, the given list of numbers does not form an AP. +EXERCISE 5.1 +1. In which of the following situations, does the list of numbers involved make an arithmetic +progression, and why? +(i) The taxi fare after each km when the fare is ` 15 for the first km and ` 8 for each +additional km. +(ii) The amount of air present in a cylinder when a vacuum pump removes 1 +4 of the +air remaining in the cylinder at a time. +(iii) The cost of digging a well after every metre of digging, when it costs ` 150 for the +first metre and rises by ` 50 for each subsequent metre. +(iv) The amount of money in the account every year, when ` 10000 is deposited at +compound interest at 8 % per annum. +2. Write first four terms of the AP, when the first term a and the common difference d are +given as follows: +(i) a = 10, +d = 10 +(ii) a = –2, +d = 0 +(iii) a = 4, +d = – 3 +(iv) a = – 1, d = 1 +2 +(v) a = – 1.25, d = – 0.25 +3. For the following APs, write the first term and the common difference: +(i) 3, 1, – 1, – 3, . . . +(ii) – 5, – 1, 3, 7, . . . +(iii) 1 +5 +9 +13 +, +, +, +, +3 +3 +3 +3 + . . . +(iv) 0.6, 1.7, 2.8, 3.9, . . . +4. Which of the following are APs ? If they form an AP, find the common difference d and +write three more terms. +(i) 2, 4, 8, 16, . . . +(ii) +5 +7 +2, +, 3, +, +2 +2 + . . . +(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . +(iv) – 10, – 6, – 2, 2, . . . +(v) 3, 3 +2 + +, 3 +2 2 + +, 3 +3 2 , + + . . . +(vi) 0.2, 0.22, 0.222, 0.2222, . . . +(vii) 0, – 4, – 8, –12, . . . +(viii) – 1 +2 , – 1 +2 , – 1 +2 , – 1 +2 , . . . +Reprint 2025-26 + +56 +MATHEMATICS +(ix) 1, 3, 9, 27, . . . +(x) a, 2a, 3a, 4a, . . . +(xi) a, a2, a3, a4, . . . +(xii) +2, +8, +18 , 32, . . . +(xiii) +3, +6, +9 , 12 , . . . +(xiv) 12, 32, 52, 72, . . . +(xv) 12, 52, 72, 73, . . . +5.3 nth Term of an AP +Let us consider the situation again, given in Section 5.1 in which Reena applied for a +job and got selected. She has been offered the job with a starting monthly salary of +` 8000, with an annual increment of ` 500. What would be her monthly salary for the +fifth year? +To answer this, let us first see what her monthly salary for the second year +would be. +It would be ` (8000 + 500) = ` 8500. In the same way, we can find the monthly +salary for the 3rd, 4th and 5th year by adding ` 500 to the salary of the previous year. +So, the salary for the 3rd year = ` (8500 + 500) += ` (8000 + 500 + 500) += ` (8000 + 2 × 500) += ` [8000 + (3 – 1) × 500] +(for the 3rd year) += ` 9000 +Salary for the 4th year = ` (9000 + 500) += ` (8000 + 500 + 500 + 500) += ` (8000 + 3 × 500) += ` [8000 + (4 – 1) × 500] +(for the 4th year) += ` 9500 +Salary for the 5th year = ` (9500 + 500) += ` (8000+500+500+500 + 500) += ` (8000 + 4 × 500) += ` [8000 + (5 – 1) × 500] +(for the 5th year) += ` 10000 +Observe that we are getting a list of numbers +8000, 8500, 9000, 9500, 10000, . . . +These numbers are in AP. (Why?) +Reprint 2025-26 + +ARITHMETIC PROGRESSIONS +57 +Now, looking at the pattern formed above, can you find her monthly salary for +the 6th year? The 15th year? And, assuming that she will still be working in the job, +what about the monthly salary for the 25th year? You would calculate this by adding +` 500 each time to the salary of the previous year to give the answer. Can we make +this process shorter? Let us see. You may have already got some idea from the way +we have obtained the salaries above. +Salary for the 15th year += Salary for the 14th year + ` 500 += += ` [8000 + 14 × 500] += ` [8000 + (15 – 1) × 500] = ` 15000 +i.e., +First salary + (15 – 1) × Annual increment. +In the same way, her monthly salary for the 25th year would be +` [8000 + (25 – 1) × 500] = ` 20000 += First salary + (25 – 1) × Annual increment +This example would have given you some idea about how to write the 15th term, +or the 25th term, and more generally, the nth term of the AP. +Let a1, a2, a3, . . . be an AP whose first term a1 is a and the common +difference is d. +Then, +the second term a2 = a + d = a + (2 – 1) d +the third term +a3 = a2 + d = (a + d) + d = a + 2d = a + (3 – 1) d +the fourth term +a4 = a3 + d = (a + 2d) + d = a + 3d = a + (4 – 1) d +. . . . . . . . +. . . . . . . . +Looking at the pattern, we can say that the nth term an = a + (n – 1) d. +So, the nth term an of the AP with first term a and common difference d is +given by an = a + (n – 1) d. + +Reprint 2025-26 + +58 +MATHEMATICS +an is also called the general term of the AP. If there are m terms in the AP, then +am represents the last term which is sometimes also denoted by l. +Let us consider some examples. +Example 3 : Find the 10th term of the AP : 2, 7, 12, . . . +Solution : Here, a = 2, +d = 7 – 2 = 5 +and +n = 10. +We have +an = a + (n – 1) d +So, +a10 = 2 + (10 – 1) × 5 = 2 + 45 = 47 +Therefore, the 10th term of the given AP is 47. +Example 4 : Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any term 0? Give +reason for your answer. +Solution : Here, a = 21, d = 18 – 21 = – 3 and an = – 81, and we have to find n. +As +an = a + ( n – 1) d, +we have +– 81 = 21 + (n – 1)(– 3) +– 81 = 24 – 3n +– 105 = – 3n +So, +n = 35 +Therefore, the 35th term of the given AP is – 81. +Next, we want to know if there is any n for which an = 0. If such an n is there, then +21 + (n – 1) (–3) = 0, +i.e., +3(n – 1) = 21 +i.e., +n = 8 +So, the eighth term is 0. +Example 5 : Determine the AP whose 3rd term is 5 and the 7th term is 9. +Solution : We have +a3 = a + (3 – 1) d = a + 2d = 5 +(1) +and +a7 = a + (7 – 1) d = a + 6d = 9 +(2) +Solving the pair of linear equations (1) and (2), we get +a = 3, +d = 1 +Hence, the required AP is 3, 4, 5, 6, 7, . . . +Reprint 2025-26 + +ARITHMETIC PROGRESSIONS +59 +Example 6 : Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . . +Solution : We have : +a2 – a1 = 11 – 5 = 6, +a3 – a2 = 17 – 11 = 6, +a4 – a3 = 23 – 17 = 6 +As ak + 1 – ak is the same for k = 1, 2, 3, etc., the given list of numbers is an AP. +Now, +a = 5 +and +d = 6. +Let 301 be a term, say, the nth term of this AP. +We know that +an = a + (n – 1) d +So, +301 = 5 + (n – 1) × 6 +i.e., +301 = 6n – 1 +So, +n = 302 +151 +6 +3 + +But n should be a positive integer (Why?). So, 301 is not a term of the given list of +numbers. +Example 7 : How many two-digit numbers are divisible by 3? +Solution : The list of two-digit numbers divisible by 3 is : +12, 15, 18, . . . , 99 +Is this an AP? Yes it is. Here, +a = 12, d = 3, an = 99. +As +an = a + (n – 1) d, +we have +99 = 12 + (n – 1) × 3 +i.e., +87 = (n – 1) × 3 +i.e., +n – 1 = 87 +3 = 29 +i.e., +n = 29 + 1 = 30 +So, there are 30 two-digit numbers divisible by 3. +Example 8 : Find the 11th term from the last term (towards the first term) of the +AP : 10, 7, 4, . . ., – 62. +Solution : Here, +a = 10, d = 7 – 10 = – 3, l = – 62, +where +l = a + (n – 1) d +Reprint 2025-26 + +60 +MATHEMATICS +To find the 11th term from the last term, we will find the total number of terms in +the AP. +So, +– 62 = 10 + (n – 1)(–3) +i.e., +– 72 = (n – 1)(–3) +i.e., +n – 1 = 24 +or +n = 25 +So, there are 25 terms in the given AP. +The 11th term from the last term will be the 15th term. (Note that it will not be +the 14th term. Why?) +So, +a15 = 10 + (15 – 1)(–3) = 10 – 42 = – 32 +i.e., the 11th term from the last term is – 32. +Alternative Solution : +If we write the given AP in the reverse order, then a = – 62 and d = 3 (Why?) +So, the question now becomes finding the 11th term with these a and d. +So, +a11 = – 62 + (11 – 1) × 3 = – 62 + 30 = – 32 +So, the 11th term, which is now the required term, is – 32. +Example 9 : A sum of ` 1000 is invested at 8% simple interest per year. Calculate the +interest at the end of each year. Do these interests form an AP? If so, find the interest +at the end of 30 years making use of this fact. +Solution : We know that the formula to calculate simple interest is given by +Simple Interest = P×R×T +100 +So, the interest at the end of the 1st year = +1000×8×1 +100 +` + = ` 80 +The interest at the end of the 2nd year = +1000×8× 2 +100 +` + = ` 160 +The interest at the end of the 3rd year = +1000×8×3 +100 +` + = ` 240 +Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on. +So, the interest (in `) at the end of the 1st, 2nd, 3rd, . . . years, respectively are +80, 160, 240, . . . +Reprint 2025-26 + +ARITHMETIC PROGRESSIONS +61 +It is an AP as the difference between the consecutive terms in the list is 80, i.e., +d = 80. Also, a = 80. +So, to find the interest at the end of 30 years, we shall find a30. +Now, +a30 = a + (30 – 1) d = 80 + 29 × 80 = 2400 +So, the interest at the end of 30 years will be ` 2400. +Example 10 : In a flower bed, there are 23 rose plants in the first row, 21 in the +second, 19 in the third, and so on. There are 5 rose plants in the last row. How many +rows are there in the flower bed? +Solution : The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are : +23, 21, 19, . . ., 5 +It forms an AP (Why?). Let the number of rows in the flower bed be n. +Then +a = 23, +d = 21 – 23 = – 2, an = 5 +As, +an = a + (n – 1) d +We have, +5 = 23 + (n – 1)(– 2) +i.e., +– 18 = (n – 1)(– 2) +i.e., +n = 10 +So, there are 10 rows in the flower bed. +EXERCISE 5.2 +1. Fill in the blanks in the following table, given that a is the first term, d the common +difference and an the nth term of the AP: +a +d +n +an +(i) +7 +3 +8 +. . . +(ii) +– 18 +. . . +10 +0 +(iii) +. . . +– 3 +18 +– 5 +(iv) +– 18.9 +2.5 +. . . +3.6 +(v) +3.5 +0 +105 +. . . +Reprint 2025-26 + +62 +MATHEMATICS +2. Choose the correct choice in the following and justify : +(i) 30th term of the AP: 10, 7, 4, . . . , is +(A) 97 +(B) 77 +(C) –77 +(D) – 87 +(ii) 11th term of the AP: – 3, +1 +2 + +, 2, . . ., is +(A) 28 +(B) 22 +(C) –38 +(D) – 48 1 +2 +3. In the following APs, find the missing terms in the boxes : +(i) 2, + , +26 +(ii) + , +13, + , +3 +(iii) 5, + , + , +1 +9 2 +(iv) – 4, + , + , + , + , +6 +(v) + , +38, + , + , + , +– 22 +4. Which term of the AP : 3, 8, 13, 18, . . . ,is 78? +5. Find the number of terms in each of the following APs : +(i) 7, 13, 19, . . . , 205 +(ii) 18, +1 +15 2 , 13, . . . , – 47 +6. Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . . +7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73. +8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th +term. +9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is +zero? +10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference. +11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term? +12. Two APs have the same common difference. The difference between their 100th terms is +100, what is the difference between their 1000th terms? +13. How many three-digit numbers are divisible by 7? +14. How many multiples of 4 lie between 10 and 250? +15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal? +16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. +Reprint 2025-26 + +ARITHMETIC PROGRESSIONS +63 +17. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253. +18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is +44. Find the first three terms of the AP. +19. Subba Rao started work in 1995 at an annual salary of ` 5000 and received an increment +of ` 200 each year. In which year did his income reach ` 7000? +20. Ramkali saved ` 5 in the first week of a year and then increased her weekly savings by +` 1.75. If in the nth week, her weekly savings become ` 20.75, find n. +5.4 Sum of First n Terms of an AP +Let us consider the situation again +given in Section 5.1 in which Shakila +put ` 100 into her daughter’s money +box when she was one year old, +` 150 on her second birthday, +` 200 on her third birthday and will +continue in the same way. How much +money will be collected in the money +box by the time her daughter is 21 +years old? +Here, the amount of money (in `) put in the money box on her first, second, third, +fourth . . . birthday were respectively 100, 150, 200, 250, . . . till her 21st birthday. To +find the total amount in the money box on her 21st birthday, we will have to write each +of the 21 numbers in the list above and then add them up. Don’t you think it would be +a tedious and time consuming process? Can we make the process shorter? This would +be possible if we can find a method for getting this sum. Let us see. +We consider the problem given to Gauss (about whom you read in +Chapter 1), to solve when he was just 10 years old. He was asked to find the sum of +the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can +you guess how did he do? He wrote : +S = 1 + 2 + 3 + . . . + 99 + 100 +And then, reversed the numbers to write +S = 100 + 99 + . . . + 3 + 2 + 1 +Adding these two, he got +2S = (100 + 1) + (99 + 2) + . . . + (3 + 98) + (2 + 99) + (1 + 100) += 101 + 101 + . . . + 101 + 101 +(100 times) +So, +S = 100 +101 +5050 +2 + + +, i.e., the sum = 5050. +Reprint 2025-26 + +64 +MATHEMATICS +We will now use the same technique to find the sum of the first n terms of an AP : +a, a + d, a + 2d, . . . +The nth term of this AP is a + (n – 1) d. Let S denote the sum of the first n terms +of the AP. We have +S = a + (a + d) + (a + 2d) + . . . + [a + (n – 1) d] +(1) +Rewriting the terms in reverse order, we have +S = [a + (n – 1) d] + [a + (n – 2) d] + . . . + (a + d) + a +(2) +On adding (1) and (2), term-wise. we get +2S = +[2 +( +1) ] +[2 +( +1) ] +... +[2 +( +1) ] +[2 +( +1) ] +times + + + + + + + + + + + + + +a +n +d +a +n +d +a +n +d +a +n +d +n +or, +2S = n [2a + (n – 1) d] +(Since, there are n terms) +or, +S = 2 +n [2a + (n – 1) d] +So, the sum of the first n terms of an AP is given by +S = 2 +n [2a + (n – 1) d] +We can also write this as +S = 2 +n [a + a + (n – 1) d] +i.e., +S = 2 +n (a + an) +(3) +Now, if there are only n terms in an AP, then an = l, the last term. +From (3), we see that +S = 2 +n (a + l ) +(4) +This form of the result is useful when the first and the last terms of an AP are +given and the common difference is not given. +Now we return to the question that was posed to us in the beginning. The amount +of money (in Rs) in the money box of Shakila’s daughter on 1st, 2nd, 3rd, 4th birthday, +. . ., were 100, 150, 200, 250, . . ., respectively. +This is an AP. We have to find the total money collected on her 21st birthday, i.e., +the sum of the first 21 terms of this AP. +Reprint 2025-26 + +ARITHMETIC PROGRESSIONS +65 +Here, a = 100, d = 50 and n = 21. Using the formula : +S = + + +2 +( +1) +2 +n +a +n +d + + +, +we have +S = + + +21 2 100 +(21 1) +50 +2 + + + + + = + + +21 200 +1000 +2 + += 21 +1200 +2  + = 12600 +So, the amount of money collected on her 21st birthday is ` 12600. +Hasn’t the use of the formula made it much easier to solve the problem? +We also use Sn in place of S to denote the sum of first n terms of the AP. We +write S20 to denote the sum of the first 20 terms of an AP. The formula for the sum of +the first n terms involves four quantities S, a, d and n. If we know any three of them, +we can find the fourth. +Remark : The nth term of an AP is the difference of the sum to first n terms and the +sum to first (n – 1) terms of it, i.e., an = Sn – Sn – 1. +Let us consider some examples. +Example 11 : Find the sum of the first 22 terms of the AP : 8, 3, –2, . . . +Solution : Here, a = 8, d = 3 – 8 = –5, n = 22. +We know that +S = + + +2 +( +1) +2 +n +a +n +d + + +Therefore, +S = + + +22 16 +21( 5) +2 + + + = 11(16 – 105) = 11(–89) = – 979 +So, the sum of the first 22 terms of the AP is – 979. +Example 12 : If the sum of the first 14 terms of an AP is 1050 and its first term is 10, +find the 20th term. +Solution : Here, S14 = 1050, n = 14, a = 10. +As +Sn = + + +2 +( +1) +2 + + +n +a +n +d , +so, +1050 = + + +14 20 13 +2 + +d = 140 + 91d +Reprint 2025-26 + +66 +MATHEMATICS +i.e., +910 = 91d +or, +d = 10 +Therefore, +a20 = 10 + (20 – 1) × 10 = 200, i.e. 20th term is 200. +Example 13 : How many terms of the AP : 24, 21, 18, . . . must be taken so that their +sum is 78? +Solution : Here, a = 24, d = 21 – 24 = –3, Sn = 78. We need to find n. +We know that +Sn = + + +2 +( +1) +2 + + +n +a +n +d +So, +78 = + + +48 +( +1)( 3) +2 + + + +n +n + =  + +51 +3 +2 + +n +n +or +3n2 – 51n + 156 = 0 +or +n2 – 17n + 52 = 0 +or +(n – 4)(n – 13) = 0 +or +n = 4 or 13 +Both values of n are admissible. So, the number of terms is either 4 or 13. +Remarks: +1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78. +2. Two answers are possible because the sum of the terms from 5th to 13th will be +zero. This is because a is positive and d is negative, so that some terms will be +positive and some others negative, and will cancel out each other. +Example 14 : Find the sum of : +(i) the first 1000 positive integers +(ii) the first n positive integers +Solution : +(i) Let S = 1 + 2 + 3 + . . . + 1000 +Using the formula Sn = +( +) +2 + +n a +l for the sum of the first n terms of an AP, we +have +S1000 = 1000 (1 +1000) +2 + + = 500 × 1001 = 500500 +So, the sum of the first 1000 positive integers is 500500. +(ii) Let Sn = 1 + 2 + 3 + . . . + n +Here a = 1 and the last term l is n. +Reprint 2025-26 + +ARITHMETIC PROGRESSIONS +67 +Therefore, +Sn = +(1 +) +2 +n +n + +or + Sn = +( +1) +2 +n n  +So, the sum of first n positive integers is given by +Sn = +( ++ 1) +2 +n n +Example 15 : Find the sum of first 24 terms of the list of numbers whose nth term is +given by +an = 3 + 2n +Solution : +As +an = 3 + 2n, +so, +a1 = 3 + 2 = 5 +a2 = 3 + 2 × 2 = 7 +a3 = 3 + 2 × 3 = 9 + +List of numbers becomes 5, 7, 9, 11, . . . +Here, +7 – 5 = 9 – 7 = 11 – 9 = 2 and so on. +So, it forms an AP with common difference d = 2. +To find S24, we have n = 24, +a = 5, +d = 2. +Therefore, +S24 = + + +24 2 +5 +(24 +1) +2 +2 + + + + + = + + +12 10 +46 + + = 672 +So, sum of first 24 terms of the list of numbers is 672. +Example 16 : A manufacturer of TV sets produced 600 sets in the third year and 700 +sets in the seventh year. Assuming that the production increases uniformly by a fixed +number every year, find : +(i) the production in the 1st year +(ii) the production in the 10th year +(iii) the total production in first 7 years +Solution : (i) Since the production increases uniformly by a fixed number every year, +the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP. +Let us denote the number of TV sets manufactured in the nth year by an. +Then, +a3 = 600 and a7 = 700 +Reprint 2025-26 + +68 +MATHEMATICS +or, +a + 2d = 600 +and +a + 6d = 700 +Solving these equations, we get d = 25 +and +a = 550. +Therefore, production of TV sets in the first year is 550. +(ii) Now +a10 = a + 9d = 550 + 9 × 25 = 775 +So, production of TV sets in the 10th year is 775. +(iii) Also, +S7 = + + +7 2 +550 +(7 +1) +25 +2 + + + + += + + +7 1100 +150 +2 + + = 4375 +Thus, the total production of TV sets in first 7 years is 4375. +EXERCISE 5.3 +1. Find the sum of the following APs: +(i) 2, 7, 12, . . ., to 10 terms. +(ii) –37, –33, –29, . . ., to 12 terms. +(iii) 0.6, 1.7, 2.8, . . ., to 100 terms. +(iv) +1 +1 +1 +, +, +15 12 10 + , . . ., to 11 terms. +2. Find the sums given below : +(i) 7 + +1 +10 2 + 14 + . . . + 84 +(ii) 34 + 32 + 30 + . . . + 10 +(iii) –5 + (–8) + (–11) + . . . + (–230) +3. In an AP: +(i) given a = 5, d = 3, an = 50, find n and Sn. +(ii) given a = 7, a13 = 35, find d and S13. +(iii) given a12 = 37, d = 3, find a and S12. +(iv) given a3 = 15, S10 = 125, find d and a10. +(v) given d = 5, S9 = 75, find a and a9. +(vi) given a = 2, d = 8, Sn = 90, find n and an. +(vii) given a = 8, an = 62, Sn = 210, find n and d. +(viii) given an = 4, d = 2, Sn = –14, find n and a. +(ix) given a = 3, n = 8, S = 192, find d. +(x) given l = 28, S = 144, and there are total 9 terms. Find a. +Reprint 2025-26 + +ARITHMETIC PROGRESSIONS +69 +4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636? +5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms +and the common difference. +6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference +is 9, how many terms are there and what is their sum? +7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. +8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 +respectively. +9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of +first n terms. +10. Show that a1, a2, . . ., an, . . . form an AP where an is defined as below : +(i) an = 3 + 4n +(ii) an = 9 – 5n +Also find the sum of the first 15 terms in each case. +11. If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What +is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and +the nth terms. +12. Find the sum of the first 40 positive integers divisible by 6. +13. Find the sum of the first 15 multiples of 8. +14. Find the sum of the odd numbers between 0 and 50. +15. A contract on construction job specifies a penalty for delay of completion beyond a +certain date as follows: ` 200 for the first day, ` 250 for the second day, ` 300 for the third +day, etc., the penalty for each succeeding day being ` 50 more than for the preceding day. +How much money the contractor has to pay as penalty, if he has delayed the work by 30 +days? +16. A sum of ` 700 is to be used to give seven cash prizes to students of a school for their +overall academic performance. If each prize is ` 20 less than its preceding prize, find the +value of each of the prizes. +17. In a school, students thought of planting trees in and around the school to reduce air +pollution. It was decided that the number of trees, that each section of each class will +plant, will be the same as the class, in which they are studying, e.g., a section of Class I +will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are +three sections of each class. How many trees will be planted by the students? +18. A spiral is made up of successive semicircles, with centres alternately at A and B, +starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in +Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive +semicircles? (Take  = 22 +7 ) +Reprint 2025-26 + +70 +MATHEMATICS +Fig. 5.4 +[Hint : Length of successive semicircles is l1, l2, l3, l4, . . . with centres at A, B, A, B, . . ., +respectively.] +19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, +18 in the row next to it and so on (see Fig. 5.5). In how many rows are the 200 logs placed +and how many logs are in the top row? +Fig. 5.5 +20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, +and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the +line (see Fig. 5.6). +Fig. 5.6 +A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops +it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and +she continues in the same way until all the potatoes are in the bucket. What is the total +distance the competitor has to run? +[Hint : To pick up the first potato and the second potato, the total distance (in metres) +run by a competitor is 2 × 5 + 2 × (5 + 3)] +Reprint 2025-26 + +ARITHMETIC PROGRESSIONS +71 +EXERCISE 5.4 (Optional)* +1. Which term of the AP : 121, 117, 113, . . ., is +its first negative term? +[Hint : Find n for an < 0] +2. The sum of the third and the seventh terms +of an AP is 6 and their product is 8. Find +the sum of first sixteen terms of the AP. +3. A ladder has rungs 25 cm apart. +(see Fig. 5.7). The rungs decrease +uniformly in length from 45 cm at the +bottom to 25 cm at the top. If the top and +the bottom rungs are +1 +2 2 m apart, what is +the length of the wood required for the +rungs? +[Hint : Number of rungs = 250 +1 +25 ] +4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value +of x such that the sum of the numbers of the houses preceding the house numbered x is +equal to the sum of the numbers of the houses following it. Find this value of x. +[Hint : Sx – 1 = S49 – Sx] +5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and +built of solid concrete. +Each step has a rise of 1 +4 m and a tread of 1 +2 m. (see Fig. 5.8). Calculate the total volume +of concrete required to build the terrace. +[Hint : Volume of concrete required to build the first step = +3 +1 +1 +50 m +4 +2 + + +] +Fig. 5.8 +Fig. 5.7 +* These exercises are not from the examination point of view. +Reprint 2025-26 + +72 +MATHEMATICS +5.5 Summary +In this chapter, you have studied the following points : +1. An arithmetic progression (AP) is a list of numbers in which each term is obtained by +adding a fixed number d to the preceding term, except the first term. The fixed number d +is called the common difference. +The general form of an AP is a, a + d, a + 2d, a + 3d, . . . +2. A given list of numbers a1, a2, a3, . . . is an AP, if the differences a2 – a1, a3 – a2, +a4 – a3, . . ., give the same value, i.e., if ak + 1 – ak is the same for different values of k. +3. In an AP with first term a and common difference d, the nth term (or the general term) is +given by + an = a + (n – 1) d. +4. The sum of the first n terms of an AP is given by : +S = + + +2 +( +1) +2 +n +a +n +d + + +5. If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP +is given by : +S = +( +) +2 +n a +l + +A NOTE TO THE READER +If a, b, c are in AP, then b = +2 +a +c + + and b is called the arithmetic +mean of a and c. +Reprint 2025-26" +class_10,6,Triangles,ncert_books/class_10/jemh1dd/jemh106.pdf,"TRIANGLES +73 +6 +6.1 Introduction +You are familiar with triangles and many of their properties from your earlier classes. +In Class IX, you have studied congruence of triangles in detail. Recall that two figures +are said to be congruent, if they have the same shape and the same size. In this +chapter, we shall study about those figures which have the same shape but not necessarily +the same size. Two figures having the same shape (and not necessarily the same size) +are called similar figures. In particular, we shall discuss the similarity of triangles and +apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier. +Can you guess how heights of mountains (say Mount Everest) or distances of +some long distant objects (say moon) have been found out? Do you think these have +TRIANGLES +Reprint 2025-26 + +74 +MATHEMATICS +been measured directly with the help of a measuring tape? In fact, all these heights +and distances have been found out using the idea of indirect measurements, which is +based on the principle of similarity of figures (see Example 7, Q.15 of Exercise 6.3 +and also Chapters 8 and 9 of this book). +6.2 Similar Figures +In Class IX, you have seen that all circles with the same radii are congruent, all +squares with the same side lengths are congruent and all equilateral triangles with the +same side lengths are congruent. +Now consider any two (or more) +circles [see Fig. 6.1 (i)]. Are they +congruent? Since all of them do not +have the same radius, they are not +congruent to each other. Note that +some are congruent and some are not, +but all of them have the same shape. +So they all are, what we call, similar. +Two similar figures have the same +shape but not necessarily the same +size. Therefore, all circles are similar. +What about two (or more) squares or +two (or more) equilateral triangles +[see Fig. 6.1 (ii) and (iii)]? As observed +in the case of circles, here also all +squares are similar and all equilateral +triangles are similar. +From the above, we can say +that all congruent figures are +similar but the similar figures need +not be congruent. +Can a circle and a square be +similar? Can a triangle and a square +be similar? These questions can be +answered by just looking at the +figures (see Fig. 6.1). Evidently +these figures are not similar. (Why?) +Fig. 6.1 +Fig. 6.2 +Reprint 2025-26 + +TRIANGLES +75 +What can you say about the two quadrilaterals ABCD and PQRS +(see Fig 6.2)?Are they similar? These figures appear to be similar but we cannot be +certain about it.Therefore, we must have some definition of similarity of figures and +based on this definition some rules to decide whether the two given figures are similar +or not. For this, let us look at the photographs given in Fig. 6.3: +Fig. 6.3 +You will at once say that they are the photographs of the same monument +(Taj Mahal) but are in different sizes. Would you say that the three photographs are +similar? Yes,they are. +What can you say about the two photographs of the same size of the same +person one at the age of 10 years and the other at the age of 40 years? Are these +photographs similar? These photographs are of the same size but certainly they are +not of the same shape. So, they are not similar. +What does the photographer do when she prints photographs of different sizes +from the same negative? You must have heard about the stamp size, passport size and +postcard size photographs. She generally takes a photograph on a small size film, say +of 35mm size and then enlarges it into a bigger size, say 45mm (or 55mm). Thus, if we +consider any line segment in the smaller photograph (figure), its corresponding line +segment in the bigger photograph (figure) will be 45 +35 +55 +or 35 + + + + + + of that of the line segment. +This really means that every line segment of the smaller photograph is enlarged +(increased) in the ratio 35:45 (or 35:55). It can also be said that every line segment +of the bigger photograph is reduced (decreased) in the ratio 45:35 (or 55:35). Further, +if you consider inclinations (or angles) between any pair of corresponding line segments +in the two photographs of different sizes, you shall see that these inclinations(or angles) +are always equal. This is the essence of the similarity of two figures and in particular +of two polygons. We say that: +Two polygons of the same number of sides are similar, if (i) their +corresponding angles are equal and (ii) their corresponding sides are in the +same ratio (or proportion). +Reprint 2025-26 + +76 +MATHEMATICS +Note that the same ratio of the corresponding sides is referred to as the scale +factor (or the Representative Fraction) for the polygons. You must have heard that +world maps (i.e., global maps) and blue prints for the construction of a building are +prepared using a suitable scale factor and observing certain conventions. +In order to understand similarity of figures more clearly, let us perform the following +activity: +Activity 1 : Place a lighted bulb at a +point O on the ceiling and directly below +it a table in your classroom. Let us cut a +polygon, say a quadrilateral ABCD, from +a plane cardboard and place this +cardboard parallel to the ground between +the lighted bulb and the table. Then a +shadow of ABCD is cast on the table. +Mark the outline of this shadow as +ABCD (see Fig.6.4). +Note that the quadrilateral ABCD is +an enlargement (or magnification) of the +quadrilateral ABCD. This is because of +the property of light that light propogates +in a straight line. You may also note that +A lies on ray OA, B lies on ray OB, C +lies on OC and D lies on OD. Thus, quadrilaterals ABCD and ABCD are of the +same shape but of different sizes. +So, quadrilateral ABCD is similiar to quadrilateral ABCD. We can also say +that quadrilateral ABCD is similar to the quadrilateral ABCD. +Here, you can also note that vertex A corresponds to vertex A, vertex B +corresponds to vertex B, vertex C corresponds to vertex C and vertex D corresponds +to vertex D. Symbolically, these correspondences are represented as A A, B B, +C C and D D. By actually measuring the angles and the sides of the two +quadrilaterals, you may verify that +(i)  A =  A,  B =  B,  C =  C,  D =  D and +(ii) AB +BC +CD +DA +A B +B C +C D +D A + + + + + + + + + +. +This again emphasises that two polygons of the same number of sides are +similar, if (i) all the corresponding angles are equal and (ii) all the corresponding +sides are in the same ratio (or proportion). +Fig. 6.4 +Reprint 2025-26 + +TRIANGLES +77 +From the above, you can easily say that quadrilaterals ABCD and PQRS of +Fig. 6.5 are similar. +Fig. 6.5 +Remark : You can verify that if one polygon is similar to another polygon and this +second polygon is similar to a third polygon, then the first polygon is similar to the third +polygon. +You may note that in the two quadrilaterals (a square and a rectangle) of +Fig. 6.6, corresponding angles are equal, but their corresponding sides are not in the +same ratio. +Fig. 6.6 +So, the two quadrilaterals are not similar. Similarly, you may note that in the two +quadrilaterals (a square and a rhombus) of Fig. 6.7, corresponding sides are in the +same ratio, but their corresponding angles are not equal. Again, the two polygons +(quadrilaterals) are not similar. +Reprint 2025-26 + +78 +MATHEMATICS +Fig. 6.7 +Thus, either of the above two conditions (i) and (ii) of similarity of two +polygons is not sufficient for them to be similar. +EXERCISE 6.1 +1. Fill in the blanks using the correct word given in brackets : +(i) All circles are +. (congruent, similar) +(ii) All squares are +. (similar, congruent) +(iii) All + triangles are similar. (isosceles, equilateral) +(iv) Two polygons of the same number of sides are similar, if (a) their corresponding +angles are + and (b) their corresponding sides are +.(equal, +proportional) +2. Give two different examples of pair of +(i) similar figures. +(ii) non-similar figures. +3. State whether the following quadrilaterals are similar or not: +Fig. 6.8 +Reprint 2025-26 + +TRIANGLES +79 +6.3 Similarity of Triangles +What can you say about the similarity of two triangles? +You may recall that triangle is also a polygon. So, we can state the same conditions +for the similarity of two triangles. That is: +Two triangles are similiar, if +(i) their corresponding angles are equal and +(ii) their corresponding sides are in the same ratio (or proportion). +Note that if corresponding angles of two +triangles are equal, then they are known as +equiangular triangles. A famous Greek +mathematician Thales gave an important truth relating +to two equiangular triangles which is as follows: +The ratio of any two corresponding sides in +two equiangular triangles is always the same. +It is believed that he had used a result called +the Basic Proportionality Theorem (now known as +the Thales Theorem) for the same. +To understand the Basic Proportionality +Theorem, let us perform the following activity: +Activity 2 : Draw any angle XAY and on its one +arm AX, mark points (say five points) P, Q, D, R and +B such that AP = PQ = QD = DR = RB. +Now, through B, draw any line intersecting arm +AY at C (see Fig. 6.9). +Also, through the point D, draw a line parallel +to BC to intersect AC at E. Do you observe from +your constructions that AD +3 +DB +2 + +? Measure AE and +EC. What about AE +EC ? Observe that AE +EC is also equal to 3 +2 . Thus, you can see that +in  ABC, DE || BC and AD +AE +DB +EC + +. Is it a coincidence? No, it is due to the following +theorem (known as the Basic Proportionality Theorem): +Thales +(640 – 546 B.C.) +Fig. 6.9 +Reprint 2025-26 + +80 +MATHEMATICS +Theorem 6.1 : If a line is drawn parallel to one side of a triangle to intersect the +other two sides in distinct points, the other two sides are divided in the same +ratio. +Proof : We are given a triangle ABC in which a line +parallel to side BC intersects other two sides AB and +AC at D and E respectively (see Fig. 6.10). +We need to prove that AD +AE +DB +EC + +. +Let us join BE and CD and then draw DM  AC and +EN  AB. +Now, area of  ADE (= 1 +2 base × height) = 1 +2 AD × EN. +Recall from Class IX, that area of  ADE is denoted as ar(ADE). +So, +ar(ADE) = 1 +2 AD × EN +Similarly, +ar(BDE) = 1 +2 DB × EN, +ar(ADE) = 1 +2 AE × DM and ar(DEC) = 1 +2 EC × DM. +Therefore, +ar(ADE) +ar(BDE) = +1 AD × EN +AD +2 +1 +DB +DB × EN +2 + +(1) +and +ar(ADE) +ar(DEC) = +1 AE × DM +AE +2 +1 +EC +EC × DM +2 + +(2) +Note that  BDE and DEC are on the same base DE and between the same parallels +BC and DE. +So, +ar(BDE) = ar(DEC) +(3) +Fig. 6.10 +Reprint 2025-26 + +TRIANGLES +81 +Therefore, from (1), (2) and (3), we have : +AD +DB = AE +EC +Is the converse of this theorem also true (For the meaning of converse, see +Appendix 1)? To examine this, let us perform the following activity: +Activity 3 : Draw an angle XAY on your +notebook and on ray AX, mark points B1, B2, +B3, B4 and B such that AB1 = B1B2 = B2B3 = +B3B4 = B4B. +Similarly, on ray AY, mark points +C1, C2, C3, C4 and C such that AC1 = C1C2 = +C2C3 = C3C4 = C4C. Then join B1C1 and BC +(see Fig. 6.11). +Note that +1 +1 +AB +B B = +1 +1 +AC +C C +(Each equal to 1 +4 ) +You can also see that lines B1C1 and BC are parallel to each other, i.e., +B1C1 || BC +(1) +Similarly, by joining B2C2, B3C3 and B4C4, you can see that: +2 +2 +AB +B B = +2 +2 +AC +C C +2 +3 + + + + + + +and B2C2 || BC +(2) +3 +3 +AB +B B = +3 +3 +AC +C C +3 +2 + + + + + + +and B3C3 || BC +(3) +4 +4 +AB +B B = +4 +4 +AC +C C +4 +1 + + + + + + +and B4C4 || BC +(4) +From (1), (2), (3) and (4), it can be observed that if a line divides two sides of a +triangle in the same ratio, then the line is parallel to the third side. +You can repeat this activity by drawing any angle XAY of different measure and +taking any number of equal parts on arms AX and AY . Each time, you will arrive at +the same result. Thus, we obtain the following theorem, which is the converse of +Theorem 6.1: +Fig. 6.11 +Reprint 2025-26 + +82 +MATHEMATICS +Fig. 6.13 +Theorem 6.2 : If a line divides any two sides of a +triangle in the same ratio, then the line is parallel +to the third side. +This theorem can be proved by taking a line DE such +that AD +AE +DB +EC + + and assuming that DE is not parallel +to BC (see Fig. 6.12). +If DE is not parallel to BC, draw a line DE +parallel to BC. +So, +AD +DB = AE +E C + + +(Why ?) +Therefore, +AE +EC = AE +E C + + +(Why ?) +Adding 1 to both sides of above, you can see that E and E must coincide. +(Why ?) +Let us take some examples to illustrate the use of the above theorems. +Example 1 : If a line intersects sides AB and AC of a  ABC at D and E respectively +and is parallel to BC, prove that AD +AB = AE +AC (see Fig. 6.13). +Solution : +DE || BC +(Given) +So, +AD +DB = AE +EC +(Theorem 6.1) +or, +DB +AD = EC +AE +or, +DB +1 +AD  = EC +1 +AE  +or, +AB +AD = AC +AE +So, +AD +AB = AE +AC +Fig. 6.12 +Reprint 2025-26 + +TRIANGLES +83 +Example 2 : ABCD is a trapezium with AB || DC. E +and F are points on non-parallel sides AD and BC +respectively such that EF is parallel to AB +(see Fig. 6.14). Show that AE +BF +ED +FC + +. +Solution : Let us join AC to intersect EF at G +(see Fig. 6.15). +AB || DC and EF || AB +(Given) +So, +EF || DC +(Lines parallel to the same line are +parallel to each other) +Now, in  ADC, +EG || DC +(As EF || DC) +So, AE +ED = AG +GC +(Theorem 6.1) + (1) +Similarly, from  CAB, +CG +AG = CF +BF +i.e., +AG +GC = BF +FC +(2) +Therefore, from (1) and (2), +AE +ED = BF +FC +Example 3 : In Fig. 6.16, +PS +SQ = PT +TR and  PST = + PRQ. Prove that PQR is an isosceles triangle. +Solution : It is given that +PS +PT +SQ +TR + + +So, +ST || QR +(Theorem 6.2) +Therefore, + PST =  PQR +(Corresponding angles) +(1) +Fig. 6.14 +Fig. 6.15 +Fig. 6.16 +Reprint 2025-26 + +84 +MATHEMATICS +Also, it is given that + PST =  PRQ +(2) +So, + PRQ =  PQR [From (1) and (2)] +Therefore, +PQ = PR +(Sides opposite the equal angles) +i.e., +PQR is an isosceles triangle. +EXERCISE 6.2 +1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). +Fig. 6.17 +2. E and F are points on the sides PQ and PR +respectively of a  PQR. For each of the following +cases, state whether EF || QR : +(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm +(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm +(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm +3. In Fig. 6.18, if LM || CB and LN || CD, prove that +AM +AN +AB +AD + + +4. In Fig. 6.19, DE || AC and DF || AE. Prove that +BF +BE +FE +EC + + +Fig. 6.18 +Fig. 6.19 +Reprint 2025-26 + +TRIANGLES +85 +5. In Fig. 6.20, DE || OQ and DF || OR. Show that +EF || QR. +6. In Fig. 6.21, A, B and C are points on OP, OQ and +OR respectively such that AB || PQ and AC || PR. +Show that BC || QR. +7. Using Theorem 6.1, prove that a line drawn through +the mid-point of one side of a triangle parallel to +another side bisects the third side. (Recall that you +have proved it in Class IX). +8. Using Theorem 6.2, prove that the line joining the +mid-points of any two sides of a triangle is parallel +to the third side. (Recall that you have done it in +Class IX). +9. ABCD is a trapezium in which AB || DC and its +diagonals intersect each other at the point O. Show +that AO +CO +BO +DO + + +10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that +AO +CO +BO +DO + + Show that ABCD is a trapezium. +6.4 Criteria for Similarity of Triangles +In the previous section, we stated that two triangles are similar, if (i) their corresponding +angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). +That is, in  ABC and  DEF, if +(i)  A =  D,  B =  E,  C =  F and +(ii) AB +BC +CA , +DE +EF +FD + + + then the two triangles are similar (see Fig. 6.22). +Fig. 6.22 +Fig. 6.20 +Fig. 6.21 +Reprint 2025-26 + +86 +MATHEMATICS +Here, you can see that A corresponds to D, B corresponds to E and C +corresponds to F. Symbolically, we write the similarity of these two triangles as +‘ ABC ~  DEF’ and read it as ‘triangle ABC is similar to triangle DEF’. The +symbol ‘~’ stands for ‘is similar to’. Recall that you have used the symbol ‘’ for +‘is congruent to’ in Class IX. +It must be noted that as done in the case of congruency of two triangles, the +similarity of two triangles should also be expressed symbolically, using correct +correspondence of their vertices. For example, for the triangles ABC and DEF of +Fig. 6.22, we cannot write  ABC ~  EDF or  ABC ~  FED. However, we +can write  BAC ~  EDF. +Now a natural question arises : For checking the similarity of two triangles, say +ABC and DEF, should we always look for all the equality relations of their corresponding +angles ( A =  D,  B =  E,  C =  F) and all the equality relations of the ratios +of their corresponding sides +AB +BC +CA +DE +EF +FD + + + + + + + +? Let us examine. You may recall that +in Class IX, you have obtained some criteria for congruency of two triangles involving +only three pairs of corresponding parts (or elements) of the two triangles. Here also, +let us make an attempt to arrive at certain criteria for similarity of two triangles involving +relationship between less number of pairs of corresponding parts of the two triangles, +instead of all the six pairs of corresponding parts. For this, let us perform the following +activity: +Activity 4 : Draw two line segments BC and EF of two different lengths, say 3 cm +and 5 cm respectively. Then, at the points B and C respectively, construct angles PBC +and QCB of some measures, say, 60° and 40°. Also, at the points E and F, construct +angles REF and SFE of 60° and 40° respectively (see Fig. 6.23). +Fig. 6.23 +Reprint 2025-26 + +TRIANGLES +87 +Let rays BP and CQ intersect each other at A and rays ER and FS intersect each +other at D. In the two triangles ABC and DEF, you can see that + B =  E,  C =  F and  A =  D. That is, corresponding angles of these two +triangles are equal. What can you say about their corresponding sides ? Note that +BC +3 +0.6. +EF +5 + + + What about AB +DE and CA +FD ? On measuring AB, DE, CA and FD, you +will find that AB +DE and CA +FD are also equal to 0.6 (or nearly equal to 0.6, if there is some +error in the measurement). Thus, AB +BC +CA +DE +EF +FD + + + You can repeat this activity by +constructing several pairs of triangles having their corresponding angles equal. Every +time, you will find that their corresponding sides are in the same ratio (or proportion). +This activity leads us to the following criterion for similarity of two triangles. +Theorem 6.3 : If in two triangles, corresponding angles are equal, then their +corresponding sides are in the same ratio (or proportion) and hence the two +triangles are similar. +This criterion is referred to as the AAA +(Angle–Angle–Angle) criterion of +similarity of two triangles. +This theorem can be proved by taking two +triangles ABC and DEF such that + A =  D,  B =  E and  C =  F +(see Fig. 6.24) +Cut DP = AB and DQ = AC and join PQ. +So, + ABC  DPQ +(Why ?) +This gives + B =  P =  E and PQ || EF +(How?) +Therefore, +DP +PE = +DQ +QF +(Why?) +i.e., +AB +DE = +AC +DF +(Why?) +Similarly, +AB +DE = BC +EF +and so AB +BC +AC +DE +EF +DF + + +. +Remark : If two angles of a triangle are respectively equal to two angles of another +triangle, then by the angle sum property of a triangle their third angles will also be +equal. Therefore, AAA similarity criterion can also be stated as follows: +Fig. 6.24 +Reprint 2025-26 + +88 +MATHEMATICS +If two angles of one triangle are respectively equal to two angles of another +triangle, then the two triangles are similar. +This may be referred to as the AA similarity criterion for two triangles. +You have seen above that if the three angles of one triangle are respectively +equal to the three angles of another triangle, then their corresponding sides are +proportional (i.e., in the same ratio). What about the converse of this statement? Is the +converse true? In other words, if the sides of a triangle are respectively proportional to +the sides of another triangle, is it true that their corresponding angles are equal? Let us +examine it through an activity : +Activity 5 : Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm, +CA = 8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm (see Fig. 6.25). +Fig. 6.25 +So, you have : +AB +BC +CA +DE +EF +FD + + +(each equal to 2 +3 ) +Now measure  A,  B,  C,  D,  E and  F. You will observe that + A =  D,  B =  E and  C =  F, i.e., the corresponding angles of the two +triangles are equal. +You can repeat this activity by drawing several such triangles (having their sides +in the same ratio). Everytime you shall see that their corresponding angles are equal. +It is due to the following criterion of similarity of two triangles: +Theorem 6.4 : If in two triangles, sides of one triangle are proportional to +(i.e., in the same ratio of ) the sides of the other triangle, then their corresponding +angles are equal and hence the two triangles are similiar. +This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for +two triangles. +This theorem can be proved by taking two triangles ABC and DEF such that +AB +BC +CA +DE +EF +FD + + + (< 1) (see Fig. 6.26): +Reprint 2025-26 + +TRIANGLES +89 +Fig. 6.26 +Cut DP = AB and DQ = AC and join PQ. +It can be seen that +DP +PE = +DQ +QF +and PQ || EF +(How?) +So, + P = E +and + Q =  F. +Therefore, +DP +DE = DQ +DF = PQ +EF +So, +DP +DE = DQ +DF = BC +EF +(Why?) +So, +BC = PQ +(Why?) +Thus, + ABC  DPQ +(Why ?) +So, + A =  D,  B =  E +and + C =  F +(How ?) +Remark : You may recall that either of the two conditions namely, (i) corresponding +angles are equal and (ii) corresponding sides are in the same ratio is not sufficient for +two polygons to be similar. However, on the basis of Theorems 6.3 and 6.4, you can +now say that in case of similarity of the two triangles, it is not necessary to check both +the conditions as one condition implies the other. +Let us now recall the various criteria for congruency of two triangles learnt in +Class IX. You may observe that SSS similarity criterion can be compared with the SSS +congruency criterion.This suggests us to look for a similarity criterion comparable to +SAS congruency criterion of triangles. For this, let us perform an activity. +Activity 6 : Draw two triangles ABC and DEF such that AB = 2 cm,  A = 50°, +AC = 4 cm, DE = 3 cm,  D = 50° and DF = 6 cm (see Fig.6.27). +Reprint 2025-26 + +90 +MATHEMATICS +Fig. 6.27 +Here, you may observe that AB +DE = AC +DF (each equal to 2 +3 ) and  A (included +between the sides AB and AC) =  D (included between the sides DE and DF). That +is, one angle of a triangle is equal to one angle of another triangle and sides including +these angles are in the same ratio (i.e., proportion). Now let us measure  B,  C, + E and  F. +You will find that  B =  E and  C =  F. That is,  A =  D,  B =  E and + C =  F. So, by AAA similarity criterion,  ABC ~  DEF. You may repeat this +activity by drawing several pairs of such triangles with one angle of a triangle equal to +one angle of another triangle and the sides including these angles are proportional. +Everytime, you will find that the triangles are similar. It is due to the following criterion +of similarity of triangles: +Theorem 6.5 : If one angle of a triangle is equal to one angle of the other +triangle and the sides including these angles are proportional, then the two +triangles are similar. +This criterion is referred to as +the SAS (Side–Angle–Side) +similarity criterion for two +triangles. +As before, this theorem can +be proved by taking two triangles +ABC and DEF such that +AB +AC +DE +DF + + ( 1) and  A =  D +(see Fig. 6.28). Cut DP = AB, DQ += AC and join PQ. +Fig. 6.28 +Reprint 2025-26 + +TRIANGLES +91 +Now, +PQ || EF and  ABC  DPQ +(How ?) +So, + A =  D,  B =  P and  C =  Q +Therefore, + ABC ~  DEF +(Why?) +We now take some examples to illustrate the use of these criteria. +Example 4 : In Fig. 6.29, if PQ || RS, prove that  POQ ~  SOR. +Fig. 6.29 +Solution : +PQ || RS +(Given) +So, + P =  S +(Alternate angles) +and + Q =  R +Also, + POQ =  SOR +(Vertically opposite angles) +Therefore, + POQ ~  SOR +(AAA similarity criterion) +Example 5 : Observe Fig. 6.30 and then find  P. +Fig. 6.30 +Solution : In  ABC and  PQR, +Reprint 2025-26 + +92 +MATHEMATICS +AB +3.8 +1 , +RQ +7.6 +2 + + +BC +6 +1 +QP +12 +2 + + + and +CA +3 3 +1 +PR +2 +6 3 + + +That is, +AB +BC +CA +RQ +QP +PR + + +So, + ABC ~  RQP +(SSS similarity) +Therefore, + C =  P +(Corresponding angles of similar triangles) +But + C = 180° –  A –  B +(Angle sum property) += 180° – 80° – 60° = 40° +So, + P = 40° +Example 6 : In Fig. 6.31, +OA . OB = OC . OD. +Show that + A =  C and  B =  D. +Solution : +OA . OB = OC . OD +(Given) +So, +OA +OC = OD +OB +(1) +Also, we have + AOD =  COB +(Vertically opposite angles) (2) +Therefore, from (1) and (2), + AOD ~  COB (SAS similarity criterion) +So, + A =  C and  D =  B +(Corresponding angles of similar triangles) +Example 7 : A girl of height 90 cm is +walking away from the base of a +lamp-post at a speed of 1.2 m/s. If the lamp +is 3.6 m above the ground, find the length +of her shadow after 4 seconds. +Solution : Let AB denote the lamp-post +and CD the girl after walking for 4 seconds +away from the lamp-post (see Fig. 6.32). +From the figure, you can see that DE is the +shadow of the girl. Let DE be x metres. +Fig. 6.31 +Fig. 6.32 +Reprint 2025-26 + +TRIANGLES +93 +Fig. 6.33 +Now, BD = 1.2 m × 4 = 4.8 m. +Note that in  ABE and  CDE, + B =  D +(Each is of 90° because lamp-post +as well as the girl are standing +vertical to the ground) +and + E =  E +(Same angle) +So, + ABE ~  CDE +(AA similarity criterion) +Therefore, +BE +DE = AB +CD +i.e., +4.8 + x +x + = 3.6 +0.9 +(90 cm = 90 +100 m = 0.9 m) +i.e., +4.8 + x = 4x +i.e., +3x = 4.8 +i.e., +x = 1.6 +So, the shadow of the girl after walking for 4 seconds is 1.6 m long. +Example 8 : In Fig. 6.33, CM and RN are +respectively the medians of  ABC and + PQR. If  ABC ~  PQR, prove that : +(i)  AMC ~  PNR +(ii) +CM +AB +RN +PQ + +(iii)  CMB ~  RNQ +Solution : (i) + ABC ~  PQR +(Given) +So, +AB +PQ = +BC +CA +QR +RP + +(1) +and + A =  P,  B =  Q and  C =  R +(2) +But +AB = 2 AM and PQ = 2 PN +(As CM and RN are medians) +So, from (1), +2AM +2PN = CA +RP +Reprint 2025-26 + +94 +MATHEMATICS +i.e., +AM +PN = CA +RP +(3) +Also, + MAC =  NPR +[From (2)] +(4) +So, from (3) and (4), + AMC ~  PNR +(SAS similarity) +(5) +(ii) From (5), +CM +RN = CA +RP +(6) +But +CA +RP = +AB +PQ +[From (1)] +(7) +Therefore, +CM +RN = +AB +PQ +[From (6) and (7)] +(8) +(iii) Again, +AB +PQ = +BC +QR +[From (1)] +Therefore, +CM +RN = BC +QR +[From (8)] +(9) +Also, +CM +RN = AB +2 BM +PQ +2 QN + +i.e., +CM +RN = +BM +QN +(10) +i.e., +CM +RN = +BC +BM +QR +QN + +[From (9) and (10)] +Therefore, + CMB ~  RNQ +(SSS similarity) +[Note : You can also prove part (iii) by following the same method as used for proving +part (i).] +EXERCISE 6.3 +1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by +you for answering the question and also write the pairs of similar triangles in the symbolic +form : +Reprint 2025-26 + +TRIANGLES +95 +Fig. 6.34 +2. In Fig. 6.35,  ODC ~  OBA,  BOC = 125° +and  CDO = 70°. Find  DOC,  DCO and + OAB. +3. Diagonals AC and BD of a trapezium ABCD +with AB || DC intersect each other at the +point O. Using a similarity criterion for two +triangles, show that +OA +OB +OC +OD + + +Fig. 6.35 +Reprint 2025-26 + +96 +MATHEMATICS +4. In Fig. 6.36, +QR +QT +QS +PR + + and  1 =  2. Show +that  PQS ~  TQR. +5. S and T are points on sides PR and QR of + PQR such that  P =  RTS. Show that + RPQ ~  RTS. +6. In Fig. 6.37, if  ABE  ACD, show that + ADE ~  ABC. +7. In Fig. 6.38, altitudes AD and CE of  ABC +intersect each other at the point P. Show +that: +(i)  AEP ~  CDP +(ii)  ABD ~  CBE +(iii)  AEP ~  ADB +(iv)  PDC ~  BEC +8. E is a point on the side AD produced of a +parallelogram ABCD and BE intersects CD +at F. Show that  ABE ~  CFB. +9. In Fig. 6.39, ABC and AMP are two right +triangles, right angled at B and M +respectively. Prove that: +(i)  ABC ~  AMP +(ii) CA +BC +PA +MP + +10. CD and GH are respectively the bisectors +of  ACB and  EGF such that D and H lie +on sides AB and FE of  ABC and  EFG +respectively. If  ABC ~  FEG, show that: +(i) CD +AC +GH +FG + +(ii)  DCB ~  HGE +(iii)  DCA ~  HGF +Fig. 6.36 +Fig. 6.37 +Fig. 6.38 +Fig. 6.39 +Reprint 2025-26 + +TRIANGLES +97 +11. In Fig. 6.40, E is a point on side CB +produced of an isosceles triangle ABC +with AB = AC. If AD  BC and EF  AC, +prove that  ABD ~  ECF. +12. Sides AB and BC and median AD of a +triangle ABC are respectively propor- +tional to sides PQ and QR and median +PM of  PQR (see Fig. 6.41). Show that + ABC ~  PQR. +13. D is a point on the side BC of a triangle +ABC such that  ADC =  BAC. Show +that CA2 = CB.CD. +14. Sides AB and AC and median AD of a +triangle ABC are respectively +proportional to sides PQ and PR and +median PM of another triangle PQR. +Show that  ABC ~  PQR. +15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time +a tower casts a shadow 28 m long. Find the height of the tower. +16. If AD and PM are medians of triangles ABC and PQR, respectively where + ABC ~  PQR, prove that +AB +AD +PQ +PM + + +6.5 Summary +In this chapter you have studied the following points : +1. Two figures having the same shape but not necessarily the same size are called similar +figures. +2. All the congruent figures are similar but the converse is not true. +3. Two polygons of the same number of sides are similar, if (i) their corresponding angles +are equal and (ii) their corresponding sides are in the same ratio (i.e., proportion). +4. If a line is drawn parallel to one side of a triangle to intersect the other two sides in +distinct points, then the other two sides are divided in the same ratio. +5. If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the +third side. +6. If in two triangles, corresponding angles are equal, then their corresponding sides are in +the same ratio and hence the two triangles are similar (AAA similarity criterion). +7. If in two triangles, two angles of one triangle are respectively equal to the two angles of +the other triangle, then the two triangles are similar (AA similarity criterion). +Fig. 6.40 +Fig. 6.41 +Reprint 2025-26 + +98 +MATHEMATICS +8. If in two triangles, corresponding sides are in the same ratio, then their corresponding +angles are equal and hence the triangles are similar (SSS similarity criterion). +9. If one angle of a triangle is equal to one angle of another triangle and the sides including +these angles are in the same ratio (proportional), then the triangles are similar +(SAS similarity criterion). +A NOTE TO THE READER +If in two right triangles, hypotenuse and one side of one triangle are +proportional to the hypotenuse and one side of the other triangle, +then the two triangles are similar. This may be referred to as the +RHS Similarity Criterion. +If you use this criterion in Example 2, Chapter 8, the proof will become +simpler. +Reprint 2025-26" +class_10,7,Coordinate Geometry,ncert_books/class_10/jemh1dd/jemh107.pdf,"COORDINATE GEOMETRY +99 +7 +7.1 Introduction +In Class IX, you have studied that to locate the position of a point on a plane, we +require a pair of coordinate axes. The distance of a point from the y-axis is called its +x-coordinate, or abscissa. The distance of a point from the x-axis is called its +y-coordinate, or ordinate. The coordinates of a point on the x-axis are of the form +(x, 0), and of a point on the y-axis are of the form (0, y). +Here is a play for you. Draw a set of a pair of perpendicular axes on a graph +paper. Now plot the following points and join them as directed: Join the point A(4, 8) to +B(3, 9) to C(3, 8) to D(1, 6) to E(1, 5) to F(3, 3) to G(6, 3) to H(8, 5) to I(8, 6) to +J(6, 8) to K(6, 9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to +form a triangle. Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle. +Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle. Lastly join S to the points +(0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6). What picture have you got? +Also, you have seen that a linear equation in two variables of the form +ax + by + c = 0, (a, b are not simultaneously zero), when represented graphically, +gives a straight line. Further, in Chapter 2, you have seen the graph of +y = ax2 + bx + c (a ¹ 0), is a parabola. In fact, coordinate geometry has been developed +as an algebraic tool for studying geometry of figures. It helps us to study geometry +using algebra, and understand algebra with the help of geometry. Because of this, +coordinate geometry is widely applied in various fields such as physics, engineering, +navigation, seismology and art! +In this chapter, you will learn how to find the distance between the two points +whose coordinates are given. You will also study how to find the coordinates of the +point which divides a line segment joining two given points in a given ratio. +COORDINATE GEOMETRY +Reprint 2025-26 + +100 +MATHEMATICS +7.2 Distance Formula +Let us consider the following situation: +A town B is located 36 km east and 15 +km north of the town A. How would you find +the distance from town A to town B without +actually measuring it. Let us see. This situation +can be represented graphically as shown in +Fig. 7.1. You may use the Pythagoras Theorem +to calculate this distance. +Now, suppose two points lie on the x-axis. +Can we find the distance between them? For +instance, consider two points A(4, 0) and B(6, 0) +in Fig. 7.2. The points A and B lie on the x-axis. +From the figure you can see that OA = 4 +units and OB = 6 units. +Therefore, the distance of B from A, i.e., +AB = OB – OA = 6 – 4 = 2 units. +So, if two points lie on the x-axis, we can +easily find the distance between them. +Now, suppose we take two points lying on +the y-axis. Can you find the distance between +them. If the points C(0, 3) and D(0, 8) lie on the +y-axis, similarly we find that CD = 8 – 3 = 5 units +(see Fig. 7.2). +Next, can you find the distance of A from C (in Fig. 7.2)? Since OA = 4 units and +OC = 3 units, the distance of A from C, i.e., AC = +2 +2 +3 +4 + + = 5 units. Similarly, you can +find the distance of B from D = BD = 10 units. +Now, if we consider two points not lying on coordinate axis, can we find the +distance between them? Yes! We shall use Pythagoras theorem to do so. Let us see +an example. +In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant. How do we use +Pythagoras theorem to find the distance between them? Let us draw PR and QS +perpendicular to the x-axis from P and Q respectively. Also, draw a perpendicular +from P on QS to meet QS at T. Then the coordinates of R and S are (4, 0) and (6, 0), +respectively. So, RS = 2 units. Also, QS = 8 units and TS = PR = 6 units. +Fig. 7.1 +Fig. 7.2 +Reprint 2025-26 + +COORDINATE GEOMETRY +101 +Therefore, QT = 2 units and PT = RS = 2 units. +Now, using the Pythagoras theorem, we +have + PQ2 = PT2 + QT2 += 22 + 22 = 8 +So, +PQ = 2 2 units +How will we find the distance between two +points in two different quadrants? +Consider the points P(6, 4) and Q(–5, –3) +(see Fig. 7.4). Draw QS perpendicular to the +x-axis. Also draw a perpendicular PT from the +point P on QS (extended) to meet y-axis at the +point R. +Fig. 7.4 +Then PT = 11 units and QT = 7 units. (Why?) +Using the Pythagoras Theorem to the right triangle PTQ, we get +PQ = +2 +2 +11 +7 + + = 170 units. +Fig. 7.3 +Reprint 2025-26 + +102 +MATHEMATICS +Let us now find the distance between any two +points P(x1, y1) and Q(x2, y2). Draw PR and QS +perpendicular to the x-axis. A perpendicular from the +point P on QS is drawn to meet it at the point +T (see Fig. 7.5). +Then, +OR = x1, OS = x2. +So, RS = x2 – x1 = PT. +Also, +SQ = y2, ST = PR = y1. +So, QT = y2 – y1. +Now, applying the Pythagoras theorem in  PTQ, we get +PQ2 = PT2 + QT2 += (x2 – x1)2 + (y2 – y1)2 +Therefore, +PQ = + + + + +2 +2 +2 +1 +2 +1 +x +x +y +y + + + +Note that since distance is always non-negative, we take only the positive square +root. So, the distance between the points P(x1, y1) and Q(x2, y2) is +PQ = + + + + +2 +2 +2 +1 +2 +1 +– ++ +– +x +x +y +y +, +which is called the distance formula. +Remarks : +1. In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by +OP = +2 +2 +x +y + +. +2. We can also write, PQ = + + + + +2 +2 +1 +2 +1 +2 +x +x +y +y + + + +. (Why?) +Example 1 : Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the +type of triangle formed. +Solution : Let us apply the distance formula to find the distances PQ, QR and PR, +where P(3, 2), Q(–2, –3) and R(2, 3) are the given points. We have +PQ = +2 +2 +2 +2 +(3 +2) +(2 +3) +5 +5 +50 + + + + + + + = 7.07 (approx.) +QR = +2 +2 +2 +2 +(–2 – 2) +(–3 – 3) +(– 4) +(–6) +52 + + + + + = 7.21 (approx.) +PR = +2 +2 +2 +2 +(3 – 2) +(2 – 3) +1 +( 1) +2 + + + + + = 1.41 (approx.) +Since the sum of any two of these distances is greater than the third distance, therefore, +the points P, Q and R form a triangle. +Fig. 7.5 +Reprint 2025-26 + +COORDINATE GEOMETRY +103 +Also, PQ2 + PR2 = QR2, by the converse of Pythagoras theorem, we have  P = 90°. +Therefore, PQR is a right triangle. +Example 2 : Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices +of a square. +Solution : Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points. One way +of showing that ABCD is a square is to use the property that all its sides should be +equal and both its digonals should also be equal. Now, +AB = +2 +2 +(1 – 4) +(7 +2) +9 +25 +34 + + + + + +BC = +2 +2 +(4 +1) +(2 +1) +25 +9 +34 + + + + + + +CD = +2 +2 +(–1 +4) +(–1 – 4) +9 +25 +34 + + + + + +DA = +2 +2 +(1 +4) +(7 – 4) +25 +9 +34 + + + + + +AC = +2 +2 +(1 1) +(7 +1) +4 +64 +68 + + + + + + +BD = +2 +2 +(4 +4) +(2 +4) +64 +4 +68 + + + + + + +Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral +ABCD are equal and its diagonals AC and BD are also equal. Thereore, ABCD is a +square. +Alternative Solution : We find +the four sides and one diagonal, say, +AC as above. Here AD2 + DC2 = +34 + 34 = 68 = AC2. Therefore, by +the converse of Pythagoras +theorem,  D = 90°. A quadrilateral +with all four sides equal and one +angle 90° is a square. So, ABCD +is a square. +Example 3 : Fig. 7.6 shows the +arrangement of desks in a +classroom. Ashima, Bharti and +Camella are seated at A(3, 1), +B(6, 4) and C(8, 6) respectively. +Do you think they are seated in a +line? Give reasons for your +answer. +Fig. 7.6 +Reprint 2025-26 + +104 +MATHEMATICS +Solution : Using the distance formula, we have +AB = +2 +2 +(6 +3) +(4 +1) +9 +9 +18 +3 2 + + + + + + + +BC = +2 +2 +(8 – 6) +(6 – 4) +4 +4 +8 +2 2 + + + + + +AC = +2 +2 +(8 – 3) +(6 – 1) +25 +25 +50 +5 2 + + + + + +Since, AB + BC = 3 2 +2 2 +5 2 +AC, + + + + we can say that the points A, B and C +are collinear. Therefore, they are seated in a line. +Example 4 : Find a relation between x and y such that the point (x , y) is equidistant +from the points (7, 1) and (3, 5). +Solution : Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5). +We are given that AP = BP. So, AP2 = BP2 +i.e., +(x – 7)2 + (y – 1)2 = (x – 3)2 + (y – 5)2 +i.e., +x2 – 14x + 49 + y2 – 2y + 1 = x2 – 6x + 9 + y2 – 10y + 25 +i.e., +x – y = 2 +which is the required relation. +Remark : Note that the graph of the equation +x – y = 2 is a line. From your earlier studies, +you know that a point which is equidistant +from A and B lies on the perpendicular +bisector of AB. Therefore, the graph of +x – y = 2 is the perpendicular bisector of AB +(see Fig. 7.7). +Example 5 : Find a point on the y-axis which +is equidistant from the points A(6, 5) and +B(– 4, 3). +Solution : We know that a point on the +y-axis is of the form (0, y). So, let the point +P(0, y) be equidistant from A and B. Then +(6 – 0)2 + (5 – y)2 = (– 4 – 0)2 + (3 – y)2 +i.e., +36 + 25 + y2 – 10y = 16 + 9 + y2 – 6y +i.e., + 4y = 36 +i.e., +y = 9 +Fig. 7.7 +Reprint 2025-26 + +COORDINATE GEOMETRY +105 +Fig. 7.8 +So, the required point is (0, 9). +Let us check our solution : AP = +2 +2 +(6 – 0) +(5 – 9) +36 +16 +52 + + + + +BP = +2 +2 +(– 4 – 0) +(3 – 9) +16 +36 +52 + + + + +Note : Using the remark above, we see that (0, 9) is the intersection of the y-axis and +the perpendicular bisector of AB. +EXERCISE 7.1 +1. Find the distance between the following pairs of points : +(i) (2, 3), (4, 1) +(ii) (– 5, 7), (– 1, 3) +(iii) (a, b), (– a, – b) +2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance +between the two towns A and B discussed in Section 7.2. +3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear. +4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle. +5. In a classroom, 4 friends are +seated at the points A, B, C and +D as shown in Fig. 7.8. Champa +and Chameli walk into the class +and after observing for a few +minutes Champa asks Chameli, +“Don’t you think ABCD is a +square?” Chameli disagrees. +Using distance formula, find +which of them is correct. +6. Name the type of quadrilateral +formed, if any, by the following +points, and give reasons for +your answer: +(i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0) +(ii) (–3, 5), (3, 1), (0, 3), (–1, – 4) +(iii) (4, 5), (7, 6), (4, 3), (1, 2) +7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9). +8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is +10 units. +Reprint 2025-26 + +106 +MATHEMATICS +9. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the +distances QR and PR. +10. Find a relation between x and y such that the point (x, y) is equidistant from the point +(3, 6) and (– 3, 4). +7.3 Section Formula +Let us recall the situation in Section 7.2. +Suppose a telephone company wants to +position a relay tower at P between A and B +is such a way that the distance of the tower +from B is twice its distance from A. If P lies +on AB, it will divide AB in the ratio 1 : 2 +(see Fig. 7.9). If we take A as the origin O, +and 1 km as one unit on both the axis, the +coordinates of B will be (36, 15). In order to +know the position of the tower, we must know +the coordinates of P. How do we find these +coordinates? +Let the coordinates of P be (x, y). Draw perpendiculars from P and B to the +x-axis, meeting it in D and E, respectively. Draw PC perpendicular to BE. Then, by +the AA similarity criterion, studied in Chapter 6,  POD and  BPC are similar. +Therefore , +OD +OP +1 +PC +PB +2 + + +, and PD +OP +1 +BC +PB +2 + + +So, +1 +36 +2 +x +x  + +and +1 +15 +2 +y +y  + + +These equations give x = 12 and y = 5. +You can check that P(12, 5) meets the +condition that OP : PB = 1 : 2. +Now let us use the understanding that +you may have developed through this +example to obtain the general formula. +Consider any two points A(x1, y1) and +B(x2, y2) and assume that P (x, y) divides +AB internally in the ratio m1 : m2, i.e., +1 +2 +PA +PB +m +m + + (see Fig. 7.10). +Fig. 7.9 +Fig. 7.10 +Reprint 2025-26 + +COORDINATE GEOMETRY +107 +Draw AR, PS and BT perpendicular to the x-axis. Draw AQ and PC parallel to +the x-axis. Then, by the AA similarity criterion, + PAQ ~  BPC +Therefore, +PA +AQ +BP +PC + + = PQ +BC +(1) +Now, +AQ = RS = OS – OR = x – x1 +PC = ST = OT – OS = x2 – x +PQ = PS – QS = PS – AR = y – y1 +BC = BT– CT = BT – PS = y2 – y +Substituting these values in (1), we get +1 +2 +m +m = +1 +1 +2 +2 +x +x +y +y +x +x +y +y + + + + + +Taking +1 +2 +m +m = +1 +2 +x +x +x +x + + +, we get x = +1 +2 +2 +1 +1 +2 +m x +m x +m +m + + +Similarly, taking +1 +2 +m +m = +1 +2 +y +y +y +y + + +, we get y = +1 +2 +2 1 +1 +2 +m y +m y +m +m + + +So, the coordinates of the point P(x, y) which divides the line segment joining the +points A(x1, y1) and B(x2, y2), internally, in the ratio m1 : m2 are +1 2 +2 1 +1 +2 +2 +1 +1 +2 +1 +2 +, +m x +m x +m y +m y +m +m +m +m + + + + + + + + + + +(2) +This is known as the section formula. +This can also be derived by drawing perpendiculars from A, P and B on the +y-axis and proceeding as above. +If the ratio in which P divides AB is k : 1, then the coordinates of the point P will be + +2 +1 +2 +1 +, +1 +1 +kx +x +ky +y +k +k + + + + + + + + + + + +Special Case : The mid-point of a line segment divides the line segment in the ratio +1 : 1. Therefore, the coordinates of the mid-point P of the join of the points A(x1, y1) +and B(x2, y2) is +1 +2 +1 +2 +1 +2 +1 +2 +1 +1 +1 +1 +, +, +1 +1 +1 +1 +2 +2 +x +x +y +y +x +x +y +y + + + + + + + + + + + + + + + + + + + + + +. +Let us solve a few examples based on the section formula. +Reprint 2025-26 + +108 +MATHEMATICS +Example 6 : Find the coordinates of the point which divides the line segment joining +the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally. +Solution : Let P(x, y) be the required point. Using the section formula, we get +x = 3(8) +1(4) +7 +3 +1 + + + +, y = 3(5) +1(–3) +3 +3 +1 + + + +Therefore, (7, 3) is the required point. +Example 7 : In what ratio does the point (– 4, 6) divide the line segment joining the +points A(– 6, 10) and B(3, – 8)? +Solution : Let (– 4, 6) divide AB internally in the ratio m1 : m2. Using the section +formula, we get +(– 4, 6) = +1 +2 +1 +2 +1 +2 +1 +2 +3 +6 +–8 +10 +, +m +m +m +m +m +m +m +m + + + + + + + + + + +(1) +Recall that if (x, y) = (a, b) then x = a and y = b. +So, +– 4 = +1 +2 +1 +2 +3 +6 +m +m +m +m + + +and +1 +2 +1 +2 +8 +10 +6 +m +m +m +m + + + + +Now, +– 4 = +1 +2 +1 +2 +3 +6 +m +m +m +m + + +gives us +– 4m1 – 4m2 = 3m1 – 6m2 +i.e., + 7m1 = 2m2 +i.e., +m1 : m2 = 2 : 7 +You should verify that the ratio satisfies the y-coordinate also. +Now, +1 +2 +1 +2 +8 +10 +m +m +m +m + + + + = +1 +2 +1 +2 +8 +10 +1 +m +m +m +m + + + + (Dividing throughout by m2) += +2 +8 +10 +7 +6 +2 +1 +7 + + + + +Reprint 2025-26 + +COORDINATE GEOMETRY +109 +Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and +B(3, – 8) in the ratio 2 : 7. +Alternatively : The ratio m1 : m2 can also be written as +1 +2 +:1, +m +m + or k : 1. Let (– 4, 6) +divide AB internally in the ratio k : 1. Using the section formula, we get +(– 4, 6) = +3 +6 +8 +10 +, +1 +1 +k +k +k +k + + + + + + + + + + + +(2) +So, +– 4 = 3 +6 +1 +k +k + + +i.e., +– 4k – 4 = 3k – 6 +i.e., +7k = 2 +i.e., +k : 1 = 2 : 7 +You can check for the y-coordinate also. +So, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and +B(3, – 8) in the ratio 2 : 7. +Note : You can also find this ratio by calculating the distances PA and PB and taking +their ratios provided you know that A, P and B are collinear. +Example 8 : Find the coordinates of the points of trisection (i.e., points dividing in +three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4). +Solution : Let P and Q be the points of +trisection of AB i.e., AP = PQ = QB +(see Fig. 7.11). +Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by +applying the section formula, are +1( 7) +2(2) 1(4) +2( 2) +, +1 +2 +1 +2 + + + + + + + + + + + +, i.e., (–1, 0) +Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are +2( 7) +1(2) 2(4) +1( 2) +, +2 +1 +2 +1 + + + + + + + + + + + +, i.e., (– 4, 2) +Fig. 7.11 +Reprint 2025-26 + +110 +MATHEMATICS +Therefore, the coordinates of the points of trisection of the line segment joining A and +B are (–1, 0) and (– 4, 2). +Note : We could also have obtained Q by noting that it is the mid-point of PB. So, we +could have obtained its coordinates using the mid-point formula. +Example 9 : Find the ratio in which the y-axis divides the line segment joining the +points (5, – 6) and (–1, – 4). Also find the point of intersection. +Solution : Let the ratio be k : 1. Then by the section formula, the coordinates of the +point which divides AB in the ratio k : 1 are +5 +4 +6 +, +1 +1 +k +k +k +k + + + + + + + + + + + + + +This point lies on the y-axis, and we know that on the y-axis the abscissa is 0. +Therefore, +5 +1 +k +k + + + + = 0 +So, +k = 5 +That is, the ratio is 5 : 1. Putting the value of k = 5, we get the point of intersection as +13 +0, +3 + + + + + + +. +Example 10 : If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a +parallelogram, taken in order, find the value of p. +Solution : We know that diagonals of a parallelogram bisect each other. +So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD +i.e., +6 +9 1 +4 +, +2 +2 + + + + + + + + + = +8 +2 +3 +, +2 +2 +p + + + + + + + + +i.e., +15 5 +, +2 +2 + + + + + + = +8 +5 +, +2 +2 +p + + + + + + + +so, +15 +2 = 8 +2 +p + +i.e., +p = 7 +Reprint 2025-26 + +COORDINATE GEOMETRY +111 +EXERCISE 7.2 +1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the +ratio 2 : 3. +2. Find the coordinates of the points of trisection of the line segment joining (4, –1) +and (–2, –3). +3. To conduct Sports Day activities, in +your rectangular shaped school +ground ABCD, lines have been +drawn with chalk powder at a +distance of 1m each. 100 flower pots +have been placed at a distance of 1m +from each other along AD, as shown +in Fig. 7.12. Niharika runs 1 +4 th the +distance AD on the 2nd line and +posts a green flag. Preet runs 1 +5 th +the distance AD on the eighth line +and posts a red flag. What is the +distance between both the flags? If +Rashmi has to post a blue flag exactly +halfway between the line segment +joining the two flags, where should +she post her flag? +4. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided +by (– 1, 6). +5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the +x-axis. Also find the coordinates of the point of division. +6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find +x and y. +7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is +(2, – 3) and B is (1, 4). +8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that +AP = 3 AB +7 +and P lies on the line segment AB. +9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and +B(2, 8) into four equal parts. +10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in +order. [Hint : Area of a rhombus = 1 +2 (product of its diagonals)] +Fig. 7.12 +Reprint 2025-26 + +112 +MATHEMATICS +7.4 Summary +In this chapter, you have studied the following points : +1. The distance between P(x1, y1) and Q(x2, y2) is +2 +2 +2 +1 +2 +1 +( +) +( +) . +x +x +y +y + + + +2. The distance of a point P(x, y) from the origin is +2 +2 . +x +y + +3. The coordinates of the point P(x, y) which divides the line segment joining the +points A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2 are +1 2 +2 1 +1 +2 +2 +1 +1 +2 +1 +2 +, +m x +m x +m y +m y +m +m +m +m + + + + + + + + + + + +4. The mid-point of the line segment joining the points P(x1, y1) and Q(x2, y2) is +1 +2 +1 +2 +, +2 +2 +x +x +y +y + + + + + + + +. +A NOTE TO THE READER +Section 7.3 discusses the Section Formula for the coordinates (x, y) of a +point P which divides internally the line segment joining the points +A(x1, y1) and B(x2, y2) in the ratio m1 : m2 as follows : +x = +1 2 +2 1 +1 +2 +m x +m x +m +m + + + , +y = +1 +2 +2 +1 +1 +2 +m y +m y +m +m + + +Note that, here, PA : PB = m1 : m2. +However, if P does not lie between A and B but lies on the line AB, +outside the line segment AB, and PA : PB = m1 : m2, we say that P divides +externally the line segment joining the points A and B. You will study +Section Formula for such case in higher classes. +Reprint 2025-26" +class_10,8,Introduction to Trigonometry,ncert_books/class_10/jemh1dd/jemh108.pdf,"INTRODUCTION TO TRIGONOMETRY +113 +8 +There is perhaps nothing which so occupies the +middle position of mathematics as trigonometry. +– J.F. Herbart (1890) +8.1 Introduction +You have already studied about triangles, and in particular, right triangles, in your +earlier classes. Let us take some examples from our surroundings where right triangles +can be imagined to be formed. For instance : +1. Suppose the students of a school are +visiting Qutub Minar. Now, if a student +is looking at the top of the Minar, a right +triangle can be imagined to be made, +as shown in Fig 8.1. Can the student +find out the height of the Minar, without +actually measuring it? + 2. Suppose a girl is sitting on the balcony +of her house located on the bank of a +river. She is looking down at a flower +pot placed on a stair of a temple situated +nearby on the other bank of the river. +A right triangle is imagined to be made +in this situation as shown in Fig.8.2. If +you know the height at which the +person is sitting, can you find the width +of the river? +INTRODUCTION TO +TRIGONOMETRY +Fig. 8.1 +Fig. 8.2 +Reprint 2025-26 + +114 +MATHEMATICS +3. Suppose a hot air balloon is flying in +the air. A girl happens to spot the +balloon in the sky and runs to her +mother to tell her about it. Her mother +rushes out of the house to look at the +balloon.Now when the girl had spotted +the balloon intially it was at point A. +When both the mother and daughter +came out to see it, it had already +travelled to another point B. Can you +find the altitude of B from the ground? +In all the situations given above, the distances or heights can be found by using +some mathematical techniques, which come under a branch of mathematics called +‘trigonometry’. The word ‘trigonometry’ is derived from the Greek words ‘tri’ +(meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact, +trigonometry is the study of relationships between the sides and angles of a triangle. +The earliest known work on trigonometry was recorded in Egypt and Babylon. Early +astronomers used it to find out the distances of the stars and planets from the Earth. +Even today, most of the technologically advanced methods used in Engineering and +Physical Sciences are based on trigonometrical concepts. +In this chapter, we will study some ratios of the sides of a right triangle with +respect to its acute angles, called trigonometric ratios of the angle. We will restrict +our discussion to acute angles only. However, these ratios can be extended to other +angles also. We will also define the trigonometric ratios for angles of measure 0° and +90°. We will calculate trigonometric ratios for some specific angles and establish +some identities involving these ratios, called trigonometric identities. +8.2 Trigonometric Ratios +In Section 8.1, you have seen some right triangles +imagined to be formed in different situations. +Let us take a right triangle ABC as shown +in Fig. 8.4. +Here,  CAB (or, in brief, angle A) is an +acute angle. Note the position of the side BC +with respect to angle A. It faces  A. We call it +the side opposite to angle A. AC is the +hypotenuse of the right triangle and the side AB +is a part of  A. So, we call it the side +adjacent to angle A. +Fig. 8.4 +Fig. 8.3 +Reprint 2025-26 + +INTRODUCTION TO TRIGONOMETRY +115 +Note that the position of sides change +when you consider angle C in place of A +(see Fig. 8.5). +You have studied the concept of ‘ratio’ in +your earlier classes. We now define certain ratios +involving the sides of a right triangle, and call +them trigonometric ratios. +The trigonometric ratios of the angle A +in right triangle ABC (see Fig. 8.4) are defined +as follows : +sine of  A = side opposite to angle A +BC +hypotenuse +AC + +cosine of  A = side adjacent to angle A +AB +hypotenuse +AC + +tangent of  A = side opposite to angle A +BC +side adjacent to angle A +AB + +cosecant of  A = +1 +hypotenuse +AC +sine of +A +side opposite to angle A +BC + + + +secant of  A = +1 +hypotenuse +AC +cosine of +A +side adjacent to angle A +AB + + + +cotangent of  A = +1 +side adjacent to angle A +AB +tangent of +A +side opposite to angle A +BC + + + +The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A +and cot A respectively. Note that the ratios cosec A, sec A and cot A are respectively, +the reciprocals of the ratios sin A, cos A and tan A. +Also, observe that tan A = +BC +BC +sin A +AC +AB +AB +cos A +AC + + + and cot A = cosA +sin A +. +So, the trigonometric ratios of an acute angle in a right triangle express the +relationship between the angle and the length of its sides. +Why don’t you try to define the trigonometric ratios for angle C in the right +triangle? (See Fig. 8.5) +Fig. 8.5 +Reprint 2025-26 + +116 +MATHEMATICS +The first use of the idea of ‘sine’ in the way we use +it today was in the work Aryabhatiyam by Aryabhata, +in A.D. 500. Aryabhata used the word ardha-jya +for the half-chord, which was shortened to jya or +jiva in due course. When the Aryabhatiyam was +translated into Arabic, the word jiva was retained as +it is. The word jiva was translated into sinus, which +means curve, when the Arabic version was translated +into Latin. Soon the word sinus, also used as sine, +became common in mathematical texts throughout +Europe. An English Professor of astronomy Edmund +Gunter (1581–1626), first used the abbreviated +notation ‘sin’. +The origin of the terms ‘cosine’ and ‘tangent’ was much later. The cosine function +arose from the need to compute the sine of the complementary angle. Aryabhatta +called it kotijya. The name cosinus originated with Edmund Gunter. In 1674, the +English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’. +Remark : Note that the symbol sin A is used as an +abbreviation for ‘the sine of the angle A’. sin A is not +the product of ‘sin’ and A. ‘sin’ separated from A +has no meaning. Similarly, cos A is not the product of +‘cos’ and A. Similar interpretations follow for other +trigonometric ratios also. +Now, if we take a point P on the hypotenuse +AC or a point Q on AC extended, of the right triangle +ABC and draw PM perpendicular to AB and QN +perpendicular to AB extended (see Fig. 8.6), how +will the trigonometric ratios of  A in  PAM differ +from those of  A in  CAB or from those of  A in + QAN? +To answer this, first look at these triangles. Is  PAM similar to  CAB? From +Chapter 6, recall the AA similarity criterion. Using the criterion, you will see that the +triangles PAM and CAB are similar. Therefore, by the property of similar triangles, +the corresponding sides of the triangles are proportional. +So, we have +AM +AB = AP +MP +AC +BC + + +Aryabhata + C.E. 476 – 550 +Fig. 8.6 +Reprint 2025-26 + +INTRODUCTION TO TRIGONOMETRY +117 +From this, we find +MP +AP = BC +sin A +AC  +. +Similarly, +AM +AB +AP +AC + + = cos A, MP +BC +tan A +AM +AB + + + and so on. +This shows that the trigonometric ratios of angle A in  PAM not differ from +those of angle A in  CAB. +In the same way, you should check that the value of sin A (and also of other +trigonometric ratios) remains the same in  QAN also. +From our observations, it is now clear that the values of the trigonometric +ratios of an angle do not vary with the lengths of the sides of the triangle, if +the angle remains the same. +Note : For the sake of convenience, we may write sin2A, cos2A, etc., in place of +(sin A)2, (cos A)2, etc., respectively. But cosec A = (sin A)–1  sin–1 A (it is called sine +inverse A). sin–1 A has a different meaning, which will be discussed in higher classes. +Similar conventions hold for the other trigonometric ratios as well. Sometimes, the +Greek letter  (theta) is also used to denote an angle. +We have defined six trigonometric ratios of an acute angle. If we know any one +of the ratios, can we obtain the other ratios? Let us see. +If in a right triangle ABC, sin A = 1, +3 +then this means that BC +1 +AC +3 + +, i.e., the +lengths of the sides BC and AC of the triangle +ABC are in the ratio 1 : 3 (see Fig. 8.7). So if +BC is equal to k, then AC will be 3k, where +k is any positive number. To determine other +trigonometric ratios for the angle A, we need to find the length of the third side +AB. Do you remember the Pythagoras theorem? Let us use it to determine the +required length AB. +AB2 = AC2 – BC2 = (3k)2 – (k)2 = 8k2 = (2 +2 k)2 +Therefore, +AB = +2 2 k + +So, we get +AB = 2 2 k +(Why is AB not – 2 2 k ?) +Now, +cos A = AB +2 2 +2 2 +AC +3 +3 +k +k + + +Similarly, you can obtain the other trigonometric ratios of the angle A. +Fig. 8.7 +Reprint 2025-26 + +118 +MATHEMATICS +Remark : Since the hypotenuse is the longest side in a right triangle, the value of +sin A or cos A is always less than 1 (or, in particular, equal to 1). +Let us consider some examples. +Example 1 : Given tan A = 4 +3 , find the other +trigonometric ratios of the angle A. +Solution : Let us first draw a right  ABC +(see Fig 8.8). +Now, we know that tan A = BC +4 +AB +3 + +. +Therefore, if BC = 4k, then AB = 3k, where k is a +positive number. +Now, by using the Pythagoras Theorem, we have +AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2 +So, +AC = 5k +Now, we can write all the trigonometric ratios using their definitions. +sin A = BC +4 +4 +AC +5 +5 +k +k + + +cos A = AB +3 +3 +AC +5 +5 +k +k + + +Therefore, cot A = +1 +3 +1 +5 +, cosec A = +tanA +4 +sin A +4 + + + and sec A = +1 +5 +cos A +3 + + +Example 2 : If  B and  Q are +acute angles such that sin B = sin Q, +then prove that  B =  Q. +Solution : Let us consider two right +triangles ABC and PQR where +sin B = sin Q (see Fig. 8.9). +We have +sin B = AC +AB +and +sin Q = PR +PQ +Fig. 8.8 +Fig. 8.9 +Reprint 2025-26 + +INTRODUCTION TO TRIGONOMETRY +119 +Then +AC +AB = PR +PQ +Therefore, +AC +PR = +AB +, say +PQ +k + +(1) +Now, using Pythagoras theorem, +BC = +2 +2 +AB +AC + +and +QR = +2 +2 +PQ – PR +So, +BC +QR = +2 +2 +2 +2 +2 +2 +2 +2 +2 +2 +2 +2 +2 +2 +AB +AC +PQ +PR +PQ +PR +PQ +PR +PQ +PR +PQ +PR +k +k +k +k + + + + + + + + + +(2) +From (1) and (2), we have +AC +PR = AB +BC +PQ +QR + +Then, by using Theorem 6.4,  ACB ~  PRQ and therefore,  B =  Q. +Example 3 : Consider  ACB, right-angled at C, in +which AB = 29 units, BC = 21 units and  ABC =  +(see Fig. 8.10). Determine the values of +(i) cos2  + sin2 , +(ii) cos2  – sin2  +Solution : In  ACB, we have +AC = +2 +2 +AB +BC + + = +2 +2 +(29) +(21) + += +(29 +21)(29 +21) +(8)(50) +400 +20units + + + + + +So, +sin  = AC +20 +BC +21 +, cos += +AB +29 +AB +29 + + + + +Now, (i) cos2 + sin2 = +2 +2 +2 +2 +2 +20 +21 +20 +21 +400 +441 +1, +29 +29 +841 +29 + + + + + + + + + + + + + + + + + + +and (ii) cos2  – sin2  = +2 +2 +2 +21 +20 +(21 +20)(21 +20) +41 +29 +29 +841 +29 + + + + + + + + + + + + + + + + + +. +Fig. 8.10 +Reprint 2025-26 + +120 +MATHEMATICS +Example 4 : In a right triangle ABC, right-angled at B, +if tan A = 1, then verify that +2 sin A cos A = 1. +Solution : In  ABC, tan A = BC +AB = 1 +(see Fig 8.11) +i.e., +BC = AB +Let AB = BC = k, where k is a positive number. +Now, +AC = +2 +2 +AB +BC + += +2 +2 +( ) +( ) +2 +k +k +k + + +Therefore, +sin A = +BC +1 +AC +2 + +and +cos A = +AB +1 +AC +2 + +So, +2 sin A cos A = +1 +1 +2 +1, +2 +2 + + + + + + + + + + which is the required value. +Example 5 : In  OPQ, right-angled at P, +OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12). +Determine the values of sin Q and cos Q. +Solution : In  OPQ, we have +OQ2 = OP2 + PQ2 +i.e., +(1 + PQ)2 = OP2 + PQ2 +(Why?) +i.e., +1 + PQ2 + 2PQ = OP2 + PQ2 +i.e., +1 + 2PQ = 72 +(Why?) +i.e., +PQ = 24 cm and OQ = 1 + PQ = 25 cm +So, +sin Q = +7 +25 and cos Q = 24 +25 +Fig. 8.12 +Fig. 8.11 +Reprint 2025-26 + +INTRODUCTION TO TRIGONOMETRY +121 +EXERCISE 8.1 +1. In  ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine : +(i) sin A, cos A +(ii) sin C, cos C +2. In Fig. 8.13, find tan P – cot R. +3. If sin A = 3 , +4 + calculate cos A and tan A. +4. Given 15 cot A = 8, find sin A and sec A. +5. Given sec  = 13 , +12 calculate all other trigonometric ratios. +6. If  A and  B are acute angles such that cos A = cos B, then show that  A =  B. +7. If cot  = 7 , +8 evaluate : (i) +(1 +sin +)(1 +sin +) , +(1 +cos +)(1 +cos ) + + + + + + + + +(ii) cot2  +8. If 3 cot A = 4, check whether +2 +2 +1 +tan A +1 + tan A + + = cos2 A – sin2A or not. +9. In triangle ABC, right-angled at B, if tan A = 1 , +3 find the value of: +(i) sin A cos C + cos A sin C +(ii) cos A cos C – sin A sin C +10. In  PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of +sin P, cos P and tan P. +11. State whether the following are true or false. Justify your answer. +(i) The value of tan A is always less than 1. +(ii) sec A = 12 +5 for some value of angle A. +(iii) cos A is the abbreviation used for the cosecant of angle A. +(iv) cot A is the product of cot and A. +(v) sin  = 4 +3 for some angle . +8.3 Trigonometric Ratios of Some Specific Angles +From geometry, you are already familiar with the construction of angles of 30°, 45°, +60° and 90°. In this section, we will find the values of the trigonometric ratios for these +angles and, of course, for 0°. +Fig. 8.13 +Reprint 2025-26 + +122 +MATHEMATICS +Trigonometric Ratios of 45° +In  ABC, right-angled at B, if one angle is 45°, then +the other angle is also 45°, i.e.,  A =  C = 45° +(see Fig. 8.14). +So, +BC = AB +(Why?) +Now, Suppose BC = AB = a. +Then by Pythagoras Theorem, AC2 = AB2 + BC2 = a2 + a2 = 2a2, +and, therefore, +AC = +2 +a + +Using the definitions of the trigonometric ratios, we have : +sin 45° = side opposite to angle 45° +BC +1 +hypotenuse +AC +2 +2 +a +a + + + +cos 45° = side adjacent toangle 45° +AB +1 +hypotenuse +AC +2 +2 +a +a + + + +tan 45° = side opposite to angle 45° +BC +1 +side adjacent to angle 45° +AB +a +a + + + +Also, cosec 45° = +1 +2 +sin 45  + +, sec 45° = +1 +2 +cos 45  + +, cot 45° = +1 +1 +tan 45  + +. +Trigonometric Ratios of 30° and 60° +Let us now calculate the trigonometric ratios of 30° +and 60°. Consider an equilateral triangle ABC. Since +each angle in an equilateral triangle is 60°, therefore, + A =  B =  C = 60°. +Draw the perpendicular AD from A to the side BC +(see Fig. 8.15). +Now + ABD  ACD +(Why?) +Therefore, +BD = DC +and + BAD =  CAD (CPCT) +Now observe that: + ABD is a right triangle, right-angled at D with  BAD = 30° and  ABD = 60° +(see Fig. 8.15). +Fig. 8.15 +Fig. 8.14 +Reprint 2025-26 + +INTRODUCTION TO TRIGONOMETRY +123 +As you know, for finding the trigonometric ratios, we need to know the lengths of the +sides of the triangle. So, let us suppose that AB = 2a. +Then, +BD = 1 BC = +2 +a +and +AD2 = AB2 – BD2 = (2a)2 – (a)2 = 3a2, +Therefore, +AD = +3 +a +Now, we have : +sin 30° = BD +1 +AB +2 +2 +a +a + + +, cos 30° = AD +3 +3 +AB +2 +2 +a +a + + +tan 30° = +BD +1 +AD +3 +3 +a +a + + +. +Also, +cosec 30° = +1 +2, +sin 30 + + + sec 30° = +1 +2 +cos 30 +3 + + +cot 30° = +1 +3 +tan 30 + + +. +Similarly, +sin 60° = AD +3 +3 +AB +2 +2 +a +a + + +, cos 60° = 1 +2 , tan 60° = +3 , +cosec 60° = +2 , +3 sec 60° = 2 and cot 60° = +1 +3 + +Trigonometric Ratios of 0° and 90° +Let us see what happens to the trigonometric ratios of angle +A, if it is made smaller and smaller in the right triangle ABC +(see Fig. 8.16), till it becomes zero. As  A gets smaller and +smaller, the length of the side BC decreases.The point C gets +closer to point B, and finally when  A becomes very close +to 0°, AC becomes almost the same as AB (see Fig. 8.17). +Fig. 8.17 +Fig. 8.16 +Reprint 2025-26 + +124 +MATHEMATICS +When  A is very close to 0°, BC gets very close to 0 and so the value of +sin A = BC +AC is very close to 0. Also, when  A is very close to 0°, AC is nearly the +same as AB and so the value of cos A = AB +AC is very close to 1. +This helps us to see how we can define the values of sin A and cos A when +A = 0°. We define : sin 0° = 0 and cos 0° = 1. +Using these, we have : +tan 0° = sin 0° +cos 0° = 0, cot 0° = +1 +, +tan 0° which is not defined. (Why?) +sec 0° = +1 +cos 0 = 1 and cosec 0° = +1 +, +sin 0 which is again not defined.(Why?) +Now, let us see what happens to the trigonometric ratios of  A, when it is made +larger and larger in  ABC till it becomes 90°. As  A gets larger and larger,  C gets +smaller and smaller. Therefore, as in the case above, the length of the side AB goes on +decreasing. The point A gets closer to point B. Finally when  A is very close to 90°, + C becomes very close to 0° and the side AC almost coincides with side BC +(see Fig. 8.18). +Fig. 8.18 +When  C is very close to 0°,  A is very close to 90°, side AC is nearly the +same as side BC, and so sin A is very close to 1. Also when  A is very close to 90°, + C is very close to 0°, and the side AB is nearly zero, so cos A is very close to 0. +So, we define : +sin 90° = 1 and cos 90° = 0. +Now, why don’t you find the other trigonometric ratios of 90°? +We shall now give the values of all the trigonometric ratios of 0°, 30°, 45°, 60° +and 90° in Table 8.1, for ready reference. +Reprint 2025-26 + +INTRODUCTION TO TRIGONOMETRY +125 +Table 8.1 + A +0° +30° +45° +60° +90° +sin A +0 +1 +2 +1 +2 +3 +2 +1 +cos A +1 +3 +2 +1 +2 +1 +2 +0 +tan A +0 +1 +3 +1 +3 +Not defined +cosec A +Not defined +2 +2 +2 +3 +1 +sec A +1 +2 +3 +2 +2 +Not defined +cot A +Not defined +3 +1 +1 +3 +0 +Remark : From the table above you can observe that as  A increases from 0° to +90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0. +Let us illustrate the use of the values in the table above through some examples. +Example 6 : In  ABC, right-angled at B, +AB = 5 cm and  ACB = 30° (see Fig. 8.19). +Determine the lengths of the sides BC and AC. +Solution : To find the length of the side BC, we will +choose the trigonometric ratio involving BC and the +given side AB. Since BC is the side adjacent to angle +C and AB is the side opposite to angle C, therefore +AB +BC = tan C +i.e., +5 +BC = tan 30° = +1 +3 +which gives +BC = 5 3 cm +Fig. 8.19 +Reprint 2025-26 + +126 +MATHEMATICS +To find the length of the side AC, we consider +sin 30° = AB +AC +(Why?) +i.e., +1 +2 = +5 +AC +i.e., +AC = 10 cm +Note that alternatively we could have used Pythagoras theorem to determine the third +side in the example above, +i.e., +AC = +2 +2 +2 +2 +AB +BC +5 +(5 3) cm = 10cm. + + + +Example 7 : In  PQR, right-angled at +Q (see Fig. 8.20), PQ = 3 cm and PR = 6 cm. +Determine  QPR and  PRQ. +Solution : Given PQ = 3 cm and PR = 6 cm. +Therefore, +PQ +PR = sin R +or +sin R = 3 +1 +6 +2 + +So, + PRQ = 30° +and therefore, + QPR = 60°. +(Why?) +You may note that if one of the sides and any other part (either an acute angle or any +side) of a right triangle is known, the remaining sides and angles of the triangle can be +determined. +Example 8 : If sin (A – B) = 1 , +2 cos (A + B) = 1 , +2 0° < A + B  90°, A > B, find A +and B. +Solution : Since, sin (A – B) = 1 +2 , therefore, A – B = 30° +(Why?) +(1) +Also, since cos (A + B) = 1 +2 , therefore, A + B = 60° +(Why?) +(2) +Solving (1) and (2), we get : A = 45° and B = 15°. +Fig. 8.20 +Reprint 2025-26 + +INTRODUCTION TO TRIGONOMETRY +127 +EXERCISE 8.2 +1. Evaluate the following : +(i) sin 60° cos 30° + sin 30° cos 60° +(ii) 2 tan2 45° + cos2 30° – sin2 60° +(iii) +cos 45° +sec 30° + cosec 30° +(iv) sin 30° + tan 45° – cosec 60° +sec 30° + cos 60° + cot 45° +(v) +2 +2 +2 +2 +2 +5 cos +60 +4 sec 30 +tan +45 +sin +30 +cos 30 + + + + + +2. Choose the correct option and justify your choice : +(i) +2 +2 tan 30 +1 +tan +30 + + + + +(A) +sin 60° +(B) cos 60° +(C) tan 60° +(D) sin 30° +(ii) +2 +2 +1 +tan +45 +1 +tan +45 + + + + +(A) +tan 90° +(B) 1 +(C) sin 45° +(D) 0 +(iii) sin 2A = 2 sin A is true when A = +(A) +0° +(B) 30° +(C) 45° +(D) 60° +(iv) +2 +2 tan 30 +1 +tan +30 + + + + +(A) +cos 60° +(B) sin 60° +(C) tan 60° +(D) sin 30° +3. If tan (A + B) = 3 and tan (A – B) = +1 +3 ; 0° < A + B  90°; A > B, find A and B. +4. State whether the following are true or false. Justify your answer. +(i) sin (A + B) = sin A + sin B. +(ii) The value of sin  increases as  increases. +(iii) The value of cos  increases as  increases. +(iv) sin  = cos  for all values of . +(v) cot A is not defined for A = 0°. +Reprint 2025-26 + +128 +MATHEMATICS +8.4 Trigonometric Identities +You may recall that an equation is called an identity +when it is true for all values of the variables involved. +Similarly, an equation involving trigonometric ratios +of an angle is called a trigonometric identity, if it is +true for all values of the angle(s) involved. +In this section, we will prove one trigonometric +identity, and use it further to prove other useful +trigonometric identities. +In  ABC, right-angled at B (see Fig. 8.21), we have: +AB2 + BC2 = AC2 +(1) +Dividing each term of (1) by AC2, we get +2 +2 +2 +2 +AB +BC +AC +AC + + = +2 +2 +AC +AC +i.e., +2 +2 +AB +BC +AC +AC + + + + + + + + + + + + + = +2 +AC +AC + + + + + + +i.e., +(cos A)2 + (sin A)2 = 1 +i.e., +cos2 A + sin2 A = 1 +(2) +This is true for all A such that 0°  A  90°. So, this is a trigonometric identity. +Let us now divide (1) by AB2. We get +2 +2 +2 +2 +AB +BC +AB +AB + + = +2 +2 +AC +AB +or, +2 +2 +AB +BC +AB +AB + + + + + + + + + + + + + = +2 +AC +AB + + + + + + +i.e., +1 + tan2 A = sec2 A +(3) +Is this equation true for A = 0°? Yes, it is. What about A = 90°? Well, tan A and +sec A are not defined for A = 90°. So, (3) is true for all A such that 0°  A  90°. +Let us see what we get on dividing (1) by BC2. We get +2 +2 +2 +2 +AB +BC +BC +BC + + = +2 +2 +AC +BC +Fig. 8.21 +Reprint 2025-26 + +INTRODUCTION TO TRIGONOMETRY +129 +i.e., +2 +2 +AB +BC +BC +BC + + + + + + + + + + + + + = +2 +AC +BC + + + + + + +i.e., +cot2 A + 1 = cosec2 A +(4) +Note that cosec A and cot A are not defined for A = 0°. Therefore (4) is true for +all A such that 0° < A  90°. +Using these identities, we can express each trigonometric ratio in terms of other +trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the +values of other trigonometric ratios. +Let us see how we can do this using these identities. Suppose we know that +tan A = +1 +3 + Then, cot A = +3 . +Since, sec2 A = 1 + tan2 A = +1 +4, +1 +3 +3 + + + sec A = 2 +3 , and cos A = +3 +2  +Again, sin A = +2 +3 +1 +1 +cos A +1 +4 +2 + + + + +. Therefore, cosec A = 2. +Example 9 : Express the ratios cos A, tan A and sec A in terms of sin A. +Solution : Since +cos2 A + sin2 A = 1, therefore, +cos2 A = 1 – sin2 A, i.e., cos A = +2 +1 +sin A + + +This gives +cos A = +2 +1 +sin A + +(Why?) +Hence, +tan A = sin A +cos A = +2 +2 +sin A +1 +1 +and sec A = cos A +1 – sin A +1 +sin A + + +Example 10 : Prove that sec A (1 – sin A)(sec A + tan A) = 1. +Solution : +LHS = sec A (1 – sin A)(sec A + tan A) = +1 +1 +sin A +(1 +sin A) +cos A +cos A +cos A + + + + + + + + + + + + + + +Reprint 2025-26 + +130 +MATHEMATICS += +2 +2 +2 +(1 +sin A)(1 + sin A) +1 +sin A +cos A +cos A + + + += +2 +2 +cos A +1 +cos A + = RHS +Example 11 : Prove that cot A – cos A +cosec A – 1 +cot A + cos A +cosec A + 1 + +Solution : LHS = +cos A +cos A +cot A – cos A +sin A +cos A +cot A + cos A +cos A +sin A + + + += +1 +1 +cos A +1 +1 +sin A +sin A +cosec A – 1 +cosec A + 1 +1 +1 +cos A +1 +1 +sin A +sin A + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + = RHS +Example 12 : Prove that sin +cos +1 +1 +, +sin +cos +1 +sec +tan + + + + + + + using the identity +sec2  = 1 + tan2 . +Solution : Since we will apply the identity involving sec  and tan , let us first +convert the LHS (of the identity we need to prove) in terms of sec  and tan  by +dividing numerator and denominator by cos  +LHS = sin + – cos ++ 1 +tan +1 +sec +sin ++ cos +– 1 +tan +1 +sec + + + + + + + + + + + += (tan +sec ) +1 +{(tan +sec ) +1} (tan +sec ) +(tan +sec ) +1 +{(tan +sec ) +1} (tan +sec ) + + + + + + + + + + + + + += +2 +2 +(tan +sec +) +(tan +sec ) +{tan +sec +1} (tan +sec ) + + + + + + + + += +– 1 +tan +sec +(tan +sec +1) (tan +sec ) + + + + + + + +Reprint 2025-26 + +INTRODUCTION TO TRIGONOMETRY +131 += +–1 +1 +, +tan +sec +sec +tan + + + + + +which is the RHS of the identity, we are required to prove. +EXERCISE 8.3 +1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. +2. Write all the other trigonometric ratios of  A in terms of sec A. +3. Choose the correct option. Justify your choice. +(i) 9 sec2 A – 9 tan2 A = +(A) 1 +(B) 9 +(C) 8 +(D) 0 +(ii) (1 + tan  + sec ) (1 + cot  – cosec ) = +(A) 0 +(B) 1 +(C) 2 +(D) –1 +(iii) (sec A + tan A) (1 – sin A) = +(A) sec A +(B) sin A +(C) cosec A +(D) cos A +(iv) +2 +2 +1 +tan +A +1 + cot +A + + +(A) sec2 A +(B) –1 +(C) cot2 A +(D) tan2 A +4. Prove the following identities, where the angles involved are acute angles for which the +expressions are defined. +(i) (cosec  – cot )2 = +1 +cos +1 +cos + + + + +(ii) +cos A +1 +sin A +2 sec A +1 + sin A +cos A + + + +(iii) +tan +cot +1 +sec +cosec +1 +cot +1 +tan + + + + + + + + + + + +[Hint : Write the expression in terms of sin  and cos ] +(iv) +2 +1 +sec A +sin A +sec A +1 – cos A + + +[Hint : Simplify LHS and RHS separately] +(v) +cos A – sin A + 1 +cosec A + cot A, +cos A + sin A – 1  + using the identity cosec2 A = 1 + cot2 A. +(vi) +1 +sin A +sec A + tan A +1 – sin A + + +(vii) +3 +3 +sin +2 sin +tan +2 cos +cos + + + + + +(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A +Reprint 2025-26 + +132 +MATHEMATICS +(ix) +1 +(cosec A – sin A)(sec A – cos A) +tan A + cot A + +[Hint : Simplify LHS and RHS separately] +(x) +2 +2 +2 +1 +tan A +1 +tan A +1 – cot A +1 + cot A + + + + + + + + + + + + + + + + = tan2 A +8.5 Summary +In this chapter, you have studied the following points : +1. In a right triangle ABC, right-angled at B, +sin A = side opposite to angle A +side adjacent to angle A +, cos A = +hypotenuse +hypotenuse +tan A = side opposite toangle A +side adjacent to angle A . +2. +1 +1 +1 +sin A +, +cosec A = +; sec A = +; tan A = +tan A = +sin A +cos A +cot A +cos A . +3. If one of the trigonometric ratios of an acute angle is known, the remaining trigonometric +ratios of the angle can be easily determined. +4. The values of trigonometric ratios for angles 0°, 30°, 45°, 60° and 90°. +5. The value of sin A or cos A never exceeds 1, whereas the value of sec A (0° £ A < 90°) or +cosec A (0° < A £ 90º) is always greater than or equal to 1. +6. sin2 A + cos2 A = 1, +sec2 A – tan2 A = 1 for 0° £ A < 90°, +cosec2 A = 1 + cot2 A for 0° < A £ 90º. +Reprint 2025-26" +class_10,9,Some Applications of Trigonometry,ncert_books/class_10/jemh1dd/jemh109.pdf,"SOME APPLICATIONS OF TRIGONOMETRY +133 +9 +9.1 Heights and Distances +In the previous chapter, you have studied about trigonometric ratios. In this chapter, +you will be studying about some ways in which trigonometry is used in the life around +you. +Let us consider Fig. 8.1 of prvious chapter, which is redrawn below in Fig. 9.1. +Fig. 9.1 +In this figure, the line AC drawn from the eye of the student to the top of the +minar is called the line of sight. The student is looking at the top of the minar. The +angle BAC, so formed by the line of sight with the horizontal, is called the angle of +elevation of the top of the minar from the eye of the student. +Thus, the line of sight is the line drawn from the eye of an observer to the point +in the object viewed by the observer. The angle of elevation of the point viewed is +SOME APPLICATIONS OF +TRIGONOMETRY +Reprint 2025-26 + +134 +MATHEMATICS +the angle formed by the line of sight with the horizontal when the point being viewed is +above the horizontal level, i.e., the case when we raise our head to look at the object +(see Fig. 9.2). +Fig. 9.2 +Now, consider the situation given in Fig. 8.2. The girl sitting on the balcony is +looking down at a flower pot placed on a stair of the temple. In this case, the line of +sight is below the horizontal level. The angle so formed by the line of sight with the +horizontal is called the angle of depression. +Thus, the angle of depression of a point on the object being viewed is the angle +formed by the line of sight with the horizontal when the point is below the horizontal +level, i.e., the case when we lower our head to look at the point being viewed +(see Fig. 9.3). +Fig. 9.3 +Now, you may identify the lines of sight, and the angles so formed in Fig. 8.3. +Are they angles of elevation or angles of depression? +Let us refer to Fig. 9.1 again. If you want to find the height CD of the minar +without actually measuring it, what information do you need? You would need to know +the following: +(i) the distance DE at which the student is standing from the foot of the minar +Reprint 2025-26 + +SOME APPLICATIONS OF TRIGONOMETRY +135 +(ii) the angle of elevation,  BAC, of the top of the minar +(iii) the height AE of the student. +Assuming that the above three conditions are known, how can we determine the +height of the minar? +In the figure, CD = CB + BD. Here, BD = AE, which is the height of the student. +To find BC, we will use trigonometric ratios of  BAC or  A. +In  ABC, the side BC is the opposite side in relation to the known  A. Now, +which of the trigonometric ratios can we use? Which one of them has the two values +that we have and the one we need to determine? Our search narrows down to using +either tan A or cot A, as these ratios involve AB and BC. +Therefore, +tan A = BC +AB or cot A = AB, +BC which on solving would give us BC. +By adding AE to BC, you will get the height of the minar. +Now let us explain the process, we have just discussed, by solving some problems. +Example 1 : A tower stands vertically on the ground. From a point on the ground, +which is 15 m away from the foot of the tower, the angle of elevation of the top of the +tower is found to be 60°. Find the height of the tower. +Solution : First let us draw a simple diagram to +represent the problem (see Fig. 9.4). Here AB +represents the tower, CB is the distance of the point +from the tower and  ACB is the angle of elevation. +We need to determine the height of the tower, i.e., +AB. Also, ACB is a triangle, right-angled at B. +To solve the problem, we choose the trigonometric +ratio tan 60° (or cot 60°), as the ratio involves AB +and BC. +Now, +tan 60° = AB +BC +i.e., +3 = AB +15 +i.e., +AB = 15 3 +Hence, the height of the tower is 15 3 m. +Fig. 9.4 +Reprint 2025-26 + +136 +MATHEMATICS +Example 2 : An electrician has to repair an +electric fault on a pole of height 5 m. She needs +to reach a point 1.3m below the top of the pole +to undertake the repair work (see Fig. 9.5). What +should be the length of the ladder that she should +use which, when inclined at an angle of 60° to +the horizontal, would enable her to reach the +required position? Also, how far from the foot +of the pole should she place the foot of the +ladder? (You may take +3 = 1.73) +Solution : In Fig. 9.5, the electrician is required to +reach the point B on the pole AD. +So, +BD = AD – AB = (5 – 1.3)m = 3.7 m. +Here, BC represents the ladder. We need to find +its length, i.e., the hypotenuse of the right triangle BDC. +Now, can you think which trigonometic ratio should we consider? +It should be sin 60°. +So, +BD +BC = sin 60° or 3.7 +BC = +3 +2 +Therefore, +BC = 3.7 +2 +3 + + = 4.28 m (approx.) +i.e., the length of the ladder should be 4.28 m. +Now, +DC +BD = cot 60° = 1 +3 +i.e., +DC = +3.7 +3 = 2.14 m (approx.) +Therefore, she should place the foot of the ladder at a distance of 2.14 m from +the pole. +Fig. 9.5 +Reprint 2025-26 + +SOME APPLICATIONS OF TRIGONOMETRY +137 +Example 3 : An observer 1.5 m tall is 28.5 m away +from a chimney. The angle of elevation of the top +of the chimney from her eyes is 45°. What is the +height of the chimney? +Solution : Here, AB is the chimney, CD the +observer and  ADE the angle of elevation (see +Fig. 9.6). In this case, ADE is a triangle, right-angled +at E and we are required to find the height of the +chimney. +We have +AB = AE + BE = AE + 1.5 +and + DE = CB = 28.5 m +To determine AE, we choose a trigonometric ratio, which involves both AE and +DE. Let us choose the tangent of the angle of elevation. +Now, +tan 45° = AE +DE +i.e., +1 = +AE +28.5 +Therefore, +AE = 28.5 +So the height of the chimney (AB) = (28.5 + 1.5) m = 30 m. +Example 4 : From a point P on the ground the angle of elevation of the top of a 10 m +tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation +of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the +distance of the building from the point P. (You may take +3 = 1.732) +Solution : In Fig. 9.7, AB denotes the height of the building, BD the flagstaff and P +the given point. Note that there are two right triangles PAB and PAD. We are required +to find the length of the flagstaff, i.e., DB and the distance of the building from the +point P, i.e., PA. +Fig. 9.6 +Reprint 2025-26 + +138 +MATHEMATICS +Since, we know the height of the building AB, we +will first consider the right  PAB. +We have +tan 30° = AB +AP +i.e., +1 +3 = 10 +AP +Therefore, +AP = 10 3 +i.e., the distance of the building from P is 10 3 m = 17.32 m. +Next, let us suppose DB = x m. Then AD = (10 + x) m. +Now, in right  PAD, +tan 45° = +AD +10 +AP +10 3 +x + + +Therefore, +1 = +10 +10 3 +x + +i.e., +x = 10  + +3 +1 + + = 7.32 +So, the length of the flagstaff is 7.32 m. +Example 5 : The shadow of a tower standing +on a level ground is found to be 40 m longer +when the Sun’s altitude is 30° than when it is +60°. Find the height of the tower. +Solution : In Fig. 9.8, AB is the tower and +BC is the length of the shadow when the +Sun’s altitude is 60°, i.e., the angle of +elevation of the top of the tower from the tip +of the shadow is 60° and DB is the length of +the shadow, when the angle of elevation +is 30°. +Now, let AB be h m and BC be x m. According to the question, DB is 40 m longer +than BC. +Fig. 9.7 +Fig. 9.8 +Reprint 2025-26 + +SOME APPLICATIONS OF TRIGONOMETRY +139 +So, +DB = (40 + x) m +Now, we have two right triangles ABC and ABD. +In  ABC, +tan 60° = AB +BC +or, +3 = h +x +(1) +In  ABD, +tan 30° = AB +BD +i.e., +1 +3 = +40 +h +x +(2) +From (1), we have +h = +3 +x +Putting this value in (2), we get  + +3 +3 +x + = x + 40, i.e., 3x = x + 40 +i.e., +x = 20 +So, +h = 20 3 +[From (1)] +Therefore, the height of the tower is 20 3 m. +Example 6 : The angles of depression of the top and the bottom of an 8 m tall building +from the top of a multi-storeyed building are 30° +and 45°, respectively. Find the height of the multi- +storeyed building and the distance between the +two buildings. +Solution : In Fig. 9.9, PC denotes the multi- +storyed building and AB denotes the 8 m tall +building. We are interested to determine the height +of the multi-storeyed building, i.e., PC and the +distance between the two buildings, i.e., AC. +Look at the figure carefully. Observe that PB is +a transversal to the parallel lines PQ and BD. +Therefore, QPB and PBD are alternate +angles, +and +so +are +equal. +So PBD = 30°. Similarly, +  PAC = 45°. +In right  PBD, we have +Fig. 9.9 +Reprint 2025-26 + +140 +MATHEMATICS +PD +BD = tan 30° = 1 +3 +or BD = PD +3 +In right  PAC, we have +PC +AC = tan 45° = 1 +i.e., +PC = AC +Also, +PC = PD + DC, therefore, PD + DC = AC. +Since, AC = BD and DC = AB = 8 m, we get PD + 8 = BD = PD 3 (Why?) +This gives +PD = + + + + + + + +8 +3 +1 +8 +4 +3 +1 m. +3 +1 +3 +1 +3 +1 + + + + + + + +So, the height of the multi-storeyed building is + + + + + + +4 +3 +1 +8 m = 4 3 + +3 m + + +and the distance between the two buildings is also  + +4 3 +3 m. + +Example 7 : From a point on a bridge across a river, the angles of depression of +the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge +is at a height of 3 m from the banks, find the width of the river. +Solution : In Fig 9.10, A and B +represent points on the bank on +opposite sides of the river, so that +AB is the width of the river. P is a +point on the bridge at a height of 3 +m, i.e., DP = 3 m. We are +interested to determine the width +of the river, which is the length of +the side AB of the D APB. +Now, +AB = AD + DB +In right  APD,  A = 30°. +So, +tan 30° = PD +AD +Fig. 9.10 +Reprint 2025-26 + +SOME APPLICATIONS OF TRIGONOMETRY +141 +i.e., +1 +3 = +3 +AD or AD = 3 3 m +Also, in right  PBD,  B = 45°. So, BD = PD = 3 m. +Now, +AB = BD + AD = 3 + 3 3 = 3 (1 + +3 ) m. +Therefore, the width of the river is 3 + +3 +1 m + +. +EXERCISE 9.1 +1. A circus artist is climbing a 20 m long rope, which is +tightly stretched and tied from the top of a vertical +pole to the ground. Find the height of the pole, if +the angle made by the rope with the ground level is +30° (see Fig. 9.11). +2. A tree breaks due to storm and the broken part +bends so that the top of the tree touches the ground +making an angle 30° with it. The distance between +the foot of the tree to the point where the top +touches the ground is 8 m. Find the height of +the tree. +3. A contractor plans to install two slides for the children to play in a park. For the children +below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and +is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have +a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What +should be the length of the slide in each case? +4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m +away from the foot of the tower, is 30°. Find the height of the tower. +5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is +temporarily tied to a point on the ground. The inclination of the string with the ground +is 60°. Find the length of the string, assuming that there is no slack in the string. +6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of +elevation from his eyes to the top of the building increases from 30° to 60° as he walks +towards the building. Find the distance he walked towards the building. +7. From a point on the ground, the angles of elevation of the bottom and the top of a +transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. +Find the height of the tower. +Fig. 9.11 +Reprint 2025-26 + +142 +MATHEMATICS +8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the +angle of elevation of the top of the statue is 60° and from the same point the angle of +elevation of the top of the pedestal is 45°. Find the height of the pedestal. +9. The angle of elevation of the top of a building from the foot of the tower is 30° and the +angle of elevation of the top of the tower from the foot of the building is 60°. If the tower +is 50 m high, find the height of the building. +10. Two poles of equal heights are standing opposite each other on either side of the road, +which is 80 m wide. From a point between them on the road, the angles of elevation of +the top of the poles are 60° and 30°, respectively. Find the height of the poles and the +distances of the point from the poles. +11. A TV tower stands vertically on a bank +of a canal. From a point on the other +bank directly opposite the tower, the +angle of elevation of the top of the +tower is 60°. From another point 20 m +away from this point on the line joing +this point to the foot of the tower, the +angle of elevation of the top of the +tower is 30° (see Fig. 9.12). Find the +height of the tower and the width of +the canal. +12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is +60° and the angle of depression of its foot is 45°. Determine the height of the tower. +13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of +depression of two ships are 30° and 45°. If one ship is exactly behind the other on the +same side of the lighthouse, find the distance between the two ships. +14. A 1.2 m tall girl spots a balloon moving +with the wind in a horizontal line at a +height of 88.2 m from the ground. The +angle of elevation of the balloon from +the eyes of the girl at any instant is +60°. After some time, the angle of +elevation reduces to 30° (see Fig. 9.13). +Find the distance travelled by the +balloon during the interval. +15. A straight highway leads to the foot of a tower. A man standing at the top of the tower +observes a car at an angle of depression of 30°, which is approaching the foot of the +Fig. 9.12 +Fig. 9.13 +Reprint 2025-26 + +SOME APPLICATIONS OF TRIGONOMETRY +143 +tower with a uniform speed. Six seconds later, the angle of depression of the car is found +to be 60°. Find the time taken by the car to reach the foot of the tower from this point. +9.2 Summary +In this chapter, you have studied the following points : +1. +(i) The line of sight is the line drawn from the eye of an observer to the point in the +object viewed by the observer. +(ii) The angle of elevation of an object viewed, is the angle formed by the line of sight +with the horizontal when it is above the horizontal level, i.e., the case when we raise +our head to look at the object. +(iii) The angle of depression of an object viewed, is the angle formed by the line of sight +with the horizontal when it is below the horizontal level, i.e., the case when we lower +our head to look at the object. +2. The height or length of an object or the distance between two distant objects can be +determined with the help of trigonometric ratios. +Reprint 2025-26" +class_10,10,Circles,ncert_books/class_10/jemh1dd/jemh110.pdf,"144 +MATHEMATICS +10 +10.1 Introduction +You have studied in Class IX that a circle is a collection of all points in a plane +which are at a constant distance (radius) from a fixed point (centre). You have +also studied various terms related to a circle like chord, segment, sector, arc etc. +Let us now examine the different situations that can arise when a circle and a line +are given in a plane. +So, let us consider a circle and a line PQ. There can be three possibilities given +in Fig. 10.1 below: +Fig. 10.1 +In Fig. 10.1 (i), the line PQ and the circle have no common point. In this case, +PQ is called a non-intersecting line with respect to the circle. In Fig. 10.1 (ii), there +are two common points A and B that the line PQ and the circle have. In this case, we +call the line PQ a secant of the circle. In Fig. 10.1 (iii), there is only one point A which +is common to the line PQ and the circle. In this case, the line is called a tangent to the +circle. +CIRCLES +Reprint 2025-26 + +CIRCLES +145 +You might have seen a pulley fitted over a well which is used +in taking out water from the well. Look at Fig. 10.2. Here the rope +on both sides of the pulley, if considered as a ray, is like a tangent +to the circle representing the pulley. +Is there any position of the line with respect to the circle +other than the types given above? You can see that there cannot +be any other type of position of the line with respect to the circle. +In this chapter, we will study about the existence of the tangents +to a circle and also study some of their properties. +10.2 Tangent to a Circle +In the previous section, you have seen that a tangent* to a circle is a line that +intersects the circle at only one point. +To understand the existence of the tangent to a circle at a point, let us perform +the following activities: +Activity 1 : Take a circular wire and attach a straight wire AB at a point P of the +circular wire so that it can rotate about the point P in a plane. Put the system on a table +and gently rotate the wire AB about the point P to get different positions of the straight +wire [see Fig. 10.3(i)]. +In various positions, the wire intersects the +circular wire at P and at another point Q1 or Q2 or +Q3, etc. In one position, you will see that it will +intersect the circle at the point P only (see position +AB of AB). This shows that a tangent exists at +the point P of the circle. On rotating further, you +can observe that in all other positions of AB, it will +intersect the circle at P and at another point, say R1 +or R2 or R3, etc. So, you can observe that there is +only one tangent at a point of the circle. +While doing activity above, you must have observed that as the position AB +moves towards the position A B, the common point, say Q1, of the line AB and the +circle gradually comes nearer and nearer to the common point P. Ultimately, it coincides +with the point P in the position AB of AB. Again note, what happens if ‘AB’ is +rotated rightwards about P? The common point R3 gradually comes nearer and nearer +to P and ultimately coincides with P. So, what we see is: +The tangent to a circle is a special case of the secant, when the two end +points of its corresponding chord coincide. +Fig. 10.3 (i) +Fig. 10.2 +*The word ‘tangent’ comes from the Latin word ‘tangere’, which means to touch and was +introduced by the Danish mathematician Thomas Fineke in 1583. +Reprint 2025-26 + +146 +MATHEMATICS +Activity 2 : On a paper, draw a circle and a +secant PQ of the circle. Draw various lines +parallel to the secant on both sides of it. You +will find that after some steps, the length of +the chord cut by the lines will gradually +decrease, i.e., the two points of intersection of +the line and the circle are coming closer and +closer [see Fig. 10.3(ii)]. In one case, it +becomes zero on one side of the secant and in +another case, it becomes zero on the other side +of the secant. See the positions PQ and PQ +of the secant in Fig. 10.3 (ii). These are the +tangents to the circle parallel to the given secant +PQ. This also helps you to see that there cannot +be more than two tangents parallel to a given +secant. +This activity also establishes, what you must have observed, while doing +Activity 1, namely, a tangent is the secant when both of the end points of the +corresponding chord coincide. +The common point of the tangent and the circle is called the point of contact +[the point A in Fig. 10.1 (iii)]and the tangent is said to touch the circle at the +common point. +Now look around you. Have you seen a bicycle +or a cart moving? Look at its wheels. All the spokes +of a wheel are along its radii. Now note the position +of the wheel with respect to its movement on the +ground. Do you see any tangent anywhere? +(See Fig. 10.4). In fact, the wheel moves along a line +which is a tangent to the circle representing the wheel. +Also, notice that in all positions, the radius through +the point of contact with the ground appears to be at +right angles to the tangent (see Fig. 10.4). We shall +now prove this property of the tangent. +Theorem 10.1 : The tangent at any point of a circle is perpendicular to the +radius through the point of contact. +Proof : We are given a circle with centre O and a tangent XY to the circle at a +point P. We need to prove that OP is perpendicular to XY. +Fig. 10.4 +Fig. 10.3 (ii) +Reprint 2025-26 + +CIRCLES +147 +Take a point Q on XY other than P and join OQ (see Fig. 10.5). +The point Q must lie outside the circle. +(Why? Note that if Q lies inside the circle, XY +will become a secant and not a tangent to the +circle). Therefore, OQ is longer than the radius +OP of the circle. That is, +OQ > OP. +Since this happens for every point on the +line XY except the point P, OP is the +shortest of all the distances of the point O to the +points of XY. So OP is perpendicular to XY. +(as shown in Theorem A1.7.) +Remarks +1. By theorem above, we can also conclude that at any point on a circle there can be +one and only one tangent. +2. The line containing the radius through the point of contact is also sometimes called +the ‘normal’ to the circle at the point. +EXERCISE 10.1 +1. How many tangents can a circle have? +2. Fill in the blanks : +(i) A tangent to a circle intersects it in + point (s). +(ii) A line intersecting a circle in two points is called a + . +(iii) A circle can have + parallel tangents at the most. +(iv) The common point of a tangent to a circle and the circle is called + . +3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at +a point Q so that OQ = 12 cm. Length PQ is : +(A) 12 cm +(B) 13 cm +(C) 8.5 cm +(D) +119 cm. +4. Draw a circle and two lines parallel to a given line such that one is a tangent and the +other, a secant to the circle. +10.3 Number of Tangents from a Point on a Circle +To get an idea of the number of tangents from a point on a circle, let us perform the +following activity: +Fig. 10.5 +Reprint 2025-26 + +148 +MATHEMATICS +Activity 3 : Draw a circle on a paper. Take a +point P inside it. Can you draw a tangent to the +circle through this point? You will find that all +the lines through this point intersect the circle in +two points. So, it is not possible to draw any +tangent to a circle through a point inside it +[see Fig. 10.6 (i)]. +Next take a point P on the circle and draw +tangents through this point. You have already +observed that there is only one tangent to the +circle at such a point [see Fig. 10.6 (ii)]. +Finally, take a point P outside the circle and +try to draw tangents to the circle from this point. +What do you observe? You will find that you +can draw exactly two tangents to the circle +through this point [see Fig. 10.6 (iii)]. +We can summarise these facts as follows: +Case 1 : There is no tangent to a circle passing +through a point lying inside the circle. +Case 2 : There is one and only one tangent to a +circle passing through a point lying on the circle. +Case 3 : There are exactly two tangents to a +circle through a point lying outside the circle. +In Fig. 10.6 (iii), T1and T2 are the points of +contact of the tangents PT1 and PT2 +respectively. +The length of the segment of the tangent +from the external point P and the point of contact +with the circle is called the length of the tangent +from the point P to the circle. +Note that in Fig. 10.6 (iii), PT1 and PT2 are the lengths of the tangents from P to +the circle. The lengths PT1 and PT2 have a common property. Can you find this? +Measure PT1 and PT2. Are these equal? In fact, this is always so. Let us give a proof +of this fact in the following theorem. +(i) +(ii) +Fig. 10.6 +(iii) +Reprint 2025-26 + +CIRCLES +149 +Theorem 10.2 : The lengths of tangents drawn +from an external point to a circle are equal. +Proof : We are given a circle with centre O, a +point P lying outside the circle and two tangents +PQ, PR on the circle from P (see Fig. 10.7). We +are required to prove that PQ = PR. +For this, we join OP, OQ and OR. Then +∠ OQP and ∠ ORP are right angles, because +these are angles between the radii and tangents, +and according to Theorem 10.1 they are right +angles. Now in right triangles OQP and ORP, +OQ = OR +(Radii of the same circle) +OP = OP +(Common) +Therefore, +∆ OQP ≅∆ ORP +(RHS) +This gives +PQ = PR +(CPCT) +Remarks +1. The theorem can also be proved by using the Pythagoras Theorem as follows: +PQ2 = OP2 – OQ2 = OP2 – OR2 = PR2 (As OQ = OR) +which gives PQ = PR. +2. Note also that ∠ OPQ = ∠ OPR. Therefore, OP is the angle bisector of ∠ QPR, +i.e., the centre lies on the bisector of the angle between the two tangents. +Let us take some examples. +Example 1 : Prove that in two concentric circles, +the chord of the larger circle, which touches the +smaller circle, is bisected at the point of contact. +Solution : We are given two concentric circles +C1 and C2 with centre O and a chord AB of the +larger circle C1 which touches the smaller circle +C2 at the point P (see Fig. 10.8). We need to prove +that AP = BP. +Let us join OP. Then, AB is a tangent to C2 at P +and OP is its radius. Therefore, by Theorem 10.1, +OP ⊥AB +Fig. 10.7 +Fig. 10.8 +Reprint 2025-26 + +150 +MATHEMATICS +Now AB is a chord of the circle C1 and OP  AB. Therefore, OP is the bisector of the +chord AB, as the perpendicular from the centre bisects the chord, +i.e., +AP = BP +Example 2 : Two tangents TP and TQ are drawn +to a circle with centre O from an external point T. +Prove that  PTQ = 2  OPQ. +Solution : We are given a circle with centre O, +an external point T and two tangents TP and TQ +to the circle, where P, Q are the points of contact +(see Fig. 10.9). We need to prove that + PTQ = 2  OPQ +Let + PTQ =  +Now, by Theorem 10.2, TP = TQ. So, TPQ is an isosceles triangle. +Therefore, + TPQ =  TQP = 1 +1 +(180° +) +90° +2 +2 + + + +Also, by Theorem 10.1, + OPT = 90° +So, + OPQ =  OPT –  TPQ = +1 +90° +90° – 2 + + + + + + + + += 1 +1 +PTQ +2 +2 + + +This gives + PTQ = 2  OPQ +Example 3 : PQ is a chord of length 8 cm of a +circle of radius 5 cm. The tangents at P and Q +intersect at a point T (see Fig. 10.10). Find the +length TP. +Solution : Join OT. Let it intersect PQ at the +point R. Then  TPQ is isosceles and TO is the +angle bisector of  PTQ. So, OT  PQ +and therefore, OT bisects PQ which gives +PR = RQ = 4 cm. +Also, +OR = +2 +2 +2 +2 +OP +PR +5 +4 cm +3 cm + + + + +. +Fig. 10.9 +Fig. 10.10 +Reprint 2025-26 + +CIRCLES +151 +Now, + TPR +  RPO = 90° =  TPR +  PTR +(Why?) +So, + RPO =  PTR +Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity. +This gives +TP +PO = RP +RO , i.e., TP +5 = 4 +3 or TP = 20 +3 cm. +Note : TP can also be found by using the Pythagoras Theorem, as follows: +Let +TP = x and TR = y. +Then +x2 = y2 + 16 +(Taking right  PRT) +(1) +x2 + 52 = (y + 3)2 +(Taking right  OPT) +(2) +Subtracting (1) from (2), we get +25 = 6y – 7 +or +y = 32 +16 +6 +3 + +Therefore, +x2 = +2 +16 +16 +16 +25 +16 +(16 +9) +3 +9 +9 + + + + + + + + + + +[From (1)] +or +x = 20 +3 +EXERCISE +10.2 +In Q.1 to 3, choose the correct option and give justification. +1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from +the centre is 25 cm. The radius of the circle is +(A) 7 cm +(B) 12 cm +(C) +15 cm +(D) 24.5 cm +2. In Fig. 10.11, if TP and TQ are the two tangents +to a circle with centre O so that  POQ = 110°, +then PTQ is equal to +(A) 60° +(B) 70° +(C) +80° +(D) 90° +3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other +at angle of 80°, then  POA is equal to +(A) 50° +(B) 60° +(C) +70° +(D) 80° +Fig. 10.11 +Reprint 2025-26 + +152 +MATHEMATICS +4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel. +5. Prove that the perpendicular at the point of contact to the tangent to a circle passes +through the centre. +6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 +cm. Find the radius of the circle. +7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the +larger circle which touches the smaller circle. +8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that +AB + CD = AD + BC +Fig. 10.12 +Fig. 10.13 +9. In Fig. 10.13, XY and XY are two parallel tangents to a circle with centre O and +another tangent AB with point of contact C intersecting XY at A and XY at B. Prove +that  AOB = 90°. +10. Prove that the angle between the two tangents drawn from an external point to a circle +is supplementary to the angle subtended by the line-segment joining the points of +contact at the centre. +11. Prove that the parallelogram circumscribing a +circle is a rhombus. +12. A triangle ABC is drawn to circumscribe a circle +of radius 4 cm such that the segments BD and +DC into which BC is divided by the point of +contact D are of lengths 8 cm and 6 cm +respectively (see Fig. 10.14). Find the sides AB +and AC. +13. Prove that opposite sides of a quadrilateral +circumscribing a circle subtend supplementary +angles at the centre of the circle. +Fig. 10.14 +Reprint 2025-26 + +CIRCLES +153 +10.4 +Summary +In this chapter, you have studied the following points : +1. +The meaning of a tangent to a circle. +2. +The tangent to a circle is perpendicular to the radius through the point of contact. +3. +The lengths of the two tangents from an external point to a circle are equal. +Reprint 2025-26" +class_10,11,Areas Related to Circles,ncert_books/class_10/jemh1dd/jemh111.pdf,"154 +MATHEMATICS +11 +11.1 Areas of Sector and Segment of a Circle +You have already come across the terms sector and +segment of a circle in your earlier classes. Recall +that the portion (or part) of the circular region enclosed +by two radii and the corresponding arc is called a +sector of the circle and the portion (or part) of the +circular region enclosed between a chord and the +corresponding arc is called a segment of the circle. +Thus, in Fig. 11.1, shaded region OAPB is a sector +of the circle with centre O. ∠ AOB is called the +angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of +the circle. For obvious reasons, OAPB is called the minor sector and +OAQB is called the major sector. You can also see that angle of the major sector is +360° – ∠ AOB. +Now, look at Fig. 11.2 in which AB is a chord +of the circle with centre O. So, shaded region APB is +a segment of the circle. You can also note that +unshaded region AQB is another segment of the circle +formed by the chord AB. For obvious reasons, APB +is called the minor segment and AQB is called the +major segment. +Remark : When we write ‘segment’ and ‘sector’ +we will mean the ‘minor segment’ and the ‘minor +sector’ respectively, unless stated otherwise. +AREAS RELATED TO CIRCLES +Fig. 11.2 +Fig. 11.1 +Reprint 2025-26 + +AREAS RELATED TO CIRCLES +155 +Now with this knowledge, let us try to find some +relations (or formulae) to calculate their areas. +Let OAPB be a sector of a circle with centre +O and radius r (see Fig. 11.3). Let the degree +measure of Ð AOB be q. +You know that area of a circle (in fact of a +circular region or disc) is pr2. +In a way, we can consider this circular region to +be a sector forming an angle of 360° (i.e., of degree +measure 360) at the centre O. Now by applying the +Unitary Method, we can arrive at the area of the +sector OAPB as follows: +When degree measure of the angle at the centre is 360, area of the +sector = pr2 +So, when the degree measure of the angle at the centre is 1, area of the +sector = +2 +360 +r + + +Therefore, when the degree measure of the angle at the centre is q, area of the +sector = +2 +360 +r + + = +2 +360 +r + +. +Thus, we obtain the following relation (or formula) for area of a sector of a +circle: +Area of the sector of angle q = + +r + + +2 +360 +, +where r is the radius of the circle and q the angle of the sector in degrees. +Now, a natural question arises : Can we find +the length of the arc APB corresponding to this +sector? Yes. Again, by applying the Unitary +Method and taking the whole length of the circle +(of angle 360°) as 2pr, we can obtain the required +length of the arc APB as +2 +360 +r +. +So, length of an arc of a sector of angle q = + + + +2 +360 +r . +Fig. 11.3 +Fig. 11.4 +Reprint 2025-26 + +156 +MATHEMATICS +Now let us take the case of the area of the +segment APB of a circle with centre O and radius r +(see Fig. 11.4). You can see that : +Area of the segment APB = Area of the sector OAPB – Area of  OAB + = +2 – area of +OAB +360 +r + + +Note : From Fig. 11.3 and Fig. 11.4 respectively, you can observe that: +Area of the major sector OAQB = r2 – Area of the minor sector OAPB +and Area of major segment AQB = r2 – Area of the minor segment APB +Let us now take some examples to understand these concepts (or results). +Example 1 : Find the area of the sector of a circle +with radius 4 cm and of angle 30°. Also, find the area +of the corresponding major sector (Use  = 3.14). +Solution : Given sector is OAPB (see Fig. 11.5). +Area of the sector = +2 +360 +r + += +2 +30 +3.14 +4 +4 cm +360  + + += +2 +2 +12.56 cm +4.19cm +3 + + (approx.) +Area of the corresponding major sector += r2 – area of sector OAPB += (3.14 × 16 – 4.19) cm2 += 46.05 cm2 = 46.1 cm2(approx.) +Alternatively, area of the major sector = +2 +(360 – +) +360 +r + += +2 +360 +30 +3.14 +16 cm +360 + + + + + + + + += +2 +2 +330 +3.14 +16cm +46.05 cm +360  + + += 46.1 cm2 (approx.) +Fig. 11.5 +Reprint 2025-26 + +AREAS RELATED TO CIRCLES +157 +Example 2 : Find the area of the segment AYB +shown in Fig. 11.6, if radius of the circle is 21 cm and + AOB = 120°. (Use  = 22 +7 ) +Solution : Area of the segment AYB += Area of sector OAYB – Area of  OAB +(1) +Now, +area of the sector OAYB = 120 +22 +21 +21 +360 +7 + + + +cm2 = 462 cm2 +(2) +For finding the area of  OAB, draw OM  AB as shown in Fig. 11.7. +Note that OA = OB. Therefore, by RHS congruence,  AMO  BMO. +So, M is the mid-point of AB and  AOM =  BOM = 1 +120 +60 +2  + +. +Let +OM = x cm +So, from  OMA, +OM +OA = cos 60° +or, +21 +x = +1 +1 +cos 60° = +2 +2 + + + + + + +or, +x = 21 +2 +So, +OM = 21 +2 cm +Also, +AM +OA = sin 60° = +3 +2 +So, +AM = 21 3 +2 + cm +Therefore, +AB = 2 AM = 2 +21 3 cm = 21 3cm +2 + +Fig. 11.6 +Fig. 11.7 +Reprint 2025-26 + +158 +MATHEMATICS +So, +area of  OAB = 1 AB × OM +2 + = +2 +1 +21 +21 3 +cm +2 +2 + + += +2 +441 3cm +4 +(3) +Therefore, area of the segment AYB = +2 +441 +462 +3 +cm +4 + + + + + + + +[From (1), (2) and (3)] += +2 +21 (88 – 21 3)cm +4 +EXERCISE 11.1 +Unless stated otherwise, use  = 22 +7 . +1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. +2. Find the area of a quadrant of a circle whose circumference is 22 cm. +3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute +hand in 5 minutes. +4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of +the corresponding : (i) minor segment +(ii) major sector. (Use  = 3.14) +5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: +(i) the length of the arc +(ii) area of the sector formed by the arc +(iii) area of the segment formed by the corresponding chord +6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find +the areas of the corresponding minor and major +segments of the circle. +(Use  = 3.14 and 3 = 1.73) +7. A chord of a circle of radius 12 cm subtends an +angle of 120° at the centre. Find the area of the +corresponding segment of the circle. +(Use  = 3.14 and 3 = 1.73) +8. A horse is tied to a peg at one corner of a square +shaped grass field of side 15 m by means of a 5 m +long rope (see Fig. 11.8). Find +Fig. 11.8 +Reprint 2025-26 + +AREAS RELATED TO CIRCLES +159 +(i) the area of that part of the field in which the +horse can graze. +(ii) the increase in the grazing area if the rope +were 10 m long instead of 5 m. (Use  = 3.14) +9. A brooch is made with silver wire in the form +of a circle with diameter 35 mm. The wire is also +used in making 5 diameters which divide the +circle into 10 equal sectors as shown in +Fig. 11.9. Find : +(i) the total length of the silver wire required. +(ii) the area of each sector of the brooch. +10. An umbrella has 8 ribs which are equally spaced +(see Fig. 11.10). Assuming umbrella to be a flat +circle of radius 45 cm, find the area between the +two consecutive ribs of the umbrella. +11. A car has two wipers which do not overlap. Each +wiper has a blade of length 25 cm sweeping +through an angle of 115°. Find the total area +cleaned at each sweep of the blades. +12. To warn ships for underwater rocks, a lighthouse +spreads a red coloured light over a sector of +angle 80° to a distance of 16.5 km. Find the area +of the sea over which the ships are warned. +(Use  = 3.14) +13. A round table cover has six equal designs as +shown in Fig. 11.11. If the radius of the cover is +28 cm, find the cost of making the designs at the +rate of ` 0.35 per cm2. (Use 3 = 1.7) +14. Tick the correct answer in the following : +Area of a sector of angle p (in degrees) of a circle +with radius R is +(A) +2 +R +180 +p  + +(B) +2 +R +180 +p  +(C) +2 +R +360 +p  + +(D) +2 +2 +R +720 +p  + +Fig. 11.9 +Fig. 11.10 +Fig. 11.11 +Reprint 2025-26 + +160 +MATHEMATICS +11.2 Summary +In this chapter, you have studied the following points : +1. Length of an arc of a sector of a circle with radius r and angle with degree measure  is +2 +360 +r + + + +2. Area of a sector of a circle with radius r and angle with degree measure  is +2 +360 +r + + + +3. Area of segment of a circle += Area of the corresponding sector – Area of the corresponding triangle. +Reprint 2025-26" +class_10,12,Surface Areas and Volumes,ncert_books/class_10/jemh1dd/jemh112.pdf,"SURFACE AREAS AND VOLUMES +161 +12 +12.1 Introduction +From Class IX, you are familiar with some of the solids like cuboid, cone, cylinder, and +sphere (see Fig. 12.1). You have also learnt how to find their surface areas and volumes. +Fig. 12.1 +In our day-to-day life, we come across a number of solids made up of combinations +of two or more of the basic solids as shown above. +You must have seen a truck with a +container fitted on its back (see Fig. 12.2), +carrying oil or water from one place to +another. Is it in the shape of any of the four +basic solids mentioned above? You may +guess that it is made of a cylinder with two +hemispheres as its ends. +SURFACE AREAS AND +VOLUMES +Fig. 12.2 +Reprint 2025-26 + +162 +MATHEMATICS +Again, you may have seen an object like the +one in Fig. 12.3. Can you name it? A test tube, right! +You would have used one in your science laboratory. +This tube is also a combination of a cylinder and a +hemisphere. Similarly, while travelling, you may have +seen some big and beautiful buildings or monuments +made up of a combination of solids mentioned above. +If for some reason you wanted to find the +surface areas, or volumes, or capacities of such +objects, how would you do it? We cannot classify +these under any of the solids you have already studied. +In this chapter, you will see how to find surface areas and volumes of such +objects. +12.2 Surface Area of a Combination of Solids +Let us consider the container seen in Fig. 12.2. How do we find the surface area of +such a solid? Now, whenever we come across a new problem, we first try to see, if +we can break it down into smaller problems, we have earlier solved. We can see that +this solid is made up of a cylinder with two hemispheres stuck at either end. It would +look like what we have in Fig. 12.4, after we put the pieces all together. +Fig. 12.4 +If we consider the surface of the newly formed object, we would be able to see +only the curved surfaces of the two hemispheres and the curved surface of the cylinder. +So, the total surface area of the new solid is the sum of the curved surface +areas of each of the individual parts. This gives, +TSA of new solid = CSA of one hemisphere + CSA of cylinder + + CSA of other hemisphere +where TSA, CSA stand for ‘Total Surface Area’ and ‘Curved Surface Area’ +respectively. +Let us now consider another situation. Suppose we are making a toy by putting +together a hemisphere and a cone. Let us see the steps that we would be going +through. +Fig. 12.3 +Reprint 2025-26 + +SURFACE AREAS AND VOLUMES +163 +First, we would take a cone and a hemisphere and bring their flat faces together. +Here, of course, we would take the base radius of the cone equal to the radius of the +hemisphere, for the toy is to have a smooth surface. So, the steps would be as shown +in Fig. 12.5. +Fig. 12.5 +At the end of our trial, we have got ourselves a nice round-bottomed toy. Now if +we want to find how much paint we would require to colour the surface of this toy, +what would we need to know? We would need to know the surface area of the toy, +which consists of the CSA of the hemisphere and the CSA of the cone. +So, we can say: +Total surface area of the toy = CSA of hemisphere + CSA of cone +Now, let us consider some examples. +Example 1 : Rasheed got a playing top (lattu) as his +birthday present, which surprisingly had no colour on +it. He wanted to colour it with his crayons. The top is +shaped like a cone surmounted by a hemisphere +(see Fig 12.6). The entire top is 5 cm in height and +the diameter of the top is 3.5 cm. Find the area he +has to colour. (Take  = 22 +7 ) +Solution : This top is exactly like the object we have discussed in Fig. 12.5. So, we +can conveniently use the result we have arrived at there. That is : +TSA of the toy = CSA of hemisphere + CSA of cone +Now, the curved surface area of the hemisphere = +2 +2 +1 (4 +) +2 +2 +r +r + + + += +2 +22 +3.5 +3.5 +2 +cm +7 +2 +2 + + + + + + + + + +Fig. 12.6 +Reprint 2025-26 + +164 +MATHEMATICS +Also, the height of the cone = height of the top – height (radius) of the hemispherical part += +3.5 +5 +cm +2 + + + + + + + + = 3.25 cm +So, the slant height of the cone (l ) = +2 +2 +2 +2 +3.5 +(3.25) cm +2 +r +h + + + + + + + + + + = 3.7 cm (approx.) +Therefore, CSA of cone = rl = +2 +22 +3.5 +3.7 cm +7 +2 + + + + + + + + +This gives the surface area of the top as += +2 +2 +22 +3.5 +3.5 +22 +3.5 +2 +cm +3.7 cm +7 +2 +2 +7 +2 + + + + + + + + + + + + + + + + + + += + + +2 +22 +3.5 3.5 +3.7 cm +7 +2 + + + = +2 +2 +11 +(3.5 +3.7) cm +39.6 cm (approx.) +2  + + +You may note that ‘total surface area of the top’ is not the sum of the total +surface areas of the cone and hemisphere. +Example 2 : The decorative block shown +in Fig. 12.7 is made of two solids — a cube +and a hemisphere. The base of the block is a +cube with edge 5 cm, and the hemisphere +fixed on the top has a diameter of 4.2 cm. +Find the total surface area of the block. +(Take  = 22 +7 ) +Solution : The total surface area of the cube = 6 × (edge)2 = 6 × 5 × 5 cm2 = 150 cm2. +Note that the part of the cube where the hemisphere is attached is not included in the +surface area. +So, +the surface area of the block = TSA of cube – base area of hemisphere ++ CSA of hemisphere += 150 – r2 + 2 r2 = (150 + r2) cm2 += +2 +2 +22 +4.2 +4.2 +150 cm +cm +7 +2 +2 + + + + + + + + + += (150 + 13.86) cm2 = 163.86 cm2 +. +Fig. 12.7 +Reprint 2025-26 + +SURFACE AREAS AND VOLUMES +165 +Example 3 : A wooden toy rocket is in the +shape of a cone mounted on a cylinder, as +shown in Fig. 12.8. The height of the entire +rocket is 26 cm, while the height of the conical +part is 6 cm. The base of the conical portion +has a diameter of 5 cm, while the base +diameter of the cylindrical portion is 3 cm. If +the conical portion is to be painted orange +and the cylindrical portion yellow, find the +area of the rocket painted with each of these +colours. (Take  = 3.14) +Solution : Denote radius of cone by r, slant +height of cone by l, height of cone by h, radius +of cylinder by r and height of cylinder by h. +Then r = 2.5 cm, h = 6 cm, r = 1.5 cm, +h = 26 – 6 = 20 cm and +l = +2 +2 +r +h + + = +2 +2 +2.5 +6 +cm + + = 6.5 cm +Here, the conical portion has its circular base resting on the base of the cylinder, but +the base of the cone is larger than the base of the cylinder. So, a part of the base of the +cone (a ring) is to be painted. +So, +the area to be painted orange = CSA of the cone + base area of the cone + – base area of the cylinder += rl + r2 – r2 += [(2.5 × 6.5) + (2.5)2 – (1.5)2] cm2 += [20.25] cm2 = 3.14 × 20.25 cm2 += 63.585 cm2 +Now, +the area to be painted yellow = CSA of the cylinder ++ area of one base of the cylinder += 2rh + (r)2 += r (2h + r) += (3.14 × 1.5) (2 × 20 + 1.5) cm2 += 4.71 × 41.5 cm2 += 195.465 cm2 +Fig. 12.8 +Reprint 2025-26 + +166 +MATHEMATICS +Example 4 : Mayank made a bird-bath for his garden +in the shape of a cylinder with a hemispherical +depression at one end (see Fig. 12.9). The height of +the cylinder is 1.45 m and its radius is 30 cm. Find the +total surface area of the bird-bath. (Take  = 22 +7 ) +Solution : Let h be height of the cylinder, and r the +common radius of the cylinder and hemisphere. Then, +the total surface area of the bird-bath = CSA of cylinder + CSA of hemisphere += 2rh + 2r2 = 2r(h + r) += +2 +22 +2 +30(145 +30) cm +7 + + + += 33000 cm2 = 3.3 m2 +EXERCISE 12.1 +Unless stated otherwise, take  = +22 +7  +1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the +resulting cuboid. +2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The +diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the +inner surface area of the vessel. +3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. +The total height of the toy is 15.5 cm. Find the total surface area of the toy. +4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest +diameter the hemisphere can have? Find the surface area of the solid. +5. A hemispherical depression is cut out from one face of a cubical wooden block such +that the diameter l of the hemisphere is equal to the edge of the cube. Determine the +surface area of the remaining solid. +6. A medicine capsule is in the shape of a +cylinder with two hemispheres stuck to each +of its ends (see Fig. 12.10). The length of +the entire capsule is 14 mm and the diameter +of the capsule is 5 mm. Find its surface area. +Fig. 12.10 +Fig. 12.9 +Reprint 2025-26 + +SURFACE AREAS AND VOLUMES +167 +7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and +diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the +top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of +the canvas of the tent at the rate of ` 500 per m2. (Note that the base of the tent will not +be covered with canvas.) +8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the +same height and same diameter is hollowed out. Find the total surface area of the +remaining solid to the nearest cm2. +9. A wooden article was made by scooping +out a hemisphere from each end of a solid +cylinder, as shown in Fig. 12.11. If the +height of the cylinder is 10 cm, and its +base is of radius 3.5 cm, find the total +surface area of the article. +12.3 Volume of a Combination of Solids +In the previous section, we have discussed how to find the surface area of solids made +up of a combination of two basic solids. Here, we shall see how to calculate their +volumes. It may be noted that in calculating the surface area, we have not added the +surface areas of the two constituents, because some part of the surface area disappeared +in the process of joining them. However, this will not be the case when we calculate +the volume. The volume of the solid formed by joining two basic solids will actually be +the sum of the volumes of the constituents, as we see in the examples below. +Example 5 : Shanta runs an industry in +a shed which is in the shape of a cuboid +surmounted by a half cylinder (see Fig. +12.12). If the base of the shed is of +dimension 7 m × 15 m, and the height of +the cuboidal portion is 8 m, find the volume +of air that the shed can hold. Further, +suppose the machinery in the shed +occupies a total space of 300 m3, and +there are 20 workers, each of whom +occupy about 0.08 m3 space on an +average. Then, how much air is in the +shed? (Take  = 22 +7 ) +Fig. 12.12 +Fig. 12.11 +Reprint 2025-26 + +168 +MATHEMATICS +Solution : The volume of air inside the shed (when there are no people or machinery) +is given by the volume of air inside the cuboid and inside the half cylinder, taken +together. +Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively. +Also, the diameter of the half cylinder is 7 m and its height is 15 m. +So, the required volume = volume of the cuboid + 1 +2 volume of the cylinder += +3 +1 +22 +7 +7 +15 +7 +8 +15 m +2 +7 +2 +2 + + + + + + + + + + + + + + = 1128.75 m3 +Next, the total space occupied by the machinery = 300 m3 +And the total space occupied by the workers = 20 × 0.08 m3 = 1.6 m3 +Therefore, the volume of the air, when there are machinery and workers += 1128.75 – (300.00 + 1.60) = 827.15 m3 +Example 6 : A juice seller was serving his +customers using glasses as shown in Fig. 12.13. +The inner diameter of the cylindrical glass was +5 cm, but the bottom of the glass had a +hemispherical raised portion which reduced the +capacity of the glass. If the height of a glass +was 10 cm, find the apparent capacity of the +glass and its actual capacity. (Use  = 3.14.) +Solution : Since the inner diameter of the glass = 5 cm and height = 10 cm, +the apparent capacity of the glass = r2h += 3.14 × 2.5 × 2.5 × 10 cm3 = 196.25 cm3 +But the actual capacity of the glass is less by the volume of the hemisphere at the +base of the glass. +i.e., +it is less by 2 +3 r3 = +3 +2 +3.14 +2.5 +2.5 +2.5 cm +3  + + + + = 32.71 cm3 +So, the actual capacity of the glass = apparent capacity of glass – volume of the +hemisphere += (196.25 – 32.71) cm3 += 163.54 cm3 +Fig. 12.13 +Reprint 2025-26 + +SURFACE AREAS AND VOLUMES +169 +Example 7 : A solid toy is in the form of a +hemisphere surmounted by a right circular cone. The +height of the cone is 2 cm and the diameter of the +base is 4 cm. Determine the volume of the toy. If a +right circular cylinder circumscribes the toy, find the +difference of the volumes of the cylinder and the toy. +(Take  = 3.14) +Solution : Let BPC be the hemisphere and ABC be the cone standing on the base +of the hemisphere (see Fig. 12.14). The radius BO of the hemisphere (as well as +of the cone) = 1 +2 × 4 cm = 2 cm. +So, +volume of the toy = +3 +2 +2 +1 +3 +3 + + + +r +r h += +3 +2 +3 +2 +1 +3.14 +(2) +3.14 +(2) +2 cm +3 +3 + + + + + + + + + + + + + = 25.12 cm3 +Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of +the base of the right circular cylinder = HP = BO = 2 cm, and its height is +EH = AO + OP = (2 + 2) cm = 4 cm +So, the volume required = volume of the right circular cylinder – volume of the toy + = (3.14 × 22 × 4 – 25.12) cm3 + = 25.12 cm3 +Hence, the required difference of the two volumes = 25.12 cm3. +EXERCISE 12.2 +Unless stated otherwise, take  = 22 +7 . + + + + +A solid is in the shape of a cone standing on a hemisphere with both their radii being +equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid +in terms of . +2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with +two cones attached at its two ends by using a thin aluminium sheet. The diameter of the +model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume +of air contained in the model that Rachel made. (Assume the outer and inner dimensions +of the model to be nearly the same.) +Fig. 12.14 +Reprint 2025-26 + +170 +MATHEMATICS +3. A gulab jamun, contains sugar syrup up to about +30% of its volume. Find approximately how much +syrup would be found in 45 gulab jamuns, each +shaped like a cylinder with two hemispherical ends +with length 5 cm and diameter 2.8 cm (see Fig. 12.15). +4. A pen stand made of wood is in the shape of a +cuboid with four conical depressions to hold pens. +The dimensions of the cuboid are 15 cm by 10 cm by +3.5 cm. The radius of each of the depressions is 0.5 +cm and the depth is 1.4 cm. Find the volume of +wood in the entire stand (see Fig. 12.16). +5. A vessel is in the form of an inverted cone. Its +height is 8 cm and the radius of its top, which is +open, is 5 cm. It is filled with water up to the brim. +When lead shots, each of which is a sphere of radius +0.5 cm are dropped into the vessel, one-fourth of +the water flows out. Find the number of lead shots +dropped in the vessel. +6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which +is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the +pole, given that 1 cm3 of iron has approximately 8g mass. (Use = 3.14) +7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on +a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water +such that it touches the bottom. Find the volume of water left in the cylinder, if the radius +of the cylinder is 60 cm and its height is 180 cm. +8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter +of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its +volume to be 345 cm3. Check whether she is correct, taking the above as the inside +measurements, and  = 3.14. +12.4 Summary +In this chapter, you have studied the following points: +1. To determine the surface area of an object formed by combining any two of the basic +solids, namely, cuboid, cone, cylinder, sphere and hemisphere. +2. To find the volume of objects formed by combining any two of a cuboid, cone, cylinder, +sphere and hemisphere. +Fig. 12.16 +Fig. 12.15 +Reprint 2025-26" +class_10,13,Statistics,ncert_books/class_10/jemh1dd/jemh113.pdf,"STATISTICS +171 +13 +13.1 Introduction +In Class IX, you have studied the classification of given data into ungrouped as well as +grouped frequency distributions. You have also learnt to represent the data pictorially +in the form of various graphs such as bar graphs, histograms (including those of varying +widths) and frequency polygons. In fact, you went a step further by studying certain +numerical representatives of the ungrouped data, also called measures of central +tendency, namely, mean, median and mode. In this chapter, we shall extend the study +of these three measures, i.e., mean, median and mode from ungrouped data to that of +grouped data. We shall also discuss the concept of cumulative frequency, the +cumulative frequency distribution and how to draw cumulative frequency curves, called +ogives. +13.2 Mean of Grouped Data +The mean (or average) of observations, as we know, is the sum of the values of all the +observations divided by the total number of observations. From Class IX, recall that if +x1, x2,. . ., xn are observations with respective frequencies f1, f2, . . ., fn, then this +means observation x1 occurs f1 times, x2 occurs f2 times, and so on. +Now, the sum of the values of all the observations = f1x1 + f2x2 + . . . + fnxn, and +the number of observations = f1 + f2 + . . . + fn. +So, the mean x of the data is given by +x = +1 1 +2 +2 +1 +2 +n +n +n +f x +f x +f x +f +f +f + + + + + + + + +Recall that we can write this in short form by using the Greek letter  (capital +sigma) which means summation. That is, +STATISTICS +Reprint 2025-26 + +172 +MATHEMATICS +x = +1 +1 +n +i +i +i +n +i +i +f x +f + + + + +which, more briefly, is written as x =  + +i +i +i +f x +f +, if it is understood that i varies from +1 to n. +Let us apply this formula to find the mean in the following example. +Example 1 : The marks obtained by 30 students of Class X of a certain school in a +Mathematics paper consisting of 100 marks are presented in table below. Find the +mean of the marks obtained by the students. +Marks obtained +10 +20 +36 +40 +50 +56 +60 +70 +72 +80 +88 +92 +95 +(xi) + Number of +1 +1 +3 +4 +3 +2 +4 +4 +1 +1 +2 +3 +1 +students ( fi) +Solution: Recall that to find the mean marks, we require the product of each xi with +the corresponding frequency fi. So, let us put them in a column as shown in Table 13.1. +Table 13.1 +Marks obtained (xi) +Number of students ( fi) +fixi +10 +1 +10 +20 +1 +20 +. +36 +3 +108 +40 +4 +160 +50 +3 +150 +56 +2 +112 +60 +4 +240 +70 +4 +280 +72 +1 +72 +80 +1 +80 +88 +2 +176 +92 +3 +276 +95 +1 +95 +Total +fi = 30 +fixi = 1779 +Reprint 2025-26 + +STATISTICS +173 +Now, + + + + +i +i +i +f x +x +f + = 1779 +30 = 59.3 +Therefore, the mean marks obtained is 59.3. +In most of our real life situations, data is usually so large that to make a meaningful +study it needs to be condensed as grouped data. So, we need to convert given ungrouped +data into grouped data and devise some method to find its mean. +Let us convert the ungrouped data of Example 1 into grouped data by forming +class-intervals of width, say 15. Remember that, while allocating frequencies to each +class-interval, students falling in any upper class-limit would be considered in the next +class, e.g., 4 students who have obtained 40 marks would be considered in the class- +interval 40-55 and not in 25-40. With this convention in our mind, let us form a grouped +frequency distribution table (see Table 13.2). +Table 13.2 +Class interval +10 - 25 +25 - 40 +40 - 55 +55 - 70 +70 - 85 +85 - 100 +Number of students +2 +3 +7 +6 +6 +6 +Now, for each class-interval, we require a point which would serve as the +representative of the whole class. It is assumed that the frequency of each class- +interval is centred around its mid-point. So the mid-point (or class mark) of each +class can be chosen to represent the observations falling in the class. Recall that we +find the mid-point of a class (or its class mark) by finding the average of its upper and +lower limits. That is, +Class mark = Upper class limit + Lower class limit +2 +With reference to Table 13.2, for the class 10-25, the class mark is 10 +25 +2 ++ +, i.e., +17.5. Similarly, we can find the class marks of the remaining class intervals. We put +them in Table 13.3. These class marks serve as our xi’s. Now, in general, for the ith +class interval, we have the frequency fi corresponding to the class mark xi. We can +now proceed to compute the mean in the same manner as in Example 1. +Reprint 2025-26 + +174 +MATHEMATICS +Table 13.3 +Class interval +Number of students ( fi) +Class mark (xi) +fixi +10 - 25 +2 +17.5 +35.0 +25 - 40 +3 +32.5 +97.5 +40 - 55 +7 +47.5 +332.5 +55 - 70 +6 +62.5 +375.0 +70 - 85 +6 +77.5 +465.0 +85 - 100 +6 +92.5 +555.0 +Total +Σ fi = 30 +Σ fixi = 1860.0 +The sum of the values in the last column gives us Σ fixi. So, the mean x of the +given data is given by +x = +1860.0 +62 +30 +i +i +i +f x +f +Σ += += +Σ +This new method of finding the mean is known as the Direct Method. +We observe that Tables 13.1 and 13.3 are using the same data and employing the +same formula for the calculation of the mean but the results obtained are different. +Can you think why this is so, and which one is more accurate? The difference in the +two values is because of the mid-point assumption in Table 13.3, 59.3 being the exact +mean, while 62 an approximate mean. +Sometimes when the numerical values of xi and fi are large, finding the product +of xi and fi becomes tedious and time consuming. So, for such situations, let us think of +a method of reducing these calculations. +We can do nothing with the fi’s, but we can change each xi to a smaller number +so that our calculations become easy. How do we do this? What about subtracting a +fixed number from each of these xi’s? Let us try this method. +The first step is to choose one among the xi’s as the assumed mean, and denote +it by ‘a’. Also, to further reduce our calculation work, we may take ‘a’ to be that xi +which lies in the centre of x1, x2, . . ., xn. So, we can choose a = 47.5 or a = 62.5. Let +us choose a = 47.5. +The next step is to find the difference di between a and each of the xi’s, that is, +the deviation of ‘a’ from each of the xi’s. +i.e., +di = xi – a = xi – 47.5 +Reprint 2025-26 + +STATISTICS +175 +The third step is to find the product of di with the corresponding fi, and take the sum +of all the fi di’s. The calculations are shown in Table 13.4. +Table 13.4 +Class interval +Number of +Class mark +di = xi – 47.5 +fidi + students ( fi) + (xi) +10 - 25 +2 +17.5 +–30 +–60 +25 - 40 +3 +32.5 +–15 +–45 +40 - 55 +7 +47.5 +0 +0 +55 - 70 +6 +62.5 +15 +90 +70 - 85 +6 +77.5 +30 +180 +85 - 100 +6 +92.5 +45 +270 +Total +Σfi = 30 +Σfidi = 435 +So, from Table 13.4, the mean of the deviations, d = +i +i +i +f d +f +Σ +Σ +. +Now, let us find the relation between d and x . +Since in obtaining di, we subtracted ‘a’ from each xi, so, in order to get the mean +x , we need to add ‘a’ to d . This can be explained mathematically as: +Mean of deviations, +d = +i +i +i +f d +f +Σ +Σ + So, +d = +( +) +i +i +i +f +x +a +f +Σ +− +Σ + = +i +i +i +i +i +f x +f a +f +f +Σ +Σ +− +Σ +Σ + = +i +i +f +x +a +f +Σ +− +Σ + = x +a +− +So, +x = a + d +i.e., +x = +i +i +i +f d +a +f +Σ ++ +Σ +Reprint 2025-26 + +176 +MATHEMATICS +Substituting the values of a, Σfidi and Σfi from Table 13.4, we get +x = +435 +47.5 +47.5 +14.5 +62 +30 ++ += ++ += +. +Therefore, the mean of the marks obtained by the students is 62. +The method discussed above is called the Assumed Mean Method. +Activity 1 : From the Table 13.3 find the mean by taking each of xi (i.e., 17.5, 32.5, +and so on) as ‘a’. What do you observe? You will find that the mean determined in +each case is the same, i.e., 62. (Why ?) +So, we can say that the value of the mean obtained does not depend on the +choice of ‘a’. +Observe that in Table 13.4, the values in Column 4 are all multiples of 15. So, if +we divide the values in the entire Column 4 by 15, we would get smaller numbers to +multiply with fi. (Here, 15 is the class size of each class interval.) +So, let ui = +ix +a +h +− +, where a is the assumed mean and h is the class size. +Now, we calculate ui in this way and continue as before (i.e., find fi ui and +then Σfiui). Taking h = 15, let us form Table 13.5. +Table 13.5 +Class interval +fi +x i +di = xi – a +ui = +ix – a +h +fiui +10 - 25 +2 +17.5 +–30 +–2 +–4 +25 - 40 +3 +32.5 +–15 +–1 +–3 +40 - 55 +7 +47.5 +0 +0 +0 +55 - 70 +6 +62.5 +15 +1 +6 +70 - 85 +6 +77.5 +30 +2 +12 +85 - 100 +6 +92.5 +45 +3 +18 +Total +Σfi = 30 +Σfiui = 29 +Let +u = +i +i +i +f u +f +Σ +Σ +Here, again let us find the relation between u and x . +Reprint 2025-26 + +STATISTICS +177 +We have, +ui = +ix +a +h + +Therefore, +u = +( +) +1 +i +i +i +i +i +i +i +x +a +f +f x +a f +h +f +h +f + + + + + + + + + + + + + + + = +1 +i +i +i +i +i +f x +f +a +h +f +f + + + + + + + + + + + + = + + +1 x +a +h + +So, +hu = x +a + +i.e., +x = a + hu +So, +x = +i +i +i +f u +a +h +f + + + + + + + + + +Now, substituting the values of a, h, fiui and fi from Table 14.5, we get +x = +29 +47.5 +15 +30 + + + + + + + += 47.5 + 14.5 = 62 +So, the mean marks obtained by a student is 62. +The method discussed above is called the Step-deviation method. +We note that : +the step-deviation method will be convenient to apply if all the di’s have a +common factor. +The mean obtained by all the three methods is the same. +The assumed mean method and step-deviation method are just simplified +forms of the direct method. +The formula x = a + hu still holds if a and h are not as given above, but are +any non-zero numbers such that ui = +ix +a +h + +. +Let us apply these methods in another example. +Reprint 2025-26 + +178 +MATHEMATICS +Example 2 : The table below gives the percentage distribution of female teachers in +the primary schools of rural areas of various states and union territories (U.T.) of +India. Find the mean percentage of female teachers by all the three methods discussed +in this section. +Percentage of +15 - 25 +25 - 35 +35 - 45 +45 - 55 +55 - 65 +65 - 75 +75 - 85 +female teachers +Number of +6 +11 +7 +4 +4 +2 +1 +States/U.T. +Source : Seventh All India School Education Survey conducted by NCERT +Solution : Let us find the class marks, xi, of each class, and put them in a column +(see Table 13.6): +Table 13.6 +Percentage of female +Number of +x i +teachers +States /U.T. ( fi) +15 - 25 +6 +20 +25 - 35 +11 +30 +35 - 45 +7 +40 +45 - 55 +4 +50 +55 - 65 +4 +60 +65 - 75 +2 +70 +75 - 85 +1 +80 +Here we take a = 50, h = 10, then di = xi – 50 and +50 +10 +i +i +x +u + + +. +We now find di and ui and put them in Table 13.7. +Reprint 2025-26 + +STATISTICS +179 +Table 13.7 +Percentage of +Number of +xi +di = xi – 50 +−50 += +10 +i +i +x +u +fixi +fidi +fiui +female +states/U.T. +teachers + ( fi) +15 - 25 +6 +20 +–30 +–3 +120 +–180 +–18 +25 - 35 +11 +30 +–20 +–2 +330 +–220 +–22 +35 - 45 +7 +40 +–10 +–1 +280 +–70 +–7 +45 - 55 +4 +50 +0 +0 +200 +0 +0 +55 - 65 +4 +60 +10 +1 +240 +40 +4 +65 - 75 +2 +70 +20 +2 +140 +40 +4 +75 - 85 +1 +80 +30 +3 +80 +30 +3 +Total +35 +1390 –360 +–36 +From the table above, we obtain Σfi = 35, +Σfixi = 1390, +Σfidi = – 360, +Σfiui = –36. +Using the direct method, +1390 +39.71 +35 +Σ += += += +Σ +i +i +i +f x +x +f +Using the assumed mean method, +x = +i +i +i +f d +a +f +Σ ++ +Σ + = +( 360) +50 +39.71 +35 +− ++ += +Using the step-deviation method, +x = +– 36 +50 +10 +39.71 +35 +i +i +i +f u +a +h +f + + +Σ + + ++ +× += ++ +× += + + + + +Σ + + + + +Therefore, the mean percentage of female teachers in the primary schools of +rural areas is 39.71. +Remark : The result obtained by all the three methods is the same. So the choice of +method to be used depends on the numerical values of xi and fi. If xi and fi are +sufficiently small, then the direct method is an appropriate choice. If xi and fi are +numerically large numbers, then we can go for the assumed mean method or +step-deviation method. If the class sizes are unequal, and xi are large numerically, we +can still apply the step-deviation method by taking h to be a suitable divisor of all the di’s. +Reprint 2025-26 + +180 +MATHEMATICS +Example 3 : The distribution below shows the number of wickets taken by bowlers in +one-day cricket matches. Find the mean number of wickets by choosing a suitable +method. What does the mean signify? +Number of +20 - 60 +60 - 100 +100 - 150 150 - 250 250 - 350 350 - 450 +wickets +Number of +7 +5 +16 +12 +2 +3 +bowlers +Solution : Here, the class size varies, and the xi +,s are large. Let us still apply the step- +deviation method with a = 200 and h = 20. Then, we obtain the data as in Table 13.8. +Table 13.8 +Number of +Number of +xi +di = xi – 200 += 20 +i +i +d +u +ui fi +wickets +bowlers +taken + ( fi) +20 - 60 +7 +40 +–160 +–8 +–56 +60 - 100 +5 +80 +–120 +–6 +–30 +100 - 150 +16 +125 +–75 +–3.75 +–60 +150 - 250 +12 +200 +0 +0 +0 +250 - 350 +2 +300 +100 +5 +10 +350 - 450 +3 +400 +200 +10 +30 +Total +45 +–106 +So, +106 +45 +− += +⋅ +u + Therefore, x = 200 + +106 +20 +45 +− + + + + + + + = 200 – 47.11 = 152.89. +This tells us that, on an average, the number of wickets taken by these 45 bowlers +in one-day cricket is 152.89. +Now, let us see how well you can apply the concepts discussed in this section! +Reprint 2025-26 + +STATISTICS +181 +Activity 2 : +Divide the students of your class into three groups and ask each group to do one of the +following activities. +1. Collect the marks obtained by all the students of your class in Mathematics in the +latest examination conducted by your school. Form a grouped frequency distribution +of the data obtained. +2. Collect the daily maximum temperatures recorded for a period of 30 days in your +city. Present this data as a grouped frequency table. +3. Measure the heights of all the students of your class (in cm) and form a grouped +frequency distribution table of this data. +After all the groups have collected the data and formed grouped frequency +distribution tables, the groups should find the mean in each case by the method which +they find appropriate. +EXERCISE 13.1 +1. A survey was conducted by a group of students as a part of their environment awareness +programme, in which they collected the following data regarding the number of plants in +20 houses in a locality. Find the mean number of plants per house. +Number of plants +0 - 2 +2 - 4 +4 - 6 +6 - 8 +8 - 10 +10 - 12 +12 - 14 +Number of houses +1 +2 +1 +5 +6 +2 +3 +Which method did you use for finding the mean, and why? +2. Consider the following distribution of daily wages of 50 workers of a factory. +Daily wages (in `) +500 - 520 +520 -540 +540 - 560 +560 - 580 +580 -600 +Number of workers +12 +14 +8 +6 +10 +Find the mean daily wages of the workers of the factory by using an appropriate method. +3. The following distribution shows the daily pocket allowance of children of a locality. +The mean pocket allowance is Rs 18. Find the missing frequency f. +Daily pocket +11 - 13 +13 - 15 +15 - 17 +17 - 19 +19 - 21 +21 - 23 +23 - 25 +allowance (in `) +Number of children +7 +6 +9 +13 +f +5 +4 +Reprint 2025-26 + +182 +MATHEMATICS +4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per +minute were recorded and summarised as follows. Find the mean heartbeats per minute +for these women, choosing a suitable method. +Number of heartbeats 65 - 68 +68 - 71 +71 - 74 +74 - 77 +77 - 80 +80 - 83 +83 - 86 +per minute +Number of women +2 +4 +3 +8 +7 +4 +2 +5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These +boxes contained varying number of mangoes. The following was the distribution of +mangoes according to the number of boxes. +Number of mangoes +50 - 52 +53 - 55 +56 - 58 +59 - 61 +62 - 64 +Number of boxes +15 +110 +135 +115 +25 +Find the mean number of mangoes kept in a packing box. Which method of finding +the mean did you choose? +6. The table below shows the daily expenditure on food of 25 households in a locality. +Daily expenditure +100 - 150 +150 - 200 +200 - 250 +250 - 300 +300 - 350 +(in `) +Number of +4 +5 +12 +2 +2 +households +Find the mean daily expenditure on food by a suitable method. +7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data +was collected for 30 localities in a certain city and is presented below: +Concentration of SO2 (in ppm) +Frequency +0.00 - 0.04 +4 +0.04 - 0.08 +9 +0.08 - 0.12 +9 +0.12 - 0.16 +2 +0.16 - 0.20 +4 +0.20 - 0.24 +2 +Find the mean concentration of SO2 in the air. +Reprint 2025-26 + +STATISTICS +183 +8. A class teacher has the following absentee record of 40 students of a class for the whole +term. Find the mean number of days a student was absent. +Number of +0 - 6 +6 - 10 10 - 14 14 - 20 20 - 28 28 - 38 38 - 40 +days +Number of +11 +10 +7 +4 +4 +3 +1 +students +9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean +literacy rate. +Literacy rate (in %) +45 - 55 +55 - 65 +65 - 75 +75 - 85 +85 - 95 +Number of cities +3 +10 +11 +8 +3 +13.3 Mode of Grouped Data +Recall from Class IX, a mode is that value among the observations which occurs most +often, that is, the value of the observation having the maximum frequency. Further, we +discussed finding the mode of ungrouped data. Here, we shall discuss ways of obtaining +a mode of grouped data. It is possible that more than one value may have the same +maximum frequency. In such situations, the data is said to be multimodal. Though +grouped data can also be multimodal, we shall restrict ourselves to problems having a +single mode only. +Let us first recall how we found the mode for ungrouped data through the following +example. +Example 4 : The wickets taken by a bowler in 10 cricket matches are as follows: +2 +6 +4 +5 +0 +2 +1 +3 +2 +3 +Find the mode of the data. +Solution : Let us form the frequency distribution table of the given data as follows: +Number of +0 +1 +2 +3 +4 +5 +6 +wickets +Number of +1 +1 +3 +2 +1 +1 +1 +matches +Reprint 2025-26 + +184 +MATHEMATICS +Clearly, 2 is the number of wickets taken by the bowler in the maximum number +(i.e., 3) of matches. So, the mode of this data is 2. +In a grouped frequency distribution, it is not possible to determine the mode by +looking at the frequencies. Here, we can only locate a class with the maximum +frequency, called the modal class. The mode is a value inside the modal class, and is +given by the formula: +Mode = +1 +0 +1 +0 +2 +2 +f +f +l +h +f +f +f + + + + + + + + + + + +where l = lower limit of the modal class, +h = size of the class interval (assuming all class sizes to be equal), +f1 = frequency of the modal class, +f0 = frequency of the class preceding the modal class, +f2 = frequency of the class succeeding the modal class. +Let us consider the following examples to illustrate the use of this formula. +Example 5 : A survey conducted on 20 households in a locality by a group of students +resulted in the following frequency table for the number of family members in a +household: +Family size +1 - 3 +3 - 5 +5 - 7 +7 - 9 +9 - 11 +Number of +7 +8 +2 +2 +1 +families +Find the mode of this data. +Solution : Here the maximum class frequency is 8, and the class corresponding to this +frequency is 3 – 5. So, the modal class is 3 – 5. +Now +modal class = 3 – 5, lower limit (l) of modal class = 3, class size (h) = 2 +frequency ( f1) of the modal class = 8, +frequency ( f0) of class preceding the modal class = 7, +frequency ( f2) of class succeeding the modal class = 2. +Now, let us substitute these values in the formula : +Reprint 2025-26 + +STATISTICS +185 +Mode = +1 +0 +1 +0 +2 +2 +f +f +l +h +f +f +f + + + + + + + + + + + += +8 +7 +2 +3 +2 +3 +3.286 +2 +8 +7 +2 +7 + + + + + + + + + + + + + + + +Therefore, the mode of the data above is 3.286. +Example 6 : The marks distribution of 30 students in a mathematics examination are +given in Table 13.3 of Example 1. Find the mode of this data. Also compare and +interpret the mode and the mean. +Solution : Refer to Table 13.3 of Example 1. Since the maximum number of students +(i.e., 7) have got marks in the interval 40 - 55, the modal class is 40 - 55. Therefore, +the lower limit (l) of the modal class = 40, +the class size (h) = 15, +the frequency ( f1) of modal class = 7, +the frequency ( f0) of the class preceding the modal class = 3, +the frequency ( f2) of the class succeeding the modal class = 6. +Now, using the formula: +Mode = +1 +0 +1 +0 +2 +2 +f +f +l +h +f +f +f + + + + + + + + + + + +, +we get +Mode = +7 +3 +40 +15 +14 +6 +3 + + + + + + + + + + + + = 52 +So, the mode marks is 52. +Now, from Example 1, you know that the mean marks is 62. +So, the maximum number of students obtained 52 marks, while on an average a +student obtained 62 marks. +Remarks : +1. In Example 6, the mode is less than the mean. But for some other problems it may +be equal or more than the mean also. +2. It depends upon the demand of the situation whether we are interested in finding the +average marks obtained by the students or the average of the marks obtained by most +Reprint 2025-26 + +186 +MATHEMATICS +of the students. In the first situation, the mean is required and in the second situation, +the mode is required. +Activity 3 : Continuing with the same groups as formed in Activity 2 and the situations +assigned to the groups. Ask each group to find the mode of the data. They should also +compare this with the mean, and interpret the meaning of both. +Remark : The mode can also be calculated for grouped data with unequal class sizes. +However, we shall not be discussing it. +EXERCISE 13.2 +1. The following table shows the ages of the patients admitted in a hospital during a year: +Age (in years) +5 - 15 +15 - 25 +25 - 35 +35 - 45 +45 - 55 +55 - 65 +Number of patients +6 +11 +21 +23 +14 +5 +Find the mode and the mean of the data given above. Compare and interpret the two +measures of central tendency. +2. The following data gives the information on the observed lifetimes (in hours) of 225 +electrical components : +Lifetimes (in hours) +0 - 20 +20 - 40 +40 - 60 +60 - 80 +80 - 100 +100 - 120 +Frequency +10 +35 +52 +61 +38 +29 +Determine the modal lifetimes of the components. +3. The following data gives the distribution of total monthly household expenditure of 200 +families of a village. Find the modal monthly expenditure of the families. Also, find the +mean monthly expenditure : +Expenditure (in `) +Number of families +1000 - 1500 +24 +1500 - 2000 +40 +2000 - 2500 +33 +2500 - 3000 +28 +3000 - 3500 +30 +3500 - 4000 +22 +4000 - 4500 +16 +4500 - 5000 +7 +Reprint 2025-26 + +STATISTICS +187 +4. The following distribution gives the state-wise teacher-student ratio in higher +secondary schools of India. Find the mode and mean of this data. Interpret the two +measures. +Number of students per teacher +Number of states / U.T. +15 - 20 +3 +20 - 25 +8 +25 - 30 +9 +30 - 35 +10 +35 - 40 +3 +40 - 45 +0 +45 - 50 +0 +50 - 55 +2 +5. The given distribution shows the number of runs scored by some top batsmen of the +world in one-day international cricket matches. +Runs scored +Number of batsmen +3000 - 4000 +4 +4000 - 5000 +18 +5000 - 6000 +9 +6000 - 7000 +7 +7000 - 8000 +6 +8000 - 9000 +3 +9000 - 10000 +1 +10000 - 11000 +1 +Find the mode of the data. +6. A student noted the number of cars passing through a spot on a road for 100 +periods each of 3 minutes and summarised it in the table given below. Find the mode +of the data : +Number of cars +0 - 10 +10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 +Frequency +7 +14 +13 +12 +20 +11 +15 +8 +Reprint 2025-26 + +188 +MATHEMATICS +13.4 Median of Grouped Data +As you have studied in Class IX, the median is a measure of central tendency which +gives the value of the middle-most observation in the data. Recall that for finding the +median of ungrouped data, we first arrange the data values of the observations in +ascending order. Then, if n is odd, the median is the +1 +2 +n  + + + + + +th observation. And, if n +is even, then the median will be the average of the +th +2 +n +and the +1 th +2 +n + + + + + + + + observations. +Suppose, we have to find the median of the following data, which gives the +marks, out of 50, obtained by 100 students in a test : +Marks obtained +20 +29 +28 +33 +42 +38 +43 +25 +Number of students +6 +28 +24 +15 +2 +4 +1 +20 +First, we arrange the marks in ascending order and prepare a frequency table as +follows : +Table 13.9 +Marks obtained +Number of students +(Frequency) +20 +6 +25 +20 +28 +24 +29 +28 +33 +15 +38 +4 +42 +2 +43 +1 +Total +100 +Reprint 2025-26 + +STATISTICS +189 +Here n = 100, which is even. The median will be the average of the +2 +n th and the +1 +2 +n + + + + + + + +th observations, i.e., the 50th and 51st observations. To find these +observations, we proceed as follows: +Table 13.10 +Marks obtained +Number of students +20 +6 +upto 25 +6 + 20 = 26 +upto 28 +26 + 24 = 50 +upto 29 +50 + 28 = 78 +upto 33 +78 + 15 = 93 +upto 38 +93 + 4 = 97 +upto 42 +97 + 2 = 99 +upto 43 +99 + 1 = 100 +Now we add another column depicting this information to the frequency table +above and name it as cumulative frequency column. +Table 13.11 +Marks obtained +Number of students +Cumulative frequency +20 +6 +6 +25 +20 +26 +28 +24 +50 +29 +28 +78 +33 +15 +93 +38 +4 +97 +42 +2 +99 +43 +1 +100 +Reprint 2025-26 + +190 +MATHEMATICS +From the table above, we see that: +50th observaton is 28 +(Why?) +51st observation is 29 +So, +Median = 28 +29 +28.5 +2 ++ += +Remark : The part of Table 13.11 consisting Column 1 and Column 3 is known as +Cumulative Frequency Table. The median marks 28.5 conveys the information that +about 50% students obtained marks less than 28.5 and another 50% students obtained +marks more than 28.5. +Now, let us see how to obtain the median of grouped data, through the following +situation. +Consider a grouped frequency distribution of marks obtained, out of 100, by 53 +students, in a certain examination, as follows: +Table 13.12 +Marks +Number of students +0 - 10 +5 +10 - 20 +3 +20 - 30 +4 +30 - 40 +3 +40 - 50 +3 +50 - 60 +4 +60 - 70 +7 +70 - 80 +9 +80 - 90 +7 +90 - 100 +8 +From the table above, try to answer the following questions: +How many students have scored marks less than 10? The answer is clearly 5. +Reprint 2025-26 + +STATISTICS +191 +How many students have scored less than 20 marks? Observe that the number +of students who have scored less than 20 include the number of students who have +scored marks from 0 - 10 as well as the number of students who have scored marks +from 10 - 20. So, the total number of students with marks less than 20 is 5 + 3, i.e., 8. +We say that the cumulative frequency of the class 10 -20 is 8. +Similarly, we can compute the cumulative frequencies of the other classes, i.e., +the number of students with marks less than 30, less than 40, . . ., less than 100. We +give them in Table 13.13 given below: +Table 13.13 +Marks obtained +Number of students +(Cumulative frequency) +Less than 10 +5 +Less than 20 +5 + 3 = 8 +Less than 30 +8 + 4 = 12 +Less than 40 +12 + 3 = 15 +Less than 50 +15 + 3 = 18 +Less than 60 +18 + 4 = 22 +Less than 70 +22 + 7 = 29 +Less than 80 +29 + 9 = 38 +Less than 90 +38 + 7 = 45 +Less than 100 +45 + 8 = 53 +The distribution given above is called the cumulative frequency distribution of +the less than type. Here 10, 20, 30, . . . 100, are the upper limits of the respective +class intervals. +We can similarly make the table for the number of students with scores, more +than or equal to 0, more than or equal to 10, more than or equal to 20, and so on. From +Table 13.12, we observe that all 53 students have scored marks more than or equal to +0. Since there are 5 students scoring marks in the interval 0 - 10, this means that there +are 53 – 5 = 48 students getting more than or equal to 10 marks. Continuing in the +same manner, we get the number of students scoring 20 or above as 48 – 3 = 45, 30 or +above as 45 – 4 = 41, and so on, as shown in Table 13.14. +Reprint 2025-26 + +192 +MATHEMATICS +Table 13.14 +Marks obtained +Number of students +(Cumulative frequency) +More than or equal to 0 +53 +More than or equal to 10 +53 – 5 = 48 +More than or equal to 20 +48 – 3 = 45 +More than or equal to 30 +45 – 4 = 41 +More than or equal to 40 +41 – 3 = 38 +More than or equal to 50 +38 – 3 = 35 +More than or equal to 60 +35 – 4 = 31 +More than or equal to 70 +31 – 7 = 24 +More than or equal to 80 +24 – 9 = 15 +More than or equal to 90 +15 – 7 = 8 +The table above is called a cumulative frequency distribution of the more +than type. Here 0, 10, 20, . . ., 90 give the lower limits of the respective class intervals. +Now, to find the median of grouped data, we can make use of any of these +cumulative frequency distributions. +Let us combine Tables 13.12 and 13.13 to get Table 13.15 given below: +Table 13.15 +Marks +Number of students ( f ) +Cumulative frequency (cf) +0 - 10 +5 +5 +10 - 20 +3 +8 +20 - 30 +4 +12 +30 - 40 +3 +15 +40 - 50 +3 +18 +50 - 60 +4 +22 +60 - 70 +7 +29 +70 - 80 +9 +38 +80 - 90 +7 +45 +90 - 100 +8 +53 +Now in a grouped data, we may not be able to find the middle observation by +looking at the cumulative frequencies as the middle observation will be some value in +Reprint 2025-26 + +STATISTICS +193 +a class interval. It is, therefore, necessary to find the value inside a class that divides +the whole distribution into two halves. But which class should this be? +To find this class, we find the cumulative frequencies of all the classes and 2 +n . +We now locate the class whose cumulative frequency is greater than (and nearest to) +2 +n  This is called the median class. In the distribution above, n = 53. So, 2 +n = 26.5. +Now 60 – 70 is the class whose cumulative frequency 29 is greater than (and nearest +to) 2 +n , i.e., 26.5. +Therefore, 60 – 70 is the median class. +After finding the median class, we use the following formula for calculating the +median. +Median = +cf +2 ++ +, +n +l +h +f + + + + + + + + + + + + + + +where +l = lower limit of median class, +n = number of observations, +cf = cumulative frequency of class preceding the median class, +f = frequency of median class, +h = class size (assuming class size to be equal). +Substituting the values +26.5, +2 +n  + l = 60, cf = 22, f = 7, h = 10 +in the formula above, we get +Median = +26.5 +22 +60 +10 +7 + + + + + + + + + += 60 + 45 +7 += 66.4 +So, about half the students have scored marks less than 66.4, and the other half have +scored marks more than 66.4. +Reprint 2025-26 + +194 +MATHEMATICS +Example 7 : A survey regarding the heights (in cm) of 51 girls of Class X of a school +was conducted and the following data was obtained: +Height (in cm) +Number of girls +Less than 140 +4 +Less than 145 +11 +Less than 150 +29 +Less than 155 +40 +Less than 160 +46 +Less than 165 +51 +Find the median height. +Solution : To calculate the median height, we need to find the class intervals and their +corresponding frequencies. +The given distribution being of the less than type, 140, 145, 150, . . ., 165 give the +upper limits of the corresponding class intervals. So, the classes should be below 140, +140 - 145, 145 - 150, . . ., 160 - 165. Observe that from the given distribution, we find +that there are 4 girls with height less than 140, i.e., the frequency of class interval +below 140 is 4. Now, there are 11 girls with heights less than 145 and 4 girls with +height less than 140. Therefore, the number of girls with height in the interval +140 - 145 is 11 – 4 = 7. Similarly, the frequency of 145 - 150 is 29 – 11 = 18, for +150 - 155, it is 40 – 29 = 11, and so on. So, our frequency distribution table with the +given cumulative frequencies becomes: +Table 13.16 +Class intervals +Frequency +Cumulative frequency +Below 140 +4 +4 +140 - 145 +7 +11 +145 - 150 +18 +29 +150 - 155 +11 +40 +155 - 160 +6 +46 +160 - 165 +5 +51 +Reprint 2025-26 + +STATISTICS +195 +Now +n = 51. So, +51 +25.5 +2 +2 +n = += +. This observation lies in the class 145 - 150. Then, +l (the lower limit) = 145, +cf (the cumulative frequency of the class preceding 145 - 150) = 11, +f (the frequency of the median class 145 - 150) = 18, +h (the class size) = 5. +Using the formula, Median = l + +cf +2 +n +h +f + + +− + +× + + + + + + +, we have +Median = +25.5 +11 +145 +5 +18 +− + + ++ +× + + + + += 145 + 72.5 +18 = 149.03. +So, the median height of the girls is 149.03 cm. +This means that the height of about 50% of the girls is less than this height, and +50% are taller than this height. +Example 8 : The median of the following data is 525. Find the values of x and y, if the +total frequency is 100. +Class intervals +Frequency +0 - 100 +2 +100 - 200 +5 +200 - 300 +x +300 - 400 +12 +400 - 500 +17 +500 - 600 +20 +600 - 700 +y +700 - 800 +9 +800 - 900 +7 +900 - 1000 +4 +Reprint 2025-26 + +196 +MATHEMATICS +Solution : +Class intervals +Frequency +Cumulative frequency +0 - 100 +2 +2 +100 - 200 +5 +7 +200 - 300 +x +7 + x +300 - 400 +12 +19 + x +400 - 500 +17 +36 + x +500 - 600 +20 +56 + x +600 - 700 +y +56 + x + y +700 - 800 +9 +65 + x + y +800 - 900 +7 +72 + x + y +900 - 1000 +4 +76 + x + y +It is given that n = 100 +So, +76 + x + y = 100, +i.e., +x + y = 24 +(1) +The median is 525, which lies in the class 500 – 600 +So, +l = 500, +f = 20, +cf = 36 + x, +h = 100 +Using the formula : +Median = +cf +2 +, +n +l +h +f + + + + + + + + + + + + we get + 525 = +50 +36 +500 +100 +20 +x + + + + + + + + + + +i.e., +525 – 500 = (14 – x) × 5 +i.e., +25 = 70 – 5x +i.e., +5x = 70 – 25 = 45 +So, +x = 9 +Therefore, from (1), we get +9 + y = 24 +i.e., +y = 15 +Reprint 2025-26 + +STATISTICS +197 +Now, that you have studied about all the three measures of central tendency, let +us discuss which measure would be best suited for a particular requirement. +The mean is the most frequently used measure of central tendency because it +takes into account all the observations, and lies between the extremes, i.e., the largest +and the smallest observations of the entire data. It also enables us to compare two or +more distributions. For example, by comparing the average (mean) results of students +of different schools of a particular examination, we can conclude which school has a +better performance. +However, extreme values in the data affect the mean. For example, the mean of +classes having frequencies more or less the same is a good representative of the data. +But, if one class has frequency, say 2, and the five others have frequency 20, 25, 20, +21, 18, then the mean will certainly not reflect the way the data behaves. So, in such +cases, the mean is not a good representative of the data. +In problems where individual observations are not important, and we wish to find +out a ‘typical’ observation, the median is more appropriate, e.g., finding the typical +productivity rate of workers, average wage in a country, etc. These are situations +where extreme values may be there. So, rather than the mean, we take the median as +a better measure of central tendency. +In situations which require establishing the most frequent value or most popular +item, the mode is the best choice, e.g., to find the most popular T.V. programme being +watched, the consumer item in greatest demand, the colour of the vehicle used by +most of the people, etc. +Remarks : +1. There is a empirical relationship between the three measures of central tendency : +3 Median = Mode + 2 Mean +2. The median of grouped data with unequal class sizes can also be calculated. However, +we shall not discuss it here. +Reprint 2025-26 + +198 +MATHEMATICS +EXERCISE 13.3 +1. The following frequency distribution gives the monthly consumption of electricity of +68 consumers of a locality. Find the median, mean and mode of the data and compare +them. +Monthly consumption (in units) +Number of consumers +65 - 85 +4 +85 - 105 +5 +105 - 125 +13 +125 - 145 +20 +145 - 165 +14 +165 - 185 +8 +185 - 205 +4 +2. If the median of the distribution given below is 28.5, find the values of x and y. +Class interval +Frequency +0 - 10 +5 +10 - 20 +x +20 - 30 +20 +30 - 40 +15 +40 - 50 +y +50 - 60 +5 +Total +60 +3. A life insurance agent found the following data for distribution of ages of 100 policy +holders. Calculate the median age, if policies are given only to persons having age 18 +years onwards but less than 60 year. +Reprint 2025-26 + +STATISTICS +199 +Age (in years) +Number of policy holders +Below 20 +2 +Below 25 +6 +Below 30 +24 +Below 35 +45 +Below 40 +78 +Below 45 +89 +Below 50 +92 +Below 55 +98 +Below 60 +100 +4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and +the data obtained is represented in the following table : +Length (in mm) +Number of leaves +118 - 126 +3 +127 - 135 +5 +136 - 144 +9 +145 - 153 +12 +154 - 162 +5 +163 - 171 +4 +172 - 180 +2 +Find the median length of the leaves. +(Hint : The data needs to be converted to continuous classes for finding the median, +since the formula assumes continuous classes. The classes then change to +117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.) +Reprint 2025-26 + +200 +MATHEMATICS +5. The following table gives the distribution of the life time of 400 neon lamps : +Life time (in hours) +Number of lamps +1500 - 2000 +14 +2000 - 2500 +56 +2500 - 3000 +60 +3000 - 3500 +86 +3500 - 4000 +74 +4000 - 4500 +62 +4500 - 5000 +48 +Find the median life time of a lamp. +6. 100 surnames were randomly picked up from a local telephone directory and the +frequency distribution of the number of letters in the English alphabets in the surnames +was obtained as follows: +Number of letters +1 - 4 +4 - 7 +7 - 10 +10 - 13 +13 - 16 +16 - 19 +Number of surnames +6 +30 +40 +16 +4 +4 +Determine the median number of letters in the surnames. Find the mean number of +letters in the surnames? Also, find the modal size of the surnames. +7. The distribution below gives the weights of 30 students of a class. Find the median +weight of the students. +Weight (in kg) +40 - 45 +45 - 50 +50 - 55 +55 - 60 +60 - 65 +65 - 70 +70 - 75 +Number of students +2 +3 +8 +6 +6 +3 +2 +13.5 Summary +In this chapter, you have studied the following points: +1. The mean for grouped data can be found by : +(i) the direct method : +i +i +i +f x +x +f +Σ += Σ +(ii) the assumed mean method : +i +i +i +f d +x +a +f +Σ += ++ Σ +Reprint 2025-26 + +STATISTICS +201 +(iii) the step deviation method : +i +i +i +f u +x +a +h +f + + + + + + + + + + + +, +with the assumption that the frequency of a class is centred at its mid-point, called its +class mark. +2. The mode for grouped data can be found by using the formula: +Mode = +1 +0 +1 +0 +2 +2 +f +f +l +h +f +f +f + + + + + + + + + + + +where symbols have their usual meanings. +3. The cumulative frequency of a class is the frequency obtained by adding the frequencies +of all the classes preceding the given class. +4. The median for grouped data is formed by using the formula: +Median = +cf +2 +n +l +h +f + + + + + + + + + + + + + + + +, +where symbols have their usual meanings. +A NOTE TO THE READER +For calculating mode and median for grouped data, it should be +ensured that the class intervals are continuous before applying the +formulae. Same condition also apply for construction of an ogive. +Further, in case of ogives, the scale may not be the same on both the axes. +Reprint 2025-26" +class_10,14,Probability,ncert_books/class_10/jemh1dd/jemh114.pdf,"202 +MATHEMATICS +14 +The theory of probabilities and the theory of errors now constitute +a formidable body of great mathematical interest and of great +practical importance. +– R.S. Woodward +14.1 Probability — A Theoretical Approach +Let us consider the following situation : +Suppose a coin is tossed at random. +When we speak of a coin, we assume it to be ‘fair’, that is, it is symmetrical so +that there is no reason for it to come down more often on one side than the other. +We call this property of the coin as being ‘unbiased’. By the phrase ‘random toss’, +we mean that the coin is allowed to fall freely without any bias or interference. +We know, in advance, that the coin can only land in one of two possible ways — +either head up or tail up (we dismiss the possibility of its ‘landing’ on its edge, which +may be possible, for example, if it falls on sand). We can reasonably assume that each +outcome, head or tail, is as likely to occur as the other. We refer to this by saying that +the outcomes head and tail, are equally likely. +For another example of equally likely outcomes, suppose we throw a die +once. For us, a die will always mean a fair die. What are the possible outcomes? +They are 1, 2, 3, 4, 5, 6. Each number has the same possibility of showing up. So +the equally likely outcomes of throwing a die are 1, 2, 3, 4, 5 and 6. +PROBABILITY +Reprint 2025-26 + +PROBABILITY +203 +Are the outcomes of every experiment equally likely? Let us see. +Suppose that a bag contains 4 red balls and 1 blue ball, and you draw a ball +without looking into the bag. What are the outcomes? Are the outcomes — a red ball +and a blue ball equally likely? Since there are 4 red balls and only one blue ball, you +would agree that you are more likely to get a red ball than a blue ball. So, the outcomes +(a red ball or a blue ball) are not equally likely. However, the outcome of drawing a +ball of any colour from the bag is equally likely. So, all experiments do not necessarily +have equally likely outcomes. +However, in this chapter, from now on, we will assume that all the experiments +have equally likely outcomes. +In Class IX, we defined the experimental or empirical probability P(E) of an +event E as +P(E) = Number of trials in which the event happened +Total number of trials +The empirical interpretation of probability can be applied to every event associated +with an experiment which can be repeated a large number of times. The requirement +of repeating an experiment has some limitations, as it may be very expensive or +unfeasible in many situations. Of course, it worked well in coin tossing or die throwing +experiments. But how about repeating the experiment of launching a satellite in order +to compute the empirical probability of its failure during launching, or the repetition of +the phenomenon of an earthquake to compute the empirical probability of a multi- +storeyed building getting destroyed in an earthquake? +In experiments where we are prepared to make certain assumptions, the repetition +of an experiment can be avoided, as the assumptions help in directly calculating the +exact (theoretical) probability. The assumption of equally likely outcomes (which is +valid in many experiments, as in the two examples above, of a coin and of a die) is one +such assumption that leads us to the following definition of probability of an event. +The theoretical probability (also called classical probability) of an event E, +written as P(E), is defined as +P(E) = +Number of outcomes favourable to E +Number of all possible outcomes of the experiment +, +where we assume that the outcomes of the experiment are equally likely. +We will briefly refer to theoretical probability as probability. +This definition of probability was given by Pierre Simon Laplace in 1795. +Reprint 2025-26 + +204 +MATHEMATICS +Probability theory had its origin in the 16th century when +an Italian physician and mathematician J.Cardan wrote the +first book on the subject, The Book on Games of Chance. +Since its inception, the study of probability has attracted +the attention of great mathematicians. James Bernoulli +(1654 – 1705), A. de Moivre (1667 – 1754), and +Pierre Simon Laplace are among those who made significant +contributions to this field. Laplace’s Theorie Analytique +des Probabilités, 1812, is considered to be the greatest +contribution by a single person to the theory of probability. +In recent years, probability has been used extensively in +many areas such as biology, economics, genetics, physics, +sociology etc. +Let us find the probability for some of the events associated with experiments +where the equally likely assumption holds. +Example 1 : Find the probability of getting a head when a coin is tossed once. Also +find the probability of getting a tail. +Solution : In the experiment of tossing a coin once, the number of possible outcomes +is two — Head (H) and Tail (T). Let E be the event ‘getting a head’. The number of +outcomes favourable to E, (i.e., of getting a head) is 1. Therefore, +P(E) = P (head) = Number of outcomes favourable to E +Number of all possible outcomes + = 1 +2 +Similarly, if F is the event ‘getting a tail’, then +P(F) = P(tail) = 1 +2 +(Why ?) +Example 2 : A bag contains a red ball, a blue ball and a yellow ball, all the balls being +of the same size. Kritika takes out a ball from the bag without looking into it. What is +the probability that she takes out the +(i) yellow ball? +(ii) red ball? +(iii) blue ball? +Solution : Kritika takes out a ball from the bag without looking into it. So, it is equally +likely that she takes out any one of them. +Pierre Simon Laplace +(1749 – 1827) +Reprint 2025-26 + +PROBABILITY +205 +Let Y be the event ‘the ball taken out is yellow’, B be the event ‘the ball taken +out is blue’, and R be the event ‘the ball taken out is red’. +Now, the number of possible outcomes = 3. +(i) The number of outcomes favourable to the event Y = 1. +So, +P(Y) = 1 +3 +Similarly, +(ii) P(R) = 1 +3 and (iii) P(B) = 1 +3 +Remarks : +1. An event having only one outcome of the experiment is called an elementary +event. In Example 1, both the events E and F are elementary events. Similarly, in +Example 2, all the three events, Y, B and R are elementary events. +2. In Example 1, we note that : P(E) + P(F) = 1 +In Example 2, we note that : P(Y) + P(R) + P(B) = 1 +Observe that the sum of the probabilities of all the elementary events of +an experiment is 1. This is true in general also. +Example 3 : Suppose we throw a die once. (i) What is the probability of getting a +number greater than 4 ? (ii) What is the probability of getting a number less than or +equal to 4 ? +Solution : (i) Here, let E be the event ‘getting a number greater than 4’. The number +of possible outcomes is six : 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5 +and 6. Therefore, the number of outcomes favourable to E is 2. So, +P(E) = P(number greater than 4) = 2 +6 = 1 +3 +(ii) Let F be the event ‘getting a number less than or equal to 4’. +Number of possible outcomes = 6 +Outcomes favourable to the event F are 1, 2, 3, 4. +So, the number of outcomes favourable to F is 4. +Therefore, +P(F) = 4 +6 = 2 +3 +Reprint 2025-26 + +206 +MATHEMATICS +Are the events E and F in the example above elementary events? No, they are +not because the event E has 2 outcomes and the event F has 4 outcomes. +Remarks : From Example 1, we note that +P(E) + P(F) = 1 +1 +1 +2 +2 + + +(1) +where E is the event ‘getting a head’ and F is the event ‘getting a tail’. +From (i) and (ii) of Example 3, we also get +P(E) + P(F) = 1 +2 +1 +3 +3 + + +(2) +where E is the event ‘getting a number >4’ and F is the event ‘getting a number  4’. +Note that getting a number not greater than 4 is same as getting a number less +than or equal to 4, and vice versa. +In (1) and (2) above, is F not the same as ‘not E’? Yes, it is. We denote the event +‘not E’ by E . +So, +P(E) + P(not E) = 1 +i.e., +P(E) + P( E ) = 1, +which gives us +P( E ) = 1 – P(E). +In general, it is true that for an event E, +P( E ) = 1 – P(E) +The event E , representing ‘not E’, is called the complement of the event E. +We also say that E and E are complementary events. +Before proceeding further, let us try to find the answers to the following questions: +(i) What is the probability of getting a number 8 in a single throw of a die? +(ii) What is the probability of getting a number less than 7 in a single throw of a die? +Let us answer (i) : +We know that there are only six possible outcomes in a single throw of a die. These +outcomes are 1, 2, 3, 4, 5 and 6. Since no face of the die is marked 8, so there is no +outcome favourable to 8, i.e., the number of such outcomes is zero. In other words, +getting 8 in a single throw of a die, is impossible. +So, +P(getting 8) = 0 +6 = 0 +Reprint 2025-26 + +PROBABILITY +207 +That is, the probability of an event which is impossible to occur is 0. Such an +event is called an impossible event. +Let us answer (ii) : +Since every face of a die is marked with a number less than 7, it is sure that we +will always get a number less than 7 when it is thrown once. So, the number of +favourable outcomes is the same as the number of all possible outcomes, which is 6. +Therefore, +P(E) = P(getting a number less than 7) = 6 +6 = 1 +So, the probability of an event which is sure (or certain) to occur is 1. Such an event +is called a sure event or a certain event. +Note : From the definition of the probability P(E), we see that the numerator (number +of outcomes favourable to the event E) is always less than or equal to the denominator +(the number of all possible outcomes). Therefore, +0  P(E)  1 +Now, let us take an example related to playing cards. Have you seen a deck of +playing cards? It consists of 52 cards which are divided into 4 suits of 13 cards each— +spades (), hearts (), diamonds () and clubs (). Clubs and spades are of black +colour, while hearts and diamonds are of red colour. The cards in each suit are ace, +king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings, queens and jacks are called face +cards. +Example 4 : One card is drawn from a well-shuffled deck of 52 cards. Calculate the +probability that the card will +(i) be an ace, +(ii) not be an ace. +Solution : Well-shuffling ensures equally likely outcomes. +(i) There are 4 aces in a deck. Let E be the event ‘the card is an ace’. +The number of outcomes favourable to E = 4 +The number of possible outcomes = 52 +(Why ?) +Therefore, +P(E) = 4 +1 +52 +13 + +(ii) Let F be the event ‘card drawn is not an ace’. +The number of outcomes favourable to the event F = 52 – 4 = 48 (Why?) +Reprint 2025-26 + +208 +MATHEMATICS +The number of possible outcomes = 52 +Therefore, +P(F) = 48 +12 +52 +13 + +Remark : Note that F is nothing but E . Therefore, we can also calculate P(F) as +follows: P(F) = P( E ) = 1 – P(E) = +1 +12 +1 +13 +13 + + + +Example 5 : Two players, Sangeeta and Reshma, play a tennis match. It is known +that the probability of Sangeeta winning the match is 0.62. What is the probability of +Reshma winning the match? +Solution : Let S and R denote the events that Sangeeta wins the match and Reshma +wins the match, respectively. +The probability of Sangeeta’s winning = P(S) = 0.62 (given) +The probability of Reshma’s winning = P(R) = 1 – P(S) +[As the events R and S are complementary] += 1 – 0.62 = 0.38 +Example 6 : Savita and Hamida are friends. What is the probability that both will +have (i) different birthdays? (ii) the same birthday? (ignoring a leap year). +Solution : Out of the two friends, one girl, say, Savita’s birthday can be any day of the +year. Now, Hamida’s birthday can also be any day of 365 days in the year. +We assume that these 365 outcomes are equally likely. +(i) If Hamida’s birthday is different from Savita’s, the number of favourable outcomes +for her birthday is 365 – 1 = 364 +So, P (Hamida’s birthday is different from Savita’s birthday) = 364 +365 +(ii) P(Savita and Hamida have the same birthday) += 1 – P (both have different birthdays) += +364 +1 +365 + +[Using P( E ) = 1 – P(E)] += +1 +365 +Reprint 2025-26 + +PROBABILITY +209 +Example 7 : There are 40 students in Class X of a school of whom 25 are girls and 15 +are boys. The class teacher has to select one student as a class representative. She +writes the name of each student on a separate card, the cards being identical. Then +she puts cards in a bag and stirs them thoroughly. She then draws one card from the +bag. What is the probability that the name written on the card is the name of (i) a girl? +(ii) a boy? +Solution : There are 40 students, and only one name card has to be chosen. +(i) The number of all possible outcomes is 40 +The number of outcomes favourable for a card with the name of a girl = 25 (Why?) +Therefore, P (card with name of a girl) = P(Girl) = 25 +5 +40 +8 + +(ii) The number of outcomes favourable for a card with the name of a boy = 15 (Why?) +Therefore, P(card with name of a boy) = P(Boy) = 15 +3 +40 +8 + +Note : We can also determine P(Boy), by taking +P(Boy) = 1 – P(not Boy) = 1 – P(Girl) = +5 +3 +1 +8 +8 + + +Example 8 : A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn +at random from the box, what is the probability that it will be +(i) white? +(ii) blue? +(iii) red? +Solution : Saying that a marble is drawn at random is a short way of saying that all the +marbles are equally likely to be drawn. Therefore, the +number of possible outcomes = 3 +2 + 4 = 9 +(Why?) +Let W denote the event ‘the marble is white’, B denote the event ‘the marble is blue’ +and R denote the event ‘marble is red’. +(i) The number of outcomes favourable to the event W = 2 +So, +P(W) = 2 +9 +Similarly, +(ii) P(B) = 3 +9 = 1 +3 +and +(iii) P(R) = 4 +9 +Note that P(W) + P(B) + P(R) = 1. +Reprint 2025-26 + +210 +MATHEMATICS +Example 9 : Harpreet tosses two different coins simultaneously (say, one is of ` 1 +and other of ` 2). What is the probability that she gets at least one head? +Solution : We write H for ‘head’ and T for ‘tail’. When two coins are tossed +simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all +equally likely. Here (H, H) means head up on the first coin (say on ` 1) and head up +on the second coin (` 2). Similarly (H, T) means head up on the first coin and tail up on +the second coin and so on. +The outcomes favourable to the event E, ‘at least one head’ are (H, H), (H, T) +and (T, H). (Why?) +So, the number of outcomes favourable to E is 3. +Therefore, +P(E) = 3 +4 +i.e., the probability that Harpreet gets at least one head is 3 +4  +Note : You can also find P(E) as follows: +P (E) = +1 +3 +1 – P(E) = 1 – 4 +4 + +1 +Since P(E) = P(no head) = 4 + + + + + + +Did you observe that in all the examples discussed so far, the number of possible +outcomes in each experiment was finite? If not, check it now. +There are many experiments in which the outcome is any number between two +given numbers, or in which the outcome is every point within a circle or rectangle, etc. +Can you now count the number of all possible outcomes? As you know, this is not +possible since there are infinitely many numbers between two given numbers, or there +are infinitely many points within a circle. So, the definition of (theoretical) probability +which you have learnt so far cannot be applied in the present form. What is the way +out? To answer this, let us consider the following example : +Example 10* : In a musical chair game, the person playing the music has been +advised to stop playing the music at any time within 2 minutes after she starts +playing. What is the probability that the music will stop within the first half-minute +after starting? +Solution : Here the possible outcomes are all the numbers between 0 and 2. This is +the portion of the number line from 0 to 2 (see Fig. 14.1). +Fig. 14.1 +* Not from the examination point of view. +Reprint 2025-26 + +PROBABILITY +211 +Let E be the event that ‘the music is stopped within the first half-minute’. +The outcomes favourable to E are points on the number line from 0 to 1 +2 +. +The distance from 0 to 2 is 2, while the distance from 0 to 1 +2 is 1 +2 . +Since all the outcomes are equally likely, we can argue that, of the total distance +of 2, the distance favourable to the event E is 1 +2 . +So, +P(E) = +Distance favourable to the event E +Total distance in which outcomes can lie = +1 +1 +2 +2 +4 + +Can we now extend the idea of Example 10 for finding the probability as the ratio of +the favourable area to the total area? +Example 11* : A missing helicopter is reported to have crashed somewhere in the +rectangular region shown in Fig. 14.2. What is the probability that it crashed inside the +lake shown in the figure? +Fig. 14.2 +Solution : The helicopter is equally likely to crash anywhere in the region. +Area of the entire region where the helicopter can crash += (4.5 × 9) km2 = 40.5 km2 +* Not from the examination point of view. +Reprint 2025-26 + +212 +MATHEMATICS +Area of the lake = (2.5 × 3) km2 = 7.5 km2 +Therefore, P (helicopter crashed in the lake) = 7.5 +5 +40.5 +405 +27 + + + + +Example 12 : A carton consists of 100 shirts of which 88 are good, 8 have minor +defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which +are good, but Sujatha, another trader, will only reject the shirts which have major +defects. One shirt is drawn at random from the carton. What is the probability that +(i) it is acceptable to Jimmy? +(ii) it is acceptable to Sujatha? +Solution : One shirt is drawn at random from the carton of 100 shirts. Therefore, +there are 100 equally likely outcomes. +(i) The number of outcomes favourable (i.e., acceptable) to Jimmy = 88 (Why?) +Therefore, P (shirt is acceptable to Jimmy) = 88 +0.88 +100  +(ii) The number of outcomes favourable to Sujatha = 88 + 8 = 96 +(Why?) +So, P (shirt is acceptable to Sujatha) = 96 +0.96 +100  +Example 13 : Two dice, one blue and one grey, are thrown at the same time. Write +down all the possible outcomes. What is the probability that the sum of the two numbers +appearing on the top of the dice is +(i) 8? +(ii) 13? +(iii) less than or equal to 12? +Solution : When the blue die shows ‘1’, the grey die could show any one of the +numbers 1, 2, 3, 4, 5, 6. The same is true when the blue die shows ‘2’, ‘3’, ‘4’, ‘5’ or +‘6’. The possible outcomes of the experiment are listed in the table below; the first +number in each ordered pair is the number appearing on the blue die and the second +number is that on the grey die. +Reprint 2025-26 + +PROBABILITY +213 +1 +2 +3 +4 +5 +6 +1 +(1, 1) +(1, 2) +(1, 3) +(1, 4) +(1, 5) +(1, 6) +2 +(2, 1) +(2, 2) +(2, 3) +(2, 4) +(2, 5) +(2, 6) +3 +(3, 1) +(3, 2) +(3, 3) +(3, 4) +(3, 5) +(3, 6) +4 +(4, 1) +(4, 2) +(4, 3) +(4, 4) +(4, 5) +(4, 6) +5 +(5, 1) +(5, 2) +(5, 3) +(5, 4) +(5, 5) +(5, 6) +6 +(6, 1) +(6, 2) +(6, 3) +(6, 4) +(6, 5) +(6, 6) +Fig. 14.3 +Note that the pair (1, 4) is different from (4, 1). (Why?) +So, the number of possible outcomes = 6 × 6 = 36. +(i) The outcomes favourable to the event ‘the sum of the two numbers is 8’ denoted +by E, are: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) +(see Fig. 14.3) +i.e., the number of outcomes favourable to E = 5. +Hence, +P(E) = +5 +36 +(ii) As you can see from Fig. 14.3, there is no outcome favourable to the event F, +‘the sum of two numbers is 13’. +So, +P(F) = +0 +0 +36  +(iii) As you can see from Fig. 14.3, all the outcomes are favourable to the event G, +‘sum of two numbers £ 12’. +So, +P(G) = 36 +1 +36  +Reprint 2025-26 + +214 +MATHEMATICS +EXERCISE 14.1 +1. Complete the following statements: +(i) Probability of an event E + Probability of the event ‘not E’ = + . +(ii) The probability of an event that cannot happen is + . Such an event is +called + . +(iii) The probability of an event that is certain to happen is + . Such an event +is called + . +(iv) The sum of the probabilities of all the elementary events of an experiment is +. +(v) The probability of an event is greater than or equal to + and less than or +equal to + . +2. Which of the following experiments have equally likely outcomes? Explain. +(i) A driver attempts to start a car. The car starts or does not start. +(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot. +(iii) A trial is made to answer a true-false question. The answer is right or wrong. +(iv) A baby is born. It is a boy or a girl. +3. Why is tossing a coin considered to be a fair way of deciding which team should get the +ball at the beginning of a football game? +4. Which of the following cannot be the probability of an event? +(A) +2 +3 +(B) –1.5 +(C) 15% +(D) 0.7 +5. If P(E) = 0.05, what is the probability of ‘not E’? +6. A bag contains lemon flavoured candies only. Malini takes out one candy without +looking into the bag. What is the probability that she takes out +(i) an orange flavoured candy? +(ii) a lemon flavoured candy? +7. It is given that in a group of 3 students, the probability of 2 students not having the +same birthday is 0.992. What is the probability that the 2 students have the same +birthday? +8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. +What is the probability that the ball drawn is (i) red ? +(ii) not red? +9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken +out of the box at random. What is the probability that the marble taken out will be +(i) red ? +(ii) white ? +(iii) not green? +Reprint 2025-26 + +PROBABILITY +215 +10. A piggy bank contains hundred 50p coins, fifty ` 1 coins, twenty ` 2 coins and ten ` 5 +coins. If it is equally likely that one of the coins will fall out when the bank is turned +upside down, what is the probability that the coin (i) will be a 50 p coin ? (ii) will not be +a ` 5 coin? +11. Gopi buys a fish from a shop for his aquarium. The +shopkeeper takes out one fish at random from a +tank containing 5 male fish and 8 female fish (see +Fig. 14.4). What is the probability that the fish taken +out is a male fish? +12. A game of chance consists of spinning an arrow +which comes to rest pointing at one of the numbers +1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 14.5 ), and these are equally +likely outcomes. What is the probability that it will +point at +(i) 8 ? +(ii) an odd number? +(iii) a number greater than 2? +(iv) a number less than 9? +13. A die is thrown once. Find the probability of getting +(i) a prime number; +(ii) a number lying between 2 and 6; +(iii) an odd number. +14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting +(i) a king of red colour +(ii) a face card +(iii) a red face card +(iv) the jack of hearts +(v) a spade +(vi) the queen of diamonds +15. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their +face downwards. One card is then picked up at random. +(i) What is the probability that the card is the queen? +(ii) If the queen is drawn and put aside, what is the probability that the second card +picked up is (a) an ace? (b) a queen? +16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just +look at a pen and tell whether or not it is defective. One pen is taken out at random from +this lot. Determine the probability that the pen taken out is a good one. +17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. +What is the probability that this bulb is defective? +(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb +is drawn at random from the rest. What is the probability that this bulb is not +defective ? +18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random +from the box, find the probability that it bears (i) a two-digit number +(ii) a perfect +square number +(iii) a number divisible by 5. +Fig. 14.4 +Fig. 14.5 +Reprint 2025-26 + +216 +MATHEMATICS +19. A child has a die whose six faces show the letters as given below: +A +B +C +D +E +A +The die is thrown once. What is the probability of getting +(i) A? +(ii) D? +20*. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is +the probability that it will land inside the circle with diameter 1m? +Fig. 14.6 +21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri +will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one +pen at random and gives it to her. What is the probability that +(i) She will buy it ? +(ii) She will not buy it ? +22. Refer to Example 13. (i) Complete the following table: +Event : +‘Sum on 2 dice’ +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +Probability +1 +36 +5 +36 +1 +36 +(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and +12. Therefore, each of them has a probability 1 +11 . Do you agree with this argument? +Justify your answer. +23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. +Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses +otherwise. Calculate the probability that Hanif will lose the game. +24. A die is thrown twice. What is the probability that +(i) 5 will not come up either time? +(ii) 5 will come up at least once? +[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the +same experiment] +* Not from the examination point of view. + 3 m + 2 m +Reprint 2025-26 + +PROBABILITY +217 +A NOTE TO THE READER +The experimental or empirical probability of an event is based on +what has actually happened while the theoretical probability of the +event attempts to predict what will happen on the basis of certain +assumptions. As the number of trials in an experiment, go on +increasing we may expect the experimental and theoretical +probabilities to be nearly the same. +25. Which of the following arguments are correct and which are not correct? Give reasons +for your answer. +(i) If two coins are tossed simultaneously there are three possible outcomes—two +heads, two tails or one of each. Therefore, for each of these outcomes, the +probability is 1 +3 +(ii) If a die is thrown, there are two possible outcomes—an odd number or an even +number. Therefore, the probability of getting an odd number is 1 +2 . +14.2 Summary +In this chapter, you have studied the following points : +1. The theoretical (classical) probability of an event E, written as P(E), is defined as +P (E) = +Number of outcomes favourable to E +Number of all possible outcomes of the experiment +where we assume that the outcomes of the experiment are equally likely. +2. The probability of a sure event (or certain event) is 1. +3. The probability of an impossible event is 0. +4. The probability of an event E is a number P(E) such that +0  P (E)  1 +5. An event having only one outcome is called an elementary event. The sum of the +probabilities of all the elementary events of an experiment is 1. +6. For any event E, P (E) + P ( E ) = 1, where E stands for ‘not E’. E and E are called +complementary events. +Reprint 2025-26" +class_11,1,Sets,ncert_books/class_11/kemh1dd/kemh101.pdf,"Chapter 1 +SETS +Georg Cantor +(1845-1918) +vIn these days of conflict between ancient and modern studies; there +must surely be something to be said for a study which did not +begin with Pythagoras and will not end with Einstein; but +is the oldest and the youngest. — G.H. HARDY v +1.1 Introduction +The concept of set serves as a fundamental part of the +present day mathematics. Today this concept is being used +in almost every branch of mathematics. Sets are used to +define the concepts of relations and functions. The study of +geometry, sequences, probability, etc. requires the knowledge +of sets. +The theory of sets was developed by German +mathematician Georg Cantor (1845-1918). He first +encountered sets while working on “problems on trigonometric +series”. In this Chapter, we discuss some basic definitions +and operations involving sets. +1.2 Sets and their Representations +In everyday life, we often speak of collections of objects of a particular kind, such as, +a pack of cards, a crowd of people, a cricket team, etc. In mathematics also, we come +across collections, for example, of natural numbers, points, prime numbers, etc. More +specially, we examine the following collections: +(i) +Odd natural numbers less than 10, i.e., 1, 3, 5, 7, 9 +(ii) +The rivers of India +(iii) +The vowels in the English alphabet, namely, a, e, i, o, u +(iv) +Various kinds of triangles +(v) +Prime factors of 210, namely, 2,3,5 and 7 +(vi) +The solution of the equation: x2 – 5x + 6 = 0, viz, 2 and 3. +We note that each of the above example is a well-defined collection of objects in +Reprint 2025-26 + +2 MATHEMATICS +the sense that we can definitely decide whether a given particular object belongs to a +given collection or not. For example, we can say that the river Nile does not belong to +the collection of rivers of India. On the other hand, the river Ganga does belong to this +colleciton. +We give below a few more examples of sets used particularly in mathematics, viz. +N : the set of all natural numbers +Z : the set of all integers +Q : the set of all rational numbers +R : the set of real numbers +Z+ : the set of positive integers +Q+ : the set of positive rational numbers, and +R+ : the set of positive real numbers. +The symbols for the special sets given above will be referred to throughout +this text. +Again the collection of five most renowned mathematicians of the world is not +well-defined, because the criterion for determining a mathematician as most renowned +may vary from person to person. Thus, it is not a well-defined collection. +We shall say that a set is a well-defined collection of objects. +The following points may be noted : +(i) +Objects, elements and members of a set are synonymous terms. +(ii) +Sets are usually denoted by capital letters A, B, C, X, Y, Z, etc. +(iii) +The elements of a set are represented by small letters a, b, c, x, y, z, etc. +If a is an element of a set A, we say that “ a belongs to A” the Greek symbol ∈ +(epsilon) is used to denote the phrase ‘belongs to’. Thus, we write a ∈ A. If ‘b’ is not +an element of a set A, we write b ∉ A and read “b does not belong to A”. +Thus, in the set V of vowels in the English alphabet, a ∈ V but b ∉ V. In the set +P of prime factors of 30, 3 ∈ P but 15 ∉ P. +There are two methods of representing a set : +(i) +Roster or tabular form +(ii) +Set-builder form. +(i) +In roster form, all the elements of a set are listed, the elements are being separated +by commas and are enclosed within braces { }. For example, the set of all even +positive integers less than 7 is described in roster form as {2, 4, 6}. Some more +examples of representing a set in roster form are given below : +(a) +The set of all natural numbers which divide 42 is {1, 2, 3, 6, 7, 14, 21, 42}. +Reprint 2025-26 + +SETS 3 +ANote In roster form, the order in which the elements are listed is immaterial. +Thus, the above set can also be represented as {1, 3, 7, 21, 2, 6, 14, 42}. +(b) +The set of all vowels in the English alphabet is {a, e, i, o, u}. +(c) +The set of odd natural numbers is represented by {1, 3, 5, . . .}. The dots +tell us that the list of odd numbers continue indefinitely. +ANote It may be noted that while writing the set in roster form an element is not +generally repeated, i.e., all the elements are taken as distinct. For example, the set +of letters forming the word ‘SCHOOL’ is { S, C, H, O, L} or {H, O, L, C, S}. Here, +the order of listing elements has no relevance. +(ii) +In set-builder form, all the elements of a set possess a single common property +which is not possessed by any element outside the set. For example, in the set +{a, e, i, o, u}, all the elements possess a common property, namely, each of them +is a vowel in the English alphabet, and no other letter possess this property. Denoting +this set by V, we write +V = {x : x is a vowel in English alphabet} +It may be observed that we describe the element of the set by using a symbol x +(any other symbol like the letters y, z, etc. could be used) which is followed by a colon +“ : ”. After the sign of colon, we write the characteristic property possessed by the +elements of the set and then enclose the whole description within braces. The above +description of the set V is read as “the set of all x such that x is a vowel of the English +alphabet”. In this description the braces stand for “the set of all”, the colon stands for +“such that”. For example, the set +A = {x : x is a natural number and 3 < x < 10} is read as “the set of all x such that +x is a natural number and x lies between 3 and 10.” Hence, the numbers 4, 5, 6, +7, 8 and 9 are the elements of the set A. +If we denote the sets described in (a), (b) and (c) above in roster form by A, B, +C, respectively, then A, B, C can also be represented in set-builder form as follows: +A= {x : x is a natural number which divides 42} +B= {y : y is a vowel in the English alphabet} +C= {z : z is an odd natural number} +Example 1 Write the solution set of the equation x2 + x – 2 = 0 in roster form. +Solution The given equation can be written as +(x – 1) (x + 2) = 0, i. e., x = 1, – 2 +Therefore, the solution set of the given equation can be written in roster form as {1, – 2}. +Example 2 Write the set {x : x is a positive integer and x2 < 40} in the roster form. +Reprint 2025-26 + +4 MATHEMATICS +Solution The required numbers are 1, 2, 3, 4, 5, 6. So, the given set in the roster form +is {1, 2, 3, 4, 5, 6}. +Example 3 Write the set A = {1, 4, 9, 16, 25, . . . }in set-builder form. +Solution We may write the set A as +A = {x : x is the square of a natural number} +Alternatively, we can write +A = {x : x = n2, where n ∈ N} +Example 4 Write the set 1 2 3 4 5 6 +{ +} +2 3 4 5 6 7 +, +, +, +, +, +in the set-builder form. +Solution We see that each member in the given set has the numerator one less than +the denominator. Also, the numerator begin from 1 and do not exceed 6. Hence, in the +set-builder form the given set is +where +is a natural number and 1 +6 +1 +n +x : x +, +n +n +n + + += +≤ +≤ + + ++ + + +Example 5 Match each of the set on the left described in the roster form with the +same set on the right described in the set-builder form : +(i) +{P, R, I, N, C, A, L} (a) { x : x is a positive integer and is a divisor of 18} +(ii) +{ 0 } +(b) { x : x is an integer and x2 – 9 = 0} +(iii) +{1, 2, 3, 6, 9, 18} +(c) {x : x is an integer and x + 1= 1} +(iv) +{3, –3} +(d) {x : x is a letter of the word PRINCIPAL} +Solution Since in (d), there are 9 letters in the word PRINCIPAL and two letters P and I +are repeated, so (i) matches (d). Similarly, (ii) matches (c) as x + 1 = 1 implies +x = 0. Also, 1, 2 ,3, 6, 9, 18 are all divisors of 18 and so (iii) matches (a). Finally, x2 – 9 = 0 +implies x = 3, –3 and so (iv) matches (b). +EXERCISE 1.1 +1. +Which of the following are sets ? Justify your answer. +(i) +The collection of all the months of a year beginning with the letter J. +(ii) +The collection of ten most talented writers of India. +(iii) +A team of eleven best-cricket batsmen of the world. +(iv) +The collection of all boys in your class. +(v) +The collection of all natural numbers less than 100. +(vi) +A collection of novels written by the writer Munshi Prem Chand. +(vii) +The collection of all even integers. +Reprint 2025-26 + +SETS 5 +(viii) +The collection of questions in this Chapter. +(ix) +A collection of most dangerous animals of the world. +2. +Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈ or ∉ in the blank +spaces: +(i) + 5. . .A +(ii) +8 . . . A +(iii) +0. . .A + (iv) + 4. . . A +(v) +2. . .A +(vi) +10. . .A +3. +Write the following sets in roster form: +(i) +A = {x : x is an integer and –3 ≤ x < 7} +(ii) +B = {x : x is a natural number less than 6} +(iii) +C = {x : x is a two-digit natural number such that the sum of its digits is 8} +(iv) +D = {x : x is a prime number which is divisor of 60} +(v) +E = The set of all letters in the word TRIGONOMETRY +(vi) +F = The set of all letters in the word BETTER +4. +Write the following sets in the set-builder form : +(i) +(3, 6, 9, 12} +(ii) +{2,4,8,16,32} +(iii) +{5, 25, 125, 625} +(iv) +{2, 4, 6, . . .} +(v) +{1,4,9, . . .,100} +5. +List all the elements of the following sets : +(i) +A = {x : x is an odd natural number} +(ii) +B = {x : x is an integer, +1 +2 +– + < x < 9 +2 } +(iii) +C = {x : x is an integer, x2 ≤ 4} +(iv) +D = {x : x is a letter in the word “LOYAL”} +(v) +E = {x : x is a month of a year not having 31 days} +(vi) +F = {x : x is a consonant in the English alphabet which precedes k }. +6. +Match each of the set on the left in the roster form with the same set on the right +described in set-builder form: +(i) +{1, 2, 3, 6} +(a) +{x : x is a prime number and a divisor of 6} +(ii) +{2, 3} +(b) +{x : x is an odd natural number less than 10} +(iii) +{M,A,T,H,E,I,C,S} (c) +{x : x is natural number and divisor of 6} +(iv) +{1, 3, 5, 7, 9} +(d) +{x : x is a letter of the word MATHEMATICS}. +1.3 The Empty Set +Consider the set +A = { x : x is a student of Class XI presently studying in a school } +We can go to the school and count the number of students presently studying in +Class XI in the school. Thus, the set A contains a finite number of elements. +We now write another set B as follows: +Reprint 2025-26 + +6 MATHEMATICS +B = { x : x is a student presently studying in both Classes X and XI } +We observe that a student cannot study simultaneously in both Classes X and XI. +Thus, the set B contains no element at all. +Definition 1 A set which does not contain any element is called the empty set or the +null set or the void set. +According to this definition, B is an empty set while A is not an empty set. The +empty set is denoted by the symbol φ or { }. +We give below a few examples of empty sets. +(i) +Let A = {x : 1 < x < 2, x is a natural number}. Then A is the empty set, +because there is no natural number between 1 and 2. +(ii) +B = {x : x2 – 2 = 0 and x is rational number}. Then B is the empty set because +the equation x2 – 2 = 0 is not satisfied by any rational value of x. +(iii) +C = {x : x is an even prime number greater than 2}.Then C is the empty set, +because 2 is the only even prime number. +(iv) +D = { x : x2 = 4, x is odd }. Then D is the empty set, because the equation +x2 = 4 is not satisfied by any odd value of x. +1.4 Finite and Infinite Sets +Let +A = {1, 2, 3, 4, 5}, +B = {a, b, c, d, e, g} +and +C = { men living presently in different parts of the world} +We observe that A contains 5 elements and B contains 6 elements. How many elements +does C contain? As it is, we do not know the number of elements in C, but it is some +natural number which may be quite a big number. By number of elements of a set S, +we mean the number of distinct elements of the set and we denote it by n (S). If n (S) +is a natural number, then S is non-empty finite set. +Consider the set of natural numbers. We see that the number of elements of this +set is not finite since there are infinite number of natural numbers. We say that the set +of natural numbers is an infinite set. The sets A, B and C given above are finite sets +and n(A) = 5, n(B) = 6 and n(C) = some finite number. +Definition 2 A set which is empty or consists of a definite number of elements is +called finite otherwise, the set is called infinite. +Consider some examples : +(i) +Let W be the set of the days of the week. Then W is finite. +(ii) +Let S be the set of solutions of the equation x2 –16 = 0. Then S is finite. +(iii) +Let G be the set of points on a line. Then G is infinite. +When we represent a set in the roster form, we write all the elements of the set +within braces { }. It is not possible to write all the elements of an infinite set within +braces { } because the numbers of elements of such a set is not finite. So, we represent +Reprint 2025-26 + +SETS 7 +some infinite set in the roster form by writing a few elements which clearly indicate the +structure of the set followed ( or preceded ) by three dots. +For example, {1, 2, 3 . . .} is the set of natural numbers, {1, 3, 5, 7, . . .} is the set +of odd natural numbers, {. . .,–3, –2, –1, 0,1, 2 ,3, . . .} is the set of integers. All these +sets are infinite. +ANote All infinite sets cannot be described in the roster form. For example, the +set of real numbers cannot be described in this form, because the elements of this +set do not follow any particular pattern. +Example 6 State which of the following sets are finite or infinite : +(i) +{x : x ∈ N and (x – 1) (x –2) = 0} +(ii) +{x : x ∈ N and x2 = 4} +(iii) +{x : x ∈ N and 2x –1 = 0} +(iv) +{x : x ∈ N and x is prime} +(v) +{x : x ∈ N and x is odd} +Solution +(i) +Given set = {1, 2}. Hence, it is finite. +(ii) +Given set = {2}. Hence, it is finite. +(iii) +Given set = φ. Hence, it is finite. +(iv) +The given set is the set of all prime numbers and since set of prime +numbers is infinite. Hence the given set is infinite +(v) +Since there are infinite number of odd numbers, hence, the given set is +infinite. +1.5 Equal Sets +Given two sets A and B, if every element of A is also an element of B and if every +element of B is also an element of A, then the sets A and B are said to be equal. +Clearly, the two sets have exactly the same elements. +Definition 3 Two sets A and B are said to be equal if they have exactly the same +elements and we write A = B. Otherwise, the sets are said to be unequal and we write +A ≠ B. +We consider the following examples : +(i) +Let A = {1, 2, 3, 4} and +B = {3, 1, 4, 2}. Then A = B. +(ii) +Let A be the set of prime numbers less than 6 and P the set of prime factors +of 30. Then A and P are equal, since 2, 3 and 5 are the only prime factors of +30 and also these are less than 6. +ANote A set does not change if one or more elements of the set are repeated. +For example, the sets A = {1, 2, 3} and B = {2, 2, 1, 3, 3} are equal, since each +Reprint 2025-26 + +8 MATHEMATICS +element of A is in B and vice-versa. That is why we generally do not repeat any +element in describing a set. +Example 7 Find the pairs of equal sets, if any, give reasons: +A = {0}, +B = {x : x > 15 and x < 5}, +C = {x : x – 5 = 0 }, +D = {x: x2 = 25}, +E = {x : x is an integral positive root of the equation x2 – 2x –15 = 0}. +Solution Since 0 ∈ A and 0 does not belong to any of the sets B, C, D and E, it +follows that, A ≠ B, A ≠ C, A ≠ D, A ≠ E. +Since B = φ but none of the other sets are empty. Therefore B ≠ C, B ≠ D +and B ≠ E. Also C = {5} but –5 ∈ D, hence C ≠ D. +Since E = {5}, C = E. Further, D = {–5, 5} and E = {5}, we find that, D ≠ E. +Thus, the only pair of equal sets is C and E. +Example 8 Which of the following pairs of sets are equal? Justify your answer. +(i) +X, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”. +(ii) +A = {n : n ∈ Z and n2 ≤ 4} and B = {x : x ∈ R and x2 – 3x + 2 = 0}. +Solution (i) We have, X = {A, L, L, O, Y}, B = {L, O, Y, A, L}. Then X and B are +equal sets as repetition of elements in a set do not change a set. Thus, +X = {A, L, O, Y} = B +(ii) A = {–2, –1, 0, 1, 2}, B = {1, 2}. Since 0 ∈ A and 0 ∉ B, A and B are not equal sets. +EXERCISE 1.2 +1. +Which of the following are examples of the null set +(i) +Set of odd natural numbers divisible by 2 +(ii) +Set of even prime numbers +(iii) +{ x : x is a natural numbers, x < 5 and x > 7 } +(iv) +{ y : y is a point common to any two parallel lines} +2. +Which of the following sets are finite or infinite +(i) +The set of months of a year +(ii) +{1, 2, 3, . . .} +(iii) +{1, 2, 3, . . .99, 100} +(iv) +The set of positive integers greater than 100 +(v) +The set of prime numbers less than 99 +3. +State whether each of the following set is finite or infinite: +(i) +The set of lines which are parallel to the x-axis +(ii) +The set of letters in the English alphabet +(iii) +The set of numbers which are multiple of 5 +Reprint 2025-26 + +SETS 9 +(iv) +The set of animals living on the earth +(v) +The set of circles passing through the origin (0,0) +4. +In the following, state whether A = B or not: +(i) +A = { a, b, c, d } +B = { d, c, b, a } +(ii) +A = { 4, 8, 12, 16 } +B = { 8, 4, 16, 18} +(iii) +A = {2, 4, 6, 8, 10} +B = { x : x is positive even integer and x ≤ 10} +(iv) +A = { x : x is a multiple of 10}, +B = { 10, 15, 20, 25, 30, . . . } +5. +Are the following pair of sets equal ? Give reasons. +(i) +A = {2, 3}, +B = {x : x is solution of x2 + 5x + 6 = 0} +(ii) +A = { x : x is a letter in the word FOLLOW} +B = { y : y is a letter in the word WOLF} +6. +From the sets given below, select equal sets : +A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4}, +C = { 4, 8, 12, 14}, +D = { 3, 1, 4, 2} +E = {–1, 1}, +F = { 0, a}, +G = {1, –1}, +H = { 0, 1} +1.6 Subsets +Consider the sets : X = set of all students in your school, Y = set of all students in your +class. +We note that every element of Y is also an element of X; we say that Y is a subset +of X. The fact that Y is subset of X is expressed in symbols as Y ⊂ X. The symbol ⊂ +stands for ‘is a subset of’ or ‘is contained in’. +Definition 4 A set A is said to be a subset of a set B if every element of A is also an +element of B. +In other words, A ⊂ B if whenever a ∈ A, then a ∈ B. It is often convenient to +use the symbol “⇒” which means implies. Using this symbol, we can write the definiton +of subset as follows: +A ⊂ B if a ∈ A ⇒ a ∈ B +We read the above statement as “A is a subset of B if a is an element of A +implies that a is also an element of B”. If A is not a subset of B, we write A ⊄ B. +We may note that for A to be a subset of B, all that is needed is that every +element of A is in B. It is possible that every element of B may or may not be in A. If +it so happens that every element of B is also in A, then we shall also have B ⊂ A. In this +case, A and B are the same sets so that we have A ⊂ B and B ⊂ A ⇔ A = B, where +“⇔” is a symbol for two way implications, and is usually read as if and only if (briefly +written as “iff”). +It follows from the above definition that every set A is a subset of itself, i.e., +A ⊂ A. Since the empty set φ has no elements, we agree to say that φ is a subset of +every set. We now consider some examples : +Reprint 2025-26 + +10 MATHEMATICS +(i) +The set Q of rational numbers is a subset of the set R of real numbes, and +we write Q ⊂ R. +(ii) +If A is the set of all divisors of 56 and B the set of all prime divisors of 56, +then B is a subset of A and we write B ⊂ A. +(iii) +Let A = {1, 3, 5} and B = {x : x is an odd natural number less than 6}. Then +A ⊂ B and B ⊂ A and hence A = B. +(iv) +Let A = { a, e, i, o, u} and B = { a, b, c, d}. Then A is not a subset of B, +also B is not a subset of A. +Let A and B be two sets. If A ⊂ B and A ≠ B , then A is called a proper subset +of B and B is called superset of A. For example, +A = {1, 2, 3} is a proper subset of B = {1, 2, 3, 4}. +If a set A has only one element, we call it a singleton set. Thus,{ a } is a +singleton set. +Example 9 Consider the sets +φ, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. +Insert the symbol ⊂ or ⊄ between each of the following pair of sets: +(i) φ . . . B +(ii) A . . . B +(iii) A . . . C +(iv) B . . . C +Solution +(i) +φ ⊂ B as φ is a subset of every set. +(ii) +A ⊄ B as 3 ∈ A and 3 ∉ B +(iii) +A ⊂ C as 1, 3 ∈ A also belongs to C +(iv) +B ⊂ C as each element of B is also an element of C. +Example 10 Let A = { a, e, i, o, u} and B = { a, b, c, d}. Is A a subset of B ? No. +(Why?). Is B a subset of A? No. (Why?) +Example 11 Let A, B and C be three sets. If A ∈ B and B ⊂ C, is it true that +A ⊂ C?. If not, give an example. +Solution No. Let A = {1}, B = {{1}, 2} and C = {{1}, 2, 3}. Here A ∈ B as A = {1} +and B ⊂ C. But A ⊄ C as 1 ∈ A and 1 ∉ C. +Note that an element of a set can never be a subset of itself. +1.6.1 Subsets of set of real numbers +As noted in Section 1.6, there are many important subsets of R. We give below the +names of some of these subsets. +The set of natural numbers N = {1, 2, 3, 4, 5, . . .} +The set of integers +Z = {. . ., –3, –2, –1, 0, 1, 2, 3, . . .} +The set of rational numbers Q = { x : x = p +q , p, q ∈ Z and q ≠ 0} +Reprint 2025-26 + +SETS 11 +which is read “ Q is the set of all numbers x such that x equals the quotient +p +q , where +p and q are integers and q is not zero”. Members of Q include –5 (which can be +expressed as +5 +1 +– +) , 7 +5 , 1 +3 2 (which can be expressed as 7 +2 ) and +11 +3 +– +. +The set of irrational numbers, denoted by T, is composed of all other real numbers. +Thus +T = {x : x ∈ R and x ∉ Q}, i.e., all real numbers that are not rational. +Members of T include +2 , +5 and π . +Some of the obvious relations among these subsets are: +N ⊂ Z ⊂ Q, Q ⊂ R, T ⊂ R, N ⊄ T. +1.6.2 Intervals as subsets of R Let a, b ∈ R and a < b. Then the set of real numbers +{ y : a < y < b} is called an open interval and is denoted by (a, b). All the points +between a and b belong to the open interval (a, b) but a, b themselves do not belong to +this interval. +The interval which contains the end points also is called closed interval and is +denoted by [ a, b ]. Thus +[ a, b ] = {x : a ≤ x ≤ b} +We can also have intervals closed at one end and open at the other, i.e., +[ a, b ) = {x : a ≤ x < b} is an open interval from a to b, including a but excluding b. +( a, b ] = { x : a < x ≤ b } is an open interval from a to b including b but excluding a. +These notations provide an alternative way of designating the subsets of set of +real numbers. For example , if A = (–3, 5) and B = [–7, 9], then A ⊂ B. The set [ 0, ∞) +defines the set of non-negative real numbers, while set ( – ∞, 0 ) defines the set of +negative real numbers. The set ( – ∞, ∞ ) describes the set of real numbers in relation +to a line extending from – ∞ to ∞. +On real number line, various types of intervals described above as subsets of R, +are shown in the Fig 1.1. +Here, we note that an interval contains infinitely many points. +For example, the set {x : x ∈ R, –5 < x ≤ 7}, written in set-builder form, can be +written in the form of interval as (–5, 7] and the interval [–3, 5) can be written in set- +builder form as {x : –3 ≤ x < 5}. +Fig 1.1 +Reprint 2025-26 + +12 MATHEMATICS +The number (b – a) is called the length of any of the intervals (a, b), [a, b], +[a, b) or (a, b]. +1.7 Universal Set +Usually, in a particular context, we have to deal with the elements and subsets of a +basic set which is relevant to that particular context. For example, while studying the +system of numbers, we are interested in the set of natural numbers and its subsets such +as the set of all prime numbers, the set of all even numbers, and so forth. This basic set +is called the “Universal Set”. The universal set is usually denoted by U, and all its +subsets by the letters A, B, C, etc. +For example, for the set of all integers, the universal set can be the set of rational +numbers or, for that matter, the set R of real numbers. For another example, in human +population studies, the universal set consists of all the people in the world. +EXERCISE 1.3 +1. +Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces : +(i) +{ 2, 3, 4 } . . . { 1, 2, 3, 4,5 } (ii) { a, b, c } . . . { b, c, d } +(iii) +{x : x is a student of Class XI of your school}. . .{x : x student of your school} +(iv) +{x : x is a circle in the plane} . . .{x : x is a circle in the same plane with + radius 1 unit} +(v) +{x : x is a triangle in a plane} . . . {x : x is a rectangle in the plane} +(vi) +{x : x is an equilateral triangle in a plane} . . . {x : x is a triangle in the same plane} +(vii) +{x : x is an even natural number} . . . {x : x is an integer} +2. +Examine whether the following statements are true or false: +(i) +{ a, b } ⊄ { b, c, a } +(ii) +{ a, e } ⊂ { x : x is a vowel in the English alphabet} +(iii) +{ 1, 2, 3 } ⊂ { 1, 3, 5 } +(iv) +{ a } ⊂ { a, b, c } +(v) +{ a } ∈ { a, b, c } +(vi) +{ x : x is an even natural number less than 6} ⊂ { x : x is a natural number +which divides 36} +3. +Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why? +(i) +{3, 4} ⊂ A +(ii) +{3, 4} ∈ A +(iii) +{{3, 4}} ⊂ A +(iv) +1 ∈ A +(v) +1 ⊂ A +(vi) +{1, 2, 5} ⊂ A +(vii) +{1, 2, 5} ∈ A +(viii) +{1, 2, 3} ⊂ A +(ix) +φ ∈ A +(x) +φ ⊂ A +(xi) +{φ} ⊂ A +4. +Write down all the subsets of the following sets +(i) +{a} +(ii) +{a, b} +(iii) +{1, 2, 3} +(iv) φ +Reprint 2025-26 + +SETS 13 +5. +Write the following as intervals : +(i) +{x : x ∈ R, – 4 < x ≤ 6} +(ii) +{x : x ∈ R, – 12 < x < –10} +(iii) +{x : x ∈ R, 0 ≤ x < 7} +(iv) +{x : x ∈ R, 3 ≤ x ≤ 4} +6. +Write the following intervals in set-builder form : +(i) +(– 3, 0) +(ii) +[6 , 12] +(iii) +(6, 12] +(iv) +[–23, 5) +7. +What universal set(s) would you propose for each of the following : +(i) +The set of right triangles. +(ii) +The set of isosceles triangles. +8. +Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the +following may be considered as universal set (s) for all the three sets A, B and C +(i) +{0, 1, 2, 3, 4, 5, 6} +(ii) +φ +(iii) +{0,1,2,3,4,5,6,7,8,9,10} +(iv) +{1,2,3,4,5,6,7,8} +1.8 Venn Diagrams +Most of the relationships between sets can be +represented by means of diagrams which are known +as Venn diagrams. Venn diagrams are named after +the English logician, John Venn (1834-1883). These +diagrams consist of rectangles and closed curves +usually circles. The universal set is represented +usually by a rectangle and its subsets by circles. +In Venn diagrams, the elements of the sets are +written in their respective circles (Figs 1.2 and 1.3) +Illustration 1 In Fig 1.2, U = {1,2,3, ..., 10} is the +universal set of which +A = {2,4,6,8,10} is a subset. +Illustration 2 In Fig 1.3, U = {1,2,3, ..., 10} is the +universal set of which +A = {2,4,6,8,10} and B = {4, 6} are subsets, +and also B ⊂ A. +The reader will see an extensive use of the +Venn diagrams when we discuss the union, intersection and difference of sets. +1.9 Operations on Sets +In earlier classes, we have learnt how to perform the operations of addition, subtraction, +multiplication and division on numbers. Each one of these operations was performed +on a pair of numbers to get another number. For example, when we perform the +operation of addition on the pair of numbers 5 and 13, we get the number 18. Again, +performing the operation of multiplication on the pair of numbers 5 and 13, we get 65. +Fig 1.2 +Fig 1.3 +Reprint 2025-26 + +14 MATHEMATICS +Similarly, there are some operations which when performed on two sets give rise to +another set. We will now define certain operations on sets and examine their properties. +Henceforth, we will refer all our sets as subsets of some universal set. +1.9.1 Union of sets Let A and B be any two sets. The union of A and B is the set +which consists of all the elements of A and all the elements of B, the common elements +being taken only once. The symbol ‘∪’ is used to denote the union. Symbolically, we +write A ∪ B and usually read as ‘A union B’. +Example 12 Let A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. Find A ∪ B. +Solution We have A ∪ B = { 2, 4, 6, 8, 10, 12} +Note that the common elements 6 and 8 have been taken only once while writing +A ∪ B. +Example 13 Let A = { a, e, i, o, u } and B = { a, i, u }. Show that A ∪ B = A +Solution We have, A ∪ B = { a, e, i, o, u } = A. +This example illustrates that union of sets A and its subset B is the set A +itself, i.e., if B ⊂ A, then A ∪ B = A. +Example 14 Let X = {Ram, Geeta, Akbar} be the set of students of Class XI, who are +in school hockey team. Let Y = {Geeta, David, Ashok} be the set of students from +Class XI who are in the school football team. Find X ∪ Y and interpret the set. +Solution We have, X ∪ Y = {Ram, Geeta, Akbar, David, Ashok}. This is the set of +students from Class XI who are in the hockey team or the football team or both. +Thus, we can define the union of two sets as follows: +Definition 5 The union of two sets A and B is the set C which consists of all those +elements which are either in A or in B (including +those which are in both). In symbols, we write. +A ∪ B = { x : x ∈A or x ∈B } +The union of two sets can be represented by +a Venn diagram as shown in Fig 1.4. +The shaded portion in Fig 1.4 represents A ∪ B. +Some Properties of the Operation of Union +(i) +A ∪ B = B ∪ A (Commutative law) +(ii) +( A ∪ B ) ∪ C = A ∪ ( B ∪ C) + (Associative law ) +(iii) +A ∪ φ = A + (Law of identity element, φ is the identity of ∪) +Fig 1.4 +Reprint 2025-26 + +SETS 15 +(iv) +A ∪ A = A + (Idempotent law) +(v) +U ∪ A = U (Law of U) +1.9.2 Intersection of sets The intersection of sets A and B is the set of all elements +which are common to both A and B. The symbol ‘∩’ is used to denote the intersection. +The intersection of two sets A and B is the set of all those elements which belong to +both A and B. Symbolically, we write A ∩ B = {x : x ∈ A and x ∈ B}. +Example 15 Consider the sets A and B of Example 12. Find A ∩ B. +Solution We see that 6, 8 are the only elements which are common to both A and B. +Hence A ∩ B = { 6, 8 }. +Example 16 Consider the sets X and Y of Example 14. Find X ∩ Y. +Solution We see that element ‘Geeta’ is the only element common to both. Hence, +X ∩ Y = {Geeta}. +Example 17 Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = { 2, 3, 5, 7 }. Find A ∩ B and +hence show that A ∩ B = B. +Solution We have A ∩ B = { 2, 3, 5, 7 } = B. We +note that B ⊂ A and that A ∩ B = B. +Definition 6 The intersection of two sets A and B +is the set of all those elements which belong to both +A and B. Symbolically, we write +A ∩ B = {x : x ∈ A and x ∈ B} +The shaded portion in Fig 1.5 indicates the +intersection of A and B. +If A and B are two sets such that A ∩ B = φ, then +A and B are called disjoint sets. +For example, let A = { 2, 4, 6, 8 } and +B = { 1, 3, 5, 7 }. Then A and B are disjoint sets, +because there are no elements which are common to +A and B. The disjoint sets can be represented by +means of Venn diagram as shown in the Fig 1.6 +In the above diagram, A and B are disjoint sets. +Some Properties of Operation of Intersection +(i) A ∩ B = B ∩ A +(Commutative law). +(ii) ( A ∩ B ) ∩ C = A ∩ ( B ∩ C ) +(Associative law). +(iii) φ ∩ A = φ, U ∩ A = A +(Law of φ and U). +(iv) A ∩ A = A +(Idempotent law) +Fig 1.5 +A +B +U +Fig 1.6 +Reprint 2025-26 + +16 MATHEMATICS +(v) A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) (Distributive law ) i. e., +∩ distributes over ∪ +This can be seen easily from the following Venn diagrams [Figs 1.7 (i) to (v)]. +(i) +(iii) +(ii) +(iv) +(v) +Figs 1.7 (i) to (v) +1.9.3 Difference of sets The difference of the sets A and B in this order is the set +of elements which belong to A but not to B. Symbolically, we write A – B and read as +“A minus B”. +Example 18 Let A = { 1, 2, 3, 4, 5, 6}, B = { 2, 4, 6, 8 }. Find A – B and B – A. +Solution We have, A – B = { 1, 3, 5 }, since the elements 1, 3, 5 belong to A but +not to B and B – A = { 8 }, since the element 8 belongs to B and not to A. +We note that A – B ≠ B – A. +Reprint 2025-26 + +SETS 17 +Fig 1.8 +Fig 1.9 +Example 19 Let V = { a, e, i, o, u } and +B = { a, i, k, u}. Find V – B and B – V +Solution We have, V – B = { e, o }, since the elements +e, o belong to V but not to B and B – V = { k }, since +the element k belongs to B but not to V. +We note that V – B ≠ B – V. Using the set- +builder notation, we can rewrite the definition of +difference as +A – B = { x : x ∈ A and x ∉ B } +The difference of two sets A and B can be +represented by Venn diagram as shown in Fig 1.8. +The shaded portion represents the difference of +the two sets A and B. +Remark The sets A – B, A ∩ B and B – A are +mutually disjoint sets, i.e., the intersection of any of +these two sets is the null set as shown in Fig 1.9. +EXERCISE 1.4 +1. +Find the union of each of the following pairs of sets : +(i) +X = {1, 3, 5} +Y = {1, 2, 3} +(ii) +A = [ a, e, i, o, u} +B = {a, b, c} +(iii) +A = {x : x is a natural number and multiple of 3} +B = {x : x is a natural number less than 6} +(iv) +A = {x : x is a natural number and 1 < x ≤6 } +B = {x : x is a natural number and 6 < x < 10 } +(v) +A = {1, 2, 3}, B = φ +2. +Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ? +3. +If A and B are two sets such that A ⊂ B, then what is A ∪ B ? +4. +If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find +(i) +A ∪ B +(ii) +A ∪ C +(iii) +B ∪ C +(iv) B ∪ D +(v) +A ∪ B ∪ C +(vi) +A ∪ B ∪ D +(vii) +B ∪ C ∪ D +5. +Find the intersection of each pair of sets of question 1 above. +6. +If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find +(i) +A ∩ B +(ii) +B ∩ C +(iii) +A ∩ C ∩ D +(iv) + A ∩ C +(v) +B ∩ D +(vi) +A ∩ (B ∪ C) +(vii) +A ∩ D +(viii) +A ∩ (B ∪ D) +(ix) +( A ∩ B ) ∩ ( B ∪ C ) +(x) +( A ∪ D) ∩ ( B ∪ C) +Reprint 2025-26 + +18 MATHEMATICS +7. +If A = {x : x is a natural number }, B = {x : x is an even natural number} +C = {x : x is an odd natural number}andD = {x : x is a prime number }, find +(i) +A ∩ B +(ii) +A ∩ C +(iii) +A ∩ D +(iv) +B ∩ C +(v) +B ∩ D +(vi) +C ∩ D +8. +Which of the following pairs of sets are disjoint +(i) +{1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6 } +(ii) +{ a, e, i, o, u } and { c, d, e, f } +(iii) +{x : x is an even integer } and {x : x is an odd integer} +9. +If A = {3, 6, 9, 12, 15, 18, 21}, B = { 4, 8, 12, 16, 20 }, +C = { 2, 4, 6, 8, 10, 12, 14, 16 }, D = {5, 10, 15, 20 }; find +(i) +A – B +(ii) +A – C +(iii) +A – D +(iv) +B – A +(v) +C – A +(vi) +D – A +(vii) +B – C +(viii) +B – D +(ix) +C – B +(x) +D – B +(xi) +C – D +(xii) +D – C +10. +If X= { a, b, c, d } and Y = { f, b, d, g}, find +(i) +X – Y +(ii) +Y – X +(iii) +X ∩ Y +11. +If R is the set of real numbers and Q is the set of rational numbers, then what is +R – Q? +12. +State whether each of the following statement is true or false. Justify your answer. +(i) +{ 2, 3, 4, 5 } and { 3, 6} are disjoint sets. +(ii) +{ a, e, i, o, u } and { a, b, c, d }are disjoint sets. +(iii) +{ 2, 6, 10, 14 } and { 3, 7, 11, 15} are disjoint sets. +(iv) +{ 2, 6, 10 } and { 3, 7, 11} are disjoint sets. +1.10 Complement of a Set +Let U be the universal set which consists of all prime numbers and A be the subset of +U which consists of all those prime numbers that are not divisors of 42. Thus, + A = {x : x ∈ U and x is not a divisor of 42 }. We see that 2 ∈ U but 2 ∉ A, because +2 is divisor of 42. Similarly, 3 ∈ U but 3 ∉ A, and 7 ∈ U but 7 ∉ A. Now 2, 3 and 7 are +the only elements of U which do not belong to A. The set of these three prime numbers, +i.e., the set {2, 3, 7} is called the Complement of A with respect to U, and is denoted by +A′. So we have A′ = {2, 3, 7}. Thus, we see that +A′ = {x : x ∈ U and x ∉ A }. This leads to the following definition. +Definition 7 Let U be the universal set and A a subset of U. Then the complement of +A is the set of all elements of U which are not the elements of A. Symbolically, we +write A′ to denote the complement of A with respect to U. Thus, +A′ = {x : x ∈ U and x ∉ A }. Obviously A′ = U – A +We note that the complement of a set A can be looked upon, alternatively, as the +difference between a universal set U and the set A. +Reprint 2025-26 + +SETS 19 +Example 20 Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′. +Solution We note that 2, 4, 6, 8, 10 are the only elements of U which do not belong to +A. Hence + A′ = { 2, 4, 6, 8,10 }. +Example 21 Let U be universal set of all the students of Class XI of a coeducational +school and A be the set of all girls in Class XI. Find A′. +Solution Since A is the set of all girls, A′ is clearly the set of all boys in the class. +ANote If A is a subset of the universal set U, then its complement A′ is also a +subset of U. +Again in Example 20 above, we have A′ = { 2, 4, 6, 8, 10 } +Hence (A′ )′= {x : x ∈ U and x ∉ A′} += {1, 3, 5, 7, 9} = A +It is clear from the definition of the complement that for any subset of the universal +set U, we have ( A′)′ = A +Now, we want to find the results for ( A ∪ B )′ and A′ ∩ B′ in the followng +example. +Example 22 Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. +Find A′, B′ , A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′ ∩ B′. +Solution Clearly A′ = {1, 4, 5, 6}, B′ = { 1, 2, 6 }. Hence A′ ∩ B′ = { 1, 6 } +Also A ∪ B = { 2, 3, 4, 5 }, so that (A ∪ B)′ = { 1, 6 } + ( A ∪ B )′ = { 1, 6 } = A′ ∩ B′ +It can be shown that the above result is true in general. If A and B are any two +subsets of the universal set U, then +( A ∪ B )′ = A′ ∩ B′. Similarly, ( A ∩ B )′ = A′ ∪ B′ . These two results are stated +in words as follows : +The complement of the union of two sets is +the intersection of their complements and the +complement of the intersection of two sets is the +union of their complements. These are called De +Morgan’s laws. These are named after the +mathematician De Morgan. +The complement A′ of a set A can be represented +by a Venn diagram as shown in Fig 1.10. +The shaded portion represents the complement of the set A. +Fig 1.10 +Reprint 2025-26 + +20 MATHEMATICS +Some Properties of Complement Sets +1. Complement laws: +(i) A ∪ A′ = U + (ii) A ∩ A′ = φ +2. De Morgan’s law: +(i) (A ∪ B)´ = A′ ∩ B′ (ii) (A ∩ B)′ = A′ ∪ B′ +3. Law of double complementation : (A′ )′ = A +4. Laws of empty set and universal set φ′ = U and U′ = φ. +These laws can be verified by using Venn diagrams. +EXERCISE 1.5 +1. +Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and +C = { 3, 4, 5, 6 }. Find (i) A′ (ii) B′ (iii) (A ∪ C)′ (iv) (A ∪ B)′ (v) (A′)′ +(vi) (B – C)′ +2. +If U = { a, b, c, d, e, f, g, h}, find the complements of the following sets : +(i) A = {a, b, c} +(ii) B = {d, e, f, g} +(iii) C = {a, c, e, g} +(iv) D = { f, g, h, a} +3. +Taking the set of natural numbers as the universal set, write down the complements +of the following sets: +(i) +{x : x is an even natural number} +(ii) { x : x is an odd natural number } +(iii) +{x : x is a positive multiple of 3} +(iv) { x : x is a prime number } +(v) +{x : x is a natural number divisible by 3 and 5} +(vi) +{ x : x is a perfect square } +(vii) { x : x is a perfect cube} +(viii) { x : x + 5 = 8 } +(ix) { x : 2x + 5 = 9} +(x) +{ x : x ≥ 7 } +(xi) { x : x ∈ N and 2x + 1 > 10 } +4. +If U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}. Verify that +(i) (A ∪ B)′ = A′ ∩ B′ +(ii) (A ∩ B)′ = A′ ∪ B′ +5. +Draw appropriate Venn diagram for each of the following : +(i) (A ∪ B)′, +(ii) A′ ∩ B′, +(iii) (A ∩ B)′, +(iv) A′ ∪ B′ +6. +Let U be the set of all triangles in a plane. If A is the set of all triangles with at +least one angle different from 60°, what is A′? +7. +Fill in the blanks to make each of the following a true statement : +(i) +A ∪ A′ = . . . +(ii) +φ′ ∩ A = . . . +(iii) +A ∩ A′ = . . . +(iv) +U′ ∩ A = . . . + Miscellaneous Examples +Example 23 Show that the set of letters needed to spell “ CATARACT ” and the +set of letters needed to spell “ TRACT” are equal. +Solution Let X be the set of letters in “CATARACT”. Then +X = { C, A, T, R } +Reprint 2025-26 + +SETS 21 +Let Y be the set of letters in “ TRACT”. Then +Y = { T, R, A, C, T } = { T, R, A, C } +Since every element in X is in Y and every element in Y is in X. It follows that X = Y. +Example 24 List all the subsets of the set { –1, 0, 1 }. +Solution Let A = { –1, 0, 1 }. The subset of A having no element is the empty +set φ. The subsets of A having one element are { –1 }, { 0 }, { 1 }. The subsets of +A having two elements are {–1, 0}, {–1, 1} ,{0, 1}. The subset of A having three +elements of A is A itself. So, all the subsets of A are φ, {–1}, {0}, {1}, {–1, 0}, {–1, 1}, +{0, 1} and {–1, 0, 1}. +Example 25 Show that A ∪ B = A ∩ B implies A = B +Solution Let a ∈ A. Then a ∈ A ∪ B. Since A ∪ B = A ∩ B , a ∈ A ∩ B. So a ∈ B. +Therefore, A ⊂ B. Similarly, if b ∈ B, then b ∈ A ∪ B. Since +A ∪ B = A ∩ B, b ∈ A ∩ B. So, b ∈ A. Therefore, B ⊂ A. Thus, A = B + Miscellaneous Exercise on Chapter 1 +1. +Decide, among the following sets, which sets are subsets of one and another: +A = { x : x ∈ R and x satisfy x2 – 8x + 12 = 0 }, +B = { 2, 4, 6 }, +C = { 2, 4, 6, 8, . . . }, D = { 6 }. +2. +In each of the following, determine whether the statement is true or false. If it is +true, prove it. If it is false, give an example. +(i) +If x ∈ A and A ∈ B , then x ∈ B +(ii) +If A ⊂ B and B ∈ C , then A ∈ C +(iii) +If A ⊂ B and B ⊂ C , then A ⊂ C +(iv) +If A ⊄ B and B ⊄ C , then A ⊄ C +(v) +If x ∈ A and A ⊄ B , then x ∈ B +(vi) +If A ⊂ B and x ∉ B , then x ∉ A +3. +Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show +that B = C. +4. +Show that the following four conditions are equivalent : +(i) A ⊂ B(ii) A – B = φ +(iii) A ∪ B = B (iv) A ∩ B = A +5. +Show that if A ⊂ B, then C – B ⊂ C – A. +6. +Show that for any sets A and B, +A = ( A ∩ B ) ∪ ( A – B ) and A ∪ ( B – A ) = ( A ∪ B ) +7. +Using properties of sets, show that +(i) A ∪ ( A ∩ B ) = A (ii) A ∩ ( A ∪ B ) = A. +8. +Show that A ∩ B = A ∩ C need not imply B = C. +Reprint 2025-26 + +22 MATHEMATICS +9. +Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set +X, show that A = B. +(Hints A = A ∩ ( A ∪ X ) , B = B ∩ ( B ∪ X ) and use Distributive law ) +10. +Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty +sets and A ∩ B ∩ C = φ. +Summary +This chapter deals with some basic definitions and operations involving sets. These +are summarised below: +®A set is a well-defined collection of objects. +®A set which does not contain any element is called empty set. +®A set which consists of a definite number of elements is called finite set, +otherwise, the set is called infinite set. +®Two sets A and B are said to be equal if they have exactly the same elements. +®A set A is said to be subset of a set B, if every element of A is also an element +of B. Intervals are subsets of R. +®The union of two sets A and B is the set of all those elements which are either +in A or in B. +®The intersection of two sets A and B is the set of all elements which are +common. The difference of two sets A and B in this order is the set of elements +which belong to A but not to B. +®The complement of a subset A of universal set U is the set of all elements of U +which are not the elements of A. +®For any two sets A and B, (A ∪ B)′ = A′ ∩ B′ and ( A ∩ B )′ = A′ ∪ B′ +Historical Note +The modern theory of sets is considered to have been originated largely by the +German mathematician Georg Cantor (1845-1918). His papers on set theory +appeared sometimes during 1874 to 1897. His study of set theory came when he +was studying trigonometric series of the form a1 sin x + a2 sin 2x + a3 sin 3x + ... +He published in a paper in 1874 that the set of real numbers could not be put into +one-to-one correspondence wih the integers. From 1879 onwards, he publishd +several papers showing various properties of abstract sets. +Reprint 2025-26 + +SETS 23 +Cantor’s work was well received by another famous mathematician Richard +Dedekind (1831-1916). But Kronecker (1810-1893) castigated him for regarding +infinite set the same way as finite sets. Another German mathematician Gottlob +Frege, at the turn of the century, presented the set theory as principles of logic. +Till then the entire set theory was based on the assumption of the existence of the +set of all sets. It was the famous Englih Philosopher Bertand Russell (1872- +1970 ) who showed in 1902 that the assumption of existence of a set of all sets +leads to a contradiction. This led to the famous Russell’s Paradox. Paul R.Halmos +writes about it in his book ‘Naïve Set Theory’ that “nothing contains everything”. +The Russell’s Paradox was not the only one which arose in set theory. +Many paradoxes were produced later by several mathematicians and logicians. +As a consequence of all these paradoxes, the first axiomatisation of set theory +was published in 1908 by Ernst Zermelo. Another one was proposed by Abraham +Fraenkel in 1922. John Von Neumann in 1925 introduced explicitly the axiom of +regularity. Later in 1937 Paul Bernays gave a set of more satisfactory +axiomatisation. A modification of these axioms was done by Kurt Gödel in his +monograph in 1940. This was known as Von Neumann-Bernays (VNB) or Gödel- +Bernays (GB) set theory. +Despite all these difficulties, Cantor’s set theory is used in present day +mathematics. In fact, these days most of the concepts and results in mathematics +are expressed in the set theoretic language. + +— v — +Reprint 2025-26" +class_11,2,Relations and Functions,ncert_books/class_11/kemh1dd/kemh102.pdf,"vMathematics is the indispensable instrument of +all physical research. – BERTHELOT v +2.1 Introduction +Much of mathematics is about finding a pattern – a +recognisable link between quantities that change. In our +daily life, we come across many patterns that characterise +relations such as brother and sister, father and son, teacher +and student. In mathematics also, we come across many +relations such as number m is less than number n, line l is +parallel to line m, set A is a subset of set B. In all these, we +notice that a relation involves pairs of objects in certain +order. In this Chapter, we will learn how to link pairs of +objects from two sets and then introduce relations between +the two objects in the pair. Finally, we will learn about +special relations which will qualify to be functions. The +concept of function is very important in mathematics since it captures the idea of a +mathematically precise correspondence between one quantity with the other. +2.2 Cartesian Products of Sets +Suppose A is a set of 2 colours and B is a set of 3 objects, i.e., +A = {red, blue}and B = {b, c, s}, +where b, c and s represent a particular bag, coat and shirt, respectively. +How many pairs of coloured objects can be made from these two sets? +Proceeding in a very orderly manner, we can see that there will be 6 +distinct pairs as given below: +(red, b), (red, c), (red, s), (blue, b), (blue, c), (blue, s). +Thus, we get 6 distinct objects (Fig 2.1). +Let us recall from our earlier classes that an ordered pair of elements +taken from any two sets P and Q is a pair of elements written in small +Fig 2.1 +Chapter 2 +RELATIONS AND FUNCTIONS +G . W. Leibnitz +(1646–1716) +Reprint 2025-26 + +RELATIONS AND FUNCTIONS 25 +brackets and grouped together in a particular order, i.e., (p,q), p ∈ P and q ∈ Q . This +leads to the following definition: +Definition 1 Given two non-empty sets P and Q. The cartesian product P × Q is the +set of all ordered pairs of elements from P and Q, i.e., +P × Q = { (p,q) : p ∈ P, q ∈ Q } +If either P or Q is the null set, then P × Q will also be empty set, i.e., P × Q = φ +From the illustration given above we note that +A × B = {(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)}. +Again, consider the two sets: +A = {DL, MP, KA}, where DL, MP, KA represent Delhi, +Madhya Pradesh and Karnataka, respectively and B = {01,02, +03}representing codes for the licence plates of vehicles issued +by DL, MP and KA . +If the three states, Delhi, Madhya Pradesh and Karnataka +were making codes for the licence plates of vehicles, with the +restriction that the code begins with an element from set A, +which are the pairs available from these sets and how many such +pairs will there be (Fig 2.2)? +The available pairs are:(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), +(KA,01), (KA,02), (KA,03) and the product of set A and set B is given by +A × B = {(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), + (KA,03)}. +It can easily be seen that there will be 9 such pairs in the Cartesian product, since +there are 3 elements in each of the sets A and B. This gives us 9 possible codes. Also +note that the order in which these elements are paired is crucial. For example, the code +(DL, 01) will not be the same as the code (01, DL). +As a final illustration, consider the two sets A= {a1, a2} and +B = {b1, b2, b3, b4} (Fig 2.3). +A × B = {( a1, b1), (a1, b2), (a1, b3), (a1, b4), (a2, b1), (a2, b2), + (a2, b3), (a2, b4)}. +The 8 ordered pairs thus formed can represent the position of points in +the plane if A and B are subsets of the set of real numbers and it is +obvious that the point in the position (a1, b2) will be distinct from the point +in the position (b2, a1). +Remarks +(i) +Two ordered pairs are equal, if and only if the corresponding first elements +are equal and the second elements are also equal. +DL +MP +KA +03 +02 +01 +Fig 2.2 +Fig 2.3 +Reprint 2025-26 + +26 +MATHEMATICS +(ii) +If there are p elements in A and q elements in B, then there will be pq +elements in A × B, i.e., if n(A) = p and n(B) = q, then n(A × B) = pq. +(iii) +If A and B are non-empty sets and either A or B is an infinite set, then so is +A × B. +(iv) +A × A × A = {(a, b, c) : a, b, c ∈ A}. Here (a, b, c) is called an ordered +triplet. +Example 1 If (x + 1, y – 2) = (3,1), find the values of x and y. +Solution Since the ordered pairs are equal, the corresponding elements are equal. +Therefore +x + 1 = 3 and y – 2 = 1. +Solving we get +x = 2 and y = 3. +Example 2 If P = {a, b, c} and Q = {r}, form the sets P × Q and Q × P. +Are these two products equal? +Solution By the definition of the cartesian product, +P × Q = {(a, r), (b, r), (c, r)} and Q × P = {(r, a), (r, b), (r, c)} +Since, by the definition of equality of ordered pairs, the pair (a, r) is not equal to the pair +(r, a), we conclude that P × Q ≠ Q × P. +However, the number of elements in each set will be the same. +Example 3 Let A = {1,2,3}, B = {3,4} and C = {4,5,6}. Find +(i) +A × (B ∩ C) +(ii) +(A × B) ∩ (A × C) +(iii) +A × (B ∪ C) +(iv) +(A × B) ∪ (A × C) +Solution (i) By the definition of the intersection of two sets, (B ∩ C) = {4}. +Therefore, A × (B ∩ C) = {(1,4), (2,4), (3,4)}. + (ii) +Now (A × B) = {(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)} +and (A × C) = {(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)} +Therefore, +(A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}. +(iii) Since, +(B ∪ C) = {3, 4, 5, 6}, we have +A × (B ∪ C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,3), +(3,4), (3,5), (3,6)}. +(iv) Using the sets A × B and A × C from part (ii) above, we obtain +(A × B) ∪ (A × C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), +(3,3), (3,4), (3,5), (3,6)}. +Reprint 2025-26 + +RELATIONS AND FUNCTIONS 27 +Example 4 If P = {1, 2}, form the set P × P × P. +Solution We have, P × P × P = {(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), + (2,2,2)}. +Example 5 If R is the set of all real numbers, what do the cartesian products R × R +and R × R × R represent? +Solution The Cartesian product R × R represents the set R × R={(x, y) : x, y ∈ R} +which represents the coordinates of all the points in two dimensional space and the +cartesian product R × R × R represents the set R × R × R ={(x, y, z) : x, y, z ∈ R} +which represents the coordinates of all the points in three-dimensional space. +Example 6 If A × B ={(p, q),(p, r), (m, q), (m, r)}, find A and B. +Solution +A = set of first elements = {p, m} +B = set of second elements = {q, r}. +EXERCISE 2.1 +1. +If +2 +5 1 +1 +3 +3 +3 3 +x +,y – +, + + + + ++ += + + + + + + + +, find the values of x and y. +2. +If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of +elements in (A×B). +3. +If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G. +4. +State whether each of the following statements are true or false. If the statement +is false, rewrite the given statement correctly. +(i) +If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}. +(ii) +If A and B are non-empty sets, then A × B is a non-empty set of ordered +pairs (x, y) such that x ∈ A and y ∈ B. +(iii) +If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ. +5. +If A = {–1, 1}, find A × A × A. +6. +If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B. +7. +Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that +(i) A × (B ∩ C) = (A × B) ∩ (A × C). (ii) A × C is a subset of B × D. +8. +Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? +List them. +9. +Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) +are in A × B, find A and B, where x, y and z are distinct elements. +Reprint 2025-26 + +28 +MATHEMATICS +10. +The Cartesian product A × A has 9 elements among which are found (–1, 0) and +(0,1). Find the set A and the remaining elements of A × A. +2.3 Relations +Consider the two sets P = {a, b, c} and Q = {Ali, Bhanu, Binoy, Chandra, Divya}. +The cartesian product of +P and Q has 15 ordered pairs which +can be listed as P × Q = {(a, Ali), +(a,Bhanu), (a, Binoy), ..., (c, Divya)}. +We can now obtain a subset of +P × Q by introducing a relation R +between the first element x and the +second element y of each ordered pair +(x, y) as +R= { (x,y): x is the first letter of the name y, x ∈ P, y ∈ Q}. +Then R = {(a, Ali), (b, Bhanu), (b, Binoy), (c, Chandra)} +A visual representation of this relation R (called an arrow diagram) is shown +in Fig 2.4. +Definition 2 A relation R from a non-empty set A to a non-empty set B is a subset of +the cartesian product A × B. The subset is derived by describing a relationship between +the first element and the second element of the ordered pairs in A × B. The second +element is called the image of the first element. +Definition 3 The set of all first elements of the ordered pairs in a relation R from a set +A to a set B is called the domain of the relation R. +Definition 4 The set of all second elements in a relation R from a set A to a set B is +called the range of the relation R. The whole set B is called the codomain of the +relation R. Note that range ⊂ codomain. +Remarks +(i) A relation may be represented algebraically either by the Roster +method or by the Set-builder method. +(ii) An arrow diagram is a visual representation of a relation. +Example 7 Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by +R = {(x, y) : y = x + 1 } +(i) Depict this relation using an arrow diagram. +(ii) Write down the domain, codomain and range of R. +Solution +(i) By the definition of the relation, +R = {(1,2), (2,3), (3,4), (4,5), (5,6)}. +Fig 2.4 +Reprint 2025-26 + +RELATIONS AND FUNCTIONS 29 +The corresponding arrow diagram is +shown in Fig 2.5. +(ii) We can see that the +domain ={1, 2, 3, 4, 5,} +Similarly, the range = {2, 3, 4, 5, 6} +and the codomain = {1, 2, 3, 4, 5, 6}. +Example 8 The Fig 2.6 shows a relation +between the sets P and Q. Write this relation (i) in set-builder form, (ii) in roster form. +What is its domain and range? +Solution It is obvious that the relation R is +“x is the square of y”. +(i) In set-builder form, R = {(x, y): x + is the square of y, x ∈ P, y ∈ Q} +(ii) In roster form, R = {(9, 3), +(9, –3), (4, 2), (4, –2), (25, 5), (25, –5)} +The domain of this relation is {4, 9, 25}. +The range of this relation is {– 2, 2, –3, 3, –5, 5}. +Note that the element 1 is not related to any element in set P. +The set Q is the codomain of this relation. +ANote The total number of relations that can be defined from a set A to a set B +is the number of possible subsets of A × B. If n(A ) = p and n(B) = q, then +n (A × B) = pq and the total number of relations is 2pq. +Example 9 Let A = {1, 2} and B = {3, 4}. Find the number of relations from A to B. +Solution We have, +A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}. +Since n (A×B ) = 4, the number of subsets of A×B is 24. Therefore, the number of +relations from A into B will be 24. +Remark A relation R from A to A is also stated as a relation on A. +EXERCISE 2.2 +1. +Let A = {1, 2, 3,...,14}. Define a relation R from A to A by +R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and +range. +Fig 2.5 +Fig 2.6 +Reprint 2025-26 + +30 +MATHEMATICS +2. +Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, +x is a natural number less than 4; x, y ∈N}. Depict this relationship using roster +form. Write down the domain and the range. +3. +A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by +R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in +roster form. +4. +The Fig2.7 shows a relationship +between the sets P and Q. Write this +relation +(i) in set-builder form (ii) roster form. +What is its domain and range? +5. +Let A = {1, 2, 3, 4, 6}. Let R be the +relation on A defined by +{(a, b): a , b ∈A, b is exactly divisible by a}. +(i) +Write R in roster form +(ii) +Find the domain of R +(iii) +Find the range of R. +6. +Determine the domain and range of the relation R defined by +R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}. +7. +Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form. +8. +Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B. +9. +Let R be the relation on Z defined by R = {(a,b): a, b ∈ Z, a – b is an integer}. +Find the domain and range of R. +2.4 +Functions +In this Section, we study a special type of relation called function. It is one of the most +important concepts in mathematics. We can, visualise a function as a rule, which produces +new elements out of some given elements. There are many terms such as ‘map’ or +‘mapping’ used to denote a function. +Definition 5 A relation f from a set A to a set B is said to be a function if every +element of set A has one and only one image in set B. +In other words, a function f is a relation from a non-empty set A to a non-empty +set B such that the domain of f is A and no two distinct ordered pairs in f have the +same first element. +If f is a function from A to B and (a, b) ∈ f, then f (a) = b, where b is called the +image of a under f and a is called the preimage of b under f. +Fig 2.7 +Reprint 2025-26 + +RELATIONS AND FUNCTIONS 31 +The function f from A to B is denoted by f: A à B. +Looking at the previous examples, we can easily see that the relation in Example 7 is +not a function because the element 6 has no image. +Again, the relation in Example 8 is not a function because the elements in the +domain are connected to more than one images. Similarly, the relation in Example 9 is +also not a function. (Why?) In the examples given below, we will see many more +relations some of which are functions and others are not. +Example 10 Let N be the set of natural numbers and the relation R be defined on +N such that R = {(x, y) : y = 2x, x, y ∈ N}. +What is the domain, codomain and range of R? Is this relation a function? +Solution The domain of R is the set of natural numbers N. The codomain is also N. +The range is the set of even natural numbers. +Since every natural number n has one and only one image, this relation is a +function. +Example 11 Examine each of the following relations given below and state in each +case, giving reasons whether it is a function or not? +(i) +R = {(2,1),(3,1), (4,2)}, (ii) R = {(2,2),(2,4),(3,3), (4,4)} +(iii) +R = {(1,2),(2,3),(3,4), (4,5), (5,6), (6,7)} +Solution (i) +Since 2, 3, 4 are the elements of domain of R having their unique images, +this relation R is a function. +(ii) +Since the same first element 2 corresponds to two different images 2 +and 4, this relation is not a function. +(iii) +Since every element has one and only one image, this relation is a +function. +Definition 6 A function which has either R or one of its subsets as its range is called +a real valued function. Further, if its domain is also either R or a subset of R, it is +called a real function. +Example 12 Let N be the set of natural numbers. Define a real valued function +f : Nà N by f (x) = 2x + 1. Using this definition, complete the table given below. +x +1 +2 +3 +4 +5 +6 +7 +y +f (1) = ... +f (2) = ... +f (3) = ... +f (4) = ... +f (5) = ... +f (6) = ... +f (7) = ... +Solution The completed table is given by +x +1 +2 +3 +4 +5 +6 +7 +y +f (1) = 3 +f (2) = 5 +f (3) = 7 +f (4) = 9 +f (5) = 11 +f (6) = 13 +f (7) =15 +Reprint 2025-26 + +32 +MATHEMATICS +2.4.1 Some functions and their graphs +(i) +Identity function Let R be the set of real numbers. Define the real valued +function f : R → R by y = f(x) = x for each x ∈ R. Such a function is called the +identity function. Here the domain and range of f are R. The graph is a straight line as +shown in Fig 2.8. It passes through the origin. +Fig 2.9 +Fig 2.8 +(ii) +Constant function Define the function f: R → R by y = f (x) = c, x ∈ R where +c is a constant and each x ∈ R. Here domain of f is R and its range is {c}. +Reprint 2025-26 + +RELATIONS AND FUNCTIONS 33 +The graph is a line parallel to x-axis. For example, if f(x)=3 for each x∈R, then its +graph will be a line as shown in the Fig 2.9. +(iii) +Polynomial function A function f : R → R is said to be polynomial function if +for each x in R, y = f (x) = a0 + a1x + a2x2 + ...+ an xn, where n is a non-negative +integer and a0, a1, a2,...,an∈R. +The functions defined by f(x) = x3 – x2 + 2, and g(x) = x4 + +2 x are some examples +of polynomial functions, whereas the function h defined by h(x) = +2 +3 +x + 2x is not a +polynomial function.(Why?) +Example 13 Define the function f: R → R by y = f(x) = x2, x ∈ R. Complete the +Table given below by using this definition. What is the domain and range of this function? +Draw the graph of f. +x +– 4 +–3 +–2 +–1 +0 +1 +2 +3 +4 +y = f(x) = x2 +Solution The completed Table is given below: +x +– 4 +–3 +–2 +–1 +0 +1 +2 +3 +4 +y = f (x) = x2 +16 +9 +4 +1 +0 +1 +4 +9 +16 +Domain of f = {x : x∈R}. Range of f = {x +2: x ∈ R}. The graph of f is given +by Fig 2.10 +Fig 2.10 +Reprint 2025-26 + +34 +MATHEMATICS +Example 14 Draw the graph of the function f :R → R defined by f (x) = x3, x∈R. +Solution We have +f(0) = 0, f(1) = 1, f(–1) = –1, f(2) = 8, f(–2) = –8, f(3) = 27; f(–3) = –27, etc. +Therefore, f = {(x,x3): x∈R}. +The graph of f is given in Fig 2.11. +Fig 2.11 +(iv) +Rational functions are functions of the type +( ) +( ) +f x +g x , where f(x) and g(x) are +polynomial functions of x defined in a domain, where g(x) ≠ 0. +Example 15 Define the real valued function f : R – {0} → R defined by +1 +( ) = +f x +x , +x ∈ R –{0}. Complete the Table given below using this definition. What is the domain +and range of this function? +x +–2 +–1.5 +–1 +–0.5 +0.25 +0.5 +1 +1.5 +2 +y = 1 +x +... +... +... +... +... +... +... +... +... +Solution The completed Table is given by +x +–2 +–1.5 +–1 +–0.5 +0.25 +0.5 +1 +1.5 +2 +y = 1 +x +– 0.5 +– 0.67 –1 +– 2 +4 +2 +1 +0.67 +0.5 +Reprint 2025-26 + +RELATIONS AND FUNCTIONS 35 +The domain is all real numbers except 0 and its range is also all real numbers +except 0. The graph of f is given in Fig 2.12. +Fig 2.13 +(v) +The Modulus function The function +f: R→R defined by f(x) = |x| for each +x ∈R is called modulus function. For each +non-negative value of x, f(x) is equal to x. +But for negative values of x, the value of +f(x) is the negative of the value of x, i.e., +0 +( ) +0 +x,x +f x +x,x +≥ + += − +< + +The graph of the modulus function is given +in Fig 2.13. +(vi) +Signum function The function +f:R→R defined by +1 if +0 +( ) +0 if +0 +1 if +0 +, +x +f x +, +x +, +x +> + + += += + +− +< + +Fig 2.12 +Reprint 2025-26 + +36 +MATHEMATICS +is called the signum function. The domain of the signum function is R and the range is +the set {–1, 0, 1}. The graph of the signum function is given by the Fig 2.14. +Fig 2.14 +(vii) Greatest integer function +The function f: R → R defined +by f(x) = [x], x ∈R assumes the +value of the greatest integer, less +than or equal to x. Such a function +is called the greatest integer +function. +From the definition of [x], we +can see that +[x] = –1 for –1 ≤ x < 0 +[x] = 0 for 0 ≤ x < 1 +[x] = 1 for 1 ≤ x < 2 +[x] = 2 for 2 ≤ x < 3 and +so on. +The graph of the function is +shown in Fig 2.15. +2.4.2 Algebra of real functions In this Section, we shall learn how to add two real +functions, subtract a real function from another, multiply a real function by a scalar +(here by a scalar we mean a real number), multiply two real functions and divide one +real function by another. +(i) +Addition of two real functions Let f : X → R and g : X → R be any two real +functions, where X ⊂ R. Then, we define (f + g): X → R by +(f + g) (x) = f (x) + g (x), for all x ∈ X. +Fig 2.15 +Reprint 2025-26 + +RELATIONS AND FUNCTIONS 37 +(ii) +Subtraction of a real function from another Let f : X → R and g: X → R be +any two real functions, where X ⊂R. Then, we define (f – g) : X→R by +(f–g) (x) = f(x) –g(x), for all x ∈ X. +(iii) +Multiplication by a scalar Let f : X→R be a real valued function and α be a +scalar. Here by scalar, we mean a real number. Then the product α f is a function from +X to R defined by (α f ) (x) = α f (x), x ∈X. +(iv) +Multiplication of two real functions The product (or multiplication) of two real +functions f:X→R and g:X→R is a function fg:X→R defined by +(fg) (x) = f(x) g(x), for all x ∈ X. +This is also called pointwise multiplication. +(v) +Quotient of two real functions Let f and g be two real functions defined from +X→R, where X⊂R. The quotient of f by g denoted by +f +g is a function defined by , + +( ) +( ) +( ) +f +f x +x +g +g x + + += + + + + +, provided g(x) ≠ 0, x ∈ X +Example 16 Let f(x) = x +2and g(x) = 2x + 1 be two real functions.Find +(f + g) (x), (f –g) (x), (fg) (x), +( ) +f +x +g + + + + + + +. +Solution We have, +(f + g) (x) = x +2 + 2x + 1, (f –g) (x) = x +2 – 2x – 1, +(fg) (x) = x +2 (2x + 1) = 2x +3 + x +2, +( ) +f +x +g + + + + + + + = +2 +2 +1 +x +x + , x ≠ +1 +2 +− +Example 17 Let f(x) = +x and g(x) = x be two functions defined over the set of non- +negative real numbers. Find (f + g) (x), (f – g) (x), (fg) (x) and +f +g + + + + + + (x). +Solution We have + (f + g) (x) = +x + x, (f – g) (x) = +x – x , +(fg) x = +3 +2 +x( x ) +x += + and +( ) +f +x +g + + + + + + + +1 +2 +0 +– +x +x +, x +x += += +≠ +Reprint 2025-26 + +38 +MATHEMATICS +EXERCISE 2.3 +1. +Which of the following relations are functions? Give reasons. If it is a function, +determine its domain and range. +(i) +{(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)} +(ii) +{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)} +(iii) +{(1,3), (1,5), (2,5)}. +2. +Find the domain and range of the following real functions: +(i) +f(x) = – x +(ii) +f(x) = +2 +9 +x +− +. +3. +A function f is defined by f(x) = 2x –5. Write down the values of +(i) + f (0), +(ii) f (7), (iii) f (–3). +4. +The function ‘t’ which maps temperature in degree Celsius into temperature in +degree Fahrenheit is defined by t(C) = 9C +5 + 32. +Find +(i) +t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212. +5. +Find the range of each of the following functions. +(i) +f (x) = 2 – 3x, x ∈ R, x > 0. +(ii) + f (x) = x2 + 2, x is a real number. +(iii) + f (x) = x, x is a real number. +Miscellaneous Examples +Example 18 Let R be the set of real numbers. +Define the real function +f: R→R by f(x) = x + 10 +and sketch the graph of this function. +Solution Here f(0) = 10, f(1) = 11, f(2) = 12, ..., +f(10) = 20, etc., and +f(–1) = 9, f(–2) = 8, ..., f(–10) = 0 and so on. +Therefore, shape of the graph of the given +function assumes the form as shown in Fig 2.16. +Remark The function f defined by f(x) = mx + c , +x ∈ R, is called linear function, where m and c are +constants. Above function is an example of a linear +function. +Fig 2.16 +Reprint 2025-26 + +RELATIONS AND FUNCTIONS 39 +Example 19 Let R be a relation from Q to Q defined by R = {(a,b): a,b ∈ Q and +a – b ∈ Z}. Show that +(i) +(a,a) ∈ R for all a ∈ Q +(ii) +(a,b) ∈ R implies that (b, a) ∈ R +(iii) +(a,b) ∈ R and (b,c) ∈ R implies that (a,c) ∈R +Solution +(i) +Since, a – a = 0 ∈ Z, if follows that (a, a) ∈ R. +(ii) +(a,b) ∈ R implies that a – b ∈ Z. So, b – a ∈ Z. Therefore, +(b, a) ∈ R +(iii) +(a, b) and (b, c) ∈ R implies that a – b ∈ Z. b – c ∈ Z. So, + a – c = (a – b) + (b – c) ∈ Z. Therefore, (a,c) ∈ R +Example 20 Let f = {(1,1), (2,3), (0, –1), (–1, –3)} be a linear function from Z into Z. +Find f(x). +Solution Since f is a linear function, f (x) = mx + c. Also, since (1, 1), (0, – 1) ∈ R, +f (1) = m + c = 1 and f (0) = c = –1. This gives m = 2 and f(x) = 2x – 1. +Example 21 Find the domain of the function +2 +2 +3 +5 +( ) +5 +4 +x +x +f x +x +x ++ ++ += +− ++ +Solution Since x +2 –5x + 4 = (x – 4) (x –1), the function f is defined for all real numbers +except at x = 4 and x = 1. Hence the domain of f is R – {1, 4}. +Example 22 The function f is defined by +f (x) = +1 +0 +1 +0 +1 +0 +x, x +, x +x +, x +− +< + + += + ++ +> + +Draw the graph of f (x). +Solution Here, f(x) = 1 – x, x < 0, this gives +f(– 4) = 1 – (– 4)= 5; +f(– 3) +=1 – (– 3) = 4, +f(– 2) += 1 – (– 2)= 3 +f(–1) += 1 – (–1) = 2; etc, +and +f(1) = 2, f (2) = 3, f (3) = 4 +f(4) = 5 and so on for f(x) = x + 1, x > 0. +Thus, the graph of f is as shown in Fig 2.17 +Fig 2.17 +Reprint 2025-26 + +40 +MATHEMATICS +Miscellaneous Exercise on Chapter 2 +1. +The relation f is defined by +2 0 +3 +( ) = +3 3 +10 +x , +x +f x +x, +x + +≤≤ + +≤≤ + +The relation g is defined by +2 , 0 +2 +( ) +3 , 2 +10 +x +x +g x +x +x + +≤ +≤ + += +≤ +≤ + +Show that f is a function and g is not a function. +2. +If f (x) = x +2, find +(1 1) +(1) +(1 1 1) +f +. +– f +. – +. +3. +Find the domain of the function f (x) +2 +2 +2 +1 +8 +12 +x +x +x – x ++ ++ += ++ +. +4. +Find the domain and the range of the real function f defined by f (x) = +( +1) +x − +. +5. +Find the domain and the range of the real function f defined by f (x) = +–1 +x +. +6. +Let +2 +2 +, +: +1 +x +f +x +x +x + + + + + + += +∈ + + + ++ + + + + + + +R be a function from R into R. Determine the range +of f. +7. +Let f, g : R→R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find +f + g, f – g and +f +g . +8. +Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by +f(x) = ax + b, for some integers a, b. Determine a, b. +9. +Let R be a relation from N to N defined by R = {(a, b) : a, b ∈N and a = b +2}. Are +the following true? +(i) +(a,a) ∈ R, for all a ∈ N +(ii) +(a,b) ∈ R, implies (b,a) ∈ R +(iii) +(a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R. +Justify your answer in each case. +10. +Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} +Are the following true? +(i) +f is a relation from A to B +(ii) +f is a function from A to B. +Justify your answer in each case. +Reprint 2025-26 + +RELATIONS AND FUNCTIONS 41 +11. +Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a +function from Z to Z? Justify your answer. +12. +Let A = {9,10,11,12,13} and let f : A→N be defined by f (n) = the highest prime +factor of n. Find the range of f. +Summary +In this Chapter, we studied about relations and functions.The main features of +this Chapter are as follows: +® Ordered pair A pair of elements grouped together in a particular order. +® Cartesian product A × B of two sets A and B is given by +A × B = {(a, b): a ∈ A, b ∈ B} +In particular R × R = {(x, y): x, y ∈ R} +and R × R × R = {(x, y, z): x, y, z ∈ R} +® If (a, b) = (x, y), then a = x and b = y. +® If n(A) = p and n(B) = q, then n(A × B) = pq. +® A × φ = φ +® In general, A × B ≠ B × A. +® Relation A relation R from a set A to a set B is a subset of the cartesian +product A × B obtained by describing a relationship between the first element +x and the second element y of the ordered pairs in A × B. +® The image of an element x under a relation R is given by y, where (x, y) ∈ R, +® The domain of R is the set of all first elements of the ordered pairs in a +relation R. +® The range of the relation R is the set of all second elements of the ordered +pairs in a relation R. +® Function A function f from a set A to a set B is a specific type of relation for +which every element x of set A has one and only one image y in set B. +We write f: A→B, where f(x) = y. +® A is the domain and B is the codomain of f. +Reprint 2025-26 + +42 +MATHEMATICS +® The range of the function is the set of images. +® A real function has the set of real numbers or one of its subsets both as its +domain and as its range. +® Algebra of functions For functions f : X → R and g : X → R, we have +(f + g) (x) = f (x) + g(x), x ∈ X +(f – g) (x) = f (x) – g(x), x ∈ X +(f.g) (x) = f (x) .g (x), x ∈ X +(kf) (x) = k ( f (x) ), x ∈ X, where k is a real number. +( ) +f +x +g + + + + = +( ) +( ) +f x +g x , x ∈ X, g(x) ≠ 0 +Historical Note +The word FUNCTION first appears in a Latin manuscript “Methodus +tangentium inversa, seu de fuctionibus” written by Gottfried Wilhelm Leibnitz +(1646-1716) in 1673; Leibnitz used the word in the non-analytical sense. He +considered a function in terms of “mathematical job” – the “employee” being +just a curve. +On July 5, 1698, Johan Bernoulli, in a letter to Leibnitz, for the first time +deliberately assigned a specialised use of the term function in the analytical +sense. At the end of that month, Leibnitz replied showing his approval. +Function is found in English in 1779 in Chambers’ Cyclopaedia: “The +term function is used in algebra, for an analytical expression any way compounded +of a variable quantity, and of numbers, or constant quantities”. +— v — +Reprint 2025-26" +class_11,3,Trigonometric Functions,ncert_books/class_11/kemh1dd/kemh103.pdf,"vA mathematician knows how to solve a problem, +he can not solve it. – MILNE v +3.1 Introduction +The word ‘trigonometry’ is derived from the Greek words +‘trigon’ and ‘metron’ and it means ‘measuring the sides of +a triangle’. The subject was originally developed to solve +geometric problems involving triangles. It was studied by +sea captains for navigation, surveyor to map out the new +lands, by engineers and others. Currently, trigonometry is +used in many areas such as the science of seismology, +designing electric circuits, describing the state of an atom, +predicting the heights of tides in the ocean, analysing a +musical tone and in many other areas. +In earlier classes, we have studied the trigonometric +ratios of acute angles as the ratio of the sides of a right +angled triangle. We have also studied the trigonometric identities and application of +trigonometric ratios in solving the problems related to heights and distances. In this +Chapter, we will generalise the concept of trigonometric ratios to trigonometric functions +and study their properties. +3.2 Angles +Angle is a measure of rotation of a given ray about its initial point. The original ray is +Chapter 3 +TRIGONOMETRIC FUNCTIONS +Arya Bhatt + (476-550) +Fig 3.1 +Vertex +Reprint 2025-26 + +44 +MATHEMATICS +called the initial side and the final position of the ray after rotation is called the +terminal side of the angle. The point of rotation is called the vertex. If the direction of +rotation is anticlockwise, the angle is said to be positive and if the direction of rotation +is clockwise, then the angle is negative (Fig 3.1). +The measure of an angle is the amount of +rotation performed to get the terminal side from +the initial side. There are several units for +measuring angles. The definition of an angle +suggests a unit, viz. one complete revolution from the position of the initial side as +indicated in Fig 3.2. +This is often convenient for large angles. For example, we can say that a rapidly +spinning wheel is making an angle of say 15 revolution per second. We shall describe +two other units of measurement of an angle which are most commonly used, viz. +degree measure and radian measure. +3.2.1 Degree measure If a rotation from the initial side to terminal side is +th +1 +360 + + + + + +of +a revolution, the angle is said to have a measure of one degree, written as 1°. A degree is +divided into 60 minutes, and a minute is divided into 60 seconds . One sixtieth of a degree is +called a minute, written as 1′, and one sixtieth of a minute is called a second, written as 1″. +Thus, +1° = 60′, +1′ = 60″ +Some of the angles whose measures are 360°,180°, 270°, 420°, – 30°, – 420° are +shown in Fig 3.3. +Fig 3.2 +Fig 3.3 +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 45 +3.2.2 Radian measure There is another unit for measurement of an angle, called +the radian measure. Angle subtended at the centre by an arc of length 1 unit in a +unit circle (circle of radius 1 unit) is said to have a measure of 1 radian. In the Fig +3.4(i) to (iv), OA is the initial side and OB is the terminal side. The figures show the +angles whose measures are 1 radian, –1 radian, 1 1 +2 radian and –1 1 +2 radian. +(i) +(ii) +(iii) +Fig 3.4 (i) to (iv) +(iv) +We know that the circumference of a circle of radius 1 unit is 2π. Thus, one +complete revolution of the initial side subtends an angle of 2π radian. +More generally, in a circle of radius r, an arc of length r will subtend an angle of +1 radian. It is well-known that equal arcs of a circle subtend equal angle at the centre. +Since in a circle of radius r, an arc of length r subtends an angle whose measure is 1 +radian, an arc of length l will subtend an angle whose measure is l +r radian. Thus, if in +a circle of radius r, an arc of length l subtends an angle θ radian at the centre, we have +θ = l +r or l = r θ. +Reprint 2025-26 + +46 +MATHEMATICS +3.2.3 Relation between radian and real numbers +Consider the unit circle with centre O. Let A be any point +on the circle. Consider OA as initial side of an angle. +Then the length of an arc of the circle will give the radian +measure of the angle which the arc will subtend at the +centre of the circle. Consider the line PAQ which is +tangent to the circle at A. Let the point A represent the +real number zero, AP represents positive real number and +AQ represents negative real numbers (Fig 3.5). If we +rope the line AP in the anticlockwise direction along the +circle, and AQ in the clockwise direction, then every real +number will correspond to a radian measure and +conversely. Thus, radian measures and real numbers can +be considered as one and the same. +3.2.4 Relation between degree and radian Since a circle subtends at the centre +an angle whose radian measure is 2π and its degree measure is 360°, it follows that +2π radian = 360° + or +π radian = 180° +The above relation enables us to express a radian measure in terms of degree +measure and a degree measure in terms of radian measure. Using approximate value +of π as 22 +7 , we have +1 radian = 180 +π +° = 57° 16′ approximately. +Also +1° = π +180 radian = 0.01746 radian approximately. +The relation between degree measures and radian measure of some common angles +are given in the following table: +A +O +1 +P +1 +2 +−1 +−2 +Q +0 +Fig 3.5 +Degree +30° +45° +60° +90° +180° +270° +360° +Radian +π +6 +π +4 +π +3 +π +2 +π +3π +2 +2π +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 47 +Notational Convention +Since angles are measured either in degrees or in radians, we adopt the convention +that whenever we write angle θ°, we mean the angle whose degree measure is θ and +whenever we write angle β, we mean the angle whose radian measure is β. +Note that when an angle is expressed in radians, the word ‘radian’ is frequently +omitted. Thus, +π +π +180 and +45 +4 += +° += +° are written with the understanding that π and π +4 +are radian measures. Thus, we can say that +Radian measure = π +180 × Degree measure +Degree measure = 180 +π ×Radian measure +Example 1 Convert 40° 20′ into radian measure. +Solution We know that 180° = π radian. +Hence +40° 20′ = 40 1 +3 degree = π +180 × 121 +3 radian = 121π +540 radian. +Therefore +40° 20′ = 121π +540 radian. +Example 2 Convert 6 radians into degree measure. +Solution We know that π radian = 180°. +Hence + 6 radians = 180 +π ×6 degree = 1080 +7 +22 +× +degree += 343 7 +11degree += 343° + 7 +60 +11 +× + minute [as 1° = 60′] += 343° + 38′ + 2 +11 minute +[as 1′ = 60″] += 343° + 38′ + 10.9″ += 343°38′ 11″ approximately. +Hence + 6 radians = 343° 38′ 11″ approximately. +Example 3 Find the radius of the circle in which a central angle of 60° intercepts an +arc of length 37.4 cm (use +22 +π +7 += +). +Reprint 2025-26 + +48 +MATHEMATICS +Solution Here l = 37.4 cm and θ = 60° = 60π +π +radian = +180 +3 +Hence, +by r = θ +l , we have +r = 37.4×3 +37.4×3×7 += +π +22 + = 35.7 cm +Example 4 The minute hand of a watch is 1.5 cm long. How far does its tip move in +40 minutes? (Use π = 3.14). +Solution In 60 minutes, the minute hand of a watch completes one revolution. Therefore, +in 40 minutes, the minute hand turns through 2 +3 of a revolution. Therefore, +2 +θ = +× 360° +3 +or 4π +3 radian. Hence, the required distance travelled is given by + l = r θ = 1.5 × 4π +3 +cm = 2π cm = 2 ×3.14 cm = 6.28 cm. +Example 5 If the arcs of the same lengths in two circles subtend angles 65°and 110° +at the centre, find the ratio of their radii. +Solution Let r1 and r2 be the radii of the two circles. Given that +θ1 = 65° = π +65 +180 × + = 13π +36 radian +and +θ2 = 110° = π +110 +180 × + = 22π +36 radian +Let l be the length of each of the arc. Then l = r1θ1 = r2θ2, which gives +13π +36 ×r1 = 22π +36 × r2 , i.e., +1 +2 +r +r = 22 +13 +Hence + r1 : r2 = 22 : 13. +EXERCISE 3.1 +1. +Find the radian measures corresponding to the following degree measures: +(i) 25° +(ii) – 47°30′ +(iii) 240° + (iv) 520° +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 49 +2. +Find the degree measures corresponding to the following radian measures +(Use +22 +π +7 += +). +(i) +11 +16 +(ii) +– 4 +(iii) +5π +3 +(iv) +7π +6 +3. +A wheel makes 360 revolutions in one minute. Through how many radians does +it turn in one second? +4. +Find the degree measure of the angle subtended at the centre of a circle of +radius 100 cm by an arc of length 22 cm (Use +22 +π +7 += +). +5. +In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of +minor arc of the chord. +6. +If in two circles, arcs of the same length subtend angles 60° and 75° at the +centre, find the ratio of their radii. +7. +Find the angle in radian through which a pendulum swings if its length is 75 cm +and th e tip describes an arc of length +(i) +10 cm +(ii) +15 cm +(iii) +21 cm +3.3 Trigonometric Functions +In earlier classes, we have studied trigonometric ratios for acute angles as the ratio of +sides of a right angled triangle. We will now extend the definition of trigonometric +ratios to any angle in terms of radian measure and study them as trigonometric functions. +Consider a unit circle with centre +at origin of the coordinate axes. Let +P (a, b) be any point on the circle with +angle AOP = x radian, i.e., length of arc +AP = x (Fig 3.6). +We define cos x = a and sin x = b +Since ∆OMP is a right triangle, we have +OM2 + MP2 = OP2 or a2 + b2 = 1 +Thus, for every point on the unit circle, +we have +a2 + b2 = 1 or cos2 x + sin2 x = 1 +Since one complete revolution +subtends an angle of 2π radian at the +centre of the circle, ∠AOB = π +2 , +Fig 3.6 +Reprint 2025-26 + +50 +MATHEMATICS +∠AOC = π and ∠AOD = 3π +2 . All angles which are integral multiples of π +2 are called +quadrantal angles. The coordinates of the points A, B, C and D are, respectively, +(1, 0), (0, 1), (–1, 0) and (0, –1). Therefore, for quadrantal angles, we have +cos 0° = 1 +sin 0° = 0, +cos π +2 = 0 +sin π +2 = 1 +cosπ = − 1 +sinπ = 0 +cos 3π +2 += 0 +sin 3π +2 += –1 +cos 2π = 1 +sin 2π = 0 +Now, if we take one complete revolution from the point P, we again come back to +same point P. Thus, we also observe that if x increases (or decreases) by any integral +multiple of 2π, the values of sine and cosine functions do not change. Thus, +sin (2nπ + x) = sinx, n ∈ Z , cos (2nπ + x) = cosx, n ∈ Z +Further, sin x = 0, if x = 0, ± π, ± 2π , ± 3π, ..., i.e., when x is an integral multiple of π +and cos x = 0, if x = ± π +2 , ± 3π +2 , ± 5π +2 , ... i.e., cos x vanishes when x is an odd +multiple of π +2 . Thus +sin x = 0 implies x = nπ, +π, +π, +π, +π, where n is any integer +cos x = 0 implies x = (2n + 1) π +2 , where n is any integer. +We now define other trigonometric functions in terms of sine and cosine functions: +cosec x = +1 +sin x , x ≠ nπ, where n is any integer. +sec x = +1 +cosx , x ≠ (2n + 1) π +2 , where n is any integer. +tan x = sin +cos +x +x , x ≠ (2n +1) π +2 , where n is any integer. +cot x = cos +sin +x +x , x ≠ n π, where n is any integer. +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 51 +not +defined +not +defined +We have shown that for all real x, sin2 x + cos2 x = 1 +It follows that +1 + tan2 x = sec2 x +(why?) +1 + cot2 x = cosec2 x +(why?) +In earlier classes, we have discussed the values of trigonometric ratios for 0°, +30°, 45°, 60° and 90°. The values of trigonometric functions for these angles are same +as that of trigonometric ratios studied in earlier classes. Thus, we have the following +table: +0° +π +6 +π +4 +π +3 +π +2 +π +3π +2 +2π +sin +0 +1 +2 +1 +2 +3 +2 +1 + 0 +– 1 + 0 +cos +1 +3 +2 +1 +2 + 1 +2 +0 +– 1 + 0 + 1 +tan +0 +1 +3 + 1 +3 + 0 + 0 +The values of cosec x, sec x and cot x +are the reciprocal of the values of sin x, +cos x and tan x, respectively. +3.3.1 Sign of trigonometric functions +Let P (a, b) be a point on the unit circle +with centre at the origin such that +∠AOP = x. If ∠AOQ = – x, then the +coordinates of the point Q will be (a, –b) +(Fig 3.7). Therefore +cos (– x) = cos x +and +sin (– x) = – sin x +Since for every point P (a, b) on +the unit circle, – 1 ≤ a ≤ 1 and +Fig 3.7 +Reprint 2025-26 + +52 +MATHEMATICS +– 1 ≤ b ≤ 1, we have – 1 ≤ cos x ≤ 1 and –1 ≤ sin x ≤ 1 for all x. We have learnt in +previous classes that in the first quadrant (0 < x < π +2 ) a and b are both positive, in the +second quadrant ( π +2 < x <π) a is negative and b is positive, in the third quadrant +(π < x < 3π +2 ) a and b are both negative and in the fourth quadrant ( 3π +2 < x < 2π) a is +positive and b is negative. Therefore, sin x is positive for 0 < x < π, and negative for +π < x < 2π. Similarly, cos x is positive for 0 < x < π +2 , negative for π +2 < x < 3π +2 and also +positive for 3π +2 < x < 2π. Likewise, we can find the signs of other trigonometric +functions in different quadrants. In fact, we have the following table. +I +II +III +IV +sin x ++ ++ + – + – +cos x ++ + – + – + + +tan x ++ + – + + + – +cosec x ++ ++ + – + – +sec x ++ + – + – + + +cot x ++ + – + + + – +3.3.2 Domain and range of trigonometric functions From the definition of sine +and cosine functions, we observe that they are defined for all real numbers. Further, +we observe that for each real number x, + – 1 ≤ sin x ≤ 1 and – 1 ≤ cos x ≤ 1 +Thus, domain of y = sin x and y = cos x is the set of all real numbers and range +is the interval [–1, 1], i.e., – 1 ≤ y ≤ 1. +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 53 +Since cosec x = +1 +sin x , the domain of y = cosec x is the set { x : x ∈ R and +x ≠ n π, n ∈ Z} and range is the set {y : y ∈ R, y ≥ 1 or y ≤ – 1}. Similarly, the domain +of y = sec x is the set {x : x ∈ R and x ≠ (2n + 1) π +2 , n ∈ Z} and range is the set +{y : y ∈ R, y ≤ – 1or y ≥ 1}. The domain of y = tan x is the set {x : x ∈ R and +x ≠ (2n + 1) π +2 , n ∈ Z} and range is the set of all real numbers. The domain of +y = cot x is the set {x : x ∈ R and x ≠ n π, n ∈ Z} and the range is the set of all real +numbers. +We further observe that in the first quadrant, as x increases from 0 to π +2 , sin x +increases from 0 to 1, as x increases from π +2 to π, sin x decreases from 1 to 0. In the +third quadrant, as x increases from π to 3π +2 , sin x decreases from 0 to –1and finally, in +the fourth quadrant, sin x increases from –1 to 0 as x increases from 3π +2 to 2π. +Similarly, we can discuss the behaviour of other trigonometric functions. In fact, we +have the following table: +Remark In the above table, the statement tan x increases from 0 to ∞ (infinity) for +0 < x < π +2 simply means that tan x increases as x increases for 0 < x < π +2 and +I quadrant +II quadrant +III quadrant +IV quadrant +sin +increases from 0 to 1 +decreases from 1 to 0 +decreases from 0 to –1 increases from –1 to 0 +cos +decreases from 1 to 0 +decreases from 0 to – 1 increases from –1 to 0 +increases from 0 to 1 +tan +increases from 0 to ∞increases from –∞to 0 +increases from 0 to ∞ +increases from –∞to 0 +cot +decreases from ∞ to 0 decreases from 0 to–∞decreases from ∞ to 0 +decreases from 0to –∞ +sec +increases from 1 to ∞increases from –∞to–1 decreases from –1to–∞decreases from ∞ to 1 +cosec decreases from ∞ to 1 increases from 1 to ∞ +increases from –∞to–1 decreases from–1to–∞ +Reprint 2025-26 + +54 +MATHEMATICS +Fig 3.10 +Fig 3.11 +Fig 3.8 +Fig 3.9 +assumes arbitraily large positive values as x approaches to π +2 . Similarly, to say that +cosec x decreases from –1 to – ∞ (minus infinity) in the fourth quadrant means that +cosec x decreases for x ∈ ( 3π +2 , 2π) and assumes arbitrarily large negative values as +x approaches to 2π. The symbols ∞ and – ∞ simply specify certain types of behaviour +of functions and variables. +We have already seen that values of sin x and cos x repeats after an interval of +2π. Hence, values of cosec x and sec x will also repeat after an interval of 2π. We +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 55 +shall see in the next section that tan (π + x) = tan x. Hence, values of tan x will repeat +after an interval of π. Since cot x is reciprocal of tan x, its values will also repeat after +an interval of π. Using this knowledge and behaviour of trigonometic functions, we can +sketch the graph of these functions. The graph of these functions are given above: +Example 6 If cos x = – 3 +5 +, x lies in the third quadrant, find the values of other five +trigonometric functions. +Solution Since cos x = +3 +5 +− + , we have sec x = +5 +3 +− +Now +sin2 x + cos2 x = 1, i.e., sin2 x = 1 – cos2 x +or +sin2 x = 1 – 9 +25 = 16 +25 +Hence +sin x = ± 4 +5 +Since x lies in third quadrant, sin x is negative. Therefore +sin x = – 4 +5 +which also gives +cosec x = – 5 +4 +Fig 3.12 +Fig 3.13 +Reprint 2025-26 + +56 +MATHEMATICS +Further, we have +tan x = sin +cos +x +x = 4 +3 and +cot x = cos +sin +x +x = 3 +4 . +Example 7 If cot x = – 5 +12 , x lies in second quadrant, find the values of other five +trigonometric functions. +Solution Since cot x = – 5 +12 , we have tan x = – 12 +5 +Now +sec2 x = 1 + tan2 x = 1 + 144 +25 = 169 +25 +Hence +sec x = ± 13 +5 +Since x lies in second quadrant, sec x will be negative. Therefore +sec x = – 13 +5 , +which also gives + +5 +cos +13 +x = − +Further, we have +sin x = tan x cos x = (– 12 +5 ) ×(– 5 +13 ) = 12 +13 +and +cosec x = +1 +sin x = 13 +12 . +Example 8 Find the value of sin 31π +3 . +Solution We know that values of sin x repeats after an interval of 2π. Therefore +sin 31π +3 = sin (10π + π +3 ) = sin π +3 = +3 +2 . +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 57 +Example 9 Find the value of cos (–1710°). +Solution We know that values of cos x repeats after an interval of 2π or 360°. +Therefore, +cos (–1710°) = cos (–1710° + 5 × 360°) + = cos (–1710° + 1800°) = cos 90° = 0. +EXERCISE 3.2 +Find the values of other five trigonometric functions in Exercises 1 to 5. +1. +cos x = – 1 +2 , x lies in third quadrant. +2. +sin x = 3 +5, x lies in second quadrant. +3. +cot x = 4 +3 , x lies in third quadrant. +4. +sec x = 13 +5 , x lies in fourth quadrant. +5. +tan x = – 5 +12 , x lies in second quadrant. +Find the values of the trigonometric functions in Exercises 6 to 10. +6. +sin 765° +7. +cosec (– 1410°) +8. +tan 19π +3 +9. +sin (– 11π +3 ) +10. +cot (– 15π +4 ) +3.4 Trigonometric Functions of Sum and Difference of Two Angles +In this Section, we shall derive expressions for trigonometric functions of the sum and +difference of two numbers (angles) and related expressions. The basic results in this +connection are called trigonometric identities. We have seen that +1. +sin (– x) = – sin x +2. +cos (– x) = cos x +We shall now prove some more results: +Reprint 2025-26 + +58 +MATHEMATICS +3. +cos (x + y) = cos x cos y – sin x sin y +Consider the unit circle with centre at the origin. Letx be the angle P4OP1and y be +the angle P1OP2. Then (x + y) is the angle P4OP2. Also let (– y) be the angle P4OP3. +Therefore, P1, P2, P3 and P4 will have the coordinates P1(cos x, sin x), +P2 [cos (x + y), sin (x + y)], P3 [cos (– y), sin (– y)] and P4 (1, 0) (Fig 3.14). +Consider the triangles P1OP3 and P2OP4. They are congruent (Why?). Therefore, +P1P3 and P2P4 are equal. By using distance formula, we get +P1P3 +2 += [cos x – cos (– y)]2 + [sin x – sin(–y]2 += (cos x – cos y)2 + (sin x + sin y)2 += cos2 x + cos2 y – 2 cos x cos y + sin2 x + sin2 y + 2sin x sin y += 2 – 2 (cos x cos y – sin x sin y) +(Why?) +Also, +P2P4 +2 += [1 – cos (x + y)] 2 + [0 – sin (x + y)]2 += 1 – 2cos (x + y) + cos2 (x + y) + sin2 (x + y) += 2 – 2 cos (x + y) +Fig 3.14 +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 59 +Since +P1P3 += P2P4, we have P1P3 +2 = P2P4 +2. +Therefore, 2 –2 (cos x cos y – sin x sin y) = 2 – 2 cos (x + y). +Hence cos (x + y) = cos x cos y – sin x sin y +4. +cos (x – y) = cos x cos y + sin x sin y +Replacing y by – y in identity 3, we get +cos (x + (– y)) = cos x cos (– y) – sin x sin (– y) +or +cos (x – y) = cos x cos y + sin x sin y +5. +cos ( +x +π – +2 +) = sin x +If we replace x by π +2 and y by x in Identity (4), we get +cos ( π +2 +x +− +) = cos π +2 cos x + sin π +2 sin x = sin x. +6. +sin ( +x +π – +2 +) = cos x +Using the Identity 5, we have +sin ( π +2 +x +− +) = cos +π +π +2 +2 +x + + + + +− +− + + + + + + + + = cos x. +7. +sin (x + y) = sin x cos y + cos x sin y +We know that + sin (x + y) = cos +π +( +) +2 +x +y + + +− ++ + + + + = cos +π +( +) +2 +x +y + + +− +− + + + + + = cos ( π +2 +x +− +) cos y + sin π +( +) +2 +x +− +sin y + = sin x cos y + cos x sin y +8. +sin (x – y) = sin x cos y – cos x sin y +If we replace y by –y, in the Identity 7, we get the result. +9. +By taking suitable values of x and y in the identities 3, 4, 7 and 8, we get the +following results: +cos +x +π +( ++ +) +2 + = – sin x +sin +x +π +( ++ +) +2 + = cos x +cos (π – x) = – cos x +sin (π – x) = sin x +Reprint 2025-26 + +60 +MATHEMATICS +cos (π + x) = – cos x +sin (π + x) = – sin x +cos (2π – x) = cos x +sin (2π – x) = – sin x +Similar results for tan x, cot x, sec x and cosec x can be obtained from the results of sin +x and cos x. +10. +If none of the angles x, y and (x + y) is an odd multiple of π +2 , then + tan (x + y) = +x +y +x +y +tan ++ tan +1 – tan +tan +Since none of the x, y and (x + y) is an odd multiple of π +2 , it follows that cos x, +cos y and cos (x + y) are non-zero. Now + tan (x + y) = sin( +) +cos( +) +x +y +x +y ++ ++ + = sin cos +cos sin +cos cos +sin sin +x +y +x +y +x +y +x +y ++ +− +. +Dividing numerator and denominator by cos x cos y, we have + tan (x + y) = +y +x +y +x +y +x +y +x +y +x +y +x +y +x +y +x +cos +cos +sin +sin +cos +cos +cos +cos +cos +cos +sin +cos +cos +cos +cos +sin +− ++ += +tan +tan +1 – tan +tan +x +y +x +y ++ +11. + tan ( x – y) = +x +y +x +y +tan +– tan +1+ tan +tan +If we replace y by – y in Identity 10, we get + tan (x – y) = tan [x + (– y)] += tan +tan( +) +1 tan tan( +) +x +y +x +y ++ +− +− +− + = tan +tan +1 tan tan +x +y +x +y +− ++ +12. +If none of the angles x, y and (x + y) is a multiple of π, then + cot ( x + y) = +x +y +y +x +cot +cot +– 1 +cot ++cot +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 61 +Since, none of the x, y and (x + y) is multiple of π, we find that sin x sin y and +sin (x + y) are non-zero. Now, +cot ( x + y)= cos ( +) +cos +cos +– sin +sin +sin ( +) +sin +cos +cos +sin +x +y +x +y +x +y +x +y +x +y +x +y ++ += ++ ++ +Dividing numerator and denominator by sin x sin y, we have +cot (x + y) = cot +cot +–1 +cot +cot +x +y +y +x ++ +13. +cot (x – y)= +x +y +y +x +cot +cot ++1 +cot +– cot + if none of angles x, y and x–y is a multiple of π +If we replace y by –y in identity 12, we get the result +14. +cos 2x = cos2x – sin2 x = 2 cos2 x – 1 = 1 – 2 sin2 x = +x +x +2 +2 +1 – tan +1 + tan +We know that +cos (x + y) = cos x cos y – sin x sin y +Replacing y by x, we get + cos 2x = cos2x – sin2 x += cos2 x – (1 – cos2 x) = 2 cos2x – 1 +Again, +cos 2x = cos2 x – sin2 x += 1 – sin2 x – sin2 x = 1 – 2 sin2 x. +We have +cos 2x = cos2 x – sin 2 x = +2 +2 +2 +2 +cos +sin +cos +sin +x +x +x +x +− ++ +Dividing numerator and denominator by cos2 x, we get +cos 2x = +2 +2 +1– tan +1+ tan +x +x , +π +π +2 +≠ ++ +x +n +, where n is an integer +15. +sin 2x = 2 sinx cos x = +x +x +2 +2tan +1 + tan + +π +π +2 +≠ ++ +x +n +, where n is an integer +We have +sin (x + y) = sin x cos y + cos x sin y +Replacing y by x, we get sin 2x = 2 sin x cos x. +Again +sin 2x = +2 +2 +2sin cos +cos +sin +x +x +x +x ++ +Reprint 2025-26 + +62 +MATHEMATICS +Dividing each term by cos2 x, we get +sin 2x = +2 +2tan +1 tan +x +x ++ +16. +tan 2x = +x +x +2 +2tan +1 – tan + if +π +2 +π +2 +≠ ++ +x +n +, where n is an integer +We know that +tan (x + y) = +tan +tan +1 +tan +tan +x +y +– +x +y ++ +Replacing y by x , we get +2 +2 tan +tan 2 +1 tan +x +x +x += +− +17. +sin 3x = 3 sin x – 4 sin3 x +We have, +sin 3x = sin (2x + x) += sin 2x cos x + cos 2x sin x += 2 sin x cos x cos x + (1 – 2sin2 x) sin x += 2 sin x (1 – sin2 x) + sin x – 2 sin3 x += 2 sin x – 2 sin3 x + sin x – 2 sin3 x += 3 sin x – 4 sin3 x +18. +cos 3x= 4 cos3 x – 3 cos x +We have, + cos 3x = cos (2x +x) += cos 2x cos x – sin 2x sin x += (2cos2 x – 1) cos x – 2sin x cos x sin x += (2cos2 x – 1) cos x – 2cos x (1 – cos2 x) += 2cos3 x – cos x – 2cos x + 2 cos3 x += 4cos3 x – 3cos x. +19. += +x +x +x +x +3 +2 +3 tan +–tan +tan3 +1– 3tan + if +π +3 +π +2 +≠ ++ +x +n +, where n is an integer +We have +tan 3x =tan (2x + x) += tan 2 +tan +1 +tan 2 tan +x +x +– +x +x ++ +2 +2 +2tan +tan +1 +tan +2tan +tan +1 +1 +tan +x +x +– +x +x . +x +– +– +x ++ += +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 63 +3 +3 +2 +2 +2 +2tan +tan +tan +3 tan +tan +1 +tan +2tan +1 +3tan +x +x – +x +x – +x +– +x – +x +– +x ++ += += +20. +(i) +cos x + cos y = +x +y +x +y ++ +– +2cos +cos +2 +2 +(ii) +cos x – cos y = – +x +y +x +y ++ +– +2sin +sin +2 +2 +(iii) +sin x + sin y = +x +y +x +y ++ +– +2sin +cos +2 +2 +(iv) +sin x – sin y = +x +y +x +y ++ +– +2cos +sin +2 +2 +We know that +cos (x + y) = cos x cos y – sin x sin y +... (1) +and +cos (x – y) = cos x cos y + sin x sin y +... (2) +Adding and subtracting (1) and (2), we get +cos (x + y) + cos(x – y) = 2 cos x cos y +... (3) +and +cos (x + y) – cos (x – y) = – 2 sin x sin y +... (4) +Further +sin (x + y) = sin x cos y + cos x sin y +... (5) +and +sin (x – y) = sin x cos y – cos x sin y +... (6) +Adding and subtracting (5) and (6), we get +sin (x + y) + sin (x – y) = 2 sin x cos y +... (7) +sin (x + y) – sin (x – y) = 2cos x sin y +... (8) +Let x + y = θ and x – y = φ. Therefore +θ +θ +and +2 +2 +x +y ++φ +−φ + + + + += += + + + + + + + + +Substituting the values of x and y in (3), (4), (7) and (8), we get +cos θ + cos φ = 2 cos +θ +θ +cos +2 +2 ++φ +−φ + + + + + + + + + + + + +cos θ – cos φ = – 2 sin θ +θ +sin +2 +– ++ φ +φ + + + + + + + + +2 + + + + +sin θ + sin φ = 2 sin +θ +θ +cos +2 +2 ++φ +−φ + + + + + + + + + + + + +Reprint 2025-26 + +64 +MATHEMATICS +sin θ – sin φ = 2 cos +θ +θ +sin +2 +2 ++φ +−φ + + + + + + + + + + + + +Since θ and φ can take any real values, we can replace θ by x and φ by y. +Thus, we get +cos x + cos y = 2 cos +cos +2 +2 +x +y +x +y ++ +− +; cos x – cos y = – 2 sin +sin +2 +2 +x +y +x +y ++ +− +, +sin x + sin y = 2 sin +cos +2 +2 +x +y +x +y ++ +− +; sin x – sin y = 2 cos +sin +2 +2 +x +y +x +y ++ +− +. +Remark As a part of identities given in 20, we can prove the following results: +21. +(i) +2 cos x cos y = cos (x + y) + cos (x – y) +(ii) +–2 sin x sin y = cos (x + y) – cos (x – y) +(iii) +2 sin x cos y = sin (x + y) + sin (x – y) +(iv) +2 cos x sin y = sin (x + y) – sin (x – y). +Example 10 Prove that +5 +3sin +sec +4sin +cot +1 +6 +3 +6 +4 +π +π +π +π +− += +Solution We have +L.H.S. = +5 +3sin +sec +4sin +cot +6 +3 +6 +4 +π +π +π +π +− += 3 × 1 +2 × 2 – 4 sin +6 +π + + +π− + + + +× 1 = 3 – 4 sin 6 +π += 3 – 4 × 1 +2 = 1 = R.H.S. +Example 11 Find the value of sin 15°. +Solution We have +sin 15° += sin (45° – 30°) += sin 45° cos 30° – cos 45° sin 30° += +1 +3 +1 +1 +3 +1 +2 +2 +2 +2 +2 2 +– +× +− +× += +. +Example 12 Find the value of tan 13 +12 +π . +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 65 +Solution We have +tan 13 +12 +π = tan +12 +π + + +π + + + + + = tan +tan +12 +4 +6 +π +π +π + + += +− + + + + += +tan +tan +4 +6 +1 +tan +tan +4 +6 +π +π +− +π +π ++ += +1 +1 +3 +1 +3 +2 +3 +1 +3 +1 +1 +3 +− +− += += +− ++ ++ +Example 13 Prove that +sin( +) +tan +tan +sin( +) +tan +tan +x +y +x +y +x +y +x +y ++ ++ += +− +− +. +Solution We have + L.H.S. +sin( +) +sin cos +cos sin +sin( +) +sin cos +cos sin +x +y +x +y +x +y +x +y +x +y +x +y ++ ++ += += +− +− +Dividing the numerator and denominator by cos x cos y, we get +sin( +) +tan +tan +sin( +) +tan +tan +x +y +x +y +x +y +x +y ++ ++ += +− +− +. +Example 14 Show that + tan 3 x tan 2 x tan x = tan 3x – tan 2 x – tan x +Solution We know that 3x = 2x + x +Therefore, +tan 3x = tan (2x + x) +or +tan2 +tan +tan3 +1– tan 2 tan +x +x +x +x +x ++ += +or +tan 3x – tan 3x tan 2x tan x = tan 2x + tan x +or +tan 3x – tan 2x – tan x = tan 3x tan 2x tan x +or +tan 3x tan 2x tan x = tan 3x – tan 2x – tan x. +Example 15 Prove that +cos +cos +2 cos +4 +4 +x +x +x +π +π + + + + ++ ++ +− += + + + + + + + + +Solution Using the Identity 20(i), we have +Reprint 2025-26 + +66 +MATHEMATICS + L.H.S. +cos +cos +4 +4 +x +x +π +π + + + + += ++ ++ +− + + + + + + + + +( +) +4 +4 +4 +4 +2cos +cos +2 +2 +x +x +x – +x +π +π +π +π + + + + ++ ++ +− ++ +− + + + + += + + + + + + + + + + + + + + + + += 2 cos 4 +π cos x = 2 × +1 +2 cos x = +2 cos x = R.H.S. +Example 16 Prove that +cos 7 +cos 5 +cot +sin 7 – sin 5 +x +x +x +x +x ++ += +Solution Using the Identities 20 (i) and 20 (iv), we get + L.H.S. = +7 +5 +7 +5 +2cos +cos +2 +2 +7 +5 +7 +5 +2cos +sin +2 +2 +x +x +x +x +x +x +x +x ++ +− ++ +− + = +cos +sin +cot +x +x +x += + = R.H.S. +Example 17 Prove that +sin5 +2sin3 +sin +tan +cos5 +cos +x +x +x +x +x +x +− ++ += += +− +Solution We have +L.H.S. +sin5 +2sin3 +sin +cos5 +cos +x +x +x +x +x +− ++ += +− + +sin5 +sin +2sin3 +cos5 +cos +x +x +x +x +x ++ +− += +− + +2sin3 +cos2 +2sin3 +– 2sin3 sin 2 +x +x +x +x +x +− += +sin3 +(cos2 +1) +sin3 sin 2 +x +x +– +x +x +− += + +2 +1 cos2 +2sin +sin 2 +2sin cos +x +x +x +x +x +− += += + = tan x = R.H.S. +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 67 +EXERCISE 3.3 +Prove that: +1. +sin2 π +6 + cos2 3 +π – tan2 +1 +– +4 +2 +π = +2. 2sin2 +6 +π + cosec2 +2 +7 +3 +cos +6 +3 +2 +π +π = +3. +2 +2 +5 +cot +cosec +3tan +6 +6 +6 +6 +π +π +π ++ ++ += +4. +2 +2 +2 +3 +2sin +2cos +2sec +10 +4 +4 +3 +π +π +π ++ ++ += +5. +Find the value of: +(i) sin 75° +(ii) tan 15° +Prove the following: +6. +cos +cos +sin +sin +sin( +) +4 +4 +4 +4 +x +y +x +y +x +y +π +π +π +π + + + + + + + + +− +− +− +− +− += ++ + + + + + + + + + + + + + + + + +7. +2 +π +tan +1 +tan +4 +π +1 +tan +tan +4 +x +x +x +x + + ++ + + + + ++ + +=  + +− + + + + +− + + + + +8. +2 +cos ( +) cos ( +) +cot +sin ( +) cos +2 +x +x +x +x +x +π + +− += +π + + +π − ++ + + + + +9. +3π +3π +cos +cos (2π +) +cot +cot (2π +) +1 +2 +2 +x +x +x +x + + + + + + ++ ++ +− ++ ++ += + + + + + + + + + + + + +10. +sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x +11. +3 +3 +cos +cos +2sin +4 +4 +x +x +x +π +π + + + + ++ +− +− += − + + + + + + + + +12. +sin2 6x – sin2 4x = sin 2x sin 10x +13. cos2 2x – cos2 6x = sin 4x sin 8x +14. +sin2 x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x +15. +cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x) +16. +cos +cos +sin +sin +sin +cos +9 +5 +17 +3 +2 +10 +x +x +x +x +x +x +− +− += − +17. +sin +sin +cos +cos +tan +5 +3 +5 +3 +4 +x +x +x +x +x ++ ++ += +18. +sin +sin +cos +cos +tan +x +y +x +y +x +y +− ++ += +− +2 +19. +sin +sin +cos +cos +tan +x +x +x +x +x ++ ++ += +3 +3 +2 +20. +sin +sin +sin +cos +sin +x +x +x +x +x +− +− += +3 +2 +2 +2 +21. +cos +cos +cos +sin +sin +sin +cot +4 +3 +2 +4 +3 +2 +3 +x +x +x +x +x +x +x ++ ++ ++ ++ += +Reprint 2025-26 + +68 +MATHEMATICS +22. +cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1 +23. +2 +2 +4 +4tan +(1 +tan +) +tan 4 +1 +6 tan +tan +x +x +x +x +x +− += +− ++ +24. +cos 4x = 1 – 8sin2 x cos2 x +25. +cos 6x = 32 cos6 x – 48cos4 x + 18 cos2 x – 1 +Miscellaneous Examples +Example 18 If sin x = +3 +5 , cos y = −12 +13 , where x and y both lie in second quadrant, +find the value of sin (x + y). +Solution We know that +sin (x + y) = sin x cos y + cos x sin y +... (1) +Now +cos2 x = 1 – sin2 x = 1 – +9 +25 = +16 +25 +Therefore +cos x = ± 4 +5 . +Since x lies in second quadrant, cos x is negative. +Hence +cos x = −4 +5 +Now +sin2y = 1 – cos2y = 1 – +144 +169 +25 +169 += +i.e. +sin y = ± 5 +13. +Since y lies in second quadrant, hence sin y is positive. Therefore, sin y = +5 +13. Substituting +the values of sin x, sin y, cos x and cos y in (1), we get +sin( +) +x +y ++ += +× − + + + ++ − + + + +× +3 +5 +12 +13 +4 +5 +5 +13 += +36 +20 +56 +65 +65 +65 +− +− += − +. +Example 19 Prove that +9 +5 +cos 2 cos +cos 3 cos +sin 5 sin +2 +2 +2 +x +x +x +x +x +x +− += +. +Solution We have +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 69 +L.H.S. += +1 +9 +2cos 2 cos +2cos +cos 3 +2 +2 +2 +x +x +x +x + + +− + + + + += +1 +9 +9 +cos 2 +cos 2 +cos +3 +cos +3 +2 +2 +2 +2 +2 +x +x +x +x +x +x +x +x + + + + + + + + + + ++ ++ +− +− ++ +− +− + + + + + + + + + + + + + + + + + + + + += +1 +2 +5 +2 +3 +2 +15 +2 +3 +2 +cos +cos +cos +cos +x +x +x +x ++ +− +− + + + + = +1 +2 +5 +2 +15 +2 +cos +cos +x +x +− + + + + += +5 +15 +5 +15 +1 +2 +2 +2 +2 +2sin +sin +2 +2 +2 +x +x +x +x + + + + + + ++ +− + + + + + + +− + + + + + + + + + + + + + + + + + + + + += − +− + + + += +sin +sin +sin +sin +5 +5 +2 +5 +5 +2 +x +x +x +x + = R.H.S. +Example 20 Find the value of tan π +8 . +Solution Let +π +8 +x = +. Then +π +2 +4 +x = +. +Now +tan +tan +tan +2 +2 +1 +2 +x +x +x += +− +or +2 +π +2tan +π +8 +tan +π +4 +1 +tan 8 += +− +Let y = tan π +8 . Then 1 = +2 +1 +2 +y +y +− +or + y2 + 2y – 1 = 0 +Therefore + y = −± += −± +2 +2 2 +2 +1 +2 +Reprint 2025-26 + +70 +MATHEMATICS +Since +π +8 lies in the first quadrant, y = tan π +8 is positve. Hence +π +tan +2 +1 +8 = +−. +Example 21 If +3 +3π +tan += +, π < +< +4 +2 +x +x +, find the value of sin +x +2 , cos +x +2 and tan +x +2 . +Solution Since +3π +π +2 +x +< +< +, cosx is negative. +Also +π +3π +2 +2 +4 +x +< +< +. +Therefore, sin +x +2 is positive and cos +x +2 is negative. +Now +sec2 x = 1 + tan2 x = 1 +9 +16 +25 +16 ++ += +Therefore +cos2 x = +16 +25 or cos x = +4 +5 +– + (Why?) +Now +2 +2 +2 +sin x + = 1 – cos x = 1 +4 +5 +9 +5 ++ += +. +Therefore +sin2 +x +2 = +9 +10 +or +sin +x +2 = +3 +10 +(Why?) +Again +2cos2 +x +2 = 1+ cos x = 1 +4 +5 +1 +5 +− += +Therefore +cos2 +x +2 = +1 +10 +or +cos +x +2 = − +1 +10 (Why?) +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 71 +Hence +tan +x +2 = +sin +cos +x +x +2 +2 +3 +10 +10 +1 += +× − + + + + + + = – 3. +Example 22 Prove that cos2 x + cos2 +2 +π +π +3 +cos +3 +3 +2 +x +x + + + + ++ ++ +− += + + + + + + + + +. +Solution We have + L.H.S. = +2π +2π +1 +cos 2 +1 +cos 2 +1 +cos 2 +3 +3 +2 +2 +2 +x +x +x + + + + ++ ++ ++ +− + + + + ++ + + + + ++ ++ +. += +1 +2π +2π +3 +cos 2 +cos 2 +cos 2 +2 +3 +3 +x +x +x + + + + + + ++ ++ ++ ++ +− + + + + + + + + + + + + += +1 +2π +3 +cos 2 +2cos 2 +cos +2 +3 +x +x + + ++ ++ + + + + += +1 +π +3 +cos 2 +2cos 2 cos π +2 +3 +x +x + + + + ++ ++ +− + + + + + + + + += +1 +π +3 +cos 2 +2cos 2 cos +2 +3 +x +x + + ++ +− + + + + += +[ +] +1 +3 +3 +cos 2 +cos 2 +2 +2 +x +x ++ +− += + = R.H.S. +Miscellaneous Exercise on Chapter 3 +Prove that: +1. +0 +13 +π +5 +cos +13 +π +3 +cos +13 +π +9 +cos +13 +π +cos +2 += ++ ++ +2. +(sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0 +3. +(cos x + cos y)2 + (sin x – sin y)2 = 4 cos2 +2 +x +y ++ +Reprint 2025-26 + +72 +MATHEMATICS +4. +(cosx – cos y)2 + (sinx – sin y)2 = 4 sin2 +2 +x +y +− +5. +sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x +6 . +6 . +6 . +6 . +6 . +(sin7 +sin5 ) +(sin 9 +sin3 ) +tan 6 +(cos7 +cos5 ) +(cos 9 +cos3 ) +x +x +x +x +x +x +x +x +x ++ ++ ++ += ++ ++ ++ +7. +sin 3x + sin 2x – sin x = 4sin x cos +x +2 cos +3 +2 +x +Find sin +x +2 , cos +x +2 and tan +x +2 in each of the following : +8. +tanx = −4 +3 , x in quadrant II +9. +cos x = −1 +3, x in quadrant III +10. +sinx = 4 +1 , x in quadrant II +Summary +®If in a circle of radius r, an arc of length l subtends an angle of θ radians, then +l = r θ +®Radian measure = π +180 × Degree measure +®Degree measure = 180 +π × Radian measure +®cos2 x + sin2 x = 1 +®1 + tan2 x = sec2 x +®1 + cot2 x = cosec2 x +®cos (2nπ + x) = cos x +®sin (2nπ + x) = sin x +®sin (– x) = – sin x +®cos (– x) = cos x +®cos (x + y) = cos x cos y – sin x sin y +®cos (x – y) = cos x cos y + sin x sin y +®cos ( π +2 +x +− +) = sin x +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 73 +®sin ( π +2 +x +− +) = cos x +®sin (x + y) = sin x cos y + cos x sin y +®sin (x – y) = sin x cos y – cos x sin y +®cos +π + +2 +x + + + + + += – sin x +sin +π + +2 +x + + + + + + = cos x +cos (π – x) = – cos x +sin (π – x) = sin x +cos (π + x) = – cos x +sin (π + x) = – sin x +cos (2π – x) = cos x +sin (2π – x) = – sin x +®If none of the angles x, y and (x ± y) is an odd multiple of π +2 , then +tan (x + y) = tan +tan +tan +tan +x +y +x +y ++ +− +1 +®tan (x – y) = +tan +tan +tan +tan +x +y +x +y +− ++ +1 +®If none of the angles x, y and (x ± y) is a multiple of π, then +cot (x + y) = +cot cot +1 +cot +cot +x +y +y +x +− ++ +®cot (x – y) = +cot +cot +1 +cot +cot +x +y +y +x +− ++ +®cos 2x = cos2 x – sin2 x = 2cos2 x – 1 = 1 – 2 sin2 x +2 +2 +1 +tan +1 +tan +– +x +x += ++ +®sin 2x = 2 sinx cos x +2 +2 tan +1 +tan +x +x += ++ +®tan 2x = +2 +2tan +1 +tan +x +x +− +®sin 3x = 3sinx – 4sin3x +®cos 3x = 4cos3x – 3cosx +Reprint 2025-26 + +74 +MATHEMATICS +®tan 3x = +3 +2 +3tan +tan +1 3tan +x +x +x +− +− +® (i) +cos x + cos y = 2cos +cos +2 +2 +x +y +x +y ++ +− +(ii) +cos x – cos y = – 2sin +sin +2 +2 +x +y +x +y ++ +− +(iii) +sin x + sin y = 2 sin +cos +2 +2 +x +y +x +y ++ +− +(iv) +sin x – sin y = 2cos +sin +2 +2 +x +y +x +y ++ +− +® (i) +2cos x cos y = cos (x + y) + cos ( x – y) +(ii) +– 2sin x sin y = cos (x + y) – cos (x – y) +(iii) +2sin x cos y = sin (x + y) + sin (x – y) +(iv) +2 cos x sin y = sin (x + y) – sin (x – y). +Historical Note +The study of trigonometry was first started in India. The ancient Indian +Mathematicians, Aryabhatta (476), Brahmagupta (598), Bhaskara I (600) and +Bhaskara II (1114) got important results. All this knowledge first went from +India to middle-east and from there to Europe. The Greeks had also started the +study of trigonometry but their approach was so clumsy that when the Indian +approach became known, it was immediately adopted throughout the world. +In India, the predecessor of the modern trigonometric functions, known as +the sine of an angle, and the introduction of the sine function represents the main +contribution of the siddhantas (Sanskrit astronomical works) to the history of +mathematics. +Bhaskara I (about 600) gave formulae to find the values of sine functions +for angles more than 90°. A sixteenth century Malayalam work Yuktibhasa +(period) contains a proof for the expansion of sin (A + B). Exact expression for +sines or cosines of 18°, 36°, 54°, 72°, etc., are given by +Bhaskara II. +Reprint 2025-26 + +TRIGONOMETRIC FUNCTIONS 75 +— v — +The symbols sin–1 x, cos–1 x, etc., for arc sin x, arc cos x, etc., were +suggested by the astronomer Sir John F.W. Hersehel (1813) The names of Thales +(about 600 B.C.) is invariably associated with height and distance problems. He +is credited with the determination of the height of a great pyramid in Egypt by +measuring shadows of the pyramid and an auxiliary staff (or gnomon) of known +height, and comparing the ratios: +H +S +h +s += + = tan (sun’s altitude) +Thales is also said to have calculated the distance of a ship at sea through +the proportionality of sides of similar triangles. Problems on height and distance +using the similarity property are also found in ancient Indian works. +Reprint 2025-26" +class_11,4,Complex Numbers and Quadratic Equations,ncert_books/class_11/kemh1dd/kemh104.pdf,"76 MATHEMATICS +Chapter +COMPLEX NUMBERS AND +QUADRATIC EQUATIONS +W. R. Hamilton +(1805-1865) +vMathematics is the Queen of Sciences and Arithmetic is the Queen of +Mathematics. – GAUSS v +4.1 Introduction +In earlier classes, we have studied linear equations in one +and two variables and quadratic equations in one variable. +We have seen that the equation x2 + 1 = 0 has no real +solution as x2 + 1 = 0 gives x2 = – 1 and square of every +real number is non-negative. So, we need to extend the +real number system to a larger system so that we can +find the solution of the equation x2 = – 1. In fact, the main +objective is to solve the equation ax2 + bx + c = 0, where +D = b2 – 4ac < 0, which is not possible in the system of +real numbers. +4.2 Complex Numbers +Let us denote +1 +− by the symbol i. Then, we have +2 +1 +i = −. This means that i is a +solution of the equation x2 + 1 = 0. +A number of the form a + ib, where a and b are real numbers, is defined to be a +complex number. For example, 2 + i3, (– 1) + +3 +i +, +1 +4 +11 +i − + + ++  + + + are complex numbers. +For the complex number z = a + ib, a is called the real part, denoted by Re z and +b is called the imaginary part denoted by Im z of the complex number z. For example, +if z = 2 + i5, then Re z = 2 and Im z = 5. +Two complex numbers z1 = a + ib and z2 = c + id are equal if a = c and b = d. +4 +Reprint 2025-26 + +COMPLEX NUMBERS AND QUADRATIC EQUATIONS 77 +Example 1 If 4x + i(3x – y) = 3 + i (– 6), where x and y are real numbers, then find +the values of x and y. +Solution We have +4x + i (3x – y) = 3 + i (–6) +... (1) +Equating the real and the imaginary parts of (1), we get +4x = 3, 3x – y = – 6, +which, on solving simultaneously, give +3 +4 +x = + and +33 +4 +y = +. +4.3 Algebra of Complex Numbers +In this Section, we shall develop the algebra of complex numbers. +4.3.1 Addition of two complex numbers Let z1 = a + ib and z2 = c + id be any two +complex numbers. Then, the sum z1 + z2 is defined as follows: +z1 + z2 = (a + c) + i (b + d), which is again a complex number. +For example, (2 + i3) + (– 6 +i5) = (2 – 6) + i (3 + 5) = – 4 + i 8 +The addition of complex numbers satisfy the following properties: +(i) +The closure law The sum of two complex numbers is a complex +number, i.e., z1 + z2 is a complex number for all complex numbers +z1 and z2. +(ii) +The commutative law For any two complex numbers z1 and z2, +z1 + z2 = z2 + + z1 +(iii) +The associative law For any three complex numbers z1, z2, z3, +(z1 + z2) + z3 = z1 + (z2 + z3). +(iv) +The existence of additive identity There exists the complex number +0 + i 0 (denoted as 0), called the additive identity or the zero complex +number, such that, for every complex number z, z + 0 = z. +(v) +The existence of additive inverse To every complex number +z = a + ib, we have the complex number – a + i(– b) (denoted as – z), +called the additive inverse or negative of z. We observe that z + (–z) = 0 +(the additive identity). +4.3.2 Difference of two complex numbers Given any two complex numbers z1 and +z2, the difference z1 – z2 is defined as follows: +z1 – z2 = z1 + (– z2). +For example, +(6 + 3i) – (2 – i) = (6 + 3i) + (– 2 + i ) = 4 + 4i +and +(2 – i) – (6 + 3i) = (2 – i) + ( – 6 – 3i) = – 4 – 4i +Reprint 2025-26 + +78 MATHEMATICS +4.3.3 Multiplication of two complex numbers Let z1 = a + ib and z2 = c + id be any +two complex numbers. Then, the product z1 z2 is defined as follows: +z1 z2 = (ac – bd) + i(ad + bc) +For example, (3 + i5) (2 + i6) = (3 × 2 – 5 × 6) + i(3 × 6 + 5 × 2) = – 24 + i28 +The multiplication of complex numbers possesses the following properties, which +we state without proofs. +(i) +The closure law The product of two complex numbers is a complex number, +the product z1 z2 is a complex number for all complex numbers z1 and z2. +(ii) +The commutative law For any two complex numbers z1 and z2, +z1 z2 = z2 z1 +. +(iii) +The associative law For any three complex numbers z1, z2, z3, +(z1 z2) z3 = z1 (z2 z3). +(iv) +The existence of multiplicative identity There exists the complex number +1 + i 0 (denoted as 1), called the multiplicative identity such that z.1 = z, +for every complex number z. +(v) +The existence of multiplicative inverse For every non-zero complex +number z = a + ib or a + bi(a ≠ 0, b ≠ 0), we have the complex number +2 +2 +2 +2 +a +–b +i +a +b +a +b ++ ++ ++ + (denoted by 1 +z or z–1 ), called the multiplicative inverse +of z such that +1 +1 +z.z = (the multiplicative identity). +(vi) +The distributive law For any three complex numbers z1, z2, z3, +(a) z1 (z2 + z3) = z1 z2 + z1 z3 +(b) (z1 + z2) z3 = z1 z3 + z2 z3 +4.3.4 Division of two complex numbers Given any two complex numbers z1 and z2, +where +2 +0 +z ≠ +, the quotient +1 +2 +z +z is defined by +1 +1 +2 +2 +1 +z +z +z +z += +For example, let +z1 = 6 + 3i and z2 = 2 – i +Then +1 +2 +1 +(6 +3 ) +2 +z +i +z +i + + += ++ +× + + +− + + = ( +) +6 +3i ++ + +( +) +( +) +( +) +2 +2 +2 +2 +1 +2 +2 +1 +2 +1 +i + + +−− + + ++ + + ++ − ++ − + + +Reprint 2025-26 + +COMPLEX NUMBERS AND QUADRATIC EQUATIONS 79 += ( +) 2 +6 +3 +5 +i +i ++ + + ++ + + + + = +( +) +( +) +1 +1 +12 +3 +6 +6 +9 12 +5 +5 +i +i + +−+ ++ += ++ + + +4.3.5 Power of i we know that +( +) +3 +2 +1 +i +i i +i +i += += − += −, +( ) +( +) +2 +2 +4 +2 +1 +1 +i +i += += − += +( ) +( +) +2 +2 +5 +2 +1 +i +i +i +i +i += += − += , + +( ) +( +) +3 +3 +6 +2 +1 +1 +i +i += += − += −, etc. +Also, we have +1 +2 +2 +1 +1 +1 +, +1, +1 +1 +i +i +i +i +i +i +i +i +− +− += +× += += − += += += − +− +− + +3 +4 +3 +4 +1 +1 +1 +1 +, +1 +1 +1 +i +i +i +i +i +i +i +i +i +− +− += += +× += += += += += +− +In general, for any integer k, i4k = 1, i4k + 1 = i, i4k + 2 = –1, i4k + 3 = – i +4.3.6 The square roots of a negative real number +Note that i2 = –1 and ( – i)2 = i2 = – 1 +Therefore, the square roots of – 1 are i, – i. However, by the symbol +1 +−, we would +mean i only. +Now, we can see that i and –i both are the solutions of the equation x2 + 1 = 0 or +x2 = –1. +Similarly +( +) +( +) +2 +2 +3 +3 +i += + i2 = 3 (– 1) = – 3 +( +) +2 +3i +− + = ( +) +2 +3 +− + i2 = – 3 +Therefore, the square roots of –3 are +3 i and +3i +− +. +Again, the symbol +3 +− is meant to represent 3i only, i.e., +3 +− = +3i . +Generally, if a is a positive real number, +a +− + = +1 +a +− = +a i , +We already know that a +b +× + = +ab for all positive real number a and b. This +result also holds true when either a > 0, b < 0 or a < 0, b > 0. What if a < 0, b < 0? +Let us examine. +Note that +Reprint 2025-26 + +80 MATHEMATICS +( +) ( +) +2 +1 +1 +1 +1 +i = +− +−= +− +− + (by assuming +a +b +× + = +ab for all real numbers) += 1 = 1, which is a contradiction to the fact that += − +2 +1 +i +. +Therefore, +a +b +ab +× +≠ + if both a and b are negative real numbers. +Further, if any of a and b is zero, then, clearly, +a +b +ab +× += += 0. +4.3.7 Identities We prove the following identity +( +) +2 +2 +2 +1 +2 +1 +2 +1 2 +2 +z +z +z +z +z z ++ += ++ ++ +, for all complex numbers z1 and z2. +Proof We have, +(z1 + z2)2 = (z1 + z2) (z1 + z2), += (z1 + z2) z1 + (z1 + z2) z2 +(Distributive law) += +2 +2 +1 +2 1 +1 2 +2 +z +z z +z z +z ++ ++ ++ +(Distributive law) + = +2 +2 +1 +1 2 +1 2 +2 +z +z z +z z +z ++ ++ ++ +(Commutative law of multiplication) + = +2 +2 +1 +1 2 +2 +2 +z +z z +z ++ ++ +Similarly, we can prove the following identities: +(i) +( +) +2 +2 +2 +1 +2 +1 +1 +2 +2 +2 +z +z +z +z z +z +− += +− ++ +(ii) +( +) +3 +3 +2 +2 +3 +1 +2 +1 +1 +2 +1 2 +2 +3 +3 +z +z +z +z z +z z +z ++ += ++ ++ ++ +(iii) +( +) +3 +3 +2 +2 +3 +1 +2 +1 +1 +2 +1 2 +2 +3 +3 +z +z +z +z z +z z +z +− += +− ++ +− +(iv) +( +) ( +) +2 +2 +1 +2 +1 +2 +1 +2 +z – z +z +z +z – z += ++ +In fact, many other identities which are true for all real numbers, can be proved +to be true for all complex numbers. +Example 2 Express the following in the form of a + bi: +(i) +( +) +1 +5 +8 +i +i + + +− + + + + +(ii) +( +) ( +) +2 +i +i +− + +3 +1 +8 i + + +− + + + + +Solution +(i) +( +) +1 +5 +8 +i +i + + +− + + + + = +2 +5 +8 i +− + = +( +) +5 +1 +8 +− +− + = 5 +8 = 5 +0 +8 +i ++ +(ii) +( +) ( +) +3 +1 +2 +8 +i +i +i + + +− +− + + + + = +5 +1 +2 +8 8 8 +i +× +× +× × + = +( ) +2 +2 +1 +256 i + +1 +256 +i +i += +. +Reprint 2025-26 + +COMPLEX NUMBERS AND QUADRATIC EQUATIONS 81 +Example 3 Express (5 – 3i)3 in the form a + ib. +Solution We have, (5 – 3i)3 = 53 – 3 × 52 × (3i) + 3 × 5 (3i)2 – (3i)3 += 125 – 225i – 135 + 27i = – 10 – 198i. +Example 4 Express ( +)( +) +3 +2 +2 3 +i +− ++ +− +− +in the form of a + ib +Solution We have, ( +) ( +) +3 +2 +2 3 +i +− ++ +− +− + = ( +) ( +) +3 +2 +2 3 +i +i +− ++ +− += +2 +6 +3 +2 6 +2 +i +i +i +−+ ++ +− + = ( +) +( +) +6 +2 +3 1 +2 2 i +−+ ++ ++ +4.4 The Modulus and the Conjugate of a Complex Number +Let z = a + ib be a complex number. Then, the modulus of z, denoted by | z |, is defined +to be the non-negative real number +2 +2 +a +b ++ +, i.e., | z | = +2 +2 +a +b ++ + and the conjugate +of z, denoted as z , is the complex number a – ib, i.e., z = a – ib. +For example, +2 +2 +3 +3 +1 +10 +i ++ += ++ += +, +2 +2 +2 +5 +2 +( +5) +29 +i +− += ++ − += +, +and +3 +3 +i +i ++ = +−, 2 +5 +2 +5 +i +i +− += ++ +, 3 +5 +i +− +− + = 3i – 5 +Observe that the multiplicative inverse of the non-zero complex number z is +given by +z–1 = +1 +a +ib ++ + = +2 +2 +2 +2 +a +b +i +a +b +a +b +− ++ ++ ++ + = +2 +2 +a +ib +a +b +− ++ + = +2 +z +z +or + z +2 +z +z += +Furthermore, the following results can easily be derived. +For any two compex numbers z1 and z2 , we have +(i) +1 +2 +1 +2 +z z +z +z += +(ii) +1 +1 +2 +2 +z +z +z +z += + provided +2 +0 +z +≠ +(iii) +1 2 +1 +2 +z z +z z += +(iv) +1 +2 +1 +2 +z +z +z +z +± += +± +(v) +1 +1 +2 +2 +z +z +z +z + += + + + + + provided z2 ≠ 0. +Reprint 2025-26 + +82 MATHEMATICS +Example 5 Find the multiplicative inverse of 2 – 3i. +Solution +Let z = 2 – 3i +Then +z = 2 + 3i and +2 +2 +2 +2 +( +3) +13 +z += ++ − += +Therefore, the multiplicative inverse of 2 +3i +− +is given by +z–1 +2 +2 +3 +2 +3 +13 +13 +13 +z +i +i +z ++ += += += ++ +The above working can be reproduced in the following manner also, +z–1 = +1 +2 +3 +2 +3 +(2 +3 )(2 +3 ) +i +i +i +i ++ += +− +− ++ += +2 +2 +2 +3 +2 +3 +2 +3 +13 +13 +13 +2 +(3 ) +i +i +i +i ++ ++ += += ++ +− +Example 6 Express the following in the form a + ib +(i) +5 +2 +1 +2 +i +i ++ +− +(ii) i–35 +Solution (i) We have, +5 +2 +5 +2 +1 +2 +1 +2 +1 +2 +1 +2 +i +i +i +i +i +i ++ ++ ++ += +× +− +− ++ + +( +) +2 +5 +5 2 +2 +2 +1 +2 +i +i +i ++ ++ +− += +− += 3 +6 2 +3(1 +2 2 ) +1 +2 +3 +i +i ++ ++ += ++ + = 1 +2 2i ++ +. +(ii) +( ) +35 +35 +17 +2 +1 +1 +1 +i +i +i +i +i +i +i +− += += += +× +− + = +2 +i +i +i += +− +EXERCISE 4.1 +Express each of the complex number given in the Exercises 1 to 10 in the +form a + ib. +1. +( ) +3 +5 +5 +i +i + + +− + + + + +2. +i +i +9 +19 ++ +3. +i −39 +Reprint 2025-26 + +COMPLEX NUMBERS AND QUADRATIC EQUATIONS 83 +Fig 4.1 +4. +3(7 + i7) + i (7 + i7) +5. +(1 – i) – ( –1 + i6) +6. + +1 +2 +5 +4 +5 +5 +2 +i +i + + + + ++ +− ++ + + + + + + + + +7. +1 +7 +1 +4 +4 +3 +3 +3 +3 +i +i +i + + + + + + + + ++ ++ ++ +−− ++ + + + + + + + + + + + + + + + + +8. +(1 – i)4 +9. +3 +1 +3 +3 +i + + ++ + + + + +10. +3 +1 +2 +3i + + +−− + + + + +Find the multiplicative inverse of each of the complex numbers given in the +Exercises 11 to 13. +11. +4 – 3i +12. +5 +3i ++ +13. +– i +14. +Express the following expression in the form of a + ib : +( +) ( +) +( +) ( +) +3 +5 +3 +5 +3 +2 +3 +2 +i +i +i +i ++ +− ++ +− +− +4.5 Argand Plane and Polar Representation +We already know that corresponding to +each ordered pair of real numbers +(x, y), we get a unique point in the XY- +plane and vice-versa with reference to a +set of mutually perpendicular lines known +as the x-axis and the y-axis. The complex +number x + iy which corresponds to the +ordered pair (x, y) can be represented +geometrically as the unique point P(x, y) +in the XY-plane and vice-versa. +Some complex numbers such as +2 + 4i, – 2 + 3i, 0 + 1i, 2 + 0i, – 5 –2i and +1 – 2i which correspond to the ordered +pairs (2, 4), ( – 2, 3), (0, 1), (2, 0), ( –5, –2), and (1, – 2), respectively, have been +represented geometrically by the points A, B, C, D, E, and F, respectively in +the Fig 4.1. +The plane having a complex number assigned to each of its point is called the +complex plane or the Argand plane. +Reprint 2025-26 + +84 MATHEMATICS +Obviously, in the Argand plane, the modulus of the complex number +x + iy = +2 +2 +x +y ++ + is the distance between the point P(x, y) and the origin O (0, 0) +(Fig 4.2). The points on the x-axis corresponds to the complex numbers of the form +a + i 0 and the points on the y-axis corresponds to the complex numbers of the form +Fig 4.2 +Fig 4.3 +0 + i b. The x-axis and y-axis in the Argand plane are called, respectively, the real axis +and the imaginary axis. +The representation of a complex number z = x + iy and its conjugate +z = x – iy in the Argand plane are, respectively, the points P (x, y) and Q (x, – y). +Geometrically, the point (x, – y) is the mirror image of the point (x, y) on the real +axis (Fig 4.3). +Reprint 2025-26 + +COMPLEX NUMBERS AND QUADRATIC EQUATIONS 85 +Miscellaneous Examples +Example 7 Find the conjugate of +(3 2 )(2 +3 ) +(1 +2 )(2 +) +i +i +i +i +− ++ ++ +− +. +Solution We have , +(3 2 )(2 +3 ) +(1 +2 )(2 +) +i +i +i +i +− ++ ++ +− += 6 +9 +4 +6 +2 +4 +2 +i +i +i +i ++ +− ++ +−+ ++ + = 12 5 +4 +3 +4 +3 +4 +3 +i +i +i +i ++ +− +× ++ +− += 48 36 +20 +15 +63 16 +16 9 +25 +i +i +i +− ++ ++ +− += ++ + = 63 +16 +25 +25i +− +Therefore, conjugate of +(3 2 )(2 +3 ) +63 +16 +is +(1 +2 )(2 +) +25 +25 +i +i +i +i +i +− ++ ++ ++ +− +. +Example 8 If x + iy = +a +ib +a ib ++ +− +, prove that x2 + y2 = 1. +Solution We have, +x + iy = +( +)( +) +( +)( +) +a +ib +a +ib +a +ib +a +ib ++ ++ +− ++ + = +2 +2 +2 +2 +2 +a +b +abi +a +b +− ++ ++ + = +2 +2 +2 +2 +2 +2 +2 +a +b +ab +i +a +b +a +b +− ++ ++ ++ +So that, x – iy = +2 +2 +2 +2 +2 +2 +2 +a +b +ab i +a +b +a +b +− +− ++ ++ +Therefore, +x2 + y2 = (x + iy) (x – iy) = +2 +2 2 +2 +2 +2 +2 2 +2 +2 2 +( +) +4 +( +) +( +) +a +b +a b +a +b +a +b +− ++ ++ ++ + = +2 +2 2 +2 +2 2 +( +) +( +) +a +b +a +b ++ ++ + = 1 +Miscellaneous Exercise on Chapter 4 +1. +Evaluate: +3 +25 +18 +1 +i +i + + + ++ + + + + + + + + +. +2. +For any two complex numbers z1 and z2, prove that +Re (z1 z2) = Re z1 Re z2 – Imz1 Imz2 +. +Reprint 2025-26 + +86 MATHEMATICS +3. +Reduce +1 +2 +3 +4 +1 +4 +1 +5 +i +i +i +i +− + + + +− + + + +− ++ ++ + + + to the standard form . +4. +If +a +ib +x +iy +c +id +− +− += +− + prove that ( +) +2 +2 +2 +2 +2 +2 +2 +a +b +x +y +c +d ++ ++ += ++ +. +5. +If z1 = 2 – i, z2 = 1 + i, find +1 +2 +1 +2 +1 +– +1 +z +z +z +z ++ ++ ++ +. +6. +If a + ib = +2 +2 +( +) +2 +1 +x +i +x ++ ++ +, prove that a2 + b2 = ( +) +2 +2 +2 +2 +( +1) +2 +1 +x +x ++ ++ +. +7. +Let z1 = 2 – i, z2 = –2 + i. Find +(i) +1 2 +1 +Re z z +z + + + + + +, +(ii) +1 1 +1 +Im z z + + + + + + +. +8. +Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i. +9. +Find the modulus of +1 +1 +1 +1 +i +i +i +i ++ +− +− +− ++ . +10. +If (x + iy)3 = u + iv, then show that +2 +2 +4( +– +) +u +v +x +y +x +y ++ += +. +11. +If α and β are different complex numbers with β +1 += , then find +β +α +1 +αβ +– +– +. +12. +Find the number of non-zero integral solutions of the equation 1 +2 +x +x +– i += +. +13. +If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that +(a2 + b2) (c2 + d2) (e2 + f 2) (g2 + h2) = A2 + B2 +14. +If +1 +1 +1 +m +i +– i + + ++ += + + + + +, then find the least positive integral value of m. +Reprint 2025-26 + +COMPLEX NUMBERS AND QUADRATIC EQUATIONS 87 +Summary +®A number of the form a + ib, where a and b are real numbers, is called a +complex number, a is called the real part and b is called the imaginary part +of the complex number. +®Let z1 = a + ib and z2 = c + id. Then +(i) +z1 + z2 = (a + c) + i (b + d) +(ii) +z1 z2 = (ac – bd) + i (ad + bc) +®For any non-zero complex number z = a + ib (a ≠ 0, b ≠ 0), there exists the +complex number +2 +2 +2 +2 +a +b +i +a +b +a +b +− ++ ++ ++ +, denoted by 1 +z or z–1, called the +multiplicative inverse of z such that (a + ib) +2 +2 +2 +2 +− + + ++ + + + + ++ ++ +a +b +i +a +b +a +b += 1 + i0 +=1 +®For any integer k, i4k = 1, i4k + 1 = i, i4k + 2 = – 1, i4k + 3 = – i +®The conjugate of the complex number z = a + ib, denoted by z , is given by +z = a – ib. +Historical Note +The fact that square root of a negative number does not exist in the real number +system was recognised by the Greeks. But the credit goes to the Indian +mathematician Mahavira (850) who first stated this difficulty clearly. “He +mentions in his work ‘Ganitasara Sangraha’ as in the nature of things a negative +(quantity) is not a square (quantity)’, it has, therefore, no square root”. +Bhaskara, another Indian mathematician, also writes in his work Bijaganita, +written in 1150. “There is no square root of a negative quantity, for it is not a +square.” Cardan (1545) considered the problem of solving +x + y = 10, xy = 40. +Reprint 2025-26 + +88 MATHEMATICS +— v — +He obtained x = 5 + +15 +− + and y = 5 – +15 +− + as the solution of it, which +was discarded by him by saying that these numbers are ‘useless’. Albert +Girard (about 1625) accepted square root of negative numbers and said that +this will enable us to get as many roots as the degree of the polynomial equation. +Euler was the first to introduce the symbol i for +1 +− and W.R. Hamilton +(about 1830) regarded the complex number a + ib as an ordered pair of real +numbers (a, b) thus giving it a purely mathematical definition and avoiding use +of the so called ‘imaginary numbers’. +Reprint 2025-26" +class_11,5,Linear Inequalities,ncert_books/class_11/kemh1dd/kemh105.pdf,"5 +Chapter +vMathematics is the art of saying many things in many +different ways. – MAXWELLv +5.1 Introduction +In earlier classes, we have studied equations in one variable and two variables and also +solved some statement problems by translating them in the form of equations. Now a +natural question arises: ‘Is it always possible to translate a statement problem in the +form of an equation? For example, the height of all the students in your class is less +than 160 cm. Your classroom can occupy atmost 60 tables or chairs or both. Here we +get certain statements involving a sign ‘<’ (less than), ‘>’ (greater than), ‘≤’ (less than +or equal) and ≥ (greater than or equal) which are known as inequalities. +In this Chapter, we will study linear inequalities in one and two variables. The +study of inequalities is very useful in solving problems in the field of science, mathematics, +statistics, economics, psychology, etc. +5.2 Inequalities +Let us consider the following situations: +(i) Ravi goes to market with ` 200 to buy rice, which is available in packets of 1kg. The +price of one packet of rice is ` 30. If x denotes the number of packets of rice, which he +buys, then the total amount spent by him is ` 30x. Since, he has to buy rice in packets +only, he may not be able to spend the entire amount of ` 200. (Why?) Hence +30x < 200 +... (1) +Clearly the statement (i) is not an equation as it does not involve the sign of equality. +(ii) Reshma has ` 120 and wants to buy some registers and pens. The cost of one +register is ` 40 and that of a pen is ` 20. In this case, if x denotes the number of +registers and y, the number of pens which Reshma buys, then the total amount spent by +her is ` (40x + 20y) and we have +40x + 20y ≤ 120 +... (2) +LINEAR INEQUALITIES +Reprint 2025-26 + +90 MATHEMATICS +Since in this case the total amount spent may be upto ` 120. Note that the statement (2) +consists of two statements +40x + 20y < 120 +... (3) +and +40x + 20y = 120 +... (4) +Statement (3) is not an equation, i.e., it is an inequality while statement (4) is an equation. +Definition 1 Two real numbers or two algebraic expressions related by the symbol +‘<’, ‘>’, ‘≤’ or ‘≥’ form an inequality. +Statements such as (1), (2) and (3) above are inequalities. +3 < 5; 7 > 5 are the examples of numerical inequalities while + x < 5; y > 2; x ≥ 3, y ≤ 4 are some examples of literal inequalities. +3 < 5 < 7 (read as 5 is greater than 3 and less than 7), 3 < x < 5 (read as x is greater +than or equal to 3 and less than 5) and 2 < y < 4 are the examples of double inequalities. +Some more examples of inequalities are: +ax + b < 0 +... (5) +ax + b > 0 +... (6) +ax + b ≤ 0 +... (7) +ax + b ≥ 0 +... (8) +ax + by < c +... (9) +ax + by > c +... (10) +ax + by ≤ c +... (11) +ax + by ≥ c +... (12) +ax2 + bx + c ≤ 0 +... (13) +ax2 + bx + c > 0 +... (14) +Inequalities (5), (6), (9), (10) and (14) are strict inequalities while inequalities (7), (8), +(11), (12), and (13) are slack inequalities. Inequalities from (5) to (8) are linear +inequalities in one variable x when a ≠ 0, while inequalities from (9) to (12) are linear +inequalities in two variables x and y when a ≠ 0, b ≠ 0. +Inequalities (13) and (14) are not linear (in fact, these are quadratic inequalities +in one variable x when a ≠ 0). +In this Chapter, we shall confine ourselves to the study of linear inequalities in one +and two variables only. +Reprint 2025-26 + + LINEAR INEQUALITIES 91 +5.3 Algebraic Solutions of Linear Inequalities in One Variable and their +Graphical Representation +Let us consider the inequality (1) of Section 6.2, viz, 30x < 200 +Note that here x denotes the number of packets of rice. +Obviously, x cannot be a negative integer or a fraction. Left hand side (L.H.S.) of this +inequality is 30x and right hand side (RHS) is 200. Therefore, we have +For x = 0, L.H.S. = 30 (0) = 0 < 200 (R.H.S.), which is true. +For x = 1, L.H.S. = 30 (1) = 30 < 200 (R.H.S.), which is true. +For x = 2, L.H.S. = 30 (2) = 60 < 200, which is true. +For x = 3, L.H.S. = 30 (3) = 90 < 200, which is true. +For x = 4, L.H.S. = 30 (4) = 120 < 200, which is true. +For x = 5, L.H.S. = 30 (5) = 150 < 200, which is true. +For x = 6, L.H.S. = 30 (6) = 180 < 200, which is true. +For x = 7, L.H.S. = 30 (7) = 210 < 200, which is false. +In the above situation, we find that the values of x, which makes the above +inequality a true statement, are 0,1,2,3,4,5,6. These values of x, which make above +inequality a true statement, are called solutions of inequality and the set {0,1,2,3,4,5,6} +is called its solution set. +Thus, any solution of an inequality in one variable is a value of the variable +which makes it a true statement. +We have found the solutions of the above inequality by trial and error method +which is not very efficient. Obviously, this method is time consuming and sometimes +not feasible. We must have some better or systematic techniques for solving inequalities. +Before that we should go through some more properties of numerical inequalities and +follow them as rules while solving the inequalities. +You will recall that while solving linear equations, we followed the following rules: +Rule 1 Equal numbers may be added to (or subtracted from) both sides of an equation. +Rule 2 Both sides of an equation may be multiplied (or divided) by the same non-zero +number. +In the case of solving inequalities, we again follow the same rules except with a +difference that in Rule 2, the sign of inequality is reversed (i.e., ‘<‘ becomes ‘>’, ≤’ +becomes ‘≥’ and so on) whenever we multiply (or divide) both sides of an inequality by +a negative number. It is evident from the facts that +3 > 2 while – 3 < – 2, +– 8 < – 7 while (– 8) (– 2) > (– 7) (– 2) , i.e., 16 > 14. +Reprint 2025-26 + +92 MATHEMATICS +Thus, we state the following rules for solving an inequality: +Rule 1 Equal numbers may be added to (or subtracted from) both sides of an inequality +without affecting the sign of inequality. +Rule 2 Both sides of an inequality can be multiplied (or divided) by the same positive +number. But when both sides are multiplied or divided by a negative number, then the +sign of inequality is reversed. +Now, let us consider some examples. +Example 1 Solve 30 x < 200 when +(i) x is a natural number, +(ii) x is an integer. +Solution We are given 30 x < 200 +or +30 +200 +30 +30 +x < +(Rule 2), i.e., x < 20 / 3. +(i) +When x is a natural number, in this case the following values of x make the +statement true. + 1, 2, 3, 4, 5, 6. +The solution set of the inequality is {1,2,3,4,5,6}. +(ii) +When x is an integer, the solutions of the given inequality are +..., – 3, –2, –1, 0, 1, 2, 3, 4, 5, 6 +The solution set of the inequality is {...,–3, –2,–1, 0, 1, 2, 3, 4, 5, 6} +Example 2 Solve 5x – 3 < 3x +1 when +(i) +x is an integer, +(ii) +x is a real number. +Solution We have, 5x –3 < 3x + 1 +or +5x –3 + 3 < 3x +1 +3 +(Rule 1) +or +5x < 3x +4 +or +5x – 3x < 3x + 4 – 3x +(Rule 1) +or +2x < 4 +or +x < 2 +(Rule 2) +(i) +When x is an integer, the solutions of the given inequality are +..., – 4, – 3, – 2, – 1, 0, 1 +(ii) +When x is a real number, the solutions of the inequality are given by x < 2, +i.e., all real numbers x which are less than 2. Therefore, the solution set of +the inequality is x ∈ (– ∞, 2). +We have considered solutions of inequalities in the set of natural numbers, set of +integers and in the set of real numbers. Henceforth, unless stated otherwise, we shall +solve the inequalities in this Chapter in the set of real numbers. +Reprint 2025-26 + + LINEAR INEQUALITIES 93 +Example 3 Solve 4x + 3 < 6x +7. +Solution We have, + 4x + 3 < 6x + 7 +or +4x – 6x < 6x + 4 – 6x +or +– 2x < 4 +or x > – 2 +i.e., all the real numbers which are greater than –2, are the solutions of the given +inequality. Hence, the solution set is (–2, ∞). +Example 4 Solve 5 +2 +5 +3 +6 +– +x +x – +≤ +. +Solution We have + 5 +2 +5 +3 +6 +– x +x – +≤ +or +2 (5 – 2x) ≤ x – 30. +or +10 – 4x ≤ x – 30 +or +– 5x ≤ – 40, i.e., x ≥ 8 +Thus, all real numbers x which are greater than or equal to 8 are the solutions of the +given inequality, i.e., x ∈ [8, ∞). +Example 5 Solve 7x + 3 < 5x + 9. Show the graph of the solutions on number line. +Solution We have 7x + 3 < 5x + 9 or + 2x < 6 or x < 3 +The graphical representation of the solutions are given in Fig 5.1. +Fig 5.1 +Example 6 Solve 3 +4 +1 1 +2 +4 +x +x +− ++ +≥ +−. Show the graph of the solutions on number line. +Solution We have +3 +4 +1 1 +2 +4 +x +x +− ++ +≥ +− +or +3 +4 +3 +2 +4 +x +x +− +− +≥ +or +2 (3x – 4) ≥ (x – 3) +Reprint 2025-26 + +94 MATHEMATICS +or +6x – 8 ≥ x – 3 +or +5x ≥ 5 or x ≥ 1 +The graphical representation of solutions is given in Fig 5.2. +Fig 5.2 +Example 7 The marks obtained by a student of Class XI in first and second terminal +examination are 62 and 48, respectively. Find the minimum marks he should get in the +annual examination to have an average of at least 60 marks. +Solution Let x be the marks obtained by student in the annual examination. Then +62 48 +60 +3 +x ++ ++ +≥ +or +110 + x ≥ 180 +or +x ≥ 70 +Thus, the student must obtain a minimum of 70 marks to get an average of at least +60 marks. +Example 8 Find all pairs of consecutive odd natural numbers, both of which are larger +than 10, such that their sum is less than 40. +Solution Let x be the smaller of the two consecutive odd natural number, so that the +other one is x +2. Then, we should have +x > 10 +... (1) +and x + ( x + 2) < 40 +... (2) +Solving (2), we get +2x + 2 < 40 +i.e., x < 19 +... (3) +From (1) and (3), we get +10 < x < 19 +Since x is an odd number, x can take the values 11, 13, 15, and 17. So, the required +possible pairs will be +(11, 13), (13, 15), (15, 17), (17, 19) +Reprint 2025-26 + + LINEAR INEQUALITIES 95 +EXERCISE 5.1 +1. +Solve 24x < 100, when +(i) +x is a natural number. +(ii) +x is an integer. +2. +Solve – 12x > 30, when +(i) +x is a natural number. +(ii) +x is an integer. +3. +Solve 5x – 3 < 7, when +(i) +x is an integer. +(ii) +x is a real number. +4. +Solve 3x + 8 >2, when +(i) +x is an integer. +(ii) +x is a real number. +Solve the inequalities in Exercises 5 to 16 for real x. +5. +4x + 3 < 5x + 7 +6. +3x – 7 > 5x – 1 +7. +3(x – 1) ≤ 2 (x – 3) +8. +3 (2 – x) ≥ 2 (1 – x) +9. +11 +2 +3 +x +x +x + ++ +< +10. +1 +3 +2 +x +x +> ++ +11. +3( +2) +5(2 +) +5 +3 +x +x +− +− +≤ +12. +1 3 +1 +4 +( +6) +2 +5 +3 +x +x + + ++ +≥ +− + + + + +13. +2 (2x + 3) – 10 < 6 (x – 2) +14. +37 – (3x + 5) > 9x – 8 (x – 3) +15. +(5 +2) +(7 +3) +4 +3 +5 +x +x +x +− +− +< +− +16. +(2 +1) +(3 +2) +(2 +) +3 +4 +5 +x +x +x +− +− +− +≥ +− +Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each +case on number line +17. +3x – 2 < 2x + 1 +18. +5x – 3 > 3x – 5 +19. +3 (1 – x) < 2 (x + 4) +20. +(5 – 2) +(7 –3) +– +2 +3 +5 +x +x +x +≥ +21. +Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he +should get in the third test to have an average of at least 60 marks. +22. +To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or +more in five examinations (each of 100 marks). If Sunita’s marks in first four +examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain +in fifth examination to get grade ‘A’ in the course. +23. +Find all pairs of consecutive odd positive integers both of which are smaller than +10 such that their sum is more than 11. +24. +Find all pairs of consecutive even positive integers, both of which are larger than +5 such that their sum is less than 23. +Reprint 2025-26 + +96 MATHEMATICS +25. +The longest side of a triangle is 3 times the shortest side and the third side is 2 cm +shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find +the minimum length of the shortest side. +26. +A man wants to cut three lengths from a single piece of board of length 91cm. +The second length is to be 3cm longer than the shortest and the third length is to +be twice as long as the shortest. What are the possible lengths of the shortest +board if the third piece is to be at least 5cm longer than the second? +[Hint: If x is the length of the shortest board, then x , (x + 3) and 2x are the +lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x ≤ 91 and +2x ≥ (x + 3) + 5]. +Miscellaneous Examples +Example 9 Solve – 8 ≤ 5x – 3 < 7. +Solution In this case, we have two inequalities, – 8 ≤ 5x – 3 and 5x – 3 < 7, which we +will solve simultaneously. We have – 8 ≤ 5x –3 < 7 +or +–5 ≤ 5x < 10 +or –1 ≤ x < 2 +Example 10 Solve – 5 ≤ 5 +3 +2 +– x ≤ 8. +Solution We have + – 5 ≤ 5 +3 +2 +– x ≤ 8 +or +–10 ≤ 5 – 3x ≤ 16 + or – 15 ≤ – 3x ≤ 11 +or +5 ≥ x ≥ – 11 +3 +which can be written as –11 +3 ≤ x ≤ 5 +Example 11 Solve the system of inequalities: +3x – 7 < 5 + x +... (1) +11 – 5 x ≤ 1 +... (2) +and represent the solutions on the number line. +Solution From inequality (1), we have +3x – 7 < 5 + x +or +x < 6 +... (3) +Also, from inequality (2), we have +11 – 5 x ≤ 1 +or +– 5 x ≤ – 10 + i.e., x ≥ 2 +... (4) +Reprint 2025-26 + + LINEAR INEQUALITIES 97 +If we draw the graph of inequalities (3) and (4) on the number line, we see that the +values of x, which are common to both, are shown by bold line in Fig 5.3. +Fig 5.3 +Thus, solution of the system are real numbers x lying between 2 and 6 including 2, i.e., +2 ≤ x < 6 +Example 12 In an experiment, a solution of hydrochloric acid is to be kept between +30° and 35° Celsius. What is the range of temperature in degree Fahrenheit if conversion +formula is given by C = 5 +9 (F – 32), where C and F represent temperature in degree +Celsius and degree Fahrenheit, respectively. +Solution It is given that 30 < C < 35. +Putting +C = 5 +9 (F – 32), we get +30 < 5 +9 (F – 32) < 35, +or +9 +5 × (30) < (F – 32) < 9 +5 × (35) +or +54 < (F – 32) < 63 +or +86 < F < 95. +Thus, the required range of temperature is between 86° F and 95° F. +Example 13 A manufacturer has 600 litres of a 12% solution of acid. How many litres +of a 30% acid solution must be added to it so that acid content in the resulting mixture +will be more than 15% but less than 18%? +Solution Let x litres of 30% acid solution is required to be added. Then +Total mixture = (x + 600) litres +Therefore +30% x + 12% of 600 > 15% of (x + 600) +and +30% x + 12% of 600 < 18% of (x + 600) +or +30 +100 +x + 12 +100 (600) > 15 +100 (x + 600) +Reprint 2025-26 + +98 MATHEMATICS +and +30 +100 +x + 12 +100 (600) < 18 +100 (x + 600) +or +30x + 7200 > 15x + 9000 +and +30x + 7200 < 18x + 10800 +or +15x > 1800 and 12x < 3600 +or +x > 120 and x < 300, +i.e. +120 < x < 300 +Thus, the number of litres of the 30% solution of acid will have to be more than +120 litres but less than 300 litres. +Miscellaneous Exercise on Chapter 5 +Solve the inequalities in Exercises 1 to 6. +1. +2 ≤ 3x – 4 ≤ 5 +2. +6 ≤ – 3 (2x – 4) < 12 +3. +7 +3 +4 +18 +2 +x +– +≤ +− +≤ +4. +3 +2 +15 +0 +5 +( x +) +− +− +< +≤ +5. +3 +12 4 +2 +5 +x +− +< − +≤ +− +6. +3 +11 +7 +11 +2 +( x +) ++ +≤ +≤ +. +Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on +number line. +7. +5x + 1 > – 24, 5x – 1 < 24 +8. +2 (x – 1) < x + 5, 3 (x + 2) > 2 – x +9. +3x – 7 > 2 (x – 6) , 6 – x > 11 – 2x +10. +5 (2x – 7) – 3 (2x + 3) ≤ 0 , 2x + 19 ≤ 6x + 47 . +11. +A solution is to be kept between 68° F and 77° F. What is the range in temperature +in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by +F = 9 +5 C + 32 ? +12. +A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to +it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we +have 640 litres of the 8% solution, how many litres of the 2% solution will have to +be added? +Reprint 2025-26 + + LINEAR INEQUALITIES 99 +— v — +13. +How many litres of water will have to be added to 1125 litres of the 45% solution +of acid so that the resulting mixture will contain more than 25% but less than 30% +acid content? +14. +IQ of a person is given by the formula +IQ = +MA +CA × 100, +where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of +12 years old children, find the range of their mental age. +Summary +®Two real numbers or two algebraic expressions related by the symbols <, >, ≤ +or ≥ form an inequality. +®Equal numbers may be added to (or subtracted from ) both sides of an inequality. +®Both sides of an inequality can be multiplied (or divided ) by the same positive +number. But when both sides are multiplied (or divided) by a negative number, +then the inequality is reversed. +®The values of x, which make an inequality a true statement, are called solutions +of the inequality. +®To represent x < a (or x > a) on a number line, put a circle on the number a and +dark line to the left (or right) of the number a. +®To represent x ≤ a (or x ≥ a) on a number line, put a dark circle on the number +a and dark the line to the left (or right) of the number x. +Reprint 2025-26" +class_11,6,Permutations and Combinations,ncert_books/class_11/kemh1dd/kemh106.pdf,"100 MATHEMATICS +vEvery body of discovery is mathematical in form because there is no +other guidance we can have – DARWINv +6.1 Introduction +Suppose you have a suitcase with a number lock. The number +lock has 4 wheels each labelled with 10 digits from 0 to 9. +The lock can be opened if 4 specific digits are arranged in a +particular sequence with no repetition. Some how, you have +forgotten this specific sequence of digits. You remember only +the first digit which is 7. In order to open the lock, how +many sequences of 3-digits you may have to check with? To +answer this question, you may, immediately, start listing all +possible arrangements of 9 remaining digits taken 3 at a +time. But, this method will be tedious, because the number +of possible sequences may be large. Here, in this Chapter, +we shall learn some basic counting techniques which will +enable us to answer this question without actually listing 3-digit arrangements. In fact, +these techniques will be useful in determining the number of different ways of arranging +and selecting objects without actually listing them. As a first step, we shall examine a +principle which is most fundamental to the learning of these techniques. +6.2 Fundamental Principle of Counting +Let us consider the following problem. Mohan has 3 pants and 2 shirts. How many +different pairs of a pant and a shirt, can he dress up with? There are 3 ways in which +a pant can be chosen, because there are 3 pants available. Similarly, a shirt can be +chosen in 2 ways. For every choice of a pant, there are 2 choices of a shirt. Therefore, +there are 3 × 2 = 6 pairs of a pant and a shirt. +6 +Chapter +PERMUTATIONS AND COMBINATIONS +Jacob Bernoulli +(1654-1705) +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 101 +Let us name the three pants as P1, P2, P3 and the two shirts as S1, S2. Then, +these six possibilities can be illustrated in the Fig. 6.1. +Let us consider another problem +of the same type. +Sabnam has 2 school bags, 3 tiffin boxes +and 2 water bottles. In how many ways +can she carry these items (choosing one +each). +A school bag can be chosen in 2 +different ways. After a school bag is +chosen, a tiffin box can be chosen in 3 +different ways. Hence, there are +2 × 3 = 6 pairs of school bag and a tiffin +box. For each of these pairs a water +bottle can be chosen in 2 different ways. +Hence, there are 6 × 2 = 12 different ways in which, Sabnam can carry these items to +school. If we name the 2 school bags as B1, B2, the three tiffin boxes as T1, T2, T3 and +the two water bottles as W1, W2, these possibilities can be illustrated in the Fig. 6.2. +Fig 6.1 +Fig 6.2 +Reprint 2025-26 + +102 MATHEMATICS +In fact, the problems of the above types are solved by applying the following +principle known as the fundamental principle of counting, or, simply, the multiplication +principle, which states that +“If an event can occur in m different ways, following which another event +can occur in n different ways, then the total number of occurrence of the events +in the given order is m×n.” +The above principle can be generalised for any finite number of events. For +example, for 3 events, the principle is as follows: +‘If an event can occur in m different ways, following which another event can +occur in n different ways, following which a third event can occur in p different ways, +then the total number of occurrence to ‘the events in the given order is m × n × p.” +In the first problem, the required number of ways of wearing a pant and a shirt +was the number of different ways of the occurence of the following events in succession: +(i) +the event of choosing a pant +(ii) +the event of choosing a shirt. +In the second problem, the required number of ways was the number of different +ways of the occurence of the following events in succession: +(i) +the event of choosing a school bag +(ii) +the event of choosing a tiffin box +(iii) +the event of choosing a water bottle. +Here, in both the cases, the events in each problem could occur in various possible +orders. But, we have to choose any one of the possible orders and count the number of +different ways of the occurence of the events in this chosen order. +Example 1 Find the number of 4 letter words, with or without meaning, which can be +formed out of the letters of the word ROSE, where the repetition of the letters is not +allowed. +Solution There are as many words as there are ways of filling in 4 vacant places +by the 4 letters, keeping in mind that the repetition is not allowed. The +first place can be filled in 4 different ways by anyone of the 4 letters R,O,S,E. Following +which, the second place can be filled in by anyone of the remaining 3 letters in 3 +different ways, following which the third place can be filled in 2 different ways; following +which, the fourth place can be filled in 1 way. Thus, the number of ways in which the +4 places can be filled, by the multiplication principle, is 4 × 3 × 2 × 1 = 24. Hence, the +required number of words is 24. +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 103 +ANote If the repetition of the letters was allowed, how many words can be formed? +One can easily understand that each of the 4 vacant places can be filled in succession +in 4 different ways. Hence, the required number of words = 4 × 4 × 4 × 4 = 256. +Example 2 Given 4 flags of different colours, how many different signals can be +generated, if a signal requires the use of 2 flags one below the other? +Solution There will be as many signals as there are ways of filling in 2 vacant places + in succession by the 4 flags of different colours. The upper vacant place can +be filled in 4 different ways by anyone of the 4 flags; following which, the lower vacant +place can be filled in 3 different ways by anyone of the remaining 3 different flags. +Hence, by the multiplication principle, the required number of signals = 4 × 3 = 12. +Example 3 How many 2 digit even numbers can be formed from the digits +1, 2, 3, 4, 5 if the digits can be repeated? +Solution There will be as many ways as there are ways of filling 2 vacant places + in succession by the five given digits. Here, in this case, we start filling in unit’s +place, because the options for this place are 2 and 4 only and this can be done in 2 +ways; following which the ten’s place can be filled by any of the 5 digits in 5 different +ways as the digits can be repeated. Therefore, by the multiplication principle, the required +number of two digits even numbers is 2 × 5, i.e., 10. +Example 4 Find the number of different signals that can be generated by arranging at +least 2 flags in order (one below the other) on a vertical staff, if five different flags are +available. +Solution A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us +count the possible number of signals consisting of 2 flags, 3 flags, 4 flags and 5 flags +separately and then add the respective numbers. +There will be as many 2 flag signals as there are ways of filling in 2 vacant places + in succession by the 5 flags available. By Multiplication rule, the number of +ways is 5 × 4 = 20. +Similarly, there will be as many 3 flag signals as there are ways of filling in 3 +vacant places + in succession by the 5 flags. +Reprint 2025-26 + +104 MATHEMATICS +The number of ways is 5 × 4 × 3 = 60. +Continuing the same way, we find that +The number of 4 flag signals = 5 × 4 × 3 × 2 = 120 +and +the number of 5 flag signals = 5 × 4 × 3 × 2 × 1 = 120 +Therefore, +the required no of signals = 20 + 60 + 120 + 120 = 320. +EXERCISE 6.1 +1. +How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 +assuming that +(i) +repetition of the digits is allowed? +(ii) +repetition of the digits is not allowed? +2. +How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the +digits can be repeated? +3. +How many 4-letter code can be formed using the first 10 letters of the English +alphabet, if no letter can be repeated? +4. +How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if +each number starts with 67 and no digit appears more than once? +5. +A coin is tossed 3 times and the outcomes are recorded. How many possible +outcomes are there? +6. +Given 5 flags of different colours, how many different signals can be generated if +each signal requires the use of 2 flags, one below the other? +6.3 Permutations +In Example 1 of the previous Section, we are actually counting the different possible +arrangements of the letters such as ROSE, REOS, ..., etc. Here, in this list, each +arrangement is different from other. In other words, the order of writing the letters is +important. Each arrangement is called a permutation of 4 different letters taken all +at a time. Now, if we have to determine the number of 3-letter words, with or without +meaning, which can be formed out of the letters of the word NUMBER, where the +repetition of the letters is not allowed, we need to count the arrangements NUM, +NMU, MUN, NUB, ..., etc. Here, we are counting the permutations of 6 different +letters taken 3 at a time. The required number of words = 6 × 5 × 4 = 120 (by using +multiplication principle). +If the repetition of the letters was allowed, the required number of words would +be 6 × 6 × 6 = 216. +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 105 +Definition 1 A permutation is an arrangement in a definite order of a number of +objects taken some or all at a time. +In the following sub-section, we shall obtain the formula needed to answer these +questions immediately. +6.3.1 Permutations when all the objects are distinct +Theorem 1 The number of permutations of n different objects taken r at a time, +where 0 < r ≤ n and the objects do not repeat is n ( n – 1) ( n – 2). . .( n – r + 1), +which is denoted by nPr. +Proof There will be as many permutations as there are ways of filling in r vacant +places + + + . . . + by +←r vacant places → +the n objects. The first place can be filled in n ways; following which, the second place +can be filled in (n – 1) ways, following which the third place can be filled in (n – 2) +ways,..., the rth place can be filled in (n – (r – 1)) ways. Therefore, the number of +ways of filling in r vacant places in succession is n(n – 1) (n – 2) . . . (n – (r – 1)) or +n ( n – 1) (n – 2) ... (n – r + 1) +This expression for nPr is cumbersome and we need a notation which will help to +reduce the size of this expression. The symbol n! (read as factorial n or n factorial ) +comes to our rescue. In the following text we will learn what actually n! means. +6.3.2 Factorial notation The notation n! represents the product of first n natural +numbers, i.e., the product 1 × 2 × 3 × . . . × (n – 1) × n is denoted as n!. We read this +symbol as ‘n factorial’. Thus, 1 × 2 × 3 × 4 . . . × (n – 1) × n = n ! +1 = 1 ! +1 × 2 = 2 ! +1× 2 × 3 = 3 ! +1 × 2 × 3 × 4 = 4 ! and so on. +We define 0 ! = 1 +We can write 5 ! = 5 × 4 ! = 5 × 4 × 3 ! = 5 × 4 × 3 × 2 ! += 5 × 4 × 3 × 2 × 1! +Clearly, for a natural number n + n ! = n (n – 1) ! += n (n – 1) (n – 2) ! +[provided (n ≥ 2)] += n (n – 1) (n – 2) (n – 3) ! +[provided (n ≥ 3)] +and so on. +Reprint 2025-26 + +106 MATHEMATICS +Example 5 Evaluate (i) 5 ! +(ii) 7 ! +(iii) 7 ! – 5! +Solution +(i) 5 ! = 1 × 2 × 3 × 4 × 5 = 120 +(ii) 7 ! = 1 × 2 × 3 × 4 × 5 × 6 ×7 = 5040 +and +(iii) 7 ! – 5! = 5040 – 120 = 4920. +Example 6 Compute (i) +7! +5! +(ii) ( +) +12! +10! (2!) +Solution +(i) We have 7! +5! = +7 +6 5! +5! +× × + = 7 × 6 = 42 +and +(ii) ( +) ( ) +12! +10! +2! = +( +) +( +) ( ) +12 11 +10! +10! +2 +× +× +× + = 6 × 11 = 66. +Example 7 Evaluate +( +) +! +! +! +n +r +n +r +− +, when n = 5, r = 2. +Solution +We have to evaluate +( +) +5! +2! 5 +2 ! +− + (since n = 5, r = 2) +We have +( +) +5! +2! 5 +2 ! +− += +5! +5 +4 +10 +2! +3! +2 +× += += +× +. +Example 8 +If +1 +1 +8! +9! +10! +x ++ += +, find x. +Solution We have +1 +1 +8! +9 8! +10 9 8! +x ++ += +× +× × +Therefore +1 +1 +9 +10 9 +x ++ += +× + or +10 +9 +10 9 +x += +× +So +x = 100. +EXERCISE 6.2 +1. +Evaluate +(i) 8 ! +(ii) 4 ! – 3 ! +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 107 +2. +Is 3 ! + 4 ! = 7 ! ? +3. Compute +8! +6! 2! +× +4. If +1 +1 +6! +7! +8! +x ++ += +, find x +5. +Evaluate ( +) +! +! +n +n +r +− +, when +(i) n = 6, r = 2 +(ii) n = 9, r = 5. +6.3.3 Derivation of the formula for nPr +( +) +! +P +! +n +r +n +n +r +− += +, 0 ≤ r ≤ n +Let us now go back to the stage where we had determined the following formula: +nPr = n (n – 1) (n – 2) . . . (n – r + 1) +Multiplying numerator and denomirator by (n – r) (n – r – 1) . . . 3 × 2 × 1, we get +( +) ( +) ( +)( +)( +) +( +)( +) +1 +2 +1 +1 +3 2 1 +P +1 +3 2 1 +n +r +n n +n +... n +r +n +r +n +r +... +n +r +n +r +... +− +− +− ++ +− +− +− +× × += +− +−− +× × + = ( +) +! +! +n +n +r +− +, +Thus +( +) +! +P +! +n +r +n +n +r += +− +, where 0 < r ≤n +This is a much more convenient expression for nPr than the previous one. +In particular, when r = n, +! +P +! +0! +n +n +n +n += += +Counting permutations is merely counting the number of ways in which some or +all objects at a time are rearranged. Arranging no object at all is the same as leaving +behind all the objects and we know that there is only one way of doing so. Thus, we +can have +n P0 = 1 = +! +! +! +( +0)! += +− +n +n +n +n +... (1) +Therefore, the formula (1) is applicable for r = 0 also. +Thus +( +) +! +P +0 +! +n +r +n +, +r +n +n +r += +≤ +≤ +− +. +Reprint 2025-26 + +108 MATHEMATICS +Theorem 2 The number of permutations of n different objects taken r at a time, +where repetition is allowed, is nr. +Proof is very similar to that of Theorem 1 and is left for the reader to arrive at. +Here, we are solving some of the problems of the pervious Section using the +formula for nPr to illustrate its usefulness. +In Example 1, the required number of words = 4P4 = 4! = 24. Here repetition is +not allowed. If repetition is allowed, the required number of words would be 44 = 256. +The number of 3-letter words which can be formed by the letters of the word +NUMBER = +6 +3 +6! +P +3! += + = 4 × 5 × 6 = 120. Here, in this case also, the repetition is not +allowed. If the repetition is allowed,the required number of words would be 63 = 216. +The number of ways in which a Chairman and a Vice-Chairman can be chosen +from amongst a group of 12 persons assuming that one person can not hold more than +one position, clearly 12 +2 +12! +P +11 12 +10! += += +× + = 132. +6.3.4 Permutations when all the objects are not distinct objects Suppose we have +to find the number of ways of rearranging the letters of the word ROOT. In this case, +the letters of the word are not all different. There are 2 Os, which are of the same kind. +Let us treat, temporarily, the 2 Os as different, say, O1 and O2. The number of +permutations of 4-different letters, in this case, taken all at a time +is 4!. Consider one of these permutations say, RO1O2T. Corresponding to this +permutation,we have 2 ! permutations RO1O2T and RO2O1T which will be exactly the +same permutation if O1 and O2 are not treated as different, i.e., if O1 and O2 are the +same O at both places. +Therefore, the required number of permutations = +4! +3 4 +12 +2! = +× += +. +Permutations when O1, O2 are + Permutations when O1, O2 are +different. +the same O. +1 +2 +2 +1 +RO O T +RO O T + + +R O O T +1 +2 +2 +1 +TO O R +TO O R + + +T O O R +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 109 +1 +2 +2 +1 +RO T O +RO T O + + +R O T O +1 +2 +2 +1 +T O R O +T O R O + + +T O R O +1 +2 +2 +1 +RTO O +RTO O + + +R T O O +1 +2 +2 +1 +T R O O +T R O O + + +T R O O +1 +2 +2 +1 +O O R T +O O T R + + +O O R T +1 +2 +2 +1 +O R O T +O R O T + + +O R O T +1 +2 +2 +1 +O T O R +O T O R + + +O T O R +1 +2 +2 +1 +O R T O +O R T O + + +O R T O +1 +2 +2 +1 +O T R O +O T R O + + +O T R O +1 +2 +2 +1 +O O T R +O O T R + + +O O T R +Let us now find the number of ways of rearranging the letters of the word +INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears +3 times. +Temporarily, let us treat these letters different and name them as I1, I2, T1, T2, T3. +The number of permutations of 9 different letters, in this case, taken all at a time is 9 !. +Consider one such permutation, say, I1 NT1 SI2 T2 U E T3. Here if I1, I2 are not same +Reprint 2025-26 + +110 MATHEMATICS +and T1, T2, T3 are not same, then I1, I2 can be arranged in 2! ways and T1, T2, T3 can +be arranged in 3! ways. Therefore, 2! × 3! permutations will be just the same permutation +corresponding to this chosen permutation I1NT1SI2T2UET3. Hence, total number of +different permutations will be 9! +2!3! +We can state (without proof) the following theorems: +Theorem 3 The number of permutations of n objects, where p objects are of the +same kind and rest are all different = +! +! +n +p . +In fact, we have a more general theorem. +Theorem 4 The number of permutations of n objects, where p1 objects are of one +kind, p2 are of second kind, ..., pk are of kth kind and the rest, if any, are of different +kind is +1 +2 +! +! +! +! +k +n +p p ... p . +Example 9 Find the number of permutations of the letters of the word ALLAHABAD. +Solution Here, there are 9 objects (letters) of which there are 4A’s, 2 L’s and rest are +all different. +Therefore, the required number of arrangements = 9! +5 6 7 8 9 +4!2! +2 +× × × × += + = 7560 +Example 10 How many 4-digit numbers can be formed by using the digits 1 to 9 if +repetition of digits is not allowed? +Solution Here order matters for example 1234 and 1324 are two different numbers. +Therefore, there will be as many 4 digit numbers as there are permutations of 9 different +digits taken 4 at a time. +Therefore, the required 4 digit numbers +( +) +9 +4 +9! +9! += P = += +9 – 4 ! +5! = 9 × 8 × 7 × 6 = 3024. +Example 11 How many numbers lying between 100 and 1000 can be formed with the +digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed? +Solution Every number between 100 and 1000 is a 3-digit number. We, first, have to +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 111 +count the permutations of 6 digits taken 3 at a time. This number would be 6P3. But, +these permutations will include those also where 0 is at the 100’s place. For example, +092, 042, . . ., etc are such numbers which are actually 2-digit numbers and hence the +number of such numbers has to be subtracted from 6P3 to get the required number. To +get the number of such numbers, we fix 0 at the 100’s place and rearrange the remaining +5 digits taking 2 at a time. This number is 5P2. So +The required number +6 +5 +3 +2 +6! +5! += P +P +3! +3! +− += +− += 4 × 5 × 6 – 4 ×5 = 100 +Example 12 Find the value of n such that +(i) +5 +3 +P +42 P +4 +n +n +, n += +> +(ii) +4 +–1 +4 +P +5 += 3 +P +n +n +, n > 4 +Solution (i) Given that +5 +3 +P +42 P +n +n += +or +n (n – 1) (n – 2) (n – 3) (n – 4) = 42 n(n – 1) (n – 2) +Since +n > 4 +so n(n – 1) (n – 2) ≠ 0 +Therefore, by dividing both sides by n(n – 1) (n – 2), we get +(n – 3 (n – 4) = 42 +or +n2 – 7n – 30 = 0 +or +n2 – 10n + 3n – 30 +or +(n – 10) (n + 3) = 0 +or +n – 10 = 0or n + 3 = 0 +or +n = 10 +or n = – 3 +As n cannot be negative, so n = 10. +(ii) Given that +4 +–1 +4 +P +5 +3 +P +n +n += +Therefore +3n (n – 1) (n – 2) (n – 3) = 5(n – 1) (n – 2) (n – 3) (n – 4) +or +3n = 5 (n – 4) +[as (n – 1) (n – 2) (n – 3) ≠ 0, n > 4] +or + n = 10. +Reprint 2025-26 + +112 MATHEMATICS +Example 13 Find r, if 5 4Pr = 6 5Pr–1 . +Solution We have +4 +5 +1 +5 +P +6 P +r +r− += +or +( +) +( +) +4! +5! +5 +6 +4 +! +5 +1 ! +r +r +× += +× +− +−+ +or +( +) +( +) ( +)( +) +5! +6 +5! +4 +! +5 +1 +5 +5 +1 ! +r +r +r +r +× += +− +− ++ +− +− +− +or +(6 – r) (5 – r) = 6 +or +r2 – 11r + 24 = 0 +or +r2 – 8r – 3r + 24 = 0 +or +(r – 8) (r – 3) = 0 +or +r = 8 or r = 3. +Hence +r = 8, 3. +Example 14 Find the number of different 8-letter arrangements that can be made +from the letters of the word DAUGHTER so that +(i) +all vowels occur together +(ii) +all vowels do not occur together. +Solution (i) +There are 8 different letters in the word DAUGHTER, in which there +are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we can for +the time being, assume them as a single object (AUE). This single object together with +5 remaining letters (objects) will be counted as 6 objects. Then we count permutations +of these 6 objects taken all at a time. This number would be 6P6 = 6!. Corresponding to +each of these permutations, we shall have 3! permutations of the three vowels A, U, E +taken all at a time . Hence, by the multiplication principle the required number of +permutations = 6 ! × 3 ! = 4320. +(ii) +If we have to count those permutations in which all vowels are never +together, we first have to find all possible arrangments of 8 letters taken all at a time, +which can be done in 8! ways. Then, we have to subtract from this number, the number +of permutations in which the vowels are always together. +Therefore, the required number +8 ! – 6 ! × 3 ! = +6 ! (7×8 – 6) += +2 × 6 ! (28 – 3) += +50 × 6 ! = 50 × 720 = 36000 +Example 15 In how many ways can 4 red, 3 yellow and 2 green discs be arranged in +a row if the discs of the same colour are indistinguishable ? +Solution Total number of discs are 4 + 3 + 2 = 9. Out of 9 discs, 4 are of the first kind +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 113 +(red), 3 are of the second kind (yellow) and 2 are of the third kind (green). +Therefore, the number of arrangements +9! +=1260 +4!3! 2! +. +Example 16 Find the number of arrangements of the letters of the word +INDEPENDENCE. In how many of these arrangements, +(i) +do the words start with P +(ii) +do all the vowels always occur together +(iii) +do the vowels never occur together +(iv) +do the words begin with I and end in P? +Solution There are 12 letters, of which N appears 3 times, E appears 4 times and D +appears 2 times and the rest are all different. Therefore +The required number of arrangements +12! +1663200 +3! 4! 2! += += + (i) +Let us fix P at the extreme left position, we, then, count the arrangements of the +remaining 11 letters. Therefore, the required number of words starting with P +11! +138600 +3! 2! 4! += += +. + (ii) +There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have +to always occur together, we treat them as a single object EEEEI for the time +being. This single object together with 7 remaining objects will account for 8 +objects. These 8 objects, in which there are 3Ns and 2 Ds, can be rearranged in +8! +3! 2! ways. Corresponding to each of these arrangements, the 5 vowels E, E, E, +E and I can be rearranged in 5! +4! ways. Therefore, by multiplication principle, +the required number of arrangements +8! +5! += +16800 +3! 2! +4! +× += +(iii) +The required number of arrangements += the total number of arrangements (without any restriction) – the number + of arrangements where all the vowels occur together. +Reprint 2025-26 + +114 MATHEMATICS += 1663200 – 16800 = 1646400 +(iv) +Let us fix I and P at the extreme ends (I at the left end and P at the right end). +We are left with 10 letters. +Hence, the required number of arrangements += 10! +3! 2! 4! = 12600 +EXERCISE 6.3 +1. +How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is +repeated? +2. +How many 4-digit numbers are there with no digit repeated? +3. +How many 3-digit even numbers can be made using the digits +1, 2, 3, 4, 6, 7, if no digit is repeated? +4. +Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, +5 if no digit is repeated. How many of these will be even? +5. +From a committee of 8 persons, in how many ways can we choose a chairman +and a vice chairman assuming one person can not hold more than one position? +6. +Find n if n – 1P3 : nP4 = 1 : 9. +7. +Find r if (i) 5 +6 +1 +P +2 P +r +r− += +(ii) 5 +6 +1 +P +P +r +r− += +. +8. +How many words, with or without meaning, can be formed using all the letters of +the word EQUATION, using each letter exactly once? +9. +How many words, with or without meaning can be made from the letters of the +word MONDAY, assuming that no letter is repeated, if. +(i) +4 letters are used at a time, +(ii) +all letters are used at a time, +(iii) +all letters are used but first letter is a vowel? +10. +In how many of the distinct permutations of the letters in MISSISSIPPI do the +four I’s not come together? +11. +In how many ways can the letters of the word PERMUTATIONS be arranged if the +(i) +words start with P and end with S, +(ii) +vowels are all together, +(iii) +there are always 4 letters between P and S? +6.4 Combinations +Let us now assume that there is a group of 3 lawn tennis players X, Y, Z. A team +consisting of 2 players is to be formed. In how many ways can we do so? Is the team +of X and Y different from the team of Y and X ? Here, order is not important. +In fact, there are only 3 possible ways in which the team could be constructed. +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 115 +These are XY, YZ and ZX (Fig 6.3). +Here, each selection is called a combination of 3 different objects taken 2 at a time. +In a combination, the order is not important. +Now consider some more illustrations. +Twelve persons meet in a room and each shakes hand with all the others. How do +we determine the number of hand shakes. X shaking hands with Y and Y with X will +not be two different hand shakes. Here, order is not important. There will be as many +hand shakes as there are combinations of 12 different things taken 2 at a time. +Seven points lie on a circle. How many chords can be drawn by joining these +points pairwise? There will be as many chords as there are combinations of 7 different +things taken 2 at a time. +Now, we obtain the formula for finding the number of combinations of n different +objects taken r at a time, denoted by nCr.. +Suppose we have 4 different objects A, B, C and D. Taking 2 at a time, if we have +to make combinations, these will be AB, AC, AD, BC, BD, CD. Here, AB and BA are +the same combination as order does not alter the combination. This is why we have not +included BA, CA, DA, CB, DB and DC in this list. There are as many as 6 combinations +of 4 different objects taken 2 at a time, i.e., 4C2 = 6. +Corresponding to each combination in the list, we can arrive at 2! permutations as +2 objects in each combination can be rearranged in 2! ways. Hence, the number of +permutations = 4C2 × 2!. +On the other hand, the number of permutations of 4 different things taken 2 at +a time = 4P2. +Therefore +4P2 = 4C2 × 2! or ( +) +4 +2 +4! +C +4 +2 ! 2! = +− +Now, let us suppose that we have 5 different objects A, B, C, D, E. Taking 3 at a +time, if we have to make combinations, these will be ABC, ABD, ABE, BCD, BCE, +CDE, ACE, ACD, ADE, BDE. Corresponding to each of these 5C3 combinations, there +are 3! permutations, because, the three objects in each combination can be +Fig. 6.3 +Reprint 2025-26 + +116 MATHEMATICS +rearranged in 3 ! ways. Therefore, the total of permutations = 5 +3 +C +3! +× +Therefore 5P3 = 5C3 × 3! or +( +) +5 +3 +5! +C +5 +3 ! 3! = +− +These examples suggest the following theorem showing relationship between +permutaion and combination: +Theorem 5 P +C +! +n +n +r +r r += +, 0 < r ≤ n. +Proof Corresponding to each combination of nCr, we have r ! permutations, because +r objects in every combination can be rearranged in r ! ways. +Hence, the total number of permutations of n different things taken r at a time +is nCr × r!. On the other hand, it is P +n +r . Thus +P +C +! +n +n +r +r +r += +× +, 0 +r +n +< +≤ +. +Remarks 1. From above ( +) +! +C +! +! +n +r +n +r +n +r += +× +− +, i.e., +( +) +! +C +! +! +n +r +n +r n +r += +− +. +In particular, if r +n += +, +! +C +1 +! 0! +n +n +n +n += += . +2. +We define nC0 = 1, i.e., the number of combinations of n different things taken +nothing at all is considered to be 1. Counting combinations is merely counting the +number of ways in which some or all objects at a time are selected. Selecting +nothing at all is the same as leaving behind all the objects and we know that there +is only one way of doing so. This way we define nC0 = 1. +3. +As +( +) +0 +! +1 +C +0! +0 ! +n +n +n += = +− +, the formula +( +) +! +C +! +! +n +r +n +r +n +r += +− + is applicable for r = 0 also. +Hence +( +) +! +C +! +! +n +r +n +r +n +r += +− +, 0 ≤ r ≤ n. +4. +( +) +( +) +( +) +! +C +! +! +n +n r +n +n +r +n +n +r +−= +− +− +− + = ( +) +! +! ! +n +n +r +r +− + = C +n +r , +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 117 +i.e., selecting r objects out of n objects is same as rejecting (n – r) objects. +5. +nCa = nCb ⇒ a = b or a = n – b, i.e., n = a + b +Theorem 6 +1 +1 +C +C +C +n +n +n +r +r +r ++ +− ++ += +Proof We have + +( +) +( +) ( +) +1 +! +! +C +C +! +! +1 ! +1 ! +n +n +r +r +n +n +r n +r +r +n +r +− ++ += ++ +− +− +− ++ += +( +) ( +) +! +1 ! +! +n +r +r +n +r +× +− +− + + ( +) ( +) ( +) +! +1 ! +1 +! +n +r +n +r +n +r +− +− ++ +− += ( +) ( +) +! +1 ! +! +n +r +n +r +− +− + +1 +1 +1 +r +n +r + + ++ + + +− ++ + + += ( +) ( +) +( +) +! +1 +1 ! +! +1 +n +n +r +r +r +n +r +r n +r +− ++ + +× +− +− +− ++ + = +( +) +( +) +1 +1 ! +C +! +1 +! +n +r +n +r n +r ++ ++ += ++ − +Example 17 If +9 +8 +C +C +n +n += +, find +17 +C +n +. +Solution We have +9 +8 +C +C +n +n += + i.e., +( +) +( +) +! +! +9! +9 ! +8 !8! +n +n +n +n += +− +− +or +1 +1 +9 +8 +n += +− + or n – 8 = 9 or n = 17 +Therefore +17 +17 +17 +C +C +1 +n += += +. +Example 18 A committee of 3 persons is to be constituted from a group of 2 men and +3 women. In how many ways can this be done? How many of these committees would +consist of 1 man and 2 women? +Solution Here, order does not matter. Therefore, we need to count combinations. +There will be as many committees as there are combinations of 5 different persons +taken 3 at a time. Hence, the required number of ways = +5 +3 +5! +4 5 +C +10 +3! 2! +2 +× += += += +. +Now, 1 man can be selected from 2 men in 2C1 ways and 2 women can be +selected from 3 women in 3C2 ways. Therefore, the required number of committees +Reprint 2025-26 + +118 MATHEMATICS += +2 +3 +1 +2 +2! +3! +C +C +6 +1! 1! +2! 1! +× += +× += +. +Example 19 What is the number of ways of choosing 4 cards from a pack of 52 +playing cards? In how many of these +(i) +four cards are of the same suit, +(ii) +four cards belong to four different suits, +(iii) +are face cards, +(iv) +two are red cards and two are black cards, +(v) +cards are of the same colour? +Solution There will be as many ways of choosing 4 cards from 52 cards as there are +combinations of 52 different things, taken 4 at a time. Therefore +The required number of ways = +52 +4 +52! +49 50 51 52 +C +4! 48! +2 3 4 +× +× +× += += +× × + = 270725 + (i) There are four suits: diamond, club, spade, heart and there are 13 cards of each +suit. Therefore, there are 13C4 ways of choosing 4 diamonds. Similarly, there are +13C4 ways of choosing 4 clubs, 13C4 ways of choosing 4 spades and 13C4 ways of +choosing 4 hearts. Therefore +The required number of ways = 13C4 + 13C4 + 13C4 + 13C4. += +13! +4 +2860 +4! 9! +× += + (ii) +There are13 cards in each suit. +Therefore, there are 13C1 ways of choosing 1 card from 13 cards of diamond, +13C1 ways of choosing 1 card from 13 cards of hearts, 13C1 ways of choosing 1 +card from 13 cards of clubs, 13C1 ways of choosing 1 card from 13 cards of +spades. Hence, by multiplication principle, the required number of ways += 13C1 × +13C1 × +13C1× +13C1 = 134 +(iii) +There are 12 face cards and 4 are to be selected out of these 12 cards. This can be +done in 12C4 ways. Therefore, the required number of ways = +12! +495 +4! 8! += +. +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 119 +(iv) +There are 26 red cards and 26 black cards. Therefore, the required number of +ways = 26C2 × + 26C2 += +( +) +2 +2 +26! +325 +2! 24! + += + + + + += 105625 +(v) +4 red cards can be selected out of 26 red cards in 26C4 ways. +4 black cards can be selected out of 26 black cards in 26C4ways. +Therefore, the required number of ways = 26C4 + 26C4 + = +26! +2 +4! 22! +× + = 29900. +EXERCISE 6.4 +1. +If nC8 = nC2, find nC2. +2. +Determine n if +(i) 2nC3 : nC3 = 12 : 1 +(ii) 2nC3 : nC3 = 11 : 1 +3. +How many chords can be drawn through 21 points on a circle? +4. +In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and +4 girls? +5. +Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 +blue balls if each selection consists of 3 balls of each colour. +6. +Determine the number of 5 card combinations out of a deck of 52 cards if there +is exactly one ace in each combination. +7. +In how many ways can one select a cricket team of eleven from 17 players in +which only 5 players can bowl if each cricket team of 11 must include exactly 4 +bowlers? +8. +A bag contains 5 black and 6 red balls. Determine the number of ways in which +2 black and 3 red balls can be selected. +9. +In how many ways can a student choose a programme of 5 courses if 9 courses +are available and 2 specific courses are compulsory for every student? +Miscellaneous Examples +Example 20 How many words, with or without meaning, each of 3 vowels and 2 +consonants can be formed from the letters of the word INVOLUTE ? +Solution In the word INVOLUTE, there are 4 vowels, namely, I,O,E,Uand 4 +consonants, namely, N, V, L and T. +Reprint 2025-26 + +120 MATHEMATICS +The number of ways of selecting 3 vowels out of 4 = 4C3 = 4. +The number of ways of selecting 2 consonants out of 4 = 4C2 = 6. +Therefore, the number of combinations of 3 vowels and 2 consonants is +4 × 6 = 24. +Now, each of these 24 combinations has 5 letters which can be arranged among +themselves in 5 ! ways. Therefore, the required number of different words is +24 × 5 ! = 2880. +Example 21 A group consists of 4 girls and 7 boys. In how many ways can a team of +5 members be selected if the team has (i) no girl ? (ii) at least one boy and one girl ? +(iii) at least 3 girls ? +Solution (i) Since, the team will not include any girl, therefore, only boys are to be +selected. 5 boys out of 7 boys can be selected in 7C5 ways. Therefore, the required +number of ways = +7 +5 +7! +6 +7 +C +21 +5! 2! +2 +× += += += +(ii) +Since, at least one boy and one girl are to be there in every team. Therefore, the +team can consist of +(a) 1 boy and 4 girls +(b) 2 boys and 3 girls +(c) 3 boys and 2 girls +(d) 4 boys and 1 girl. +1 boy and 4 girls can be selected in 7C1 × 4C4 ways. +2 boys and 3 girls can be selected in 7C2 × 4C3 ways. +3 boys and 2 girls can be selected in 7C3 × 4C2 ways. +4 boys and 1 girl can be selected in 7C4 × 4C1 ways. +Therefore, the required number of ways += 7C1 × 4C4 + 7C2 × 4C3 + +7C3 × 4C2 + 7C4 × 4C1 += 7 + 84 + 210 + 140 = 441 +(iii) +Since, the team has to consist of at least 3 girls, the team can consist of +(a) 3 girls and 2 boys, or + (b) 4 girls and 1 boy. +Note that the team cannot have all 5 girls, because, the group has only 4 girls. +3 girls and 2 boys can be selected in 4C3 × 7C2 ways. +4 girls and 1 boy can be selected in 4C4 × 7C1 ways. +Therefore, the required number of ways += 4C3 × 7C2 + 4C4 × 7C1 = 84 + 7 = 91 +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 121 +Example 22 Find the number of words with or without meaning which can be made +using all the letters of the word AGAIN. If these words are written as in a dictionary, +what will be the 50th word? +Solution There are 5 letters in the word AGAIN, in which A appears 2 times. Therefore, +the required number of words = 5! +60 +2! = +. +To get the number of words starting with A, we fix the letter A at the extreme left +position, we then rearrange the remaining 4 letters taken all at a time. There will be as +many arrangements of these 4 letters taken 4 at a time as there are permutations of 4 +different things taken 4 at a time. Hence, the number of words starting with +A = 4! = 24. Then, starting with G, the number of words +4! +2! += + = 12 as after placing G +at the extreme left position, we are left with the letters A, A, I and N. Similarly, there +are 12 words starting with the next letter I. Total number of words so far obtained += 24 + 12 + 12 =48. +The 49th word is NAAGI. The 50th word is NAAIG. +Example 23 How many numbers greater than 1000000 can be formed by using the +digits 1, 2, 0, 2, 4, 2, 4? +Solution Since, 1000000 is a 7-digit number and the number of digits to be used is also +7. Therefore, the numbers to be counted will be 7-digit only. Also, the numbers have to +be greater than 1000000, so they can begin either with 1, 2 or 4. +The number of numbers beginning with 1 = +6! +4 5 +6 +3! 2! +2 +× × += + = 60, as when 1 is +fixed at the extreme left position, the remaining digits to be rearranged will be 0, 2, 2, 2, +4, 4, in which there are 3, 2s and 2, 4s. +Total numbers begining with 2 += 6! +3 4 5 6 +2! 2! +2 +× × × += + = 180 +and total numbers begining with 4 +6! +4 5 6 +3! += += +× × = 120 +Reprint 2025-26 + +122 MATHEMATICS +Therefore, the required number of numbers = 60 + 180 + 120 = 360. +Alternative Method +The number of 7-digit arrangements, clearly, +7! +420 +3! 2! = +. But, this will include those +numbers also, which have 0 at the extreme left position. The number of such +arrangements +6! +3! 2! (by fixing 0 at the extreme left position) = 60. +Therefore, the required number of numbers = 420 – 60 = 360. +ANote If one or more than one digits given in the list is repeated, it will be +understood that in any number, the digits can be used as many times as is given in +the list, e.g., in the above example 1 and 0 can be used only once whereas 2 and 4 +can be used 3 times and 2 times, respectively. +Example 24 In how many ways can 5 girls and 3 boys be seated in a row so that no +two boys are together? +Solution Let us first seat the 5 girls. This can be done in 5! ways. For each such +arrangement, the three boys can be seated only at the cross marked places. +× G × G × G × G × G ×. +There are 6 cross marked places and the three boys can be seated in 6P3 ways. +Hence, by multiplication principle, the total number of ways += 5! × 6P3 = +6! +5!× 3! += 4 × 5 × 2 × 3 × 4 × 5 × 6 = 14400. +Miscellaneous Exercise on Chapter 6 +1. +How many words, with or without meaning, each of 2 vowels and 3 consonants +can be formed from the letters of the word DAUGHTER ? +2. +How many words, with or without meaning, can be formed using all the letters of +the word EQUATION at a time so that the vowels and consonants occur together? +3. +A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways +can this be done when the committee consists of: +(i) exactly 3 girls ? (ii) atleast 3 girls ? (iii) atmost 3 girls ? +4. +If the different permutations of all the letter of the word EXAMINATION are +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 123 +listed as in a dictionary, how many words are there in this list before the first +word starting with E ? +5. +How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 +which are divisible by 10 and no digit is repeated ? +6. +The English alphabet has 5 vowels and 21 consonants. How many words with +two different vowels and 2 different consonants can be formed from the +alphabet ? +7. +In an examination, a question paper consists of 12 questions divided into two +parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student +is required to attempt 8 questions in all, selecting at least 3 from each part. In +how many ways can a student select the questions ? +8. +Determine the number of 5-card combinations out of a deck of 52 cards if each +selection of 5 cards has exactly one king. +9. +It is required to seat 5 men and 4 women in a row so that the women occupy the +even places. How many such arrangements are possible ? +10. +From a class of 25 students, 10 are to be chosen for an excursion party. There +are 3 students who decide that either all of them will join or none of them will +join. In how many ways can the excursion party be chosen ? +11. +In how many ways can the letters of the word ASSASSINATION be arranged +so that all the S’s are together ? +Summary +®Fundamental principle of counting If an event can occur in m different +ways, following which another event can occur in n different ways, then the +total number of occurrence of the events in the given order is m × n. +®The number of permutations of n different things taken r at a time, where +repetition is not allowed, is denoted by nPr and is given by nPr = +! +( +)! +n +n +r +− +, +where 0 ≤ r ≤ n. +®n! = 1 × 2 × 3 × ...×n +®n! = n × (n – 1) ! +®The number of permutations of n different things, taken r at a time, where +repeatition is allowed, is nr. +®The number of permutations of n objects taken all at a time, where p1 objects +Reprint 2025-26 + +124 MATHEMATICS +are of first kind, p2 objects are of the second kind, ..., pk objects are of the kth +kind and rest, if any, are all different is +1 +2 +! +! +! +! +k +n +p p ... p . +®The number of combinations of n different things taken r at a time, denoted by +nCr , is given by nCr = +! +! +! +n +r ( n +r ) += +− +, 0 ≤ r ≤ n. +Historical Note +The concepts of permutations and combinations can be traced back to the advent +of Jainism in India and perhaps even earlier. The credit, however, goes to the +Jains who treated its subject matter as a self-contained topic in mathematics, +under the name Vikalpa. +Among the Jains, Mahavira, (around 850) is perhaps the world’s first +mathematician credited with providing the general formulae for permutations and +combinations. +In the 6th century B.C., Sushruta, in his medicinal work, Sushruta Samhita, +asserts that 63 combinations can be made out of 6 different tastes, taken one at a +time, two at a time, etc. Pingala, a Sanskrit scholar around third century B.C., +gives the method of determining the number of combinations of a given number +of letters, taken one at a time, two at a time, etc. in his work Chhanda Sutra. +Bhaskaracharya (born 1114) treated the subject matter of permutations and +combinations under the name Anka Pasha in his famous work Lilavati. In addition +to the general formulae for nCr and nPr already provided by Mahavira, +Bhaskaracharya gives several important theorems and results concerning the +subject. +Outside India, the subject matter of permutations and combinations had its +humble beginnings in China in the famous book I–King (Book of changes). It is +difficult to give the approximate time of this work, since in 213 B.C., the emperor +had ordered all books and manuscripts in the country to be burnt which fortunately +was not completely carried out. Greeks and later Latin writers also did some +scattered work on the theory of permutations and combinations. +Some Arabic and Hebrew writers used the concepts of permutations and +combinations in studying astronomy. Rabbi ben Ezra, for instance, determined +the number of combinations of known planets taken two at a time, three at a time +and so on. This was around 1140. It appears that Rabbi ben Ezra did not know +Reprint 2025-26 + + PERMUTATIONS AND COMBINATIONS 125 +the formula for nCr. However, he was aware that nCr = nCn–r for specific values +n and r. In 1321, Levi Ben Gerson, another Hebrew writer came up with the +formulae for nPr , nPn and the general formula for nCr. +The first book which gives a complete treatment of the subject matter of +permutations and combinations is Ars Conjectandi written by a Swiss, Jacob +Bernoulli (1654 – 1705), posthumously published in 1713. This book contains +essentially the theory of permutations and combinations as is known today. +— v — +Reprint 2025-26" +class_11,7,Binomial Theorem,ncert_books/class_11/kemh1dd/kemh107.pdf,"126 MATHEMATICS +vMathematics is a most exact science and its conclusions are capable of +absolute proofs. – C.P. STEINMETZv +7.1 Introduction +In earlier classes, we have learnt how to find the squares +and cubes of binomials like a + b and a – b. Using them, we +could evaluate the numerical values of numbers like +(98)2 = (100 – 2)2, (999)3 = (1000 – 1)3, etc. However, for +higher powers like (98)5, (101)6, etc., the calculations become +difficult by using repeated multiplication. This difficulty was +overcome by a theorem known as binomial theorem. It gives +an easier way to expand (a + b)n, where n is an integer or a +rational number. In this Chapter, we study binomial theorem +for positive integral indices only. +7.2 Binomial Theorem for Positive Integral Indices +Let us have a look at the following identities done earlier: +(a+ b)0 = 1 +a + b ≠ 0 +(a+ b)1 = a + b +(a+ b)2 = a2 + 2ab + b2 +(a+ b)3 = a3 + 3a2b + 3ab2 + b3 +(a+ b)4 = (a + b)3 (a + b) = a4 + 4a3b + 6a2b2 + 4ab3 + b4 +In these expansions, we observe that +(i) +The total number of terms in the expansion is one more than the index. For +example, in the expansion of (a + b)2 , number of terms is 3 whereas the index of +(a + b)2 is 2. +(ii) +Powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the +second quantity ‘b’ increase by 1, in the successive terms. +(iii) +In each term of the expansion, the sum of the indices of a and b is the same and +is equal to the index of a + b. +7 +Chapter +Blaise Pascal +(1623-1662) +BINOMIAL THEOREM +Reprint 2025-26 + +BINOMIAL THEOREM 127 +We now arrange the coefficients in these expansions as follows (Fig 7.1): +Do we observe any pattern in this table that will help us to write the next row? Yes we +do. It can be seen that the addition of 1’s in the row for index 1 gives rise to 2 in the row +for index 2. The addition of 1, 2 and 2, 1 in the row for index 2, gives rise to 3 and 3 in +the row for index 3 and so on. Also, 1 is present at the beginning and at the end of each +row. This can be continued till any index of our interest. +We can extend the pattern given in Fig 7.2 by writing a few more rows. +Pascal’s Triangle +The structure given in Fig 7.2 looks like a triangle with 1 at the top vertex and running +down the two slanting sides. This array of numbers is known as Pascal’s triangle, +after the name of French mathematician Blaise Pascal. It is also known as Meru +Prastara by Pingla. +Expansions for the higher powers of a binomial are also possible by using Pascal’s +triangle. Let us expand (2x + 3y)5 by using Pascal’s triangle. The row for index 5 is +1 +5 +10 +10 +5 +1 +Using this row and our observations (i), (ii) and (iii), we get +(2x + 3y)5 += (2x)5 + 5(2x)4 (3y) + 10(2x)3 (3y)2 +10 (2x)2 (3y)3 + 5(2x)(3y)4 +(3y)5 += 32x5 + 240x4y + 720x3y2 + 1080x2y3 + 810xy4 + 243y5. +Fig 7.1 +Fig 7.2 +Reprint 2025-26 + +128 MATHEMATICS +Now, if we want to find the expansion of (2x + 3y)12, we are first required to get +the row for index 12. This can be done by writing all the rows of the Pascal’s triangle +till index 12. This is a slightly lengthy process. The process, as you observe, will become +more difficult, if we need the expansions involving still larger powers. +We thus try to find a rule that will help us to find the expansion of the binomial for +any power without writing all the rows of the Pascal’s triangle, that come before the +row of the desired index. +For this, we make use of the concept of combinations studied earlier to rewrite +the numbers in the Pascal’s triangle. We know that +! +C +!( +)! +n +r +n +r n – r += + , 0 ≤ r ≤ n and +n is a non-negative integer. Also, nC0 = 1 = nCn +The Pascal’s triangle can now be rewritten as (Fig 7.3) +Observing this pattern, we can now write the row of the Pascal’s triangle for any index +without writing the earlier rows. For example, for the index 7 the row would be +7C0 7C1 +7C2 +7C3 +7C4 +7C5 +7C6 +7C7. +Thus, using this row and the observations (i), (ii) and (iii), we have +(a + b)7 = 7C0 a7 + + 7C1a6b + 7C2a5b2 + 7C3a4b3 + 7C4a3b4 + 7C5a2b5 + 7C6ab6 + 7C7b7 +An expansion of a binomial to any positive integral index say n can now be visualised +using these observations. We are now in a position to write the expansion of a binomial +to any positive integral index. +Fig 7.3 Pascal’s triangle +Reprint 2025-26 + +BINOMIAL THEOREM 129 +7.2.1 + Binomial theorem for any positive integer n, +(a + b)n = nC0an + nC1an–1b + nC2an–2 b2 + ...+ nCn – 1a.bn–1 + nCnbn +Proof The proof is obtained by applying principle of mathematical induction. +Let the given statement be +P(n) : (a + b)n = nC0an + nC1an – 1b + nC2an – 2b2 + ...+ nCn–1a.bn – 1 + nCnbn +For n = 1, we have +P (1) : (a + b)1 = 1C0a1 + 1C1b1 = a + b +Thus, P (1) is true. +Suppose P (k) is true for some positive integer k, i.e. +(a + b)k = kC0ak + kC1ak – 1b + kC2ak – 2b2 + ...+ kCkbk +. . . +(1) +We shall prove that P(k + 1) is also true, i.e., +(a + b)k + 1 = k + 1C0 ak + 1 + k + 1C1 akb + k + 1C2 ak – 1b2 + ...+ k + 1Ck+1 bk + 1 +Now, (a + b)k + 1 = (a + b) (a + b)k += (a + b) (kC0 ak + kC1ak – 1 b + kC2 ak – 2 b2 +...+ kCk – 1 abk – 1 + kCk bk) +[from (1)] += kC0 ak + 1 + kC1 akb + kC2ak – 1b2 +...+ kCk – 1 a2bk – 1 + kCk abk + kC0 akb ++ kC1ak – 1b2 + kC2ak – 2b3+...+ kCk-1abk + kCkbk + 1 +[by actual multiplication] += kC0ak + 1 + (kC1+ kC0) akb + (kC2 + kC1)ak – 1b2 + ... + + (kCk+ kCk–1) abk + kCkbk + 1 + [grouping like terms] += k + 1C0a k + 1 + k + 1C1akb + k + 1C2 + ak – 1b2 +...+ k + 1Ckabk + k + 1Ck + 1 bk +1 +(by using k + 1C0=1, kCr + kCr–1 = k + 1Cr +and kCk = 1= k + 1Ck + 1) +Thus, it has been proved that P (k + 1) is true whenever P(k) is true. Therefore, by +principle of mathematical induction, P(n) is true for every positive integer n. +We illustrate this theorem by expanding (x + 2)6: +(x + 2)6 = 6C0x6 + 6C1x5.2 + 6C2x422 + 6C3x3.23 + 6C4x2.24 + 6C5x.25 + 6C6.26. += x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x + 64 +Thus (x + 2)6 = x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x + 64. +Reprint 2025-26 + +130 MATHEMATICS +Observations +1. +The notation ∑ += +− +n +k +k +k +n +k +n +b +a +0 +C +stands for +nC0anb0 + nC1an–1b1 + ...+ nCran–rbr + ...+nCnan–nbn, where b0 = 1 = an–n. +Hence the theorem can also be stated as +∑ += +− += ++ +n +k +k +k +n +k +n +n +b +a +b +a +0 +C +) +( +. +2. +The coefficients nCr occuring in the binomial theorem are known as binomial +coefficients. +3. +There are (n+1) terms in the expansion of (a+b)n, i.e., one more than the index. +4. +In the successive terms of the expansion the index of a goes on decreasing by +unity. It is n in the first term, (n–1) in the second term, and so on ending with zero +in the last term. At the same time the index of b increases by unity, starting with +zero in the first term, 1 in the second and so on ending with n in the last term. +5. +In the expansion of (a+b)n, the sum of the indices of a and b is n + 0 = n in the +first term, (n – 1) + 1 = n in the second term and so on 0 + n = n in the last term. +Thus, it can be seen that the sum of the indices of a and b is n in every term of the +expansion. +7.2.2 +Some special cases In the expansion of (a + b)n, +(i) +Taking a = x and b = – y, we obtain +(x – y)n += [x + (–y)]n += nC0xn + nC1xn – 1(–y) + nC2xn–2(–y)2 + nC3xn–3(–y)3 + ... + nCn (–y)n += nC0xn – nC1xn – 1y + nC2xn – 2y2 – nC3xn – 3y3 + ... + (–1)n nCn yn +Thus (x–y)n = nC0xn – nC1xn – 1 y + nC2xn – 2 y2 + ... + (–1)n nCn yn +Using this, we have +(x–2y)5 = 5C0x5 – 5C1x4 (2y) + 5C2x3 (2y)2 – 5C3x2 (2y)3 + + 5C4 x(2y)4 – 5C5(2y)5 + = x5 –10x4y + 40x3y2 – 80x2y3 + 80xy4 – 32y5. +(ii) +Taking a = 1, b = x, we obtain +(1 + x)n = nC0(1)n + nC1(1)n – 1x + nC2(1)n – 2 x2 + ... + nCnxn += nC0 + nC1x + nC2x2 + nC3x3 + ... + nCnxn +Thus +(1 + x)n = nC0 + nC1x + nC2x2 + nC3x3 + ... + nCnxn +Reprint 2025-26 + +BINOMIAL THEOREM 131 +In particular, for x = 1, we have +2n = nC0 + nC1 + nC2 + ... + nCn. +(iii) +Taking a = 1, b = – x, we obtain +(1– x)n = nC0 – nC1x + nC2x2 – ... + (– 1)n nCnxn +In particular, for x = 1, we get + 0 = nC0 – nC1 + nC2 – ... + (–1)n nCn +Example 1 Expand +4 +2 +3 +x +x + + ++ + + + +, x ≠ 0 +Solution By using binomial theorem, we have +4 +2 +3 +x +x + + ++ + + + + += 4C0(x2)4 + 4C1(x2)3 + + + + + + +x +3 + + 4C2(x2)2 +2 +3  + + + + + +x ++ 4C3(x2) +3 +3  + + + + + +x ++ 4C4 +4 +3  + + + + + +x += x8 + 4.x6 . x +3 ++ 6.x4 . +2 +9 +x + 4.x2. +3 +27 +x + +4 +81 +x += x8 + 12x5 + 54x2 + +4 +81 +108 +x +x ++ +. +Example 2 Compute (98)5. +Solution We express 98 as the sum or difference of two numbers whose powers are +easier to calculate, and then use Binomial Theorem. +Write 98 = 100 – 2 +Therefore, (98)5 = (100 – 2)5 += 5C0 (100)5 – 5C1 (100)4.2 + 5C2 (100)322 +– 5C3 (100)2 (2)3 + 5C4 (100) (2)4 – 5C5 (2)5 += 10000000000 – 5 × 100000000 × 2 + 10 × 1000000 × 4 – 10 ×10000 +× 8 + 5 × 100 × 16 – 32 += 10040008000 – 1000800032 = 9039207968. +Example 3 Which is larger (1.01)1000000 or 10,000? +Solution Splitting 1.01 and using binomial theorem to write the first few terms we +have +Reprint 2025-26 + +132 MATHEMATICS + (1.01)1000000 = (1 + 0.01)1000000 += 1000000C0 + 1000000C1(0.01) + other positive terms += 1 + 1000000 × 0.01 + other positive terms += 1 + 10000 + other positive terms +> 10000 +Hence +(1.01)1000000 > 10000 +Example 4 Using binomial theorem, prove that 6n–5n always leaves remainder +1 when divided by 25. +Solution For two numbers a and b if we can find numbers q and r such that +a = bq + r, then we say that b divides a with q as quotient and r as remainder. Thus, in +order to show that 6n – 5n leaves remainder 1 when divided by 25, we prove that +6n – 5n = 25k + 1, where k is some natural number. +We have +(1 + a)n = nC0 + nC1a + nC2a2 + ... + nCnan +For a = 5, we get + (1 + 5)n = nC0 + nC15 + nC252 + ... + nCn5n +i.e. +(6)n = 1 + 5n + 52.nC2 + + 53.nC3 + ... + 5n +i.e. +6n – 5n = 1+52 (nC2 + + nC35 + ... + 5n-2) +or +6n – 5n = 1+ 25 (nC2 + 5 .nC3 + ... + 5n-2) +or +6n – 5n = 25k+1 where k = nC2 + 5 .nC3 + ... + 5n–2. +This shows that when divided by 25, 6n – 5n leaves remainder 1. +EXERCISE 7.1 +Expand each of the expressions in Exercises 1 to 5. +1. +(1–2x)5 +2. +5 +2 +2 +x +– +x + + + + + + +3. +(2x – 3)6 +Reprint 2025-26 + +BINOMIAL THEOREM 133 +4. +5 +1 +3 +x +x + + ++ + + + + +5. +6 +1  + + + + ++ x +x +Using binomial theorem, evaluate each of the following: +6. +(96)3 +7. +(102)5 +8. +(101)4 +9. +(99)5 +10. +Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000. +11. +Find (a + b)4 – (a – b)4. Hence, evaluate +4) +2 +3 +( ++ +– +4) +2 +– +3 +( +. +12. +Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate ( +2 + 1)6 + ( 2 – 1)6. +13. +Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer. +14. +Prove that ∑ += += +n +r +n +n +r +r +0 +4 +C +3 +. +Miscellaneous Exercise on Chapter 7 +1. +If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever +n is a positive integer. +[Hint write an = (a – b + b)n and expand] +2. +Evaluate ( +) +( +) +6 +6 +3 +2 +3 +2 ++ +− +− +. +3. +Find the value of ( +) +( +) +4 +4 +2 +2 +2 +2 +1 +1 +a +a +a +a ++ +− ++ +− +− +. +4. +Find an approximation of (0.99)5 using the first three terms of its expansion. +5. +Expand using Binomial Theorem +4 +2 +1 +0 +2 +x +, x +x + + ++ +− +≠ + + + + +. +6. +Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem. +Summary +®The expansion of a binomial for any positive integral n is given by Binomial +Theorem, which is (a + b)n = nC0an + nC1an – 1b + nC2an – 2b2 + ...+ + nCn – 1a.bn – 1 + nCnbn. +®The coefficients of the expansions are arranged in an array. This array is +called Pascal’s triangle. +Reprint 2025-26 + +134 MATHEMATICS +Historical Note +The ancient Indian mathematicians knew about the coefficients in the +expansions of (x + y)n, 0 ≤ n ≤ 7. The arrangement of these coefficients was in +the form of a diagram called Meru-Prastara, provided by Pingla in his book +Chhanda shastra (200B.C.). This triangular arrangement is also found in the +work of Chinese mathematician Chu-shi-kie in 1303. The term binomial coefficients +was first introduced by the German mathematician, Michael Stipel (1486-1567) in +approximately 1544. Bombelli (1572) also gave the coefficients in the expansion of +(a + b)n, for n = 1,2 ...,7 and Oughtred (1631) gave them for n = 1, 2,..., 10. The +arithmetic triangle, popularly known as Pascal’s triangle and similar to the Meru- +Prastara of Pingla was constructed by the French mathematician Blaise Pascal +(1623-1662) in 1665. +The present form of the binomial theorem for integral values of n appeared in +Trate du triange arithmetic, written by Pascal and published posthumously in +1665. +— v — +Reprint 2025-26" +class_11,8,Sequences and Series,ncert_books/class_11/kemh1dd/kemh108.pdf,"vNatural numbers are the product of human spirit. – DEDEKIND v +8.1 Introduction +In mathematics, the word, “sequence” is used in much the +same way as it is in ordinary English. When we say that a +collection of objects is listed in a sequence, we usually mean +that the collection is ordered in such a way that it has an +identified first member, second member, third member and +so on. For example, population of human beings or bacteria +at different times form a sequence. The amount of money +deposited in a bank, over a number of years form a sequence. +Depreciated values of certain commodity occur in a +sequence. Sequences have important applications in several +spheres of human activities. +Sequences, following specific patterns are called progressions. In previous class, +we have studied about arithmetic progression (A.P). In this Chapter, besides discussing +more about A.P.; arithmetic mean, geometric mean, relationship between A.M. +and G.M., special series in forms of sum to n terms of consecutive natural numbers, +sum to n terms of squares of natural numbers and sum to n terms of cubes of +natural numbers will also be studied. +8.2 Sequences +Let us consider the following examples: +Assume that there is a generation gap of 30 years, we are asked to find the +number of ancestors, i.e., parents, grandparents, great grandparents, etc. that a person +might have over 300 years. +Here, the total number of generations = 300 +10 +30 = +Fibonacci +(1175-1250) +Chapter +SEQUENCES AND SERIES +8 +Reprint 2025-26 + +136 MATHEMATICS +The number of person’s ancestors for the first, second, third, …, tenth generations are +2, 4, 8, 16, 32, …, 1024. These numbers form what we call a sequence. +Consider the successive quotients that we obtain in the division of 10 by 3 at +different steps of division. In this process we get 3,3.3,3.33,3.333, ... and so on. These +quotients also form a sequence. The various numbers occurring in a sequence are +called its terms. We denote the terms of a sequence by a1, a2, a3, …, an, …, etc., the +subscripts denote the position of the term. The nth term is the number at the nth position +of the sequence and is denoted by an. The nth term is also called the general + term of the +sequence. +Thus, the terms of the sequence of person’s ancestors mentioned above are: +a1 = 2, a2 = 4, a3 = 8, …, a10 = 1024. +Similarly, in the example of successive quotients +a1 = 3, a2 = 3.3, a3 = 3.33, …, a6 = 3.33333, etc. +A sequence containing finite number of terms is called a finite sequence. For +example, sequence of ancestors is a finite sequence since it contains 10 terms (a fixed +number). +A sequence is called infinite, if it is not a finite sequence. For example, the +sequence of successive quotients mentioned above is an infinite sequence, infinite in +the sense that it never ends. +Often, it is possible to express the rule, which yields the various terms of a sequence +in terms of algebraic formula. Consider for instance, the sequence of even natural +numbers 2, 4, 6, … +Here +a1 = 2 = 2 × 1 +a2 = 4 = 2 × 2 +a3 = 6 = 2 × 3 +a4 = 8 = 2 × 4 +.... .... .... +.... .... .... +.... .... .... +.... .... .... +a23 = 46 = 2 × 23, a24 = 48 = 2 × 24, and so on. +In fact, we see that the nth term of this sequence can be written as an = 2n, +where n is a natural number. Similarly, in the sequence of odd natural numbers 1,3,5, …, +the nth term is given by the formula, an = 2n – 1, where n is a natural number. +In some cases, an arrangement of numbers such as 1, 1, 2, 3, 5, 8,.. has no visible +pattern, but the sequence is generated by the recurrence relation given by +a1 = a2 = 1 +a3 = a1 + a2 +an = an – 2 + an – 1, n > 2 +This sequence is called Fibonacci sequence. +Reprint 2025-26 + +SEQUENCES AND SERIES 137 +In the sequence of primes 2,3,5,7,…, we find that there is no formula for the nth +prime. Such sequence can only be described by verbal description. +In every sequence, we should not expect that its terms will necessarily be given +by a specific formula. However, we expect a theoretical scheme or a rule for generating +the terms a1, a2, a3,…,an,… in succession. +In view of the above, a sequence can be regarded as a function whose domain +is the set of natural numbers or some subset of it. Sometimes, we use the functional +notation a(n) for an. +8.3 Series +Let a1, a2, a3,…,an, be a given sequence. Then, the expression +a1 + a2 + a3 +,…+ an + ... +is called the series associated with the given sequence .The series is finite or infinite +according as the given sequence is finite or infinite. Series are often represented in +compact form, called sigma notation, using the Greek letter ∑(sigma) as means of +indicating the summation involved. Thus, the series a1 + a2 + a3 + ... + an is abbreviated +as +1 +n +k +k +a +=∑ +. +Remark When the series is used, it refers to the indicated sum not to the sum itself. +For example, 1 + 3 + 5 + 7 is a finite series with four terms. When we use the phrase +“sum of a series,” we will mean the number that results from adding the terms, the +sum of the series is 16. +We now consider some examples. +Example 1 Write the first three terms in each of the following sequences defined by +the following: +(i) +an = 2n + 5, +(ii) +an = +3 +4 +n − +. +Solution (i) Here an = 2n + 5 +Substituting n = 1, 2, 3, we get +a1 = 2(1) + 5 = 7, a2 = 9, a3 = 11 +Therefore, the required terms are 7, 9 and 11. +(ii) +Here an = +3 +4 +n − +. Thus, +1 +2 +3 +1 3 +1 +1 +0 +4 +2 +4 +a +, a +,a +− += += − += − += +Reprint 2025-26 + +138 MATHEMATICS +Hence, the first three terms are +1 +1 +2 +4 +– +, – + and 0. +Example 2 What is the 20th term of the sequence defined by +an = (n – 1) (2 – n) (3 + n) ? +Solution Putting n = 20 , we obtain +a20 = (20 – 1) (2 – 20) (3 + 20) + = 19 × (– 18) × (23) = – 7866. +Example 3 Let the sequence an be defined as follows: +a1 = 1, an = an – 1 + 2 for n ≥ 2. +Find first five terms and write corresponding series. +Solution We have +a1 = 1, a2 = a1 + 2 = 1 + 2 = 3, a3 = a2 + 2 = 3 + 2 = 5, +a4 = a3 + 2 = 5 + 2 = 7, a5 = a4 + 2 = 7 + 2 = 9. +Hence, the first five terms of the sequence are 1,3,5,7 and 9. The corresponding series +is 1 + 3 + 5 + 7 + 9 +... +EXERCISE 8.1 +Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth +terms are: +1. + an = n (n + 2) +2. +an = +1 +n +n+ +3. +an = 2n +4. + an = 2 +3 +6 +n − +5. + an = (–1)n–1 5n+1 +6. +an +2 +5 +4 +n +n ++ += +. +Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth +terms are: +7. + an = 4n – 3; a17, a24 +8. +an = +2 +7 +; +2n +n +a +9. +an = (–1)n – 1n3; a9 +10. +20 +( – 2) ; +3 +n +n n +a +a +n += ++ +. +Reprint 2025-26 + +SEQUENCES AND SERIES 139 +Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the +corresponding series: +11. +a1 = 3, an = 3an – 1 + 2 for all n > 1 +12. +a1 = – 1, an = +1 +na +n +−, n ≥ 2 +13. +a1 = a2 = 2, an = an – 1–1, n > 2 +14. +The Fibonacci sequence is defined by +1 = a1 = a2 and an = an – 1 + an – 2, n > 2. +Find +1 +n +n +a +a ++ , for n = 1, 2, 3, 4, 5 +8.4 Geometric Progression (G . P.) +Let us consider the following sequences: +(i) 2,4,8,16,..., (ii) +1 +1 1 +1 +9 27 81 243 +– +– +, +, +, +... (iii) . +,. +,. +,... +01 0001 000001 +In each of these sequences, how their terms progress? We note that each term, except +the first progresses in a definite order. +In (i), we have a +a +a +a +a +a +a +1 +2 +1 +3 +2 +4 +3 +2 +2 +2 +2 += += += += +, +, +, + and so on. +In (ii), we observe, a +a +a +a +a +a +a +1 +2 +1 +3 +2 +4 +3 +1 +9 +1 +3 +1 +3 +1 +3 += += += += +, +, +, + + + + and so on. +Similarly, state how do the terms in (iii) progress? It is observed that in each case, +every term except the first term bears a constant ratio to the term immediately preceding +it. In (i), this constant ratio is 2; in (ii), it is – 1 +3 and in (iii), the constant ratio is 0.01. +Such sequences are called geometric sequence or geometric progression abbreviated +as G.P. +A sequence a1, a2, a3, …, an, … is called geometric progression, if each term is +non-zero and +a +a +k +k ++ 1 = r (constant), for k ≥ 1. +By letting a1 = a, we obtain a geometric progression, a, ar, ar2, ar3,…., where a +is called the first term and r is called the common ratio of the G.P. Common ratio in +geometric progression (i), (ii) and (iii) above are 2, – 1 +3 and 0.01, respectively. +Reprint 2025-26 + +140 MATHEMATICS +As in case of arithmetic progression, the problem of finding the nth term or sum of n +terms of a geometric progression containing a large number of terms would be difficult +without the use of the formulae which we shall develop in the next Section. We shall +use the following notations with these formulae: +a = the first term, r = the common ratio, l = the last term, +n = the numbers of terms, +Sn = the sum of first n terms. +8.4.1 General term of a G .P. Let us consider a G.P. with first non-zero term ‘a’ and +common ratio ‘r’. Write a few terms of it. The second term is obtained by multiplying +a by r, thus a2 = ar. Similarly, third term is obtained by multiplying a2 by r. Thus, +a3 = a2r = ar2, and so on. +We write below these and few more terms. +1st term = a1 = a = ar1–1, 2nd term = a2 = ar = ar2–1, 3rd term = a3 = ar2 = ar3–1 +4th term = a4 = ar3 = ar4–1, 5th term = a5 = ar4 = ar5–1 +Do you see a pattern? What will be 16th term? +a16 = ar16–1 = ar15 +Therefore, the pattern suggests that the nth term of a G.P. is given by +an = arn–1. +Thus, a, G.P. can be written as a, ar, ar2, ar3, … arn – 1; a, ar, ar2,...,arn – 1... ;according +as G.P. is finite or infinite, respectively. +The series a + ar + ar2 + ... + arn–1 or a + ar + ar2 + ... + arn–1 +...are called +finite or infinite geometric series, respectively. +8.4.2. + Sum to n terms of a G .P. Let the first term of a G.P. be a and the common +ratio be r. Let us denote by Sn the sum to first n terms of G.P. Then +Sn = a + ar + ar2 +...+ arn–1 +... (1) +Case 1 +If r = 1, we have Sn = a + a + a + ... + a (n terms) = na +Case 2 +If r ≠ 1, multiplying (1) by r, we have + rSn = ar + ar2 + ar3 + ... + arn +... (2) +Subtracting (2) from (1), we get (1 – r) Sn = a – arn = a(1 – rn) +This gives +or +( +1) +S +1 +n +n +a r +r +− += +− +Example 4 Find the 10th and nth terms of the G.P. 5, 25,125,… . +Solution Here a = 5 and r = 5. Thus, a10 = 5(5)10–1 = 5(5)9 = 510 +and +an = arn–1 = 5(5)n–1 = 5n . +Reprint 2025-26 + +SEQUENCES AND SERIES 141 +Example 5 Which term of the G.P., 2,8,32, ... up to n terms is 131072? +Solution Let 131072 be the nth term of the given G.P. Here a = 2 and r = 4. +Therefore +131072 = an = 2(4)n – 1 +or +65536 = 4n – 1 +This gives +48 = 4n – 1. +So that n – 1 = 8, i.e., n = 9. Hence, 131072 is the 9th term of the G.P. +Example 6 In a G.P., the 3rd term is 24 and the 6th term is 192.Find the 10th term. +Solution Here, a +ar +3 +2 +24 += += +... (1) +and +a +ar +6 +5 +192 += += +... (2) +Dividing (2) by (1), we get r = 2. Substituting r = 2 in (1), we get a = 6. +Hence a10 = 6 (2)9 = 3072. +Example 7 Find the sum of first n terms and the sum of first 5 terms of the geometric +series +2 +4 +1 +3 +9 +... ++ ++ ++ +Solution Here a = 1 and r = 2 +3 . Therefore +Sn = +2 +1 +3 +(1 +) +2 +1 +1 +3 +n +n +a +r +r + + + + +− + + + + + + +− + + + + += +− +− + = +2 +3 1 +3 +n + + + + +− + + + + + + + + + + +In particular, +5 +5 +2 +S +3 1 +3 + + + + += +− + + + + + + + + + + + = +211 +3 +243 +× + = 211 +81 . +Example 8 How many terms of the G.P. +3 3 +3 2 4 +, , +,... are needed to give the +sum +3069 +512 ? +Solution Let n be the number of terms needed. Given that a = 3, r = 1 +2 and +3069 +S +512 +n = +Since +(1 +) +S +1 +n +n +a +– r +r += +− +Reprint 2025-26 + +142 MATHEMATICS +Therefore +1 +3(1 +) +3069 +1 +2 +6 1 +1 +512 +2 +1 +2 +n +n +− + + += += +− + + + + +− +or +3069 +3072 = +1 +1 +2n +− +or +1 +2n = +3069 +1 3072 +− +3 +1 +3072 +1024 += += +or +2n = 1024 = 210, which gives n = 10. +Example 9 The sum of first three terms of a G.P. is 13 +12 and their product is – 1. +Find the common ratio and the terms. +Solution Let a +r , a, ar be the first three terms of the G.P. Then + a +ar +a +r + ++ += 13 +12 +... (1) +and + +( ) ( +) +1 +a +a +ar +– +r + + += + + + + + ... (2) +From (2), we get a3 = – 1, i.e., a = – 1 (considering only real roots) +Substituting a = –1 in (1), we have +1 +13 +1 +12 +– +– –r +r += + or 12r2 + 25r + 12 = 0. +This is a quadratic in r, solving, we get +3 +4 +or +4 +3 +r +– +– += +. +Thus, the three terms of G.P. are : 4 +3 +–3 +3 +4 +–4 +, 1, + for = + and +, +1, + for = +3 +4 +4 +4 +3 +3 +– +r +– +r +, +Example10 Find the sum of the sequence 7, 77, 777, 7777, ... to n terms. +Solution This is not a G.P., however, we can relate it to a G.P. by writing the terms as +Sn = 7 + 77 + 777 + 7777 + ... to n terms +Reprint 2025-26 + +SEQUENCES AND SERIES 143 += 7 [9 99 999 9999 +to term] +9 +... +n ++ ++ ++ ++ += +2 +3 +4 +7 [(10 +1) +(10 +1) (10 +1) (10 +1) +terms] +9 +...n +− ++ +− ++ +− ++ +− ++ += +2 +3 +7 [(10 +10 +10 +terms) (1+1+1+... terms)] +9 +...n +– +n ++ ++ ++ += 7 10(10 +1) +7 10 (10 +1) +9 +10 +1 +9 +9 +n +n +n +n + + + + +− +− +− += +− + + + + +− + + + + + . +Example 11 A person has 2 parents, 4 grandparents, 8 great grandparents, and so on. +Find the number of his ancestors during the ten generations preceding his own. +Solution Here a = 2, r = 2 and n = 10 +Using the sum formula +Sn = +( +1) +1 +n +a r +r +− +− +We have +S10 = 2(210 – 1) = 2046 +Hence, the number of ancestors preceding the person is 2046. +8.4.3 Geometric Mean (G .M.) The geometric mean of two positive numbers a +and b is the number +ab . Therefore, the geometric mean of 2 and 8 is 4. We +observe that the three numbers 2,4,8 are consecutive terms of a G.P. This leads to a +generalisation of the concept of geometric means of two numbers. +Given any two positive numbers a and b, we can insert as many numbers as +we like between them to make the resulting sequence in a G.P. +Let G1, G2,…, Gn be n numbers between positive numbers a and b such that +a,G1,G2,G3,…,Gn,b is a G.P. Thus, b being the (n + 2)th term,we have +1 +n +b +ar ++ += +, or +1 +1 +n +b +r +a ++ + + +=  + + + +. +Hence +1 +1 +1 +G +n +b +ar +a a ++ + + += +=  + + + +, +2 +1 +2 +2 +G +n +b +ar +a a ++ + + += +=  + + + +, +3 +1 +3 +3 +G +n +b +ar +a a ++ + + += +=  + + + +, +1 +G +n +n +n +n +b +ar +a a ++ + + += +=  + + + +Reprint 2025-26 + +144 MATHEMATICS +Example12 Insert three numbers between 1 and 256 so that the resulting sequence + is a G.P. +Solution Let G1, G2,G3 be three numbers between 1 and 256 such that + 1, G1,G2,G3 ,256 is a G.P. +Therefore +256 = r4 giving r = ± 4 (Taking real roots only) +For r = 4, we have G1 = ar = 4, G2 = ar2 = 16, G3 = ar3 = 64 +Similarly, for r = – 4, numbers are – 4,16 and – 64. +Hence, we can insert 4, 16, 64 between 1 and 256 so that the resulting sequences are +in G.P. +8.5 Relationship Between A.M. and G.M. +Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively. +Then +A +andG +2 +a +b +ab ++ += += +Thus, we have +A – G = +2 +a b +ab ++ − + = +2 +2 +a +b +ab ++ − += ( +) +2 +0 +2 +a +b +− +≥ +... (1) +From (1), we obtain the relationship A ≥ G. +Example 13 If A.M. and G.M. of two positive numbers a and b are 10 and 8, +respectively, find the numbers. +Solution Given that +A.M. +10 +2 +a b ++ += += +... (1) +and +G.M. +8 +ab += += +... (2) +From (1) and (2), we get +a + b = 20 +... (3) +ab = 64 +... (4) +Putting the value of a and b from (3), (4) in the identity (a – b)2 = (a + b)2 – 4ab, +we get +(a – b)2 = 400 – 256 = 144 +or +a – b = ± 12 +... (5) +Reprint 2025-26 + +SEQUENCES AND SERIES 145 +Solving (3) and (5), we obtain +a = 4, b = 16 or a = 16, b = 4 +Thus, the numbers a and b are 4, 16 or 16, 4 respectively. +EXERCISE 8.2 +1. +Find the 20th and nth terms of the G.P. 5 5 5 +2 4 8 +, +, +, ... +2. +Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2. +3. +The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show +that q2 = ps. +4. +The 4th term of a G.P. is square of its second term, and the first term is – 3. +Determine its 7th term. +5. +Which term of the following sequences: +(a) +2 2 2 4 +is128 ? +, +, ,... +(b) +3 3 3 3 +is729 ? +, , +,... +(c) +1 1 +1 +1 +is +3 9 27 +19683 +, +, +,... +? +6. +For what values of x, the numbers +2 +7 +7 +2 +– +, x, – +are in G.P.? +Find the sum to indicated number of terms in each of the geometric progressions in +Exercises 7 to 10: +7. +0.15, 0.015, 0.0015, ... 20 terms. +8. +7 , +21 , 3 7 , ... n terms. +9. +1, – a, a2, – a3, ... n terms (if a ≠ – 1). +10. +x3, x5, x7, ... n terms (if x ≠ ± 1). +11. +Evaluate +11 +1 +(2 +3 ) +k +k= ++ +∑ +. +12. +The sum of first three terms of a G.P. is 39 +10 and their product is 1. Find the +common ratio and the terms. +13. +How many terms of G.P. 3, 32, 33, … are needed to give the sum 120? +14. +The sum of first three terms of a G.P. is 16 and the sum of the next three terms is +128. Determine the first term, the common ratio and the sum to n terms of the G.P. +15. +Given a G.P. with a = 729 and 7th term 64, determine S7. +Reprint 2025-26 + +146 MATHEMATICS +16. +Find a G.P. for which sum of the first two terms is – 4 and the fifth term is +4 times the third term. +17. +If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, +y, z are in G.P. +18. +Find the sum to n terms of the sequence, 8, 88, 888, 8888… . +19. +Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, +16, 32 and 128, 32, 8, 2, 1 +2 . +20. +Show that the products of the corresponding terms of the sequences a, ar, ar2, +…arn – 1 and A, AR, AR2, … ARn – 1 form a G.P, and find the common ratio. +21. +Find four numbers forming a geometric progression in which the third term is +greater than the first term by 9, and the second term is greater than the 4th by 18. +22. +If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that +aq – r br – pcP – q = 1. +23. +If the first and the nth term of a G.P. are a and b, respectively, and if P is the +product of n terms, prove that P2 = (ab)n. +24. +Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from +(n + 1)th to (2n)th term is 1 +nr +. +25. +If a, b, c and d are in G.P. show that +(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2 . +26. +Insert two numbers between 3 and 81 so that the resulting sequence is G.P. +27. +Find the value of n so that a +b +a +b +n +n +n +n ++ ++ ++ ++ +1 +1 + may be the geometric mean between +a and b. +28. +The sum of two numbers is 6 times their geometric mean, show that numbers +are in the ratio ( +) ( +) +3 +2 2 : 3 2 2 ++ +− +. +29. +If A and G be A.M. and G.M., respectively between two positive numbers, +prove that the numbers are A +A +G +A +G +( +)( +) +± ++ +− +. +30. +The number of bacteria in a certain culture doubles every hour. If there were 30 +bacteria present in the culture originally, how many bacteria will be present at the +end of 2nd hour, 4th hour and nth hour ? +Reprint 2025-26 + +SEQUENCES AND SERIES 147 +31. +What will Rs 500 amounts to in 10 years after its deposit in a bank which pays +annual interest rate of 10% compounded annually? +32. +If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then +obtain the quadratic equation. +Miscellaneous Examples +Example 14 If a, b, c, d and p are different real numbers such that +(a2 + b2 + c2)p2 – 2(ab + bc + cd) p + (b2 + c2 + d2) ≤ 0, then show that a, b, c and d +are in G.P. +Solution Given that +(a2 + b2 + c2) p2 – 2 (ab + bc + cd) p + (b2 + c2 + d2) ≤ 0 + ... (1) +But L.H.S. += (a2p2 – 2abp + b2) + (b2p2 – 2bcp + c2) + (c2p2 – 2cdp + d2), +which gives (ap – b)2 + (bp – c)2 + (cp – d)2 ≥ 0 + ... (2) +Since the sum of squares of real numbers is non negative, therefore, from (1) and (2), +we have, +(ap – b)2 + (bp – c)2 + (cp – d)2 = 0 +or +ap – b = 0, bp – c = 0, cp – d = 0 +This implies that b +c +d +p +a +b +c += += += +Hence a, b, c and d are in G.P. +Miscellaneous Exercise On Chapter 8 +1. +If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that +f(1) = 3 and +1 +( ) +120 +n +x +f x += += +∑ +, find the value of n. +2. +The sum of some terms of G.P. is 315 whose first term and the common ratio are +5 and 2, respectively. Find the last term and the number of terms. +3. +The first term of a G.P. is 1. The sum of the third term and fifth term is 90. +Find the common ratio of G.P. +4. +The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these +numbers in that order, we obtain an arithmetic progression. Find the numbers. + 5. + A G.P. consists of an even number of terms. If the sum of all the terms is 5 times +the sum of terms occupying odd places, then find its common ratio. +Reprint 2025-26 + +148 MATHEMATICS + 6. +If a +bx +a +bx +b +cx +b +cx +c +dx +c +dx x ++ +− += ++ +− += ++ +− +≠ +( +), +0 + then show that a, b, c and d are in G.P. + 7. +Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. +Prove that P2Rn = Sn. + 8. +If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P. + 9. +If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0, +where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15. +10. +The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show +that +( +) ( +) +2 +2 +2 +2 +: +a b +m +m – n +: m – +m – n += ++ + . +11. +Find the sum of the following series up to n terms: +(i) 5 + 55 +555 + … +(ii) .6 +. 66 +. 666+… +12. +Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms. +13. +A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to +pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid +amount. How much will the tractor cost him? +14. +Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to +pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid +amount. How much will the scooter cost him? +15. +A person writes a letter to four of his friends. He asks each one of them to copy +the letter and mail to four different persons with instruction that they move the +chain similarly. Assuming that the chain is not broken and that it costs 50 paise to +mail one letter. Find the amount spent on the postage when 8th set of letter is +mailed. +16. +A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. +Find the amount in 15th year since he deposited the amount and also calculate the +total amount after 20 years. +17. +A manufacturer reckons that the value of a machine, which costs him Rs. 15625, +will depreciate each year by 20%. Find the estimated value at the end of 5 years. +18. +150 workers were engaged to finish a job in a certain number of days. 4 workers +dropped out on second day, 4 more workers dropped out on third day and so on. +It took 8 more days to finish the work. Find the number of days in which the work +was completed. +Reprint 2025-26 + +SEQUENCES AND SERIES 149 +Summary +®By a sequence, we mean an arrangement of number in definite order according +to some rule. Also, we define a sequence as a function whose domain is the +set of natural numbers or some subsets of the type {1, 2, 3, ....k}. A sequence +containing a finite number of terms is called a finite sequence. A sequence is +called infinite if it is not a finite sequence. +®Let a1, a2, a3, ... be the sequence, then the sum expressed as a1 + a2 + a3 + ... +is called series. A series is called finite series if it has got finite number of +terms. +®A sequence is said to be a geometric progression or G.P., if the ratio of any +term to its preceding term is same throughout. This constant factor is called +the common ratio. Usually, we denote the first term of a G.P. by a and its +common ratio by r. The general or the nth term of G.P. is given by an= arn – 1. +The sum Sn of the first n terms of G.P. is given by +( +) +( +) + – 1 +1– +S +1 +1 +1 – +n +n +n +a r +a +r += +or +, if r +r – +r +≠ +®The geometric mean (G.M.) of any two positive numbers a and b is given by +ab i.e., the sequence a, G, b is G.P. +Historical Note +Evidence is found that Babylonians, some 4000 years ago, knew of arithmetic and +geometric sequences. According to Boethius (510), arithmetic and geometric +sequences were known to early Greek writers. Among the Indian mathematician, +Aryabhatta (476) was the first to give the formula for the sum of squares and cubes +of natural numbers in his famous work Aryabhatiyam, written around +499. He also gave the formula for finding the sum to n terms of an arithmetic +sequence starting with pth term. Noted Indian mathematicians Brahmgupta +Reprint 2025-26 + +150 MATHEMATICS +— v — +(598), Mahavira (850) and Bhaskara (1114-1185) also considered the sum of squares +and cubes. Another specific type of sequence having important applications in +mathematics, called Fibonacci sequence, was discovered by Italian mathematician +Leonardo Fibonacci (1170-1250). Seventeenth century witnessed the classification +of series into specific forms. In 1671 James Gregory used the term infinite series in +connection with infinite sequence. It was only through the rigorous development of +algebraic and set theoretic tools that the concepts related to sequence and series +could be formulated suitably. +Reprint 2025-26" +class_11,9,Straight Lines,ncert_books/class_11/kemh1dd/kemh109.pdf,"vG eometry, as a logical system, is a means and even the most powerful +means to make children feel the strength of the human spirit that is +of their own spirit. – H. FREUDENTHALv +9.1 Introduction +We are familiar with two-dimensional coordinate geometry +from earlier classes. Mainly, it is a combination of algebra +and geometry. A systematic study of geometry by the use +of algebra was first carried out by celebrated French +philosopher and mathematician René Descartes, in his book +‘La Géométry, published in 1637. This book introduced the +notion of the equation of a curve and related analytical +methods into the study of geometry. The resulting +combination of analysis and geometry is referred now as +analytical geometry. In the earlier classes, we initiated +the study of coordinate geometry, where we studied about +coordinate axes, coordinate plane, plotting of points in a +plane, distance between two points, section formulae, etc. All these concepts are the +basics of coordinate geometry. +Let us have a brief recall of coordinate geometry done in earlier classes. To +recapitulate, the location of the points (6, – 4) and +(3, 0) in the XY-plane is shown in Fig 9.1. +We may note that the point (6, – 4) is at 6 units +distance from the y-axis measured along the positive +x-axis and at 4 units distance from the x-axis +measured along the negative y-axis. Similarly, the +point (3, 0) is at 3 units distance from the y-axis +measured along the positive x-axis and has zero +distance from the x-axis. +We also studied there following important +formulae: +9 +Chapter +STRAIGHT LINES +René Descartes + (1596 -1650) +Fig 9.1 +Reprint 2025-26 + +152 +MATHEMATICS +I. +Distance between the points P (x1, y1) and Q (x2, y2) is +( +) +( +) +1 +2 +2 +2 +2 +1 +PQ +x – x +y – y += ++ +For example, distance between the points (6, – 4) and (3, 0) is +( +) +( +) +2 +2 +3 +6 +0 +4 +9 +16 +5 +− ++ ++ += ++ += + units. +II. +The coordinates of a point dividing the line segment joining the points (x1, y1) +and (x2, y2) internally, in the ratio m: n are + + + + + + ++ ++ ++ ++ +n +m +y +n +y +m +n +m +x +n +x +m +1 +2 +1 +2 +, +. +For example, the coordinates of the point which divides the line segment joining +A (1, –3) and B (–3, 9) internally, in the ratio 1: 3 are given by +1 ( +3) +3 1 +0 +1 3 +. +. +x +− ++ += += ++ +and +( +) +1.9 + 3. –3 += += 0. +1+ 3 +y +III. In particular, if m = n, the coordinates of the mid-point of the line segment +joining the points (x1, y1) and (x2, y2) are + + + + + + ++ ++ +2 +, +2 +2 +1 +2 +1 +y +y +x +x +. +IV. +Area of the triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is +( +) +( +) +( +) +1 +2 +3 +2 +3 +3 +1 +1 +2 +1 +2 +− ++ +− ++ +− +y +y +y +y +y +y +x +x +x +. +For example, the area of the triangle, whose vertices are (4, 4), (3, – 2) and (– 3, 16) is +54 +1 4( 2 +16) +3(16 +4) +( 3)(4 +2) +27. +2 +2 +− +−− ++ +− ++ − ++ += += +Remark If the area of the triangle ABC is zero, then three points A, B and C lie on +a line, i.e., they are collinear. +In the this Chapter, we shall continue the study of coordinate geometry to study +properties of the simplest geometric figure – straight line. Despite its simplicity, the +line is a vital concept of geometry and enters into our daily experiences in numerous +interesting and useful ways. Main focus is on representing the line algebraically, for +which slope is most essential. +9.2 Slope of a Line +A line in a coordinate plane forms two angles with the x-axis, which are supplementary. +Reprint 2025-26 + +STRAIGHT LINES 153 +The angle (say) θ made by the line l with positive +direction of x-axis and measured anti clockwise +is called the inclination of the line. Obviously +0° ≤ θ ≤ 180° (Fig 9.2). +We observe that lines parallel to x-axis, or +coinciding with x-axis, have inclination of 0°. The +inclination of a vertical line (parallel to or +coinciding with y-axis) is 90°. +Definition 1 If θ is the inclination of a line +l, then tan θ is called the slope or gradient of +the line l. +The slope of a line whose inclination is 90° is not +defined. +The slope of a line is denoted by m. +Thus, m = tan θ, θ ≠ 90° +It may be observed that the slope of x-axis is zero and slope of y-axis is not defined. +9.2.1 Slope of a line when coordinates of any two points on the line are given +We know that a line is completely determined when we are given two points on it. +Hence, we proceed to find the slope of a +line in terms of the coordinates of two points +on the line. +Let P(x1, y1) and Q(x2, y2) be two +points on non-vertical line l whose inclination +is θ. Obviously, x1 ≠ x2, otherwise the line +will become perpendicular to x-axis and its +slope will not be defined. The inclination of +the line l may be acute or obtuse. Let us +take these two cases. +Draw perpendicular QR to x-axis and +PM perpendicular to RQ as shown in +Figs. 9.3 (i) and (ii). +Case 1 When angle θ is acute: +In Fig 9.3 (i), ∠MPQ = θ. +... (1) +Therefore, slope of line l = m = tan θ. +But in ∆MPQ, we have +2 +1 +2 +1 +MQ +tanθ +. +MP +y +y +x +x +− += += +− +... (2) +Fig 9.2 +Fig 9. 3 (i) +Reprint 2025-26 + +154 +MATHEMATICS +From equations (1) and (2), we have +2 +1 +2 +1 +. +y +y +m +x +x +− += +− +Case II When angle θ is obtuse: +In Fig 9.3 (ii), we have + ∠MPQ = 180° – θ. +Therefore, θ = 180° – ∠MPQ. +Now, slope of the line l + m = tan θ += tan ( 180° – ∠MPQ) = – tan ∠MPQ += +2 +1 +1 +2 +MQ +MP +y +y +x +x +− +− += − +− + = +2 +1 +2 +1 +y +y . +x +x +− +− +Consequently, we see that in both the cases the slope m of the line through the points +(x1, y1) and (x2, y2) is given by +2 +1 +2 +1 +y +y +m +x +x +− += +− +. +9.2.2 Conditions for parallelism and perpendicularity of lines in terms of their +slopes In a coordinate plane, suppose that non-vertical lines l1 and l2 have slopes m1 +and m2, respectively. Let their inclinations be α and +β, respectively. +If the line l1 is parallel to l2 (Fig 9.4), then their +inclinations are equal, i.e., +α = β, and hence, tan α = tan β +Therefore +m1 = m2, i.e., their slopes are equal. +Conversely, if the slope of two lines l1 and l2 +is same, i.e., +m1 = m2. +Then +tan α = tan β. +By the property of tangent function (between 0° and 180°), α = β. +Therefore, the lines are parallel. +Fig 9. 3 (ii) +Fig 9. 4 +Reprint 2025-26 + +STRAIGHT LINES 155 +Hence, two non vertical lines l1 and l2 are parallel if and only if their slopes +are equal. +If the lines l1 and l2 are perpendicular (Fig 9.5), then β = α + 90°. +Therefore, tan β = tan (α + 90°) += – cot α = +1 +tanα +− +i.e., +m2 = +1 +1 +m +− +or +m1 m2 = – 1 +Conversely, if m1 m2 = – 1, i.e., tan α tan β = – 1. +Then tan α = – cot β = tan (β + 90°) or tan (β – 90°) +Therefore, +α and β differ by 90°. +Thus, lines l1 and l2 are perpendicular to each other. +Hence, two non-vertical lines are perpendicular to each other if and only if +their slopes are negative reciprocals of each other, +i.e., +m2= +1 +1 +m +− + or, m1 m2 = – 1. +Let us consider the following example. +Example 1 Find the slope of the lines: +(a) +Passing through the points (3, – 2) and (–1, 4), +(b) +Passing through the points (3, – 2) and (7, – 2), +(c) +Passing through the points (3, – 2) and (3, 4), +(d) +Making inclination of 60° with the positive direction of x-axis. +Solution (a) The slope of the line through (3, – 2) and (– 1, 4) is + +4 +( 2) +6 +3 +1 3 +4 +2 +m +−− += += += − +−− +− + . +(b) The slope of the line through the points (3, – 2) and (7, – 2) is + +–2 – (–2) +0 += += += 0 +7 – 3 +4 +m +. +(c) The slope of the line through the points (3, – 2) and (3, 4) is +Fig 9. 5 +Reprint 2025-26 + +156 +MATHEMATICS +4 – (–2) +6 += += +3 – 3 +0 +m +, which is not defined. +(d) Here inclination of the line α = 60°. Therefore, slope of the line is +m = tan 60° = +3 . +9.2.3 Angle between two lines When we think about more than one line in a plane, +then we find that these lines are either intersecting or parallel. Here we will discuss the +angle between two lines in terms of their slopes. +Let L1 and L2 be two non-vertical lines with slopes m1 and m2, respectively. If α1 +and α2 are the inclinations of lines L1 and L2, respectively. Then +α +tan +and +α +tan +2 +2 +1 +1 += += +m +m +. +We know that when two lines intersect each other, they make two pairs of +vertically opposite angles such that sum of any two adjacent angles is 180°. Let θ and +φ be the adjacent angles between the lines L1 and L2 (Fig 9.6). Then +θ = α2 – α1 and α1, α2 ≠ 90°. +Therefore tan θ = tan (α2 – α1) +2 +1 +2 +1 +1 +2 +1 +2 +tan +tan +1 tan +tan +1 +m +m +m m +α +α +α +α +− +− += += ++ ++ + (as 1 + m1m2 ≠ 0) +and φ = 180° – θ so that +tan φ = tan (180° – θ ) = – tan θ = +2 +1 +1 +2 +– +1 +m +m +m m +− ++ +, as 1 + m1m2 ≠ 0 +Fig 9. 6 +Now, there arise two cases: +Reprint 2025-26 + +STRAIGHT LINES 157 +Case I If +2 +1 +1 +2 +1 +– +m +m +m m ++ +is positive, then tan θ will be positive and tan φ will be negative, +which means θ will be acute and φ will be obtuse. +Case II If +2 +1 +1 +2 +1 +– +m +m +m m ++ +is negative, then tan θ will be negative and tan φ will be positive, +which means that θ will be obtuse and φ will be acute. +Thus, the acute angle (say θ) between lines L1 and L2 with slopes m1 and m2, +respectively, is given by +2 +1 +1 +2 +1 +2 +tan θ +, as +1 +0 +1 +m +m +m m +m m +− += ++ +≠ ++ +... (1) +The obtuse angle (say φ) can be found by using φ =1800 – θ. +Example 2 If the angle between two lines is π +4 and slope of one of the lines is 1 +2 , find +the slope of the other line. +Solution We know that the acute angle θ between two lines with slopes m1 and m2 +is given by +2 +1 +1 +2 +tanθ +1 +m +m +m m +− += ++ +... (1) +Let m1 = 2 +1 +, m2 = m and θ = π +4 . +Now, putting these values in (1), we get +1 +1 +π +2 +2 +tan +or 1 +1 +1 +4 +1 +1 +2 +2 +m +m +, +m +m +− +− += += ++ ++ +which gives +1 +1 +2 +2 +1 +or +1 +1 +1 +1 +1 +2 +2 +m +m +– . +m +m +− +− += += ++ ++ +Reprint 2025-26 + +158 +MATHEMATICS +Fig 9.7 +1 +Therefore +3 +or +3 +m +m +. += += − +Hence, slope of the other line is +3 or +1 +3 +− +. Fig 9.7 explains the +reason of two answers. +Example 3 Line through the points (–2, 6) and (4, 8) is perpendicular to the line +through the points (8, 12) and (x, 24). Find the value of x. +Solution Slope of the line through the points (– 2, 6) and (4, 8) is + +( +) +1 +8 +6 +2 +1 +4 +2 +6 +3 +m +− += += += +−− +Slope of the line through the points (8, 12) and (x, 24) is + +2 +24 12 +12 +8 +8 +m +x +x +− += += +− +− +Since two lines are perpendicular, +m1 m2 = –1, which gives + 1 +12 +1 or += 4 +3 +8 +x +x +× += − +− +. +EXERCISE 9.1 +1. +Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), +(5, – 5) and (– 4, –2). Also, find its area. +2. +The base of an equilateral triangle with side 2a lies along the y-axis such that the +mid-point of the base is at the origin. Find vertices of the triangle. +Reprint 2025-26 + +STRAIGHT LINES 159 +3. +Find the distance between P (x1, y1) and Q (x2, y2) when : (i) PQ is parallel to the +y-axis, (ii) PQ is parallel to the x-axis. +4. +Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4). +5. +Find the slope of a line, which passes through the origin, and the mid-point of the +line segment joining the points P (0, – 4) and B (8, 0). +6. +Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and +(–1, –1) are the vertices of a right angled triangle. +7. +Find the slope of the line, which makes an angle of 30° with the positive direction +of y-axis measured anticlockwise. +8. +Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and +(–3, 2) are the vertices of a parallelogram. +9. +Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2). +10. The slope of a line is double of the slope of another line. If tangent of the angle +between them is 3 +1 +, find the slopes of the lines. +11. +A line passes through (x1, y1) and (h, k). If slope of the line is m, show that +k – y1 = m (h – x1). +9.3 Various Forms of the Equation of a Line +We know that every line in a plane contains infinitely many points on it. This relationship +between line and points leads us to find the solution of the following problem: +How can we say that a given point lies on the given line? Its answer may be that +for a given line we should have a definite condition on the points lying on the line. +Suppose P (x, y) is an arbitrary point in the XY-plane and L is the given line. For the +equation of L, we wish to construct a statement or condition for the point P that is +true, when P is on L, otherwise false. Of course the statement is merely an algebraic +equation involving the variables x and y. Now, we will discuss the equation of a line +under different conditions. +9.3.1 Horizontal and vertical lines If a horizontal line L is at a distance a from the x- +axis then ordinate of every point lying on the line is either a or – a [Fig 9.8 (a)]. +Therefore, equation of the line L is either y = a or y = – a. Choice of sign will depend +upon the position of the line according as the line is above or below the y-axis. Similarly, +the equation of a vertical line at a distance b from the y-axis is either x = b or +x = – b [Fig 9.8(b)]. +Reprint 2025-26 + +160 +MATHEMATICS +Example 4 Find the equations of the lines +parallel to axes and passing through +(– 2, 3). +Solution Position of the lines is shown in the +Fig 9.9. The y-coordinate of every point on the +line parallel to x-axis is 3, therefore, equation +of the line parallel tox-axis and passing through +(– 2, 3) is y = 3. Similarly, equation of the line +parallel to y-axis and passing through (– 2, 3) +is x = – 2. +9.3.2 Point-slope form Suppose that +P0 (x0, y0) is a fixed point on a non-vertical +line L, whose slope is m. Let P (x, y) be an +arbitrary point on L (Fig 9.10). +Then, by the definition, the slope of L is given +by +( +) +x +x +m +y +y +x +x +y +y +m +0 +0 +0 +0 + , +i.e. +, +− += +− +− +− += + ...(1) +Since the point P0 (x0 , y0) along with +all points (x, y) on L satisfies (1) and no other +point in the plane satisfies (1). Equation (1) +is indeed the equation for the given line L. +Fig 9.8 +Fig 9.10 +Fig 9.9 +Reprint 2025-26 + +STRAIGHT LINES 161 +Thus, the point (x, y) lies on the line with slope m through the fixed point (x0, y0), +if and only if, its coordinates satisfy the equation +y – y0 = m (x – x0) +Example 5 Find the equation of the line through (– 2, 3) with slope – 4. +Solution Here m = – 4 and given point (x0 , y0) is (– 2, 3). +By slope-intercept form formula +(1) above, equation of the given +line is +y – 3 = – 4 (x + 2) or +4x + y + 5 = 0, which is the +required equation. +9.3.3 Two-point form Let the +line L passes through two given +points P1 (x1, y1) and P2 (x2, y2). +Let P (x, y) be a general point +on L (Fig 9.11). +The three points P1, P2 and P are +collinear, therefore, we have +slope of P1P = slope of P1P2 +i.e., +1 +2 +1 +2 +1 +1 +1 +1 +2 +1 +2 +1 +or +y +y +y +y +y +y +, +y +( x +). +y +x +x +x +x +x +x +x +− +− +− += +− += +− +− +− +− +Thus, equation of the line passing through the points (x1, y1) and (x2, y2) is given by + +) +( +1 +1 +2 +1 +2 +1 +x +x +x +x +y +y +y +y +− +− +− += +− +... (2) +Example 6 Write the equation of the line through the points (1, –1) and (3, 5). +Solution Here x1 = 1, y1 = – 1, x2 = 3 and y2 = 5. Using two-point form (2) above +for the equation of the line, we have + +( +) +( +) ( +) +5 – –1 +– –1 = +– 1 +3 – 1 +y +x +or –3x + y + 4 = 0, which is the required equation. +9.3.4 Slope-intercept form Sometimes a line is known to us with its slope and an +intercept on one of the axes. We will now find equations of such lines. +Fig 9.11 +Reprint 2025-26 + +162 +MATHEMATICS +Fig 9.12 +Case I Suppose a line L with slope m cuts +the y-axis at a distance c from the origin +(Fig 9.12). The distance c is called the y- +intercept of the line L. Obviously, +coordinates of the point where the line meet +the y-axis are (0, c). Thus, L has slope m +and passes through a fixed point (0, c). +Therefore, by point-slope form, the equation +of L is +0 +or +y +c +m( x +) +y +mx +c +−= +− += ++ +Thus, the point (x, y) on the line with slope m and y-intercept c lies on the line if and +only if +y = mx +c +...(3) +Note that the value of c will be positive or negative according as the intercept is made +on the positive or negative side of the y-axis, respectively. +Case II Suppose line L with slope m makes x-intercept d. Then equation of L is +y = m(x – d) +... (4) +Students may derive this equation themselves by the same method as in Case I. +Example 7 Write the equation of the lines for which tan θ = 2 +1 +, where θ is the +inclination of the line and (i) y-intercept is +3 +2 +– + (ii) x-intercept is 4. +Solution (i) Here, slope of the line is m = tan θ = 2 +1 +and y - intercept c = – 2 +3 +. +Therefore, by slope-intercept form (3) above, the equation of the line is +0 +3 +2 +or +2 +3 +2 +1 += ++ +− +− += +x +y +x +y +, +which is the required equation. +(ii) Here, we have m = tan θ = 2 +1 +and d = 4. +Therefore, by slope-intercept form (4) above, the equation of the line is +0 +4 +2 +or +) +4 +( +2 +1 += ++ +− +− += +x +y +x +y +, +which is the required equation. +Reprint 2025-26 + +STRAIGHT LINES 163 +9.3.5 Intercept - form Suppose a line L makes x-intercept a and y-intercept b on the +axes. Obviously L meets x-axis at the point +(a, 0) and y-axis at the point (0, b) (Fig .9.13). +By two-point form of the equation of the line, +we have +0 +0 +( +) +or +0 +b +y +x +a +ay +bx +ab +a +− +− += +− += − ++ +− +, +i.e., +1 += ++ +b +y +a +x +. +Thus, equation of the line making intercepts +a and b on x-and y-axis, respectively, is +1 += ++ +b +y +a +x +... (5) +Example 8 Find the equation of the line, which makes intercepts –3 and 2 on the +x- and y-axes respectively. +Solution Here a = –3 and b = 2. By intercept form (5) above, equation of the line is + +1 +or +2 +3 +6 +0 +3 +2 +x +y +x +y ++ += +− ++ += +− +. +Any equation of the form Ax + By + C = 0, where A and B are not zero simultaneously +is called general linear equation or general equation of a line. +EXERCISE 9.2 +In Exercises 1 to 8, find the equation of the line which satisfy the given conditions: +1. +Write the equations for the x-and y-axes. +2. +Passing through the point (– 4, 3) with slope 2 +1 +. +3. +Passing through (0, 0) with slope m. +4. +Passing through ( +) +3 +2 +,2 +and inclined with the x-axis at an angle of 75o. +5. +Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2. +6. +Intersecting the y-axis at a distance of 2 units above the origin and making an +angle of 30o with positive direction of the x-axis. +Fig 9.13 +Reprint 2025-26 + +164 +MATHEMATICS +7. +Passing through the points (–1, 1) and (2, – 4). +8. +The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the +median through the vertex R. +9. +Find the equation of the line passing through (–3, 5) and perpendicular to the line +through the points (2, 5) and (–3, 6). +10. +A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides +it in the ratio 1: n. Find the equation of the line. +11. +Find the equation of a line that cuts off equal intercepts on the coordinate axes +and passes through the point (2, 3). +12. + Find equation of the line passing through the point (2, 2) and cutting off intercepts +on the axes whose sum is 9. +13. + Find equation of the line through the point (0, 2) making an angle 2π +3 with the +positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis +at a distance of 2 units below the origin. +14. +The perpendicular from the origin to a line meets it at the point (–2, 9), find the +equation of the line. +15. +The length L (in centimetre) of a copper rod is a linear function of its Celsius +temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134 +when C = 110, express L in terms of C. +16. +The owner of a milk store finds that, he can sell 980 litres of milk each week at +Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear +relationship between selling price and demand, how many litres could he sell +weekly at Rs 17/litre? +17. +P (a, b) is the mid-point of a line segment between axes. Show that equation +of the line is +2 += ++ +b +y +a +x +. +18. +Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find +equation of the line. +19. +By using the concept of equation of a line, prove that the three points (3, 0), +(– 2, – 2) and (8, 2) are collinear. +9.4 Distance of a Point From a Line +The distance of a point from a line is the length of the perpendicular drawn from the +point to the line. Let L : Ax + By + C = 0 be a line, whose distance from the point +P (x1, y1) is d. Draw a perpendicular PM from the point P to the line L (Fig 9.14). If the +Reprint 2025-26 + +STRAIGHT LINES 165 +line meets the x-and y-axes at the points Q and R, respectively. Then, coordinates of +the points are Q +C +0 +A , + + +− + + + +and R +C +0 +B +, + + +− + + + +. Thus, the area of the triangle PQR is +given by +area +1 +( PQR) +PM.QR +2 +∆ += +, which gives +2 area (∆PQR) +PM = +QR +... (1) +Also, area +( +) +1 +1 +1 +1 +C +C +C +(∆PQR) +0 +0 +0 +2 +B +A +B +x +y +y + + + + + += ++ ++ − +− +− ++ +− + + + + + + + + + + + +2 +1 +1 +1 +C +C +C +2 +B +A +AB +y +x += ++ ++ +or +1 +1 +C +2 area (∆PQR) +A +B +C +and +AB . +, +y +x += ++ ++ +( +) +2 +2 +2 +2 +C +C +C +QR +0 +0 +A +B +AB +A +B + + += ++ += ++ ++ +− + + + + +Substituting the values of area (∆PQR) and QR in (1), we get + +1 +1 +2 +2 +A +B +C +PM +A +B +y +x + ++ += ++ +Fig 9.14 +Reprint 2025-26 + +166 +MATHEMATICS + or +1 +1 +2 +2 +A +B +C +A +B +y +x +d ++ ++ += ++ +. +Thus, the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1, y1) +is given by + +1 +1 +2 +2 +A +B +C +A +B +y +x +d ++ ++ += ++ +. +9.4.1 Distance between two +parallel lines We know that slopes of +two parallel lines are equal. +Therefore, two parallel lines can be +taken in the form +y = mx + c1 + ... (1) +and +y = mx + c2 + ... (2) +Line (1) will intersect x-axis at the point +A +1 +0 +c , +m + + +− + + + +as shown in Fig 9.15. +Distance between two lines is equal to the length of the perpendicular from point +A to line (2). Therefore, distance between the lines (1) and (2) is +( +) +( +) +1 +2 +1 +2 +2 +2 +or = +1 +1 +c +m +c +c +c +m +d +m +m + + +− +− ++ − + + +− + + ++ ++ +. +Thus, the distance d between two parallel lines +1 +y +mx +c += ++ +and +2 +y +mx +c += ++ +is given by +1 +2 +2 += +1 +c +c +d +m +− ++ + . +If lines are given in general form, i.e., Ax + By + C1 = 0 and Ax + By + C2 = 0, +then above formula will take the form +1 +2 +2 +2 +C +C +A +B +d +− += ++ +Students can derive it themselves. +Fig 9.15 +Reprint 2025-26 + +STRAIGHT LINES 167 +Example 9 Find the distance of the point (3, – 5) from the line 3x – 4y –26 = 0. +Solution Given line is +3x – 4y –26 = 0 +... (1) +Comparing (1) with general equation of line Ax + By + C = 0, we get +A = 3, B = – 4 and C = – 26. +Given point is (x1, y1) = (3, –5). The distance of the given point from given line is +( +)( +) +( +) +1 +1 +2 +2 +2 +2 +3 3 +4 +5 +26 +A +B +C +3. +5 +A +B +3 +4 +. +– +– +– +x +y +d +– ++ ++ ++ += += += ++ ++ +Example 10 Find the distance between the parallel lines 3x – 4y +7 = 0 and +3x – 4y + 5 = 0 +Solution Here A = 3, B = –4, C1 = 7 and C2 = 5. Therefore, the required distance is +( +) +2 +2 +7 +5 +2. +5 +3 +4 +– +d +– += += ++ +EXERCISE 9.3 +1. +Reduce the following equations into slope - intercept form and find their slopes +and the y - intercepts. +(i) +x + 7y = 0, +(ii) +6x + 3y – 5 = 0, +(iii) y = 0. +2. +Reduce the following equations into intercept form and find their intercepts on +the axes. +(i) +3x + 2y – 12 = 0, (ii) +4x – 3y = 6, +(iii) 3y + 2 = 0. +3. +Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2). +4. +Find the points on the x-axis, whose distances from the line +1 +3 +4 +x +y ++ += are 4 units. +5. +Find the distance between parallel lines +(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 (ii) l (x + y) + p = 0 and l (x + y) – r = 0. +6. +Find equation of the line parallel to the line 3 +4 +2 +0 +x +y +− ++ += +and passing through +the point (–2, 3). +7. +Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having +x intercept 3. +8. +Find angles between the lines +.1 +3 +and +1 +3 += ++ += ++ +y +x +y +x +9. +The line through the points (h, 3) and (4, 1) intersects the line7 +9 +19 +0 +x +y +. +− +− += +at right angle. Find the value of h. +Reprint 2025-26 + +168 +MATHEMATICS +10. +Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is + A (x –x1) + B (y – y1) = 0. +11. +Two lines passing through the point (2, 3) intersects each other at an angle of 60o. +If slope of one line is 2, find equation of the other line. +12. +Find the equation of the right bisector of the line segment joining the points (3, 4) +and (–1, 2). +13. +Find the coordinates of the foot of perpendicular from the point (–1, 3) to the +line 3x – 4y – 16 = 0. +14. +The perpendicular from the origin to the line y = mx + c meets it at the point +(–1, 2). Find the values of m and c. +15. +If p and q are the lengths of perpendiculars from the origin to the +lines +θ +2 +cos +θ +sin +θ +cos +k +y +x += +− + and x sec θ + y cosec θ = k, respectively, prove +that p2 + 4q2 = k2. +16. +In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation +and length of altitude from the vertex A. +17. +If p is the length of perpendicular from the origin to the line whose intercepts on +the axes are a and b, then show that +. +1 +1 +1 +2 +2 +2 +b +a +p ++ += +Miscellaneous Examples +Example 11 If the lines 2 +3 +0 +5 +3 +0 +x +y +, +x +ky ++ +− += ++ +− += + and 3 +2 +0 +x +y +− +− += +are +concurrent, find the value of k. +Solution Three lines are said to be concurrent, if they pass through a common point, +i.e., point of intersection of any two lines lies on the third line. Here given lines are + 2x + y – 3 = 0 +... (1) + 5x + ky – 3 = 0 +... (2) + 3x – y – 2 = 0 +... (3) +Solving (1) and (3) by cross-multiplication method, we get + +1 += += +or += 1, += 1 +–2 – 3 +–9 + 4 +–2 – 3 +x +y +x +y +. +Therefore, the point of intersection of two lines is (1, 1). Since above three lines are +concurrent, the point (1, 1) will satisfy equation (2) so that +5.1 + k .1 – 3 = 0 or k = – 2. +Reprint 2025-26 + +STRAIGHT LINES 169 +Example 12 Find the distance of the line 4x – y = 0 from the point P (4, 1) measured +along the line making an angle of 135° with the positive x-axis. +Solution Given line is 4x – y = 0 +... (1) +In order to find the distance of the +line (1) from the point P (4, 1) along +another line, we have to find the point +of intersection of both the lines. For +this purpose, we will first find the +equation of the second line +(Fig 9.16). Slope of second line is +tan 135° = –1. Equation of the line +with slope – 1 through the point +P (4, 1) is +y – 1 = – 1 (x – 4) or x + y – 5 = 0 + ... (2) +Solving (1) and (2), we get x = 1 and y = 4 so that point of intersection of the two lines +is Q (1, 4). Now, distance of line (1) from the point P (4, 1) along the line (2) += the distance between the points P (4, 1) and Q (1, 4). += ( +) +( +) +2 +2 +1 +4 +4 1 +3 2 units. +− ++ +− += +Example 13 Assuming that straight lines work as the plane mirror for a point, find +the image of the point (1, 2) in the line x – 3y + 4 = 0. +Solution Let Q (h, k) is the image of the point P (1, 2) in the line +x – 3y + 4 = 0 +... (1) +Therefore, the line (1) is the perpendicular bisector of line segment PQ (Fig 9.17). +Fig 9.16 +(1, 4) +Fig 9.17 +Reprint 2025-26 + +170 +MATHEMATICS +Hence Slope of line PQ = +1 +Slope of line +3 +4 +0 +x +y +− +− ++ += +, +so that +2 +1 +or +3 +5 +1 +1 +3 +k +h +k +h +− +− += ++ += +− +... (2) +and the mid-point of PQ, i.e., point + + + + + + ++ ++ +2 +2 +, +2 +1 k +h +will satisfy the equation (1) so that +3 +3 +or +0 +4 +2 +2 +3 +2 +1 +− += +− += ++ + + + + + + ++ +− ++ +k +h +k +h +... (3) +Solving (2) and (3), we get h = 5 +6 +and k = 5 +7 +. +Hence, the image of the point (1, 2) in the line (1) is +6 +7 +5 +5 +, + + + + + +. +Example 14 Show that the area of the triangle formed by the lines +y = m1x + c1, y = m2x + c2 and x = 0 is ( +) +2 +1 +2 +1 +2 +2 +c +c +m +m +– +− +. +Solution Given lines are +y = m1 x + c1 +... (1) +y = m2 x + c2 +... (2) +x = 0 +... (3) +We know that line y = mx + c meets +the line x = 0 (y-axis) at the point +(0, c). Therefore, two vertices of the +triangle formed by lines (1) to (3) are +P (0, c1) and Q (0, c2) (Fig 9.18). +Third vertex can be obtained by solving +equations (1) and (2). Solving (1) and +(2), we get +Fig 9.18 +Reprint 2025-26 + +STRAIGHT LINES 171 +( +) +( +) +( +) +( +) +2 +1 +1 2 +2 1 +1 +2 +1 +2 +and +c +c +m c +m c +x +y +m +m +m +m +− +− += += +− +− +Therefore, third vertex of the triangle is R +( +) +( +) +( +) +( +) +2 +1 +1 2 +2 1 +1 +2 +1 +2 +c +c +m c +m c +, +m +m +m +m + + +− +− + + + + +− +− + +. +Now, the area of the triangle is +( +) +( +) +2 +1 +2 +1 2 +2 1 +2 +1 +1 2 +2 1 +2 +2 +1 +1 +1 +2 +1 +2 +1 +2 +1 +2 +1 +0 +0 +2 +2 +c +c +m c +m c +c +c +m c +m c +c +c +c +c +m +m +m +m +m +m +m +m + + + + +− +− +− += +− ++ +− ++ +− += + + + + +− +− +− +− + + + + +− +Example 15 A line is such that its segment +between the lines +5x – y + 4 = 0 and 3x + 4y – 4 = 0 is bisected at +the point (1, 5). Obtain its equation. +Solution Given lines are +5x – y + 4 = 0 + ... (1) +3x + 4y – 4 = 0 + ... (2) +Let the required line intersects the lines (1) and +(2) at the points, (α1, β1) and (α2, β2), respectively +(Fig 9.19). Therefore +5α1 – β1 + 4 = 0 and +3 α2 + 4 β2 – 4 = 0 +or β1 = 5α1 + 4 and +2 +2 +4–3α +β +4 += +. +We are given that the mid point of the segment of +the required line between (α1, β1) and (α2, β2) is (1, 5). Therefore +1 +1 +2 +2 +β + β ++ α +α += 1 and += 5 +2 +2 +, +or +2 +1 +2 +1 +4 – 3α +5α + 4 + +4 +α + += 2 and += 5, +α +2 +or α1 + α2 = 2 and 20 α1 – 3 α2 = 20 +... (3) +Solving equations in (3) for α1 and α2, we get +Fig 9.19 +Reprint 2025-26 + +172 +MATHEMATICS +1 +26 +α = 23 and +2 +20 += +α +23 and hence, +23 +222 +4 +23 +26 +.5 +β1 += ++ += +. +Equation of the required line passing through (1, 5) and (α1, β1) is +)1 +( +1 +α +5 +β +5 +1 +1 +− +− +− += +− +x +y +or +222 +5 +23 +5 +( +1) +26 +1 +23 +y +x +− +− += +− +− +or +107x – 3y – 92 = 0, +which is the equation of required line. +Example 16 Show that the path of a moving point such that its distances from two +lines 3x – 2y = 5 and 3x + 2y = 5 are equal is a straight line. +Solution Given lines are +3x – 2y = 5 +… (1) +and +3x + 2y = 5 +… (2) +Let (h, k) is any point, whose distances from the lines (1) and (2) are equal. Therefore +5 +2 +3 +5 +2 +3 +or +4 +9 +5 +2 +3 +4 +9 +5 +2 +3 +− ++ += +− +− ++ +− ++ += ++ +− +− +k +h +k +h +k +h +k +h +, +which gives 3h – 2k – 5 = 3h + 2k – 5 or – (3h – 2k – 5) = 3h + 2k – 5. +Solving these two relations we get k = 0 or h = 3 +5 +. Thus, the point (h, k) satisfies the +equations y = 0 or x = 3 +5 +, which represent straight lines. Hence, path of the point +equidistant from the lines (1) and (2) is a straight line. +Miscellaneous Exercise on Chapter 9 +1. Find the values of k for which the line (k–3) x – (4 – k2) y + k2 –7k + 6 = 0 is +(a) +Parallel to the x-axis, +(b) +Parallel to the y-axis, +(c) +Passing through the origin. +2. +Find the equations of the lines, which cut-off intercepts on the axes whose sum +and product are 1 and – 6, respectively. +Reprint 2025-26 + +STRAIGHT LINES 173 +3. +What are the points on the y-axis whose distance from the line +1 +3 +4 +x +y ++ += is +4 units. +4. +Find perpendicular distance from the origin to the line joining the points (cosθ, sin θ) +and (cos φ, sin φ). +5. +Find the equation of the line parallel to y-axis and drawn through the point of +intersection of the lines x – 7y + 5 = 0 and 3x + y = 0. +6. +Find the equation of a line drawn perpendicular to the line +1 +6 +4 += ++ y +x +through the +point, where it meets the y-axis. +7. +Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0. +8. +Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and +2x – y – 3 = 0 may intersect at one point. +9. +If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are +concurrent, then show that m1(c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0. +10. +Find the equation of the lines through the point (3, 2) which make an angle of 45o +with the line x – 2y = 3. +11. +Find the equation of the line passing through the point of intersection of the lines +4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes. +12. +Show that the equation of the line passing through the origin and making an angle +θ with the line y +mx +c is y +x +m +m += ++ += +± tan ‚ +tan ‚ +1∓ + . +13. +In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4? +14. +Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line +2x – y = 0. +15. +Find the direction in which a straight line must be drawn through the point (–1, 2) +so that its point of intersection with the line x + y = 4 may be at a distance of +3 units from this point. +16. +The hypotenuse of a right angled triangle has its ends at the points (1, 3) and +(– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle which +are parallel to the axes. +17. +Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the +line to be a plane mirror. +18. +If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find +the value of m. +19. +If sum of the perpendicular distances of a variable point P (x, y) from the lines +x + y – 5 = 0 and 3x – 2y +7 = 0 is always 10. Show that P must move on a line. +Reprint 2025-26 + +174 +MATHEMATICS +20. +Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 +and 3x + 2y + 6 = 0. +21. +A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the +reflected ray passes through the point (5, 3). Find the coordinates of A. +22. +Prove that the product of the lengths of the perpendiculars drawn from the +points ( +) +2 +2 0 +a +b , +− +and ( +) +2 +2 0 +a +b , +− +− +to the line +2 +cosθ +sinθ +1is +x +y +b +a +b ++ += +. +23. +A person standing at the junction (crossing) of two straight paths represented by +the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose +equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he +should follow. +Summary +®Slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) +is given by +2 +1 +1 +2 +1 +2 +2 +1 +1 +2 +y +y +y +y +m +x +x +, +. +x +x +x +x +− +− += += +≠ +− +− +® If a line makes an angle á with the positive direction of x-axis, then the +slope of the line is given by m = tan α, α ≠ 90°. +®Slope of horizontal line is zero and slope of vertical line is undefined. +® An acute angle (say θ) between lines L1 and L2 with slopes m1 and m2 is +given by +2 +1 +1 +2 +1 +2 +tanθ +1 +0 +1 +m – m +, +m m +m m += ++ +≠ ++ +. +®Two lines are parallel if and only if their slopes are equal. +®Two lines are perpendicular if and only if product of their slopes is –1. +®Three points A, B and C are collinear, if and only if slope of AB = slope of BC. +®Equation of the horizontal line having distance a from the x-axis is either +y = a or y = – a. +®Equation of the vertical line having distance b from the y-axis is either +x = b or x = – b. +®The point (x, y) lies on the line with slope m and through the fixed point (xo, yo), +if and only if its coordinates satisfy the equation y – yo = m (x – xo). +® Equation of the line passing through the points (x1, y1) and (x2, y2) is given by +). +( +1 +1 +2 +1 +2 +1 +x +x +x +x +y +y +y +y +− +− +− += +− +Reprint 2025-26 + +STRAIGHT LINES 175 +®The point (x, y) on the line with slope m and y-intercept c lies on the line if and +only if y = mx + c. +®If a line with slope m makes x-intercept d. Then equation of the line is +y = m (x – d). +®Equation of a line making intercepts a and b on the x-and y-axis, +respectively, is +1 += ++ +b +y +a +x +. +®Any equation of the form Ax + By + C = 0, with A and B are not zero, +simultaneously, is called the general linear equation or general equation of +a line. +®The perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1, y1) +is given by +1 +1 +2 +2 +A +B +C +A +B +x +y +d ++ ++ += ++ +. +®Distance between the parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0, +is given by +2 +1 +2 +2 +C +C +A +B +d +− += ++ +. +Reprint 2025-26" +class_11,10,Conic Sections,ncert_books/class_11/kemh1dd/kemh110.pdf,"176 MATHEMATICS +vLet the relation of knowledge to real life be very visible to your pupils +and let them understand how by knowledge the world could be +transformed. – BERTRAND RUSSELL v +10.1 +Introduction +In the preceding Chapter 10, we have studied various forms +of the equations of a line. In this Chapter, we shall study +about some other curves, viz., circles, ellipses, parabolas +and hyperbolas. The names parabola and hyperbola are +given by Apollonius. These curves are in fact, known as +conic sections or more commonly conics because they +can be obtained as intersections of a plane with a double +napped right circular cone. These curves have a very wide +range of applications in fields such as planetary motion, +design of telescopes and antennas, reflectors in flashlights +and automobile headlights, etc. Now, in the subsequent sections we will see how the +intersection of a plane with a double napped right circular cone +results in different types of curves. +10.2 +Sections of a Cone +Let l be a fixed vertical line and m be another line intersecting it at +a fixed point V and inclined to it at an angle α (Fig10.1). +Suppose we rotate the line m around the line l in such a way +that the angle α remains constant. Then the surface generated is +a double-napped right circular hollow cone herein after referred as +Apollonius +(262 B.C. -190 B.C.) +10 +Chapter +Fig 10. 1 +CONIC SECTIONS +Reprint 2025-26 + +CONIC SECTIONS 177 +Fig 10. 2 +Fig 10. 3 +cone and extending indefinitely far in both directions (Fig10.2). +The point V is called the vertex; the line l is the axis of the cone. The rotating line +m is called a generator of the cone. The vertex separates the cone into two parts +called nappes. +If we take the intersection of a plane with a cone, the section so obtained is called +a conic section. Thus, conic sections are the curves obtained by intersecting a right +circular cone by a plane. +We obtain different kinds of conic sections depending on the position of the +intersecting plane with respect to the cone and by the angle made by it with the vertical +axis of the cone. Let β be the angle made by the intersecting plane with the vertical +axis of the cone (Fig10.3). +The intersection of the plane with the cone can take place either at the vertex of +the cone or at any other part of the nappe either below or above the vertex. +10.2.1 Circle, ellipse, parabola and hyperbola When the plane cuts the nappe (other +than the vertex) of the cone, we have the following situations: +(a) +When β = 90o, the section is a circle (Fig10.4). +(b) +When α < β < 90o, the section is an ellipse (Fig10.5). +(c) +When β = α; the section is a parabola (Fig10.6). +(In each of the above three situations, the plane cuts entirely across one nappe of +the cone). +(d) +When 0 ≤ β < α; the plane cuts through both the nappes and the curves of +intersection is a hyperbola (Fig10.7). +Reprint 2025-26 + +178 MATHEMATICS +Fig 10. 4 +10.2.2 Degenerated conic sections +When the plane cuts at the vertex of the cone, we have the following different cases: +(a) +When α < β ≤ 90o, then the section is a point (Fig10.8). +(b) +When β = α, the plane contains a generator of the cone and the section is a +straight line (Fig10.9). +It is the degenerated case of a parabola. +(c) +When 0 ≤ β < α, the section is a pair of intersecting straight lines (Fig10.10). +It is the degenerated case of a hyperbola. +Fig 10. 6 +Fig 10. 7 +Fig 10. 5 +Reprint 2025-26 + +CONIC SECTIONS 179 +In the following sections, we shall obtain the equations of each of these conic +sections in standard form by defining them based on geometric properties. +Fig 10. 8 +Fig 10. 9 +Fig 10. 10 +10.3 +Circle +Definition 1 A circle is the set of all points in a plane that are equidistant from a fixed +point in the plane. +The fixed point is called the centre of the circle and the distance from the centre +to a point on the circle is called the radius of the circle (Fig 10.11). +Reprint 2025-26 + +180 MATHEMATICS +The equation of the circle is simplest if the centre of the circle is at the origin. +However, we derive below the equation of the circle with a given centre and radius +(Fig 10.12). +Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on +the circle (Fig10.12). Then, by the definition, | CP | = r . By the distance formula, +we have +2 +2 +( +) +( +) +x – h +y – k +r ++ += +i.e. +(x – h)2 + (y – k)2 = r2 +This is the required equation of the circle with centre at (h,k) and radius r . +Example 1 Find an equation of the circle with centre at (0,0) and radius r. +Solution Here h = k = 0. Therefore, the equation of the circle is x2 + y2 = r2. +Example 2 Find the equation of the circle with centre (–3, 2) and radius 4. +Solution Here h = –3, k = 2 and r = 4. Therefore, the equation of the required circle is +(x + 3)2 + (y –2)2 = 16 +Example 3 Find the centre and the radius of the circle x2 + y2 + 8x + 10y – 8 = 0 +Solution The given equation is +(x2 + 8x) + (y2 + 10y) = 8 +Now, completing the squares within the parenthesis, we get +(x2 + 8x + 16) + (y2 + 10y + 25) = 8 + 16 + 25 +i.e. +(x + 4)2 + (y + 5)2 = 49 +i.e. +{x – (– 4)}2 + {y – (–5)}2 = 72 +Therefore, the given circle has centre at (– 4, –5) and radius 7. +Fig 10. 11 +Fig 10. 12 +Reprint 2025-26 + +CONIC SECTIONS 181 +Example 4 Find the equation of the circle which passes through the points (2, – 2), and +(3,4) and whose centre lies on the line x + y = 2. +Solution Let the equation of the circle be (x – h)2 + (y – k)2 = r2. +Since the circle passes through (2, – 2) and (3,4), we have +(2 – h)2 + (–2 – k)2 = r2 +... (1) +and +(3 – h)2 + (4 – k)2 = r2 +... (2) +Also since the centre lies on the line x + y = 2, we have +h + k = 2 +... (3) +Solving the equations (1), (2) and (3), we get +h = 0.7, k = 1.3 and r2 = 12.58 +Hence, the equation of the required circle is +(x – 0.7)2 + (y – 1.3)2 = 12.58. +EXERCISE 10.1 +In each of the following Exercises 1 to 5, find the equation of the circle with +1. +centre (0,2) and radius 2 +2. +centre (–2,3) and radius 4 +3. +centre ( +4 +1 +, +2 +1 +) and radius 12 +1 +4. +centre (1,1) and radius +2 +5. +centre (–a, –b) and radius +2 +2 +b +a +− +. +In each of the following Exercises 6 to 9, find the centre and radius of the circles. +6. +(x + 5)2 + (y – 3)2 = 36 +7. +x2 + y2 – 4x – 8y – 45 = 0 +8. +x2 + y2 – 8x + 10y – 12 = 0 +9. +2x2 + 2y2 – x = 0 +10. +Find the equation of the circle passing through the points (4,1) and (6,5) and +whose centre is on the line 4x + y = 16. +11. +Find the equation of the circle passing through the points (2,3) and (–1,1) and +whose centre is on the line x – 3y – 11 = 0. +12. +Find the equation of the circle with radius 5 whose centre lies on x-axis and +passes through the point (2,3). +13. +Find the equation of the circle passing through (0,0) and making intercepts a and +b on the coordinate axes. +14. +Find the equation of a circle with centre (2,2) and passes through the point (4,5). +15. +Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25? +Reprint 2025-26 + +182 MATHEMATICS +Fig 10. 13 +Fig 10.14 +10.4 Parabola +Definition 2 A parabola is the set of all points +in a plane that are equidistant from a fixed line +and a fixed point (not on the line) in the plane. +The fixed line is called the directrix of +the parabola and the fixed point F is called the +focus (Fig 10.13). (‘Para’ means ‘for’ and +‘bola’ means ‘throwing’, i.e., the shape +described when you throw a ball in the air). +ANote If the fixed point lies on the fixed +line, then the set of points in the plane, which +are equidistant from the fixed point and the +fixed line is the straight line through the fixed +point and perpendicular to the fixed line. We +call this straight line as degenerate case of +the parabola. +A line through the focus and perpendicular +to the directrix is called the axis of the +parabola. The point of intersection of parabola +with the axis is called the vertex of the parabola +(Fig10.14). +10.4.1 Standard equations of parabola +The equation of a parabola is simplest if the +vertex is at the origin and the axis of symmetry is along the x-axis or y-axis. The four +possible such orientations of parabola are shown below in Fig10.15 (a) to (d). +Reprint 2025-26 + +CONIC SECTIONS 183 +We will derive the equation for the parabola shown above in Fig 10.15 (a) with +focus at (a, 0) a > 0; and directricx x = – a as below: +Let F be the focus and l the directrix. Let +FM be perpendicular to the directrix and bisect +FM at the point O. Produce MO to X. By the +definition of parabola, the mid-point O is on the +parabola and is called the vertex of the parabola. +Take O as origin, OX the x-axis and OY +perpendicular to it as the y-axis. Let the distance +from the directrix to the focus be 2a. Then, the +coordinates of the focus are (a, 0), and the +equation of the directrix is x + a = 0 as in Fig10.16. +Let P(x, y) be any point on the parabola such that +PF = PB, +... (1) +where PB is perpendicular to l. The coordinates of B are (– a, y). By the distance +formula, we have +PF = +2 +2 +( +) +x – a +y ++ + and PB = +2 +( +) +x +a ++ +Since PF = PB, we have +2 +2 +2 +( +) +( +) +x – a +y +x +a ++ += ++ +i.e. +(x – a)2 + y2 = (x + a)2 +or +x2 – 2ax + a2 + y2 = x2 + 2ax + a2 +or +y2 = 4ax ( a > 0). +Fig 10.15 (a) to (d) +Fig 10.16 +Reprint 2025-26 + +184 MATHEMATICS +Hence, any point on the parabola satisfies +y2 = 4ax. +... (2) +Conversely, let P(x, y) satisfy the equation (2) +PF += +2 +2 +( +) +x – a +y ++ += +2 +( +) +4 +x – a +ax ++ += +2 +( +) +x +a ++ += PB +... (3) +and so P(x,y) lies on the parabola. +Thus, from (2) and (3) we have proved that the equation to the parabola with +vertex at the origin, focus at (a,0) and directrix x = – a is y2 = 4ax. +Discussion In equation (2), since a > 0, x can assume any positive value or zero but +no negative value and the curve extends indefinitely far into the first and the fourth +quadrants. The axis of the parabola is the positive x-axis. +Similarly, we can derive the equations of the parabolas in: +Fig 11.15 (b) as y2 = – 4ax, +Fig 11.15 (c) as x2 = 4ay, +Fig 11.15 (d) as x2 = – 4ay, +These four equations are known as standard equations of parabolas. +ANote The standard equations of parabolas have focus on one of the coordinate +axis; vertex at the origin and thereby the directrix is parallel to the other coordinate +axis. However, the study of the equations of parabolas with focus at any point and +any line as directrix is beyond the scope here. +From the standard equations of the parabolas, Fig10.15, we have the following +observations: +1. +Parabola is symmetric with respect to the axis of the parabola.If the equation +has a y2 term, then the axis of symmetry is along the x-axis and if the +equation has an x2 term, then the axis of symmetry is along the y-axis. +2. +When the axis of symmetry is along the x-axis the parabola opens to the +(a) +right if the coefficient of x is positive, +(b) +left if the coefficient of x is negative. +3. +When the axis of symmetry is along the y-axis the parabola opens +(c) +upwards if the coefficient of y is positive. +(d) +downwards if the coefficient of y is negative. +Reprint 2025-26 + +CONIC SECTIONS 185 +10.4.2 Latus rectum +Definition 3 Latus rectum of a parabola is a line segment perpendicular to the axis of +the parabola, through the focus and whose end points lie on the parabola (Fig10.17). +To find the Length of the latus rectum of the parabola y2 = 4ax (Fig 10.18). +By the definition of the parabola, AF = AC. +But +AC = FM = 2a +Hence +AF = 2a. +And since the parabola is symmetric with respect to x-axis AF = FB and so +AB = Length of the latus rectum = 4a. +Fig 10.17 +Fig 10.18 +Example 5 Find the coordinates of the focus, axis, +the equation of the directrix and latus rectum of +the parabola y2 = 8x. +Solution The given equation involves y2, so the +axis of symmetry is along the x-axis. +The coefficient of x is positive so the parabola opens +to the right. Comparing with the given equation +y2 = 4ax, we find that a = 2. +Thus, the focus of the parabola is (2, 0) and the equation of the directrix of the parabola +is x = – 2 (Fig 10.19). +Length of the latus rectum is 4a = 4 × 2 = 8. +Fig 10.19 +Reprint 2025-26 + +186 MATHEMATICS +Example 6 Find the equation of the parabola with focus (2,0) and directrix x = – 2. +Solution Since the focus (2,0) lies on the x-axis, the x-axis itself is the axis of the +parabola. Hence the equation of the parabola is of the form either +y2 = 4ax or y2 = – 4ax. Since the directrix is x = – 2 and the focus is (2,0), the parabola +is to be of the form y2 = 4ax with a = 2. Hence the required equation is + y2 = 4(2)x = 8x +Example 7 Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2). +Solution Since the vertex is at (0,0) and the focus is at (0,2) which lies on y-axis, the +y-axis is the axis of the parabola. Therefore, equation of the parabola is of the form +x2 = 4ay. thus, we have +x2 = 4(2)y, i.e., x2 = 8y. +Example 8 Find the equation of the parabola which is symmetric about the y-axis, and +passes through the point (2,–3). +Solution Since the parabola is symmetric about y-axis and has its vertex at the origin, +the equation is of the form x2 = 4ay or x2 = – 4ay, where the sign depends on whether +the parabola opens upwards or downwards. But the parabola passes through (2,–3) +which lies in the fourth quadrant, it must open downwards. Thus the equation is of +the form x2 = – 4ay. +Since the parabola passes through ( 2,–3), we have +22 = – 4a (–3), i.e., a = 1 +3 +Therefore, the equation of the parabola is +x2 = +1 +4 3 + + +− + + + + y, i.e., 3x2 = – 4y. +EXERCISE 10.2 +In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the +parabola, the equation of the directrix and the length of the latus rectum. +1. +y2 = 12x +2. +x2 = 6y +3. y2 = – 8x +4. +x2 = – 16y +5. +y2 = 10x +6. x2 = – 9y +In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the +given conditions: +Reprint 2025-26 + +CONIC SECTIONS 187 +Fig 10.20 +Fig 10.21 +Fig 10.22 +We denote the length of the major axis by 2a, the length of the minor axis by 2b +and the distance between the foci by 2c. Thus, the length of the semi major axis is a +and semi-minor axis is b (Fig10.22). +7. +Focus (6,0); directrix x = – 6 +8. +Focus (0,–3); directrix y = 3 +9. +Vertex (0,0); focus (3,0) +10. +Vertex (0,0); focus (–2,0) +11. +Vertex (0,0) passing through (2,3) and axis is along x-axis. +12. +Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis. +10. 5 +Ellipse +Definition 4 An ellipse is the set of all points in +a plane, the sum of whose distances from two +fixed points in the plane is a constant. +The two fixed points are called the foci (plural +of ‘focus’) of the ellipse (Fig10.20). +ANote The constant which is the sum of +the distances of a point on the ellipse from the +two fixed points is always greater than the +distance between the two fixed points. +The mid point of the line segment joining the foci is called the centre of the +ellipse. The line segment through the foci of the ellipse is called the major axis and the +line segment through the centre and perpendicular to the major axis is called the minor +axis. The end points of the major axis are called the vertices of the ellipse(Fig 10.21). +Reprint 2025-26 + +188 MATHEMATICS +10.5.1 Relationship between semi-major +axis, semi-minor axis and the distance of +the focus from the centre of the ellipse +(Fig 10.23). +Take a point P at one end of the major axis. +Sum of the distances of the point P to the +foci is F1 P + F2P = F1O + OP + F2P +(Since, F1P = F1O + OP) + = c + a + a – c = 2a +Take a point Q at one end of the minor axis. +Sum of the distances from the point Q to the foci is +F1Q + F2Q = +2 +2 +2 +2 +c +b +c +b ++ ++ ++ + = +2 +2 +2 +c +b ++ +Since both P and Q lies on the ellipse. +By the definition of ellipse, we have +2 +2 +2 +c +b ++ += 2a, i.e., +a = +2 +2 +c +b ++ +or +a 2 = b2 + c2 + , i.e., c = +2 +2 +b +a +− +. +10.5.2 Eccentricity +Definition 5 The eccentricity of an ellipse is the ratio of the distances from the centre +of the ellipse to one of the foci and to one of the vertices of the ellipse (eccentricity is +denoted by e) i.e., +c +e +a += +. +Then since the focus is at a distance of c from the centre, in terms of the eccentricity +the focus is at a distance of ae from the centre. +10.5.3 Standard equations of an ellipse The equation of an ellipse is simplest if the +centre of the ellipse is at the origin and the foci are on the x-axis or y-axis. The two +such possible orientations are shown in Fig 10.24. +We will derive the equation for the ellipse shown above in Fig 10.24 (a) with foci +on the x-axis. +Fig 10.23 +Reprint 2025-26 + +CONIC SECTIONS 189 +Fig 10.24 + (a) +Let F1 and F2 be the foci and O be the mid-point of the line segment F1F2. Let O +be the origin and the line from O through F2 be the positive +x-axis and that through F1as the negative x-axis. +Let, the line through O perpendicular to the +x-axis be the y-axis. Let the coordinates of F1 be +(– c, 0) and F2 be (c, 0) (Fig 10.25). +Let P(x, y) be any point on the ellipse such +that the sum of the distances from P to the two +foci be 2a so given +PF1 + PF2 = 2a. + ... (1) +Using the distance formula, we have + +2 +2 +2 +2 +) +( +) +( +y +c +x +y +c +x ++ +− ++ ++ ++ + = 2a +i.e., +2 +2) +( +y +c +x ++ ++ + = 2a – +2 +2) +( +y +c +x ++ +− +Squaring both sides, we get +(x + c)2 + y2 = 4a2 – 4a +2 +2 +2 +2 +) +( +) +( +y +c +x +y +c +x ++ +− ++ ++ +− +Fig 10.25 +2 +2 +2 +2 +1 +x +y +a +b ++ += +Reprint 2025-26 + +190 MATHEMATICS +which on simplification gives +x +a +c +a +y +c +x +− += ++ +− +2 +2) +( +Squaring again and simplifying, we get +2 +2 +2 +2 +2 +c +a +y +a +x +− ++ + = 1 +i.e., +2 +2 +2 +2 +b +y +a +x ++ + = 1 +(Since c2 = a2 – b2) +Hence any point on the ellipse satisfies +2 +2 +2 +2 +b +y +a +x ++ + = 1. +... (2) +Conversely, let P (x, y) satisfy the equation (2) with 0 < c < a. Then +y2 = b2 + + + + + + +− +2 +2 +1 +a +x +Therefore, PF1 += +2 +2 +( +) +x +c +y ++ ++ += + + + + + + +− ++ ++ +2 +2 +2 +2 +2) +( +a +x +a +b +c +x += +2 +2 +2 +2 +2 +2 +( +) +( +) a +x +x +c +a +c +a + + +− ++ ++ +− + + + + + (since b2 = a2 – c2) += +2 +cx +c +a +a +x +a +a + + ++ += + + + + + +Similarly +PF2 = +c +a +x +a +− +Hence +PF1 + PF2 = +2 +c +c +a +x +a – +x +a +a +a ++ ++ += +... (3) +Reprint 2025-26 + +CONIC SECTIONS 191 +So, any point that satisfies +2 +2 +2 +2 +b +y +a +x ++ += 1, satisfies the geometric condition and so +P(x, y) lies on the ellipse. +Hence from (2) and (3), we proved that the equation of an ellipse with centre of +the origin and major axis along the x-axis is +2 +2 +2 +2 +x +y +a +b ++ += 1. +Discussion From the equation of the ellipse obtained above, it follows that for every +point P (x, y) on the ellipse, we have +2 +2 +2 +2 +1 +b +y +a +x +− += + ≤ 1, i.e., x2 ≤ a2, so – a ≤ x ≤ a. +Therefore, the ellipse lies between the lines x = – a and x = a and touches these lines. +Similarly, the ellipse lies between the lines y = – b and y = b and touches these +lines. +Similarly, we can derive the equation of the ellipse in Fig 10.24 (b) as +2 +2 +2 +2 +1 +x +y +b +a ++ += . +These two equations are known as standard equations of the ellipses. +ANote The standard equations of ellipses have centre at the origin and the +major and minor axis are coordinate axes. However, the study of the ellipses with +centre at any other point, and any line through the centre as major and the minor +axes passing through the centre and perpendicular to major axis are beyond the +scope here. +From the standard equations of the ellipses (Fig10.24), we have the following +observations: +1. Ellipse is symmetric with respect to both the coordinate axes since if +(x, y) is a point on the ellipse, then (– x, y), (x, –y) and (– x, –y) are also points on +the ellipse. +2. The foci always lie on the major axis. The major axis can be determined by +finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis +if the coefficient of x2 has the larger denominator and it is along the y-axis if the +coefficient of y2 has the larger denominator. +Reprint 2025-26 + +192 MATHEMATICS +10.5.4 Latus rectum +Definition 6 Latus rectum of an ellipse is a +line segment perpendicular to the major axis +through any of the foci and whose end points +lie on the ellipse (Fig 10.26). +To find the length of the latus rectum +of the ellipse +1 +x +y +a +b +2 +2 +2 +2 ++ += +Let the length of AF2 be l. +Then the coordinates of A are (c, l ),i.e., +(ae, l ) +Since A lies on the ellipse +2 +2 +2 +2 +1 +x +y +a +b ++ += , we have +2 +2 +2 +2 +( +) +1 +ae +l +a +b ++ += +⇒ l2 = b2 (1 – e2) +But +2 +2 +2 +2 +2 +2 +2 +2 +1 +c +a – b +b +e +– +a +a +a += += += +Therefore +l2 = +4 +2 +b +a +, i.e., +2 +b +l +a += +Since the ellipse is symmetric with respect to y-axis (of course, it is symmetric w.r.t. +both the coordinate axes), AF2 = F2B and so length of the latus rectum is +2 +2b +a . +Example 9 Find the coordinates of the foci, the vertices, the length of major axis, the +minor axis, the eccentricity and the latus rectum of the ellipse +2 +2 +1 +25 +9 +x +y ++ += +Solution Since denominator of +2 +25 +x is larger than the denominator of +2 +9 +y , the major +Fig 10. 26 +Reprint 2025-26 + +CONIC SECTIONS 193 +axis is along the x-axis. Comparing the given equation with +2 +2 +2 +2 +1 +x +y +a +b ++ += , we get +a = 5 and b = 3. Also +2 +2 +25 +9 +4 +c +a – b +– += += += +Therefore, the coordinates of the foci are (– 4,0) and (4,0), vertices are (– 5, 0) and +(5, 0). Length of the major axis is 10 units length of the minor axis 2b is 6 units and the +eccentricity is 4 +5 and latus rectum is +2 +2 +18 +5 +b +a = + . +Example 10 Find the coordinates of the foci, the vertices, the lengths of major and +minor axes and the eccentricity of the ellipse 9x2 + 4y2 = 36. +Solution The given equation of the ellipse can be written in standard form as +2 +2 +1 +4 +9 +x +y ++ += +Since the denominator of +2 +9 +y is larger than the denominator of +2 +4 +x , the major axis is +along the y-axis. Comparing the given equation with the standard equation +2 +2 +2 +2 +1 +x +y +b +a ++ += , we have b = 2 and a = 3. +Also +c = +2 +2 +a – b = +9 +4 +5 +– += +and +5 +3 +c +e +a += += +Hence the foci are (0, 5 ) and (0, – +5 ), vertices are (0,3) and (0, –3), length of the +major axis is 6 units, the length of the minor axis is 4 units and the eccentricity of the +ellipse is +5 +3 . +Example 11 Find the equation of the ellipse whose vertices are (± 13, 0) and foci are +(± 5, 0). +Solution Since the vertices are on x-axis, the equation will be of the form +2 +2 +2 +2 +1 +x +y +a +b ++ += , where a is the semi-major axis. +Reprint 2025-26 + +194 MATHEMATICS +Given that a = 13, c = ± 5. +Therefore, from the relation c2 = a2 – b2, we get +25 = 169 – b2 , i.e., b = 12 +Hence the equation of the ellipse is +2 +2 +1 +169 +144 +x +y ++ += . +Example 12 Find the equation of the ellipse, whose length of the major axis is 20 and +foci are (0, ± 5). +Solution Since the foci are on y-axis, the major axis is along the y-axis. So, equation +of the ellipse is of the form +2 +2 +2 +2 +1 +x +y +b +a ++ += . +Given that +a = semi-major axis +20 +10 +2 += += +and the relation +c2 = a2 – b2 gives +52 = 102 – b2 i.e., b2 = 75 +Therefore, the equation of the ellipse is +2 +2 +1 +75 +100 +x +y ++ += +Example 13 Find the equation of the ellipse, with major axis along the x-axis and +passing through the points (4, 3) and (– 1,4). +Solution The standard form of the ellipse is +2 +2 +2 +2 +b +y +a +x ++ + = 1. Since the points (4, 3) +and (–1, 4) lie on the ellipse, we have +1 +9 +16 +2 +2 += ++ b +a +... (1) +and +2 +2 +16 +1 +b +a ++ + = 1 +….(2) +Solving equations (1) and (2), we find that +2 +247 +7 +a = + and +2 +247 +15 +b = +. +Hence the required equation is +Reprint 2025-26 + +CONIC SECTIONS 195 +2 +2 +1 +247 +247 +15 +7 +x +y ++ += + + + + + + +, i.e., 7x2 + 15y2 = 247. +EXERCISE 10.3 +In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length +of major axis, the minor axis, the eccentricity and the length of the latus rectum of the +ellipse. +1. +2 +2 +1 +36 +16 +x +y ++ += +2. +2 +2 +1 +4 +25 +x +y ++ += +3. +2 +2 +1 +16 +9 +x +y ++ += +4. +2 +2 +1 +25 +100 +x +y ++ += +5. +2 +2 +1 +49 +36 +x +y ++ += +6. +400 +100 +2 +2 +y +x ++ += 1 +7. +36x2 + 4y2 = 144 +8. +16x2 + y2 = 16 +9. +4x2 + 9y2 = 36 +In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies +the given conditions: +10. +Vertices (± 5, 0), foci (± 4, 0) +11. +Vertices (0, ± 13), foci (0, ± 5) +12. +Vertices (± 6, 0), foci (± 4, 0) +13. +Ends of major axis (± 3, 0), ends of minor axis (0, ± 2) +14. +Ends of major axis (0, ± +5 ), ends of minor axis (± 1, 0) +15. +Length of major axis 26, foci (± 5, 0) +16. +Length of minor axis 16, foci (0, ± 6). +17. +Foci (± 3, 0), a = 4 +18. +b = 3, c = 4, centre at the origin; foci on the x axis. +19. +Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and +(1,6). +20. +Major axis on the x-axis and passes through the points (4,3) and (6,2). +10.6 +Hyperbola +Definition 7 A hyperbola is the set of all points in a plane, the difference of whose +distances from two fixed points in the plane is a constant. +Reprint 2025-26 + +196 MATHEMATICS +The term “difference” that is used in the definition means the distance to the +farther point minus the distance to the closer point. The two fixed points are called the +foci of the hyperbola. The mid-point of the line segment joining the foci is called the +centre of the hyperbola. The line through the foci is called the transverse axis and +the line through the centre and perpendicular to the transverse axis is called the conjugate +axis. The points at which the hyperbola +intersects the transverse axis are called the +vertices of the hyperbola (Fig 10.27). +We denote the distance between the +two foci by 2c, the distance between two +vertices (the length of the transverse axis) +by 2a and we define the quantity b as +b = +2 +2 +c – a +Also 2b is the length of the conjugate axis +(Fig 10.28). +To find the constant P1F2 – P1F1 : +By taking the point P at A and B in the Fig 10.28, we have +BF1 – BF2 = AF2 – AF1 (by the definition of the hyperbola) +BA +AF1– BF2 = AB + BF2– AF1 +i.e., AF1 = BF2 +So that, BF1 – BF2 = BA + AF1– BF2 = BA = 2a +Fig 10.27 +Fig 10.28 +Reprint 2025-26 + +CONIC SECTIONS 197 +10.6.1 Eccentricity +Definition 8 Just like an ellipse, the ratio e = c +a is called the eccentricity of the +hyperbola. Since c ≥ a, the eccentricity is never less than one. In terms of the +eccentricity, the foci are at a distance of ae from the centre. +10.6.2 Standard equation of Hyperbola The equation of a hyperbola is simplest if +the centre of the hyperbola is at the origin and the foci are on the x-axis or y-axis. The +two such possible orientations are shown in Fig10.29. +We will derive the equation for the hyperbola shown in Fig 10.29(a) with foci on +the x-axis. +Let F1 and F2 be the foci and O be the mid-point of the line segment F1F2. Let O +be the origin and the line through O +through F2 be the positive x-axis and +that through F1 as the negative +x-axis. The line through O +perpendicular to the x-axis be the +y-axis. Let the coordinates of F1 be +(– c,0) and F2 be (c,0) (Fig 10.30). +Let P(x, y) be any point on the +hyperbola such that the difference +of the distances from P to the farther +point minus the closer point be 2a. +So given, PF1 – PF2 = 2a +Fig 10.29 +(a) +(b) +Fig 10.30 +Reprint 2025-26 + +198 MATHEMATICS +Using the distance formula, we have +2 +2 +2 +2 +( +) +( +) +2 +x +c +y +– +x – c +y +a ++ ++ ++ += +i.e., +2 +2 +2 +2 +( +) +2 +( +) +x +c +y +a +x – c +y ++ ++ += ++ ++ +Squaring both side, we get +(x + c)2 + y2 = 4a2 + 4a +2 +2 +( +) +x – c +y ++ + + (x – c)2 + y2 +and on simplifying, we get +a +cx – a = +2 +2 +( +) +x – c +y ++ +On squaring again and further simplifying, we get +2 +2 +2 +2 +2 +1 +x +y +– +a +c – a += +i.e., +2 +2 +2 +2 +1 +x +y +– +a +b += +(Since c2 – a2 = b2) +Hence any point on the hyperbola satisfies +2 +2 +2 +2 +1 +x +y +– +a +b += 1. +Conversely, let P(x, y) satisfy the above equation with 0 < a < c. Then +y2 + = b2 +2 +2 +2 +x – a +a + + + + + + +Therefore, +PF1 = + +2 +2 +( +) +x +c +y ++ ++ += + +2 +2 +2 +2 +2 +( +) +x – a +x +c +b +a + + ++ ++ + + + + + = a + +x +a +c +Similarly, +PF2 = a – a +c x +In hyperbola c > a; and since P is to the right of the line x = a, x > a, c +a x > a. Therefore, +a – c +a x becomes negative. Thus, PF2 = c +a x – a. +Reprint 2025-26 + +CONIC SECTIONS 199 +Therefore +PF1 – PF2 = a + c +a x – cx +a + a = 2a +Also, note that if P is to the left of the line x = – a, then +PF1 +c +– a +x +a + + += ++ + + + +, PF2 = a – c x +a +. +In that case P F2 – PF1 = 2a. So, any point that satisfies +2 +2 +2 +2 +1 +x +y +– +a +b += , lies on the +hyperbola. +Thus, we proved that the equation of hyperbola with origin (0,0) and transverse axis +along x-axis is +2 +2 +2 +2 +1 +x +y +– +a +b += . +ANote A hyperbola in which a = b is called an equilateral hyperbola. +Discussion From the equation of the hyperbola we have obtained, it follows that, we +have for every point (x, y) on the hyperbola, +2 +2 +2 +2 +1 +x +y +a +b += + + ≥ 1. +i.e, a +x ≥ 1, i.e., x ≤ – a or x ≥ a. Therefore, no portion of the curve lies between the +lines x = + a and x = – a, (i.e. no real intercept on the conjugate axis). +Similarly, we can derive the equation of the hyperbola in Fig 11.31 (b) as +2 +2 +2 +2 +y +x +a +b +− += 1 +These two equations are known as the standard equations of hyperbolas. +ANote The standard equations of hyperbolas have transverse and conjugate +axes as the coordinate axes and the centre at the origin. However, there are +hyperbolas with any two perpendicular lines as transverse and conjugate axes, but +the study of such cases will be dealt in higher classes. +From the standard equations of hyperbolas (Fig10.27), we have the following +observations: +1. +Hyperbola is symmetric with respect to both the axes, since if (x, y) is a point on +the hyperbola, then (– x, y), (x, – y) and (– x, – y) are also points on the hyperbola. +Reprint 2025-26 + +200 MATHEMATICS +2. +The foci are always on the transverse axis. It is the positive term whose +denominator gives the transverse axis. For example, +2 +2 +1 +9 +16 +x +y +– += +has transverse axis along x-axis of length 6, while +2 +2 +1 +25 +16 +y +x +– += +has transverse axis along y-axis of length 10. +10.6.3 Latus rectum +Definition 9 Latus rectum of hyperbola is a line segment perpendicular to the transverse +axis through any of the foci and whose end points lie on the hyperbola. +As in ellipse, it is easy to show that the length of the latus rectum in hyperbola is +2 +2b +a . +Example 14 Find the coordinates of the foci and the vertices, the eccentricity,the +length of the latus rectum of the hyperbolas: +(i) +2 +2 +1 +9 +16 +x +y +– += , (ii) y2 – 16x2 = 16 +Solution (i) Comparing the equation +2 +2 +1 +9 +16 +x +y +– += with the standard equation +2 +2 +2 +2 +1 +x +y +– +a +b += +Here, a = 3, b = 4 and c = +2 +2 +9 +16 +5 +a +b ++ += ++ += +Therefore, the coordinates of the foci are (± 5, 0) and that of vertices are (± 3, 0).Also, +The eccentricity e = +5 +3 +c +a = +. The latus rectum +2 +2 +32 +3 +b +a += += +(ii) Dividing the equation by 16 on both sides, we have +2 +2 +1 +16 +1 +y +x +– += +Comparing the equation with the standard equation +2 +2 +2 +2 +1 +y +x +– +a +b += , we find that +a = 4, b = 1 and +2 +2 +16 +1 +17 +c +a +b += ++ += ++ += +. +Reprint 2025-26 + +CONIC SECTIONS 201 +Therefore, the coordinates of the foci are (0, ± 17 ) and that of the vertices are +(0, ± 4). Also, +The eccentricity +17 +4 +c +e +a += += +. The latus rectum +2 +2 +1 +2 +b +a += += +. +Example 15 Find the equation of the hyperbola with foci (0, ± 3) and vertices +(0, ± +11 +2 +). +Solution Since the foci is on y-axis, the equation of the hyperbola is of the form +2 +2 +2 +2 +1 +y +x +– +a +b += +Since vertices are (0, ± +11 +2 + ), a = +11 +2 +Also, since foci are (0, ± 3); c = 3 and b2 = c2 – a2 = 25 +4 . +Therefore, the equation of the hyperbola is +2 +2 +11 +25 +4 +4 +y +x +– + + + + + + + + + + + + + = 1, i.e., 100 y2 + – 44 x2 = 275. +Example 16 Find the equation of the hyperbola where foci are (0, ±12) and the length +of the latus rectum is 36. +Solution Since foci are (0, ± 12), it follows that c = 12. +Length of the latus rectum = +36 +2 +2 += +a +b + or b2 = 18a +Therefore + c2 = a2 + b2; gives +144 = a2 + 18a +i.e., +a2 + 18a – 144 = 0, +So + a = – 24, 6. +Since a cannot be negative, we take a = 6 and so b2 = 108. +Therefore, the equation of the required hyperbola is +2 +2 +1 +36 +108 +y +x +– += , i.e., 3y2 – x2 = 108 +Reprint 2025-26 + +202 MATHEMATICS +Fig 10.31 + EXERCISE 10.4 +In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the +eccentricity and the length of the latus rectum of the hyperbolas. +1. +2 +2 +1 +16 +9 +x +y +– += +2. +2 +2 +1 +9 +27 +y +x +– += +3. 9y2 – 4x2 = 36 +4. 16x2 – 9y2 = 576 +5. +5y2 – 9x2 = 36 +6. 49y2 – 16x2 = 784. +In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given +conditions. +7. Vertices (± 2, 0), foci (± 3, 0) +8. Vertices (0, ± 5), foci (0, ± 8) +9. Vertices (0, ± 3), foci (0, ± 5) +10. Foci (± 5, 0), the transverse axis is of length 8. +11. Foci (0, ±13), the conjugate axis is of length 24. +12. Foci (± 3 +5 , 0), the latus rectum is of length 8. +13. Foci (± 4, 0), the latus rectum is of length 12 +14. vertices (± 7,0), e = 3 +4 . +15. Foci (0, ± 10 ), passing through (2,3) +Miscellaneous Examples +Example 17 The focus of a parabolic mirror as shown in Fig 10.31 is at a distance of +5 cm from its vertex. If the mirror is 45 cm deep, find +the distance AB (Fig 10.31). +Solution Since the distance from the focus to the +vertex is 5 cm. We have, a = 5. If the origin is taken at +the vertex and the axis of the mirror lies along the +positive x-axis, the equation of the parabolic section is +y2 = 4 (5) x = 20 x +Note that +x = 45. Thus +y2 = 900 +Therefore +y = ± 30 +Hence +AB = +2y = 2 × 30 = 60 cm. +Example 18 A beam is supported at its ends by +supports which are 12 metres apart. Since the load is concentrated at its centre, there +Reprint 2025-26 + +CONIC SECTIONS 203 +is a deflection of 3 cm at the centre and the deflected beam is in the shape of a +parabola. How far from the centre is the deflection 1 cm? +Solution Let the vertex be at the lowest point and the axis vertical. Let the coordinate +axis be chosen as shown in Fig 10.32. +Fig 10.32 +The equation of the parabola takes the form x2 = 4ay. Since it passes through +3 +6 100 +, + + + + + +, we have (6)2 = 4a +3 +100 + + + + + +, i.e., a = 36 100 +12 +× + = 300 m +Let AB be the deflection of the beam which is 1 +100 m. Coordinates of B are (x, 2 +100 ). +Therefore +x2 = 4 × 300 × 2 +100 = 24 +i.e. +x = +24 += 2 6 metres +Example 19 A rod AB of length 15 cm rests in between two coordinate axes in such +a way that the end point A lies on x-axis and end point B lies on +y-axis. A point P(x, y) is taken on the rod in such a way +that AP = 6 cm. Show that the locus of P is an ellipse. +Solution Let AB be the rod making an angle θ with +OX as shown in Fig 10.33 and P (x, y) the point on it +such that AP = 6 cm. +Since +AB = +15 cm, we have +PB = +9 cm. +From P draw PQ and PR perpendiculars on y-axis and +x-axis, respectively. +Fig 10.33 +Reprint 2025-26 + +204 MATHEMATICS +From +∆ PBQ, cos θ = 9 +x +From +∆ PRA, sin θ = 6 +y +Since +cos2 θ + sin2 θ = 1 +2 +2 +1 +9 +6 +x +y + + + + ++ += + + + + + + + + +or +2 +2 +1 +81 +36 +x +y ++ += +Thus the locus of P is an ellipse. +Miscellaneous Exercise on Chapter 10 +1. +If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus. +2. +An arch is in the form of a parabola with its axis vertical. The arch is 10 m high +and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola? +3. +The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. +The roadway which is horizontal and 100 m long is supported by vertical wires +attached to the cable, the longest wire being 30 m and the shortest being 6 m. +Find the length of a supporting wire attached to the roadway 18 m from the +middle. +4. +An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. +Find the height of the arch at a point 1.5 m from one end. +5. +A rod of length 12 cm moves with its ends always touching the coordinate axes. +Determine the equation of the locus of a point P on the rod, which is 3 cm from +the end in contact with the x-axis. +6. +Find the area of the triangle formed by the lines joining the vertex of the parabola +x2 = 12y to the ends of its latus rectum. +7. +A man running a racecourse notes that the sum of the distances from the two flag +posts from him is always 10 m and the distance between the flag posts is 8 m. +Find the equation of the posts traced by the man. +8. +An equilateral triangle is inscribed in the parabola y2 = 4 ax, where one vertex is +at the vertex of the parabola. Find the length of the side of the triangle. +Reprint 2025-26 + +CONIC SECTIONS 205 +Summary +In this Chapter the following concepts and generalisations are studied. +®A circle is the set of all points in a plane that are equidistant from a fixed point +in the plane. +®The equation of a circle with centre (h, k) and the radius r is +(x – h)2 + (y – k)2 = r2. +®A parabola is the set of all points in a plane that are equidistant from a fixed +line and a fixed point in the plane. +®The equation of the parabola with focus at (a, 0) a > 0 and directrix x = – a is +y2 = 4ax. +®Latus rectum of a parabola is a line segment perpendicular to the axis of the +parabola, through the focus and whose end points lie on the parabola. +®Length of the latus rectum of the parabola y2 = 4ax is 4a. +®An ellipse is the set of all points in a plane, the sum of whose distances from +two fixed points in the plane is a constant. +®The equation of an ellipse with foci on the x-axis is +2 +2 +2 +2 +1 +x +y ++ += +a +b +. +®Latus rectum of an ellipse is a line segment perpendicular to the major axis +through any of the foci and whose end points lie on the ellipse. +®Length of the latus rectum of the ellipse +2 +2 +2 +2 ++ +=1 +x +y +a +b + is +2 +2b +a . +®The eccentricity of an ellipse is the ratio between the distances from the centre +of the ellipse to one of the foci and to one of the vertices of the ellipse. +®A hyperbola is the set of all points in a plane, the difference of whose distances +from two fixed points in the plane is a constant. +®The equation of a hyperbola with foci on the x-axis is : +2 +2 +2 +2 +1 +x +y +a +b +− += +Reprint 2025-26 + +206 MATHEMATICS +®Latus rectum of hyperbola is a line segment perpendicular to the transverse +axis through any of the foci and whose end points lie on the hyperbola. +®Length of the latus rectum of the hyperbola : +2 +2 +2 +2 +1 +x +y +a +b +− += is : +2 +2b +a . +®The eccentricity of a hyperbola is the ratio of the distances from the centre of +the hyperbola to one of the foci and to one of the vertices of the hyperbola. +Historical Note +Geometry is one of the most ancient branches of mathematics. The Greek +geometers investigated the properties of many curves that have theoretical and +practical importance. Euclid wrote his treatise on geometry around 300 B.C. He +was the first who organised the geometric figures based on certain axioms +suggested by physical considerations. Geometry as initially studied by the ancient +Indians and Greeks, who made essentially no use of the process of algebra. The +synthetic approach to the subject of geometry as given by Euclid and in +Sulbasutras, etc., was continued for some 1300 years. In the 200 B.C., Apollonius +wrote a book called ‘The Conic’ which was all about conic sections with many +important discoveries that have remained unsurpassed for eighteen centuries. +Modern analytic geometry is called ‘Cartesian’ after the name of Rene +Descartes (1596-1650) whose relevant ‘La Geometrie’ was published in 1637. +But the fundamental principle and method of analytical geometry were already +discovered by Pierre de Fermat (1601-1665). Unfortunately, Fermats treatise on +the subject, entitled Ad Locus Planos et So LIDOS Isagoge (Introduction to +Plane and Solid Loci) was published only posthumously in +1679. So, Descartes came to be regarded as the unique inventor of the analytical +geometry. +Isaac Barrow avoided using cartesian method. Newton used method of +undetermined coefficients to find equations of curves. He used several types of +coordinates including polar and bipolar. Leibnitz used the terms ‘abscissa’, +‘ordinate’ and ‘coordinate’. L’ Hospital (about 1700) wrote an important textbook +on analytical geometry. +Clairaut (1729) was the first to give the distance formula although in clumsy +form. He also gave the intercept form of the linear equation. Cramer (1750) +Reprint 2025-26 + +CONIC SECTIONS 207 +made formal use of the two axes and gave the equation of a circle as +( y – a)2 + (b – x)2 = r +He gave the best exposition of the analytical geometry of his time. Monge +(1781) gave the modern ‘point-slope’ form of equation of a line as +y – y′ = a (x – x′) +and the condition of perpendicularity of two lines as aa′ + 1 = 0. +S.F. Lacroix (1765–1843) was a prolific textbook writer, but his contributions +to analytical geometry are found scattered. He gave the ‘two-point’ form of +equation of a line as + +β +β +β = +( +α) +α +α +– +y – +x – +– +′ +′ +and the length of the perpendicular from (α, β) on y = ax + b as +2 +(β +) +1 +– a – b +a ++ +. +His formula for finding angle between two lines was tan θ +1 +a – a +aa +′ + + +=  +′ ++ + +. It is, of +course, surprising that one has to wait for more than 150 years after the invention +of analytical geometry before finding such essential basic formula. In 1818, C. +Lame, a civil engineer, gave mE + m′E′ = 0 as the curve passing through the +points of intersection of two loci E = 0 and E′ = 0. +Many important discoveries, both in Mathematics and Science, have been +linked to the conic sections. The Greeks particularly Archimedes (287–212 B.C.) +and Apollonius (200 B.C.) studied conic sections for their own beauty. These +curves are important tools for present day exploration of outer space and also for +research into behaviour of atomic particles. +— v — +Reprint 2025-26" +class_11,11,Introduction to Three Dimensional Geometry,ncert_books/class_11/kemh1dd/kemh111.pdf,"208 MATHEMATICS +vMathematics is both the queen and the hand-maiden of +all sciences – E.T. BELLv +11.1 Introduction +You may recall that to locate the position of a point in a +plane, we need two intersecting mutually perpendicular lines +in the plane. These lines are called the coordinate axes +and the two numbers are called the coordinates of the +point with respect to the axes. In actual life, we do not +have to deal with points lying in a plane only. For example, +consider the position of a ball thrown in space at different +points of time or the position of an aeroplane as it flies +from one place to another at different times during its flight. +Similarly, if we were to locate the position of the +lowest tip of an electric bulb hanging from the ceiling of a +room or the position of the central tip of the ceiling fan in a room, we will not only +require the perpendicular distances of the point to be located from two perpendicular +walls of the room but also the height of the point from the floor of the room. Therefore, +we need not only two but three numbers representing the perpendicular distances of +the point from three mutually perpendicular planes, namely the floor of the room and +two adjacent walls of the room. The three numbers representing the three distances +are called the coordinates of the point with reference to the three coordinate +planes. So, a point in space has three coordinates. In this Chapter, we shall study the +basic concepts of geometry in three dimensional space.* +* +For various activities in three dimensional geometry one may refer to the Book, “A Hand Book for +designing Mathematics Laboratory in Schools”, NCERT, 2005. +Leonhard Euler +(1707-1783) +11 +Chapter +INTRODUCTION TO THREE +DIMENSIONAL GEOMETRY +Reprint 2025-26 + +INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 209 +11.2 Coordinate Axes and Coordinate Planes in Three Dimensional Space +Consider three planes intersecting at a point O +such that these three planes are mutually +perpendicular to each other (Fig 11.1). These +three planes intersect along the lines X′OX, Y′OY +and Z′OZ, called the x, y and z-axes, respectively. +We may note that these lines are mutually +perpendicular to each other. These lines constitute +the rectangular coordinate system. The planes +XOY, YOZ and ZOX, called, respectively the +XY-plane, YZ-plane and the ZX-plane, are +known as the three coordinate planes. We take +the XOY plane as the plane of the paper and the +line Z′OZ as perpendicular to the plane XOY. If the plane of the paper is considered +as horizontal, then the line Z′OZ will be vertical. The distances measured from +XY-plane upwards in the direction of OZ are taken as positive and those measured +downwards in the direction of OZ′ are taken as negative. Similarly, the distance +measured to the right of ZX-plane along OY are taken as positive, to the left of +ZX-plane and along OY′ as negative, in front of the YZ-plane along OX as positive +and to the back of it along OX′ as negative. The point O is called the origin of the +coordinate system. The three coordinate planes divide the space into eight parts known +as octants. These octants could be named as XOYZ, X′OYZ, X′OY′Z, XOY′Z, XOYZ′, +X′OYZ′, X′OY′Z′ and XOY′Z′. and denoted by I, II, III, ..., VIII , respectively. +11.3 Coordinates of a Point in Space +Having chosen a fixed coordinate system in the +space, consisting of coordinate axes, coordinate +planes and the origin, we now explain, as to how, +given a point in the space, we associate with it three +coordinates (x,y,z) and conversely, given a triplet +of three numbers (x, y, z), how, we locate a point in +the space. +Given a point P in space, we drop a +perpendicular PM on the XY-plane with M as the +foot of this perpendicular (Fig 11.2). Then, from the point M, we draw a perpendicular +ML to the x-axis, meeting it at L. Let OL be x, LM be y and MP be z. Then x,y and z +are called the x, y and z coordinates, respectively, of the point P in the space. In +Fig 11.2, we may note that the point P (x, y, z) lies in the octant XOYZ and so all x, y, +z are positive. If P was in any other octant, the signs of x, y and z would change +Fig 11.1 +Fig 11.2 +Reprint 2025-26 + +210 MATHEMATICS +accordingly. Thus, to each point P in the space there corresponds an ordered triplet +(x, y, z) of real numbers. +Conversely, given any triplet (x, y, z), we would first fix the point L on the x-axis +corresponding to x, then locate the point M in the XY-plane such that (x, y) are the +coordinates of the point M in the XY-plane. Note that LM is perpendicular to the +x-axis or is parallel to the y-axis. Having reached the point M, we draw a perpendicular +MP to the XY-plane and locate on it the point P corresponding to z. The point P so +obtained has then the coordinates (x, y, z). Thus, there is a one to one correspondence +between the points in space and ordered triplet (x, y, z) of real numbers. +Alternatively, through the point P in the +space, we draw three planes parallel to the +coordinate planes, meeting the x-axis, y-axis +and z-axis in the points A, B and C, respectively +(Fig 11.3). Let OA = x, OB = y and OC = z. +Then, the point P will have the coordinates x, y +and z and we write P (x, y, z). Conversely, given +x, y and z, we locate the three points A, B and +C on the three coordinate axes. Through the +points A, B and C we draw planes parallel to +the YZ-plane, ZX-plane and XY-plane, +respectively. The point of interesection of these three planes, namely, ADPF, BDPE +and CEPF is obviously the point P, corresponding to the ordered triplet (x, y, z). We +observe that if P (x, y, z) is any point in the space, then x, y and z are perpendicular +distances from YZ, ZX and XY planes, respectively. +ANote The coordinates of the origin O are (0,0,0). The coordinates of any point +on the x-axis will be as (x,0,0) and the coordinates of any point in the YZ-plane will +be as (0, y, z). +Remark The sign of the coordinates of a point determine the octant in which the +point lies. The following table shows the signs of the coordinates in eight octants. +Table 11.1 +Fig 11.3 +I +II +III +IV +V +VI +VII +VIII +x ++ +– +– ++ ++ +– +– ++ +y ++ ++ +– +– ++ ++ +– +– +z ++ ++ ++ ++ +– +– +– +– + Octants +Coordinates +Reprint 2025-26 + +INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 211 +Example 1 In Fig 11.3, if P is (2,4,5), find the coordinates of F. +Solution For the point F, the distance measured along OY is zero. Therefore, the +coordinates of F are (2,0,5). +Example 2 Find the octant in which the points (–3,1,2) and (–3,1,– 2) lie. +Solution From the Table 11.1, the point (–3,1, 2) lies in second octant and the point +(–3, 1, – 2) lies in octant VI. +EXERCISE 11.1 +1. +A point is on the x-axis. What are its y-coordinate and z-coordinates? +2. +A point is in the XZ-plane. What can you say about its y-coordinate? +3. +Name the octants in which the following points lie: +(1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (– 4, 2, –5), (– 4, 2, 5), +(–3, –1, 6) (– 2, – 4, –7). +4. +Fill in the blanks: +(i) +The x-axis and y-axis taken together determine a plane known as_______. +(ii) +The coordinates of points in the XY-plane are of the form _______. +(iii) +Coordinate planes divide the space into ______ octants. +11.4 Distance between Two Points +We have studied about the distance +between two points in two-dimensional +coordinate system. Let us now extend this +study to three-dimensional system. +Let P(x1, y1, z1) and Q ( x2, y2, z2) +be two points referred to a system of +rectangular axes OX, OY and OZ. +Through the points P and Q draw planes +parallel to the coordinate planes so as to +form a rectangular parallelopiped with one +diagonal PQ (Fig 11.4). +Now, since ∠PAQ is a right +angle, it follows that, in triangle PAQ, + PQ2 = PA2 + AQ2 +... (1) +Also, triangle ANQ is right angle triangle with ∠ANQ a right angle. +Fig 11.4 +Reprint 2025-26 + +212 MATHEMATICS +Therefore +AQ2 = AN2 + NQ2 +... (2) +From +(1) and (2), we have +PQ2 = PA2 + AN2 + NQ2 +Now +PA = y2 – y1, AN = x2 – x1 and NQ = z2 – z1 +Hence +PQ2 = (x2 – x1)2 + (y2 – y1) +2 + (z2 – z1)2 +Therefore +PQ = +2 +1 +2 +2 +1 +2 +2 +1 +2 +) +( +) +( +) +( +z +z +y +y +x +x +− ++ +− ++ +− +This gives us the distance between two points (x1, y1, z1) and (x2, y2, z2). +In particular, if x1 = y1 = z1 = 0, i.e., point P is origin O, then OQ = +2 +2 +2 +2 +2 +2 +z +y +x ++ ++ +, +which gives the distance between the origin O and any point Q (x2, y2, z2). +Example 3 Find the distance between the points P(1, –3, 4) and Q (– 4, 1, 2). +Solution The distance PQ between the points P (1,–3, 4) and Q (– 4, 1, 2) is +PQ = +2 +2 +2 +) +4 +2 +( +)3 +1( +)1 +4 +( +− ++ ++ ++ +− +− += +4 +16 +25 ++ ++ += 45 = 3 5 units +Example 4 Show that the points P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1) are collinear. +Solution We know that points are said to be collinear if they lie on a line. +Now, +PQ = +14 +4 +1 +9 +) +5 +3 +( +) +3 +2 +( +) +2 +1( +2 +2 +2 += ++ ++ += +− ++ +− ++ ++ +QR = +14 +2 +56 +16 +4 +36 +)3 +1 +( +) +2 +0 +( +)1 +7 +( +2 +2 +2 += += ++ ++ += +− +− ++ +− ++ +− +and +PR = +14 +3 +126 +36 +9 +81 +) +5 +1 +( +)3 +0 +( +) +2 +7 +( +2 +2 +2 += += ++ ++ += +− +− ++ +− ++ ++ +Thus, PQ + QR = PR. Hence, P, Q and R are collinear. +Example 5 Are the points A (3, 6, 9), B (10, 20, 30) and C (25, – 41, 5), the vertices +of a right angled triangle? +Solution By the distance formula, we have +AB2 = (10 – 3)2 + (20 – 6)2 + (30 – 9)2 += 49 + 196 + 441 = 686 +BC2 += (25 – 10)2 + (– 41 – 20)2 + (5 – 30)2 += 225 + 3721 + 625 = 4571 +Reprint 2025-26 + +INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 213 +CA2 += (3 – 25)2 + (6 + 41)2 + (9 – 5)2 += 484 + 2209 + 16 = 2709 +We find that CA2 + AB2 ≠ BC2. +Hence, the triangle ABC is not a right angled triangle. +Example 6 Find the equation of set of points P such that PA2 + PB2 = 2k2, where +A and B are the points (3, 4, 5) and (–1, 3, –7), respectively. +Solution Let the coordinates of point P be (x, y, z). +Here +PA2 = (x – 3)2 + (y – 4)2 + ( z – 5)2 +PB2 = (x + 1)2 + (y – 3)2 + (z + 7)2 +By the given condition PA2 + PB2 = 2k2, we have +(x – 3)2 + (y – 4)2 + (z – 5)2 + (x + 1)2 + (y – 3)2 + (z + 7)2 = 2k2 +i.e., +2x2 + 2y2 + 2z2 – 4x – 14y + 4z = 2k2 – 109. +EXERCISE 11.2 +1. +Find the distance between the following pairs of points: +(i) +(2, 3, 5) and (4, 3, 1) +(ii) +(–3, 7, 2) and (2, 4, –1) +(iii) +(–1, 3, – 4) and (1, –3, 4) +(iv) +(2, –1, 3) and (–2, 1, 3). +2. +Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear. +3. +Verify the following: +(i) +(0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle. +(ii) +(0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle. +(iii) +(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram. +4. +Find the equation of the set of points which are equidistant from the points +(1, 2, 3) and (3, 2, –1). +5. +Find the equation of the set of points P, the sum of whose distances from + A (4, 0, 0) and B (– 4, 0, 0) is equal to 10. +Miscellaneous Examples +Example 7 Show that the points A (1, 2, 3), B (–1, –2, –1), C (2, 3, 2) and +D (4, 7, 6) are the vertices of a parallelogram ABCD, but it is not a rectangle. +Solution To show ABCD is a parallelogram we need to show opposite side are equal +Note that. +Reprint 2025-26 + +214 MATHEMATICS +AB = +2 +2 +2 +) +3 +1 +( +) +2 +2 +( +)1 +1 +( +− +− ++ +− +− ++ +− +− + = +4 +16 +16 ++ ++ += 6 +BC = +2 +2 +2 +)1 +2 +( +) +2 +3 +( +)1 +2 +( ++ ++ ++ ++ ++ + = +9 +25 +9 ++ ++ + = +43 +CD = +2 +2 +2 +) +2 +6 +( +) +3 +7 +( +) +2 +4 +( +− ++ +− ++ +− + = +6 +16 +16 +4 += ++ ++ +DA = +2 +2 +2 +) +6 +3 +( +) +7 +2 +( +) +4 +1( +− ++ +− ++ +− + = +43 +9 +25 +9 += ++ ++ +Since AB = CD and BC = AD, ABCD is a parallelogram. +Now, it is required to prove that ABCD is not a rectangle. For this, we show that +diagonals AC and BD are unequal. We have +AC += +3 +1 +1 +1 +)3 +2 +( +) +2 +3 +( +)1 +2 +( +2 +2 +2 += ++ ++ += +− ++ +− ++ +− +BD += +155 +49 +81 +25 +)1 +6 +( +) +2 +7 +( +)1 +4 +( +2 +2 +2 += ++ ++ += ++ ++ ++ ++ ++ +. +Since AC ≠ BD, ABCD is not a rectangle. +ANote We can also show that ABCD is a parallelogram, using the property that +diagonals AC and BD bisect each other. +Example 8 Find the equation of the set of the points P such that its distances from the +points A (3, 4, –5) and B (– 2, 1, 4) are equal. +Solution If P (x, y, z) be any point such that PA = PB. +Now +2 +2 +2 +2 +2 +2 +) +4 +( +)1 +( +) +2 +( +) +5 +( +) +4 +( +)3 +( +− ++ +− ++ ++ += ++ ++ +− ++ +− +z +y +x +z +y +x +or + +2 +2 +2 +2 +2 +2 +) +4 +( +)1 +( +) +2 +( +) +5 +( +) +4 +( +)3 +( +− ++ +− ++ ++ += ++ ++ +− ++ +− +z +y +x +z +y +x +or +10x+ 6y – 18z – 29 = 0. +Example 9 The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates +of A and B are (3, –5, 7) and (–1, 7, – 6), respectively, find the coordinates of the +point C. +Solution Let the coordinates of C be (x, y, z) and the coordinates of the centroid G be +(1, 1, 1). Then +Reprint 2025-26 + +INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 215 +x + +− += +3 +1 +3 +1, i.e., x = 1; y −+ += +5 +7 +3 +1, i.e., y = 1; z + +− += +7 +6 +3 +1, i.e., z = 2. +Hence, coordinates of C are (1, 1, 2). +Miscellaneous Exercise on Chapter 11 +1. +Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and +C (– 1, 1, 2). Find the coordinates of the fourth vertex. +2. +Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) +and (6, 0, 0). +3. +If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), +Q (– 4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c. +4. +If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of +the set of points P such that PA2 + PB2 = k2, where k is a constant. +Summary +®In three dimensions, the coordinate axes of a rectangular Cartesian coordi- +nate system are three mutually perpendicular lines. The axes are called the x, +y and z-axes. +®The three planes determined by the pair of axes are the coordinate planes, +called XY, YZ and ZX-planes. +®The three coordinate planes divide the space into eight parts known as octants. +®The coordinates of a point P in three dimensional geometry is always written +in the form of triplet like (x, y, z). Here x, y and z are the distances from the +YZ, ZX and XY-planes. +® (i) +Any point on x-axis is of the form (x, 0, 0) + (ii) +Any point on y-axis is of the form (0, y, 0) +(iii) +Any point on z-axis is of the form (0, 0, z). +®Distance between two points P(x1, y1, z1) and Q (x2, y2, z2) is given by +2 +2 +2 +2 +1 +2 +1 +2 +1 +PQ +( x +x ) +( y +y ) +( z +z ) += +− ++ +− ++ +− +Reprint 2025-26 + +216 MATHEMATICS +— v — +Historical Note +Rene’ Descartes (1596–1650), the father of analytical geometry, essentially dealt +with plane geometry only in 1637. The same is true of his co-inventor Pierre +Fermat (1601-1665) and La Hire (1640-1718). Although suggestions for the three +dimensional coordinate geometry can be found in their works but no details. +Descartes had the idea of coordinates in three dimensions but did not develop it. +J.Bernoulli (1667-1748) in a letter of 1715 to Leibnitz introduced the three coor- +dinate planes which we use today. It was Antoinne Parent +(1666-1716), who gave a systematic development of analytical solid geometry +for the first time in a paper presented to the French Academy in 1700. +L.Euler (1707-1783) took up systematically the three dimensional coordinate ge- +ometry, in Chapter 5 of the appendix to the second volume of his “Introduction +to Geometry” in 1748. +It was not until the middle of the nineteenth century that geometry was extended +to more than three dimensions, the well-known application of which is in the +Space-Time Continuum of Einstein’s Theory of Relativity. +Reprint 2025-26" +class_11,12,Limits and Derivatives,ncert_books/class_11/kemh1dd/kemh112.pdf,"vWith the Calculus as a key, Mathematics can be successfully applied to the +explanation of the course of Nature – WHITEHEAD v +12.1 Introduction +This chapter is an introduction to Calculus. Calculus is that +branch of mathematics which mainly deals with the study +of change in the value of a function as the points in the +domain change. First, we give an intuitive idea of derivative +(without actually defining it). Then we give a naive definition +of limit and study some algebra of limits. Then we come +back to a definition of derivative and study some algebra +of derivatives. We also obtain derivatives of certain +standard functions. +12.2 Intuitive Idea of Derivatives +Physical experiments have confirmed that the body dropped +from a tall cliff covers a distance of 4.9t2 metres in t seconds, +i.e., distance s in metres covered by the body as a function of time t in seconds is given +by s = 4.9t2. +The adjoining Table 13.1 gives the distance travelled in metres at various intervals +of time in seconds of a body dropped from a tall cliff. +The objective is to find the veloctiy of the body at time t = 2 seconds from this +data. One way to approach this problem is to find the average velocity for various +intervals of time ending at t = 2 seconds and hope that these throw some light on the +velocity at t = 2 seconds. +Average velocity between t = t1 and t = t2 equals distance travelled between + t = t1 and t = t2 seconds divided by (t2 – t1). Hence the average velocity in the first +two seconds +12 +Chapter +LIMITS AND DERIVATIVES +Sir Issac Newton +(1642-1727) +Reprint 2025-26 + +218 +MATHEMATICS += +2 +1 +2 +1 +Distance travelled between +2 +0 +Time interval ( +) +t +and t +t +t += += +− += ( +) +( +) +19.6 +0 +9.8 +/ +2 +0 +m +m s +s +− += +− +. +Similarly, the average velocity between t = 1 + and t = 2 is +( +) +( +) +19.6 – 4.9 +2 1 +m +s +− += 14.7 m/s +Likewise we compute the average velocitiy +between t = t1 and t = 2 for various t1. The following +Table 13.2 gives the average velocity (v), t = t1 +seconds and t = 2 seconds. +Table 12.2 +t1 +0 +1 +1.5 +1.8 +1.9 +1.95 +1.99 +v +9.8 +14.7 +17.15 +18.62 +19.11 +19.355 +19.551 +From Table 12.2, we observe that the average velocity is gradually increasing. +As we make the time intervals ending at t = 2 smaller, we see that we get a better idea +of the velocity at t = 2. Hoping that nothing really dramatic happens between 1.99 +seconds and 2 seconds, we conclude that the average velocity at t = 2 seconds is just +above 19.551m/s. +This conclusion is somewhat strengthened by the following set of computation. +Compute the average velocities for various time intervals starting at t = 2 seconds. As +before the average velocity v between t = 2 seconds and t = t2 seconds is += +2 +2 +Distance travelled between 2 seconds and +seconds +2 +t +t − += +2 +2 +Distance travelled in +seconds + Distance travelled in 2 seconds +2 +t +t +− +− +t +s +0 +0 +1 +4.9 +1.5 +11.025 +1.8 +15.876 +1.9 +17.689 +1.95 +18.63225 +2 +19.6 +2.05 +20.59225 +2.1 +21.609 +2.2 +23.716 +2.5 +30.625 +3 +44.1 +4 +78.4 +Table 12.1 +Reprint 2025-26 + + LIMITS AND DERIVATIVES 219 += +2 +2 +Distance travelled in seconds + 19.6 +2 +t +t +− +− +The following Table 12.3 gives the average velocity v in metres per second +between t = 2 seconds and t2 seconds. +Table 12.3 +t2 +4 +3 +2.5 +2.2 +2.1 +2.05 +2.01 +v +29.4 +24.5 22.05 20.58 20.09 19.845 +19.649 +Here again we note that if we take smaller time intervals starting at t = 2, we get +better idea of the velocity at t = 2. +In the first set of computations, what we have done is to find average velocities +in increasing time intervals ending at t = 2 and then hope that nothing dramatic happens +just before t = 2. In the second set of computations, we have found the average velocities +decreasing in time intervals ending at t = 2 and then hope that nothing dramatic happens +just after t = 2. Purely on the physical grounds, both these sequences of average +velocities must approach a common limit. We can safely conclude that the velocity of +the body at t = 2 is between 19.551m/s and 19.649 m/s. Technically, we say that the +instantaneous velocity at t = 2 is between 19.551 m/s and 19.649 m/s. As is +well-known, velocity is the rate of change of displacement. Hence what we have +accomplished is the following. From the given data of distance covered at various time +instants we have estimated the rate of +change of the distance at a given instant +of time. We say that the derivative of +the distance function s = 4.9t2 at t = 2 +is between 19.551 and 19.649. +An alternate way of viewing this +limiting process is shown in Fig 12.1. +This is a plot of distance s of the body +from the top of the cliff versus the time +t elapsed. In the limit as the sequence +of time intervals h1, h2, ..., approaches +zero, the sequence of average velocities +approaches the same limit as does the +sequence of ratios +Fig 12.1 +Reprint 2025-26 + +220 +MATHEMATICS +3 +3 +1 +1 +2 +2 +1 +2 +3 +C B +C B +C B +, +, +AC +AC +AC +, ... +where C1B1 = s1 – s0 is the distance travelled by the body in the time interval h1 = AC1, +etc. From the Fig 12.1 it is safe to conclude that this latter sequence approaches the +slope of the tangent to the curve at point A. In other words, the instantaneous velocity +v(t) of a body at time t = 2 is equal to the slope of the tangent of the curve s = 4.9t2 at +t = 2. +12.3 Limits +The above discussion clearly points towards the fact that we need to understand limit- +ing process in greater clarity. We study a few illustrative examples to gain some famil- +iarity with the concept of limits. +Consider the function f(x) = x2. Observe that as x takes values very close to 0, +the value of f(x) also moves towards 0 (See Fig 2.10 Chapter 2). We say +( ) +0 +lim +0 +x +f +x +→ += +(to be read as limit of f (x) as x tends to zero equals zero). The limit of f (x) as x tends +to zero is to be thought of as the value f (x) should assume at x = 0. +In general as x → a, f (x) → l, then l is called limit of the function f (x) which is +symbolically written as +( ) +lim +x +a f +x +l +→ += . +Consider the following function g(x) = |x|, x ≠0. Observe that g(0) is not defined. +Computing the value of g(x) for values of x very +near to 0, we see that the value of g(x) moves +towards 0. So, +0 +lim +x→ g(x) = 0. This is intuitively +clear from the graph of y = |x| for x ≠0. +(See Fig 2.13, Chapter 2). +Consider the following function. +( ) +2 +4 , +2 +2 +x +h x +x +x +− += +≠ +− +. +Compute the value of h(x) for values of +x very near to 2 (but not at 2). Convince yourself +that all these values are near to 4. This is +somewhat strengthened by considering the graph +of the function y = h(x) given here (Fig 12.2). +Fig 12.2 +Reprint 2025-26 + + LIMITS AND DERIVATIVES 221 +In all these illustrations the value which the function should assume at a given +point x = a did not really depend on how is x tending to a. Note that there are essentially +two ways x could approach a number a either from left or from right, i.e., all the +values of x near a could be less than a or could be greater than a. This naturally leads +to two limits – the right hand limit and the left hand limit. Right hand limit of a +function f(x) is that value of f(x) which is dictated by the values of f(x) when x tends +to a from the right. Similarly, the left hand limit. To illustrate this, consider the function +( ) +1, +0 +2, +0 +x +f x +x +≤ + +=  +> + +Graph of this function is shown in the Fig 12.3. It is +clear that the value of f at 0 dictated by values of f(x) with +x ≤ 0 equals 1, i.e., the left hand limit of f (x) at 0 is +0 +lim +( ) 1 +x +f x +→ += . +Similarly, the value of f at 0 dictated by values of +f (x) with x > 0 equals 2, i.e., the right hand limit of f (x) +at 0 is +0 +lim +( ) +2 +x +f x ++ +→ += +. +In this case the right and left hand limits are different, and hence we say that the +limit of f (x) as x tends to zero does not exist (even though the function is defined at 0). +Summary + We say lim +x +a +→ +– f(x) is the expected value of f at x = a given the values of f near +x to the left of a. This value is called the left hand limit of f at a. +We say lim +( ) +x +a f x ++ +→ + is the expected value of f at x = a given the values of +f near x to the right of a. This value is called the right hand limit of f(x) at a. +If the right and left hand limits coincide, we call that common value as the limit +of f(x) at x = a and denote it by lim +x +a +→ f(x). +Illustration 1 Consider the function f(x) = x + 10. We want to find the limit of this +function at x = 5. Let us compute the value of the function f(x) for x very near to 5. +Some of the points near and to the left of 5 are 4.9, 4.95, 4.99, 4.995. . ., etc. Values +of the function at these points are tabulated below. Similarly, the real number 5.001, +Fig 12.3 +Reprint 2025-26 + +222 +MATHEMATICS +5.01, 5.1 are also points near and to the right of 5. Values of the function at these points +are also given in the Table 12.4. +Table 12.4 +From the Table 12.4, we deduce that value of f(x) at x = 5 should be greater than +14.995 and less than 15.001 assuming nothing dramatic happens between x = 4.995 +and 5.001. It is reasonable to assume that the value of the f(x) at x = 5 as dictated by +the numbers to the left of 5 is 15, i.e., +( ) +– +5 +lim +15 +x +f x +→ += +. +Similarly, when x approaches 5 from the right, f(x) should be taking value 15, i.e., +( ) +5 +lim +15 +x +f x ++ +→ += +. +Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are +both equal to 15. Thus, +( ) +( ) +( ) +5 +5 +5 +lim +lim +lim +15 +x +x +x +f x +f x +f x +− ++ +→ +→ +→ += += += +. +This conclusion about the limit being equal to 15 is somewhat strengthened by +seeing the graph of this function which is given in Fig 2.16, Chapter 2. In this figure, we +note that as x approaches 5 from either right or left, the graph of the function +f(x) = x +10 approaches the point (5, 15). +We observe that the value of the function at x = 5 also happens to be equal to 15. +Illustration 2 Consider the function f(x) = x3. Let us try to find the limit of this +function at x = 1. Proceeding as in the previous case, we tabulate the value of f(x) at +x near 1. This is given in the Table 12.5. +Table 12.5 +From this table, we deduce that value of f(x) at x = 1 should be greater than +0.997002999 and less than 1.003003001 assuming nothing dramatic happens between +x +0.9 +0.99 +0.999 +1.001 +1.01 +1.1 +f(x) +0.729 +0.970299 +0.997002999 +1.003003001 +1.030301 +1.331 +x +4.9 +4.95 +4.99 +4.995 +5.001 +5.01 +5.1 +f(x) +14.9 +14.95 +14.99 +14.995 +15.001 +15.01 +15.1 +Reprint 2025-26 + + LIMITS AND DERIVATIVES 223 +x = 0.999 and 1.001. It is reasonable to assume that the value of the f(x) at x = 1 as +dictated by the numbers to the left of 1 is 1, i.e., +( ) +1 +lim +1 +x +f x +− +→ += . +Similarly, when x approaches 1 from the right, f(x) should be taking value 1, i.e., +( ) +1 +lim +1 +x +f x ++ +→ += . +Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are +both equal to 1. Thus, +( ) +( ) +( ) +1 +1 +1 +lim +lim +lim +1 +x +x +x +f x +f x +f x +− ++ +→ +→ +→ += += += . +This conclusion about the limit being equal to 1 is somewhat strengthened by +seeing the graph of this function which is given in Fig 2.11, Chapter 2. In this figure, we +note that as x approaches 1 from either right or left, the graph of the function +f(x) = x3 approaches the point (1, 1). +We observe, again, that the value of the function at x = 1 also happens to be +equal to 1. +Illustration 3 Consider the function f(x) = 3x. Let us try to find the limit of this +function at x = 2. The following Table 12.6 is now self-explanatory. +Table 12.6 +x +1.9 +1.95 +1.99 +1.999 +2.001 +2.01 +2.1 +f(x) +5.7 +5.85 +5.97 +5.997 +6.003 +6.03 +6.3 +As before we observe that as x approaches 2 +from either left or right, the value of f(x) seem to +approach 6. We record this as +( ) +( ) +( ) +2 +2 +2 +lim +lim +lim +6 +x +x +x +f x +f x +f x +− ++ +→ +→ +→ += += += +Its graph shown in Fig 12.4 strengthens this +fact. +Here again we note that the value of the function +at x = 2 coincides with the limit at x = 2. +Illustration 4 Consider the constant function +f(x) = 3. Let us try to find its limit at x = 2. This +function being the constant function takes the same +Fig 12.4 +Reprint 2025-26 + +224 +MATHEMATICS +value (3, in this case) everywhere, i.e., its value at points close to 2 is 3. Hence +( ) +( ) +( ) +2 +2 +2 +lim +lim +lim +3 +x +x +x +f x +f x +f x ++ +→ +→ +→ += += += +Graph of f(x) = 3 is anyway the line parallel to x-axis passing through (0, 3) and +is shown in Fig 2.9, Chapter 2. From this also it is clear that the required limit is 3. In +fact, it is easily observed that +( ) +lim +3 +x +a f x +→ += + for any real number a. +Illustration 5 Consider the function f(x) = x2 + x. We want to find +( ) +1 +lim +x +f x +→ +. We +tabulate the values of f(x) near x = 1 in Table 12.7. +Table 12.7 +x +0.9 +0.99 +0.999 +1.01 +1.1 +1.2 +f(x) +1.71 +1.9701 +1.997001 +2.0301 +2.31 +2.64 +From this it is reasonable to deduce that +( ) +( ) +( ) +1 +1 +1 +lim +lim +lim +2 +x +x +x +f x +f x +f x +− ++ +→ +→ +→ += += += +. +From the graph of f(x) = x2 + x +shown in the Fig 12.5, it is clear that as x +approaches 1, the graph approaches (1, 2). +Here, again we observe that the +1 +lim +x→ f (x) = f (1) +Now, convince yourself of the +following three facts: +2 +1 +1 +1 +lim +1, lim +1 and lim +1 +2 +x +x +x +x +x +x +→ +→ +→ += += ++ = +Then +2 +2 +1 +1 +1 +lim +lim +1 1 +2 +lim +x +x +x +x +x +x +x +→ +→ +→ + ++ += + = += ++ + +. +Also +( +) +( +) +2 +1 +1 +1 +1 +lim . lim +1 +1.2 +2 +lim +1 +lim +x +x +x +x +x +x +x x +x +x +→ +→ +→ +→ + + + ++ += += += ++ += ++ + + + +. +Fig 12.5 +Reprint 2025-26 + + LIMITS AND DERIVATIVES 225 +Illustration 6 Consider the function f(x) = sin x. We are interested in +2 +lim sin +x +x +π +→ +, +where the angle is measured in radians. +Here, we tabulate the (approximate) value of f(x) near 2 +π (Table 12.8). From +this, we may deduce that +( ) +( ) +( ) +2 +2 +2 +lim +lim +lim +1 +x +x +x +f x +f x +f x +− ++ +π +π +π +→ +→ +→ += += += +. +Further, this is supported by the graph of f(x) = sin x which is given in the Fig 3.8 +(Chapter 3). In this case too, we observe that +2 +lim +x +π +→ + sin x = 1. +Table 12.8 +x +0.1 +2 +π − +0.01 +2 +π − +0.01 +2 +π + +0.1 +2 +π + +f(x) +0.9950 +0.9999 +0.9999 +0.9950 +Illustration 7 Consider the function f(x) = x + cos x. We want to find the +0 +lim +x→f (x). +Here we tabulate the (approximate) value of f(x) near 0 (Table 12.9). +Table 12.9 +From the Table 13.9, we may deduce that +( ) +( ) +( ) +0 +0 +0 +lim +lim +lim +1 +x +x +x +f x +f x +f x +− ++ +→ +→ +→ += += += +In this case too, we observe that +0 +lim +x→f (x) = f (0) = 1. +Now, can you convince yourself that +[ +] +0 +0 +0 +lim +cos +lim +limcos +x +x +x +x +x +x +x +→ +→ +→ ++ += ++ + is indeed true? +x +– 0.1 +– 0.01 +– 0.001 +0.001 +0.01 +0.1 +f(x) +0.9850 +0.98995 +0.9989995 +1.0009995 +1.00995 +1.0950 +Reprint 2025-26 + +226 +MATHEMATICS +Illustration 8 Consider the function +( ) +2 +1 +f x +x += + for +0 +x > +. We want to know +0 +lim +x→f (x). +Here, observe that the domain of the function is given to be all positive real +numbers. Hence, when we tabulate the values of f(x), it does not make sense to talk of +x approaching 0 from the left. Below we tabulate the values of the function for positive +x close to 0 (in this table n denotes any positive integer). +From the Table 12.10 given below, we see that as x tends to 0, f(x) becomes +larger and larger. What we mean here is that the value of f(x) may be made larger than +any given number. +Table 12.10 +x +1 +0.1 +0.01 +10–n +f(x) +1 +100 +10000 +102n +Mathematically, we say +( ) +0 +lim +x +f x +→ += +∞ +We also remark that we will not come across such limits in this course. +Illustration 9 We want to find +( ) +0 +lim +x +f x +→ +, where +( ) +2, +0 +0 +, +0 +2, +0 +x +x +f x +x +x +x +− +< + + += += + ++ +> + +As usual we make a table of x near 0 with f(x). Observe that for negative values of x +we need to evaluate x – 2 and for positive values, we need to evaluate x + 2. +Table 12.11 +From the first three entries of the Table 12.11, we deduce that the value of the +function is decreasing to –2 and hence. +( ) +0 +lim +2 +x +f x +− +→ += − +x +– 0.1 +– 0.01 +– 0.001 +0.001 +0.01 +0.1 +f(x) +– 2.1 +– 2.01 +– 2.001 +2.001 +2.01 +2.1 +Reprint 2025-26 + + LIMITS AND DERIVATIVES 227 +From the last three entires of the table we deduce that the value of the function +is increasing from 2 and hence +( ) +0 +lim +2 +x +f x ++ +→ += +Since the left and right hand limits at 0 do not coincide, +we say that the limit of the function at 0 does not exist. + Graph of this function is given in the Fig12.6. Here, +we remark that the value of the function at x = 0 is well +defined and is, indeed, equal to 0, but the limit of the function +at x = 0 is not even defined. +Illustration 10 As a final illustration, we find +( ) +1 +lim +x +f x +→ +, +where +( ) +2 +1 +0 +1 +x +x +f x +x ++ +≠ + +=  += + +Table 12.12 +x +0.9 +0.99 +0.999 +1.001 +1.01 +1.1 +f(x) +2.9 +2.99 +2.999 +3.001 +3.01 +3.1 +As usual we tabulate the values of f(x) for x near 1. From the values of f(x) for +x less than 1, it seems that the function should take value 3 at x = 1., i.e., +( ) +1 +lim +3 +x +f x +− +→ += +. +Similarly, the value of f(x) should be 3 as dic- +tated by values of f(x) at x greater than 1. i.e. +( ) +1 +lim +3 +x +f x ++ +→ += +. +But then the left and right hand limits coin- +cide and hence + +( ) +( ) +( ) +1 +1 +1 +lim +lim +lim +3 +x +x +x +f x +f x +f x +− ++ +→ +→ +→ += += += +. +Graph of function given in Fig 12.7 strength- +ens our deduction about the limit. Here, we +Fig 12.6 +Fig 12.7 +Reprint 2025-26 + +228 +MATHEMATICS +note that in general, at a given point the value of the function and its limit may be +different (even when both are defined). +12.3.1 Algebra of limits In the above illustrations, we have observed that the limiting +process respects addition, subtraction, multiplication and division as long as the limits +and functions under consideration are well defined. This is not a coincidence. In fact, +below we formalise these as a theorem without proof. +Theorem 1 Let f and g be two functions such that both lim +x +a +→ f (x) and lim +x +a +→ g(x) exist. +Then + (i) Limit of sum of two functions is sum of the limits of the functions, i.e., +lim +x +a +→[f(x) + g (x)] = lim +x +a +→ f(x) + lim +x +a +→ g(x). + (ii) +Limit of difference of two functions is difference of the limits of the functions, i.e., +lim +x +a +→[f(x) – g(x)] = lim +x +a +→ f(x) – lim +x +a +→ g(x). +(iii) +Limit of product of two functions is product of the limits of the functions, i.e., +lim +x +a +→ [f(x) . g(x)] = lim +x +a +→ f(x). lim +x +a +→ g(x). +(iv) +Limit of quotient of two functions is quotient of the limits of the functions (whenever +the denominator is non zero), i.e., +( ) +( ) +( ) +( ) +lim +lim +lim +x +a +x +a +x +a +f x +f x +g x +g x +→ +→ +→ += +ANote In particular as a special case of (iii), when g is the constant function +such that g(x) = λ , for some real number λ , we have +( +) ( ) +( ) +lim +. +.lim +x +a +x +a +f +x +f x +→ +→ + + +λ += λ + + +. +In the next two subsections, we illustrate how to exploit this theorem to evaluate +limits of special types of functions. +12.3.2 Limits of polynomials and rational functions A function f is said to be a +polynomial function of degree n f(x) = a0 + a1x + a2x2 +. . . + anxn, where ais are real +numbers such that an ≠ 0 for some natural number n. +We know that lim +x +a +→x = a. Hence +Reprint 2025-26 + + LIMITS AND DERIVATIVES 229 +( +) +2 +2 +lim +lim +. +lim .lim +. +x +a +x +a +x +a +x +a +x +x x +x +x +a a +a +→ +→ +→ +→ += += += += +An easy exercise in induction on n tells us that +lim +n +n +x +a x +a +→ += +Now, let +( ) +2 +0 +1 +2 +... +n +n +f x +a +a x +a x +a x += ++ ++ ++ ++ + be a polynomial function. Thinking +of each of +2 +0 +1 +2 +, +, +,..., +n +n +a +a x a x +a x as a function, we have +( ) +lim +x +a f +x +→ += +2 +0 +1 +2 +lim +... +n +n +x +a a +a x +a x +a x +→ + ++ ++ ++ ++ + + += +2 +0 +1 +2 +lim +lim +lim +... +lim +n +n +x +a +x +a +x +a +x +a +a +a x +a x +a x +→ +→ +→ +→ ++ ++ ++ ++ += +2 +0 +1 +2 +lim +lim +... +lim +n +n +x +a +x +a +x +a +a +a +x +a +x +a +x +→ +→ +→ ++ ++ ++ ++ += +2 +0 +1 +2 +... +n +n +a +a a +a a +a a ++ ++ ++ ++ += +( ) +f a +(Make sure that you understand the justification for each step in the above!) +A function f is said to be a rational function, if f(x) = +( ) +( ) +g x +h x , where g(x) and h(x) +are polynomials such that h(x) ≠ 0. Then +( ) +( ) +( ) +( ) +( ) +( ) +( ) +lim +lim +lim +lim +x +a +x +a +x +a +x +a +g x +g x +g a +f x +h x +h x +h a +→ +→ +→ +→ += += += +However, if h(a) = 0, there are two scenarios – (i) when g(a) ≠ 0 and (ii) when +g(a) = 0. In the former case we say that the limit does not exist. In the latter case we +can write g(x) = (x – a)k g1 (x), where k is the maximum of powers of (x – a) in g(x) +Similarly, h(x) = (x – a)l h1 (x) as h (a) = 0. Now, if k > l, we have +( ) +( ) +( ) +( +) +( ) +( +) +( ) +1 +1 +lim +lim +lim +lim +lim +k +x +a +x +a +l +x +a +x +a +x +a +g x +x +a +g +x +f x +h x +x +a +h +x +→ +→ +→ +→ +→ +− += += +− +Reprint 2025-26 + +230 +MATHEMATICS += +( +) +( +) +( ) +( ) +( ) +( ) +1 +1 +1 +1 +lim +0. +0 +lim +k l +x +a +x +a +x +a +g +x +g +a +h x +h a +− +→ +→ +− += += +If k < l, the limit is not defined. +Example 1 Find the limits: (i) +3 +2 +1 +lim +1 +x +x +x +→ + +− ++ + + + (ii) +( +) +3 +lim +1 +x +x x +→ + ++ + + +(iii) +2 +10 +1 +lim 1 +... +x +x +x +x +→− + ++ ++ ++ ++ + +. +Solution The required limits are all limits of some polynomial functions. Hence the +limits are the values of the function at the prescribed points. We have +(i) +1 +lim +x→ [x3 – x2 + 1] = 13 – 12 + 1 = 1 +(ii) +( +) +( +) +( ) +3 +lim +1 +3 3 1 +3 4 +12 +x +x x +→ + ++ += ++ += += + + +(iii) +2 +10 +1 +lim 1 +... +x +x +x +x +→− + ++ ++ ++ ++ + + +( +) +( +) +( +) +2 +10 +1 +1 +1 +... +1 += + − ++ − ++ ++ − + +1 1 1... +1 +1 += −+ ++ = . +Example 2 Find the limits: +(i) +2 +1 +1 +lim +100 +x +x +x +→ + + ++ + + ++ + + +(ii) +3 +2 +2 +2 +4 +4 +lim +4 +x +x +x +x +x +→ + + +− ++ + + +− + + +(iii) +2 +3 +2 +2 +4 +lim +4 +4 +x +x +x +x +x +→ + + +− + + +− ++ + + +(iv) +3 +2 +2 +2 +2 +lim +5 +6 +x +x +x +x +x +→ + + +− + + +− ++ + + +(v) +2 +3 +2 +1 +2 +1 +lim +3 +2 +x +x +x +x +x +x +x +→ +− + + +− + + +− +− ++ + +. +Solution All the functions under consideration are rational functions. Hence, we first +evaluate these functions at the prescribed points. If this is of the form +0 +0 , we try to +rewrite the function cancelling the factors which are causing the limit to be of +the form 0 +0 . +Reprint 2025-26 + + LIMITS AND DERIVATIVES 231 +(i) +We have +2 +2 +1 +1 +1 +1 +2 +lim +100 +1 100 +101 +x +x +x +→ ++ ++ += += ++ ++ +(ii) +Evaluating the function at 2, it is of the form +0 +0 . +Hence +3 +2 +2 +2 +4 +4 +lim +4 +x +x +x +x +x +→ +− ++ +− + = +( +) +( +)( +) +2 +2 +2 +lim +2 +2 +x +x x +x +x +→ +− ++ +− += +( +) +( +) +2 +2 +lim +as +2 +2 +x +x x +x +x +→ +− +≠ ++ + = +( +) +2 2 +2 +0 +0 +2 +2 +4 +− += += ++ +. +(iii) +Evaluating the function at 2, we get it of the form 0 +0 . +Hence +2 +3 +2 +2 +4 +lim +4 +4 +x +x +x +x +x +→ +− +− ++ + = +( +)( +) +( +) +2 +2 +2 +2 +lim +2 +x +x +x +x x +→ ++ +− +− += +( +) +( +) +( +) +2 +2 +2 +2 +4 +lim +2 +2 2 +2 +0 +x +x +x x +→ ++ ++ += += +− +− +which is not defined. +(iv) +Evaluating the function at 2, we get it of the form 0 +0 . +Hence +3 +2 +2 +2 +2 +lim +5 +6 +x +x +x +x +x +→ +− +− ++ + = +( +) +( +)( +) +2 +2 +2 +lim +2 +3 +x +x +x +x +x +→ +− +− +− += +( +) +( ) +2 +2 +2 +2 +4 +lim +4 +3 +2 +3 +1 +x +x +x +→ += += += − +− +− +− +. +Reprint 2025-26 + +232 +MATHEMATICS +(v) +First, we rewrite the function as a rational function. +2 +3 +2 +2 +1 +3 +2 +x +x +x +x +x +x +− + + +− + + +− +− ++ + + = +( +) +( +) +2 +2 +1 +1 +3 +2 +x +x x +x x +x + + +− + + +− + + +− +− ++ + + += +( +) +( +)( +) +2 +1 +1 +1 +2 +x +x x +x x +x + + +− +− + + +− +− +− + + + + += +( +)( +) +2 +4 +4 +1 +1 +2 +x +x +x x +x + + +− ++ +− + + +− +− + + + + += +( +)( +) +2 +4 +3 +1 +2 +x +x +x x +x +− ++ +− +− +Evaluating the function at 1, we get it of the form 0 +0 . +Hence +2 +2 +3 +2 +1 +2 +1 +lim +3 +2 +x +x +x +x +x +x +x +→ + + +− +− + + +− +− ++ + + += +( +)( +) +2 +1 +4 +3 +lim +1 +2 +x +x +x +x x +x +→ +− ++ +− +− += +( +)( +) +( +)( +) +1 +3 +1 +lim +1 +2 +x +x +x +x x +x +→ +− +− +− +− += +( +) +1 +3 +lim +2 +x +x +x x +→ +− +− + = ( +) +1 +3 +1 1 +2 +− +− + = 2. +We remark that we could cancel the term (x – 1) in the above evaluation because +1 +x ≠. +Evaluation of an important limit which will be used in the sequel is given as a +theorem below. +Theorem 2 For any positive integer n, +1 +lim +n +n +n +x +a +x +a +na +x +a +− +→ +− += +− +. +Remark The expression in the above theorem for the limit is true even if n is any +rational number and a is positive. +Reprint 2025-26 + + LIMITS AND DERIVATIVES 233 +Proof Dividing (xn – an) by (x – a), we see that +xn – an = (x–a) (xn–1 + xn–2 a + xn–3 a2 + ... + x an–2 + an–1) +Thus, lim +lim +n +n +x +a +x +a +x +a +x +a +→ +→ +− += +− +(xn–1 + xn–2 a + xn–3 a2 + ... + x an–2 + an–1) += an – l + a an–2 +. . . + an–2 (a) +an–l += an–1 + an – 1 +...+an–1 + an–1 (n terms) += +1 +n +na − +Example 3 Evaluate: +(i) +15 +10 +1 +1 +lim +1 +x +x +x +→ +− +− +(ii) +0 +1 +1 +lim +x +x +x +→ ++ +− +Solution (i) We have +15 +10 +1 +1 +lim +1 +x +x +x +→ +− +− += +15 +10 +1 +1 +1 +lim +1 +1 +x +x +x +x +x +→ + + +− +− +÷ + + +− +− + + += +15 +10 +1 +1 +1 +1 +lim +lim +1 +1 +x +x +x +x +x +x +→ +→ + + + + +− +− +÷ + + + + +− +− + + + + += 15 (1)14 ÷ 10(1)9 (by the theorem above) += 15 ÷ 10 +3 +2 += +(ii) Put y = 1 + x, so that +1 +y → as +0. +x → +Then +0 +1 +1 +lim +x +x +x +→ ++ +− = +1 +1 +lim +–1 +y +y +y +→ +− += +1 +1 +2 +2 +1 +1 +lim +1 +y +y +y +→ +− +− += +1 1 +2 +1 (1) +2 +− + (by the remark above) = +1 +2 +Reprint 2025-26 + +234 +MATHEMATICS +12.4 Limits of Trigonometric Functions +The following facts (stated as theorems) about functions in general come in handy in +calculating limits of some trigonometric functions. +Theorem 3 Let f and g be two real valued functions with the same domain such that +f (x) ≤ g( x) for all x in the domain of definition, For some a, if both lim +x +a +→ f(x) and +lim +x +a +→ g(x) exist, then lim +x +a +→ f(x) ≤ lim +x +a +→ g(x). This is illustrated in Fig 12.8. +Theorem 4 (Sandwich Theorem) Let f, g and h be real functions such that +f (x) ≤ g( x) ≤ h(x) for all x in the common domain of definition. For some real number +a, if lim +x +a +→ f(x) = l = lim +x +a +→ h(x), then lim +x +a +→ g(x) = l. This is illustrated in Fig 12.9. +Given below is a beautiful geometric proof of the following important +inequality relating trigonometric functions. +sin +cos +1 +x +x +x +< +< +for +π +0 +2 +x +< +< +(*) +Fig 12.8 +Fig 12.9 +Reprint 2025-26 + + LIMITS AND DERIVATIVES 235 +Proof We know that sin (– x) = – sin x and cos( – x) = cos x. Hence, it is sufficient +to prove the inequality for +π +0 +2 +x +< +< +. +In the Fig 12.10, O is the centre of the unit circle such that +the angle AOC is x radians and 0 < x < π +2 . Line segments B A and +CD are perpendiculars to OA. Further, join AC. Then +Area of +OAC +∆ +< Area of sector OAC < Area of ∆OAB. +i.e., +2 +1 +1 +OA.CD +.π.(OA) +OA.AB +2 +2π +2 +x +< +< +. +i.e., +CD < x . OA < AB. +From ∆OCD, +sin x = CD +OA (since OC = OA) and hence CD = OA sin x. Also tan x = AB +OA and +hence +AB = OA. tan x. Thus +OA sin x < OA. x < OA. tan x. +Since length OA is positive, we have +sin x < x < tan x. +Since 0 < x < π +2 , sinx is positive and thus by dividing throughout by sin x, we have +1< +1 +sin +cos +x +x +x +< +. +Taking reciprocals throughout, we have +sin +cos +1 +x +x +x +< +< +which complete the proof. +Theorem 5 The following are two important limits. +(i) +0 +sin +lim +1 +x +x +x +→ += . +(ii) +0 +1 +cos +lim +0 +x +x +x +→ +− += +. +Proof (i) The inequality in (*) says that the function sin x +x +is sandwiched between the +function cos x and the constant function which takes value 1. +Fig 12.10 +Reprint 2025-26 + +236 +MATHEMATICS +Further, since +0 +lim +x→ cos x = 1, we see that the proof of (i) of the theorem is +complete by sandwich theorem. +To prove (ii), we recall the trigonometric identity 1 – cos x = 2 sin2 +2 +x + + + + + +. +Then +0 +1 cos +lim +x +x +x +→ +− + = +2 +0 +0 +2sin +sin +2 +2 +lim +lim +.sin 2 +2 +x +x +x +x +x +x +x +→ +→ + + + + + + + + + + + + + + += + + + + += +0 +0 +sin 2 +lim +.limsin +1.0 +0 +2 +2 +x +x +x +x +x +→ +→ + + + + + + + + += += + + + + +Observe that we have implicitly used the fact that +0 +x → + is equivalent to +0 +2 +x → +. This +may be justified by putting y = 2 +x . +Example 4 Evaluate: +(i) +0 +sin4 +lim sin2 +x +x +x +→ +(ii) +0 +tan +lim +x +x +x +→ +Solution (i) +0 +sin 4 +lim sin 2 +x +x +x +→ +0 +sin4 +2 +lim +. +.2 +4 +sin2 +x +x +x +x +x +→ + + += + + + + += +0 +sin 4 +sin 2 +2.lim +4 +2 +x +x +x +x +x +→ + + + + +÷ + + + + + + + + += +4 +0 +2 +0 +sin4 +sin2 +2. lim +lim +4 +2 +x +x +x +x +x +x +→ +→ + + + + +÷ + + + + + + + + += 2.1.1 = 2 (as x → 0, 4x → 0 and 2x → 0) +Reprint 2025-26 + + LIMITS AND DERIVATIVES 237 +(ii) We have +0 +tan +lim +x +x +x +→ + = +0 +sin +lim +cos +x +x +x +x +→ + = +0 +0 +sin +1 +lim +. lim cos +x +x +x +x +x +→ +→ + = 1.1 = 1 +A general rule that needs to be kept in mind while evaluating limits is the following. +Say, given that the limit +( ) +( ) +lim +x +a +f x +g x +→ + exists and we want to evaluate this. First we check +the value of f (a) and g(a). If both are 0, then we see if we can get the factor which +is causing the terms to vanish, i.e., see if we can write f(x) = f1 (x) f2(x) so that +f1 (a) = 0 and f2 (a) ≠ 0. Similarly, we write g(x) = g1 (x) g2(x), where g1(a) = 0 and +g2(a) ≠ 0. Cancel out the common factors from f(x) and g(x) (if possible) and write +( ) +( ) +( ) +( ) +f x +p x +g x +q x += +, where q(x) ≠ 0. +Then +( ) +( ) +( ) +( ) +lim +x +a +f x +p a +g x +q a +→ += +. +EXERCISE 12.1 +Evaluate the following limits in Exercises 1 to 22. +1. +3 +lim +3 +x +x +→ ++ +2. +π +22 +lim +7 +x +x +→ + + +− + + + + +3. +2 +1 +limπ +r +r +→ +4. +4 +4 +3 +lim +2 +x +x +x +→ ++ +− +5. +10 +5 +1 +1 +lim +1 +x +x +x +x +→− ++ ++ +− +6. +( +) +5 +0 +1 +1 +lim +x +x +x +→ ++ +− +7. +2 +2 +2 +3 +10 +lim +4 +x +x +x +x +→ +− +− +− +8. +4 +2 +3 +81 +lim +2 +5 +3 +x +x +x +x +→ +− +− +− +9. +0 +lim +1 +x +ax +b +cx +→ ++ ++ +10. +1 +3 +1 +1 +6 +1 +lim +1 +z +z +z +→ +− +− +11. +2 +2 +1 +lim +, +0 +x +ax +bx +c a +b +c +cx +bx +a +→ ++ ++ ++ ++ +≠ ++ ++ +12. +2 +1 +1 +2 +lim +2 +x +x +x +→− ++ ++ +13. +0 +sin +lim +x +ax +bx +→ +14. +0 +sin +lim +, , +0 +sin +x +ax a b +bx +→ +≠ +Reprint 2025-26 + +238 +MATHEMATICS +15. +( +) +( +) +π +sin π +lim π π +x +x +x +→ +− +− +16. +0 +cos +lim π +x +x +x +→ +− +17. +0 +cos2 +1 +lim cos +1 +x +x +x +→ +− +− +18. +0 +cos +lim +sin +x +ax +x +x +b +x +→ ++ +19. +0 +lim +sec +x +x +x +→ +20. +0 +sin +lim +, , +0 +sin +x +ax +bx a b a +b +ax +bx +→ ++ ++ +≠ ++ +, +21. +0 +lim (cosec +cot ) +x +x +x +→ +− +22. +π +2 +tan 2 +lim +π +2 +x +x +x +→ +− +23. Find +( ) +0 +lim +x +f x +→ + and +( ) +1 +lim +x +f x +→ +, where +( ) +( +) +2 +3, +0 +3 +1 , +0 +x +x +f x +x +x ++ +≤ + +=  ++ +> + +24. Find +( ) +1 +lim +x +f x +→ +, where +( ) +2 +2 +1, +1 +1, +1 +x +x +f x +x +x + +− +≤ + += − +− +> + +25. Evaluate +( ) +0 +lim +x +f x +→ +, where +( ) +| +|, +0 +0, +0 +x +x +f x +x +x + +≠ + +=  + += + +26. Find +( ) +0 +lim +x +f x +→ +, where +( ) +, +0 +| +| +0, +0 +x +x +x +f x +x + +≠ + +=  + += + +27. Find +( ) +5 +lim +x +f x +→ +, where +( ) +| +| 5 +f x +x += +− +28. Suppose +( ) +, +1 +4, +1 +, +1 +a +bx +x +f x +x +b +ax +x ++ +< + + += += + +− +> + +and if +1 +lim +x→f (x) = f (1) what are possible values of a and b? +Reprint 2025-26 + + LIMITS AND DERIVATIVES 239 +29. +Let a1, a2, . . ., an be fixed real numbers and define a function +( ) +( +) ( +) ( +) +1 +2 ... +n +f x +x +a +x +a +x +a += +− +− +− +. +What is +1 +lim +x +a +→f (x) ? For some a ≠ a1, a2, ..., an, compute lim +x +a +→ f (x). +30. +If ( ) +1, +0 +0, +0 +1, +0 +x +x +f x +x +x +x + ++ +< + += += + + +− +> + +. +For what value (s) of a does lim +x +a +→f (x) exists? +31. +If the function f(x) satisfies +( ) +2 +1 +2 +lim +π +1 +x +f x +x +→ +− += +− +, evaluate +( ) +1 +lim +x +f x +→ +. +32. + If +( ) +2 +3 +, +0 +, +0 +1 +, +1 +mx +n +x +f x +nx +m +x +nx +m +x + ++ +< + += ++ +≤ +≤ + + ++ +> + +. For what integers m and n does both +( ) +0 +lim +x +f x +→ +and +( ) +1 +lim +x +f x +→ +exist? +12.5 Derivatives +We have seen in the Section 13.2, that by knowing the position of a body at various +time intervals it is possible to find the rate at which the position of the body is changing. +It is of very general interest to know a certain parameter at various instants of time and +try to finding the rate at which it is changing. There are several real life situations +where such a process needs to be carried out. For instance, people maintaining a +reservoir need to know when will a reservoir overflow knowing the depth of the water +at several instances of time, Rocket Scientists need to compute the precise velocity +with which the satellite needs to be shot out from the rocket knowing the height of the +rocket at various times. Financial institutions need to predict the changes in the value of +a particular stock knowing its present value. In these, and many such cases it is desirable +to know how a particular parameter is changing with respect to some other parameter. +The heart of the matter is derivative of a function at a given point in its domain +of definition. +Reprint 2025-26 + +240 +MATHEMATICS +Definition 1 Suppose f is a real valued function and a is a point in its domain of +definition. The derivative of f at a is defined by +( +) +( ) +0 +lim +h +f a +h +f a +h +→ ++ +− +provided this limit exists. Derivative of f (x) at a is denoted by f′(a). +Observe that f′ (a) quantifies the change in f(x) at a with respect to x. +Example 5 Find the derivative at x = 2 of the function f(x) = 3x. +Solution We have +f′ (2) = +( +) +( ) +0 +2 +2 +lim +h +f +h +f +h +→ ++ +− += +( +) +( ) +0 +3 2 +3 2 +lim +h +h +h +→ ++ +− += +0 +0 +0 +6 +3 +6 +3 +lim +lim +lim3 +3 +h +h +h +h +h +h +h +→ +→ +→ ++ +− += += += +. +The derivative of the function 3x at x = 2 is 3. +Example 6 Find the derivative of the function f(x) = 2x2 + 3x – 5 at x = –1. Also prove +that f ′ (0) + 3f ′ ( –1) = 0. +Solution We first find the derivatives of f(x) at x = –1 and at x = 0. We have +( +) +' +1 +f +− += +( +) +( +) +0 +1 +1 +lim +h +f +h +f +h +→ +−+ +− +− += +( +) +( +) +( +) +( +) +2 +2 +0 +2 +1 +3 +1 +5 +2 +1 +3 +1 +5 +lim +h +h +h +h +→ + + + + +−+ ++ +−+ +− +− +− ++ +− +− + + + + += +( +) +( ) +2 +0 +0 +2 +lim +lim 2 +1 +2 0 +1 +1 +h +h +h +h +h +h +→ +→ +− += +− += +−= − +and + +( ) +' 0 +f += +( +) +( ) +0 +0 +0 +lim +h +f +h +f +h +→ ++ +− += +( +) +( +) +( ) +( ) +2 +2 +0 +2 0 +3 0 +5 +2 0 +3 0 +5 +lim +h +h +h +h +→ + + + + ++ ++ ++ +− +− ++ +− + + + + +Reprint 2025-26 + + LIMITS AND DERIVATIVES 241 += +( +) +( ) +2 +0 +0 +2 +3 +lim +lim 2 +3 +2 0 +3 +3 +h +h +h +h +h +h +→ +→ ++ += ++ += ++ += +Clearly +( ) +( +) +' 0 +3 ' +1 +0 +f +f ++ +− += +Remark At this stage note that evaluating derivative at a point involves effective use +of various rules, limits are subjected to. The following illustrates this. +Example 7 Find the derivative of sin x at x = 0. +Solution Let f(x) = sin x. Then +f ′(0) = +( +) +( ) +0 +0 +0 +lim +h +f +h +f +h +→ ++ +− += +( +) +( ) +0 +sin 0 +sin 0 +lim +h +h +h +→ ++ +− + = +0 +sin +lim +1 +h +h +h +→ += +Example 8 Find the derivative of f(x) = 3 at x = 0 and at x = 3. +Solution Since the derivative measures the change in function, intuitively it is clear +that the derivative of the constant function must be zero at every point. This is indeed, +supported by the following computation. +( ) +' 0 +f += +( +) +( ) +0 +0 +0 +0 +0 +3 +3 +0 +lim +lim +lim +0 +h +h +h +f +h +f +h +h +h +→ +→ +→ ++ +− +− += += += +. +Similarly +( ) +' 3 +f + = +( +) +( ) +0 +0 +3 +3 +3 +3 +lim +lim +0 +h +h +f +h +f +h +h +→ +→ ++ +− +− += += +. +We now present a geomet- +ric interpretation of derivative of a +function at a point. Let y = f(x) be +a function and let P = (a, f(a)) and +Q = (a + h, f(a + h) be two points +close to each other on the graph +of this function. The Fig 12.11 is +now self explanatory. +Fig 12.11 +Reprint 2025-26 + +242 +MATHEMATICS +We know that +( ) +( +) +( ) +0 +lim +h +f a +h +f a +f +a +h +→ ++ +− +′ += +From the triangle PQR, it is clear that the ratio whose limit we are taking is +precisely equal to tan(QPR) which is the slope of the chord PQ. In the limiting process, +as h tends to 0, the point Q tends to P and we have +( +) +( ) +0 +Q +P +QR +lim +lim PR +h +f a +h +f a +h +→ +→ ++ +− += +This is equivalent to the fact that the chord PQ tends to the tangent at P of the +curve y = f(x). Thus the limit turns out to be equal to the slope of the tangent. Hence +( ) +tan ψ +f +a +′ += +. +For a given function f we can find the derivative at every point. If the derivative +exists at every point, it defines a new function called the derivative of f . Formally, we +define derivative of a function as follows. +Definition 2 Suppose f is a real valued function, the function defined by +( +) +( ) +0 +lim +h +f x +h +f x +h +→ ++ +− +wherever the limit exists is defined to be the derivative of f at x and is denoted by +f′(x). This definition of derivative is also called the first principle of derivative. +Thus +( ) +( +) +( ) +0 +' +lim +h +f +x +h +f x +f +x +h +→ ++ +− += +Clearly the domain of definition of f′ (x) is wherever the above limit exists. There +are different notations for derivative of a function. Sometimes f′ (x) is denoted by +( ) +( +) +d +f x +dx + or if y = f(x), it is denoted by dy +dx . This is referred to as derivative of f(x) +or y with respect to x. It is also denoted by D (f (x) ). Further, derivative of f at x = a +is also denoted by +( ) +or +a +a +d +df +f x +dx +dx + or even +x a +df +dx += + + + + + + +. +Example 9 Find the derivative of f(x) = 10x. +Solution Since f′ ( x) = +( +) +( ) +0 +lim +h +f x +h +f x +h +→ ++ +− +Reprint 2025-26 + + LIMITS AND DERIVATIVES 243 += +( +) +( ) +0 +10 +10 +lim +h +x +h +x +h +→ ++ +− += +0 +10 +lim +h +h +h +→ + = +( +) +0 +lim 10 +10 +h→ += +. +Example 10 Find the derivative of f(x) = x2. +Solution We have, f ′(x) = +( +) +( ) +0 +lim +h +f x +h +f x +h +→ ++ +− += +( +) +( ) +2 +2 +0 +lim +h +x +h +x +h +→ ++ +− + = +( +) +0 +lim +2 +2 +h +h +x +x +→ ++ += +Example 11 Find the derivative of the constant function f (x) = a for a fixed real +number a. +Solution We have, f ′(x) = +( +) +( ) +0 +lim +h +f x +h +f x +h +→ ++ +− += +0 +0 +0 +lim +lim +0 +h +h +a +a +h +h +→ +→ +− += += + as +0 +h ≠ +Example 12 Find the derivative of f(x) = 1 +x +Solution We have f ′(x) = +( +) +( ) +0 +lim +h +f x +h +f x +h +→ ++ +− += +0 +1 +1 +– +( +) +lim +h +x +h +x +h +→ ++ += +( +) +( +) +0 +1 +lim +h +x +x +h +h +x x +h +→ + + +− ++ + + ++ + + + + += +( +) +0 +1 +lim +h +h +h x x +h +→ + + +− + + ++ + + + + += +( +) +0 +1 +lim +h +x x +h +→ +− ++ += +2 +1 +x +− +Reprint 2025-26 + +244 +MATHEMATICS +12.5.1 Algebra of derivative of functions Since the very definition of derivatives +involve limits in a rather direct fashion, we expect the rules for derivatives to follow +closely that of limits. We collect these in the following theorem. +Theorem 5 Let f and g be two functions such that their derivatives are defined in a +common domain. Then +(i) +Derivative of sum of two functions is sum of the derivatives of the +functions. +( ) +( ) +( ) +( ) +d +d +d +f x +g x +f x +g x +dx +dx +dx + + ++ += ++ + + +. +(ii) +Derivative of difference of two functions is difference of the derivatives of +the functions. +( ) +( ) +( ) +( ) +d +d +d +f x +g x +f x +g x +dx +dx +dx + + +− += +− + + +. +(iii) +Derivative of product of two functions is given by the following product +rule. +( ) +( ) +. +( ) . ( ) +( ) . +( ) +d +d +d +f x +g x +f x +g x +f x +g x +dx +dx +dx + += ++ + + +(iv) +Derivative of quotient of two functions is given by the following quotient +rule (whenever the denominator is non–zero). +( +) +2 +( ). ( ) +( ) +( ) +( ) +( ) +( ) +d +d +f x +g x +f x +g x +d +f x +dx +dx +dx +g x +g x +− + += + + + + +The proofs of these follow essentially from the analogous theorem for limits. We +will not prove these here. As in the case of limits this theorem tells us how to compute +derivatives of special types of functions. The last two statements in the theorem may +be restated in the following fashion which aids in recalling them easily: +Let +( ) +u +f x += + and v = g (x). Then +( +) +uv ′ = u v +uv +′ +′ ++ +This is referred to a Leibnitz rule for differentiating product of functions or the +product rule. Similarly, the quotient rule is +Reprint 2025-26 + + LIMITS AND DERIVATIVES 245 +u +v +′ + + + + + + += +2 +u v +uv +v +′ +′ +− +Now, let us tackle derivatives of some standard functions. +It is easy to see that the derivative of the function f(x) = x is the constant +function 1. This is because +( ) +f +x +′ += +( +) +( ) +0 +lim +h +f x +h +f x +h +→ ++ +− + = +0 +lim +h +x +h +x +h +→ ++ +− += +0 +lim1 1 +h→ += +. +We use this and the above theorem to compute the derivative of +f(x) = 10x = x + .... + x (ten terms). By (i) of the above theorem + +( ) +df x +dx + = +d +dx ( +) +... +x +x ++ ++ + (ten terms) += +. . . +d +d +x +x +dx +dx ++ ++ + (ten terms) += 1 +... 1 ++ ++ (ten terms) = 10. +We note that this limit may be evaluated using product rule too. Write +f(x) = 10x = uv, where u is the constant function taking value 10 everywhere and +v(x) = x. Here, f(x) = 10x = uv we know that the derivative of u equals 0. Also +derivative of v(x) = x equals 1. Thus by the product rule we have +( ) +f +x +′ + = ( +) +( +) +10 +0. +10.1 10 +x +uv +u v +uv +x +′ +′ +′ +′ += += ++ += ++ += +On similar lines the derivative of f(x) = x2 may be evaluated. We have +f(x) = x2 = x .x and hence +df +dx = +( +) +( ) +( ) +. +. +. +d +d +d +x x +x x +x +x +dx +dx +dx += ++ += 1. +.1 +2 +x +x +x ++ += +. +More generally, we have the following theorem. +Theorem 6 Derivative of f(x) = xn is nxn – 1 for any positive integer n. +Proof By definition of the derivative function, we have + +( ) +( +) +( ) +0 +' +lim +h +f x +h +f +x +f +x +h +→ ++ +− += += +( +) +0 +lim +n +n +h +x +h +x +h +→ ++ +− +. +Reprint 2025-26 + +246 +MATHEMATICS +Binomial theorem tells that (x + h)n = ( +) +( +) +( +) +1 +0 +1 +C +C +... +C +n +n +n +n +n +n +n +x +x +h +h +− ++ ++ ++ +and +hence (x + h)n – xn = h(nxn – 1 +... + hn – 1). Thus +( ) +df x +dx += +( +) +0 +lim +n +n +h +x +h +x +h +→ ++ +− += +( +) +1 +1 +0 +.... +lim +n +n +h +h nx +h +h +− +− +→ ++ ++ += +( +) +1 +1 +0 +lim +... +n +n +h +nx +h +− +− +→ ++ ++ + = +1 +n +nx −. +Alternatively, we may also prove this by induction on n and the product rule as +follows. The result is true for n = 1, which has been proved earlier. We have +( +) +n +d +x +dx += +( +) +1 +. n +d +x x +dx +− += +( ) ( +) +( +) +1 +1 +. +. +n +n +d +d +x +x +x +x +dx +dx +− +− ++ +(by product rule) += +( +) +( +) +1 +2 +1. +. +1 +n +n +x +x +n +x +− +− ++ +− + (by induction hypothesis) += +( +) +1 +1 +1 +1 +n +n +n +x +n +x +nx +− +− +− ++ +− += +. +Remark The above theorem is true for all powers of x, i.e., n can be any real number +(but we will not prove it here). +12.5.2 Derivative of polynomials and trigonometric functions We start with the +following theorem which tells us the derivative of a polynomial function. +Theorem 7 Let f(x) = +1 +1 +1 +0 +.... +n +n +n +n +a x +a +x +a x +a +− +− ++ ++ ++ ++ + be a polynomial function, where +ai s are all real numbers and an ≠ 0. Then, the derivative function is given by +( +) +1 +2 +1 +( ) +1 +... +n +x +n +n +df x +na x +n +a +x +dx +− +− +− += ++ +− ++ ++ +2 +1 +2a x +a ++ +. +Proof of this theorem is just putting together part (i) of Theorem 5 and Theorem 6. +Example 13 Compute the derivative of 6x100 – x55 + x. +Solution A direct application of the above theorem tells that the derivative of the +above function is +99 +54 +600 +55 +1 +x +x +− ++ . +Reprint 2025-26 + + LIMITS AND DERIVATIVES 247 +Example 14 Find the derivative of f(x) = 1 + x + x2 + x3 +... + x50 at x = 1. +Solution A direct application of the above Theorem 6 tells that the derivative of the +above function is 1 + 2x + 3x2 + . . . + 50x49. At x = 1 the value of this function equals +1 + 2(1) + 3(1)2 + .. . + 50(1)49 = 1 + 2 + 3 + . . . + 50 = ( +)( +) +50 +51 +2 + = 1275. +Example 15 Find the derivative of f(x) = +1 +x +x ++ +Solution Clearly this function is defined everywhere except at x = 0. We use the +quotient rule with u = x + 1 and v = x. Hence u′ = 1 and v′ = 1. Therefore + +( ) +1 +df x +d +x +d +u +dx +dx +x +dx +v ++ + + + + += += + + + + + + + + +( ) ( +) +2 +2 +2 +1 +1 1 +1 +x +x +u v +uv +v +x +x +− ++ +′ +′ +− += += += − +Example 16 Compute the derivative of sin x. +Solution Let f(x) = sin x. Then +( ) +df x +dx += +( +) +( ) +( +) +( ) +0 +0 +sin +sin +lim +lim +h +h +f +x +h +f +x +x +h +x +h +h +→ +→ ++ +− ++ +− += += +0 +2 +2cos +sin +2 +2 +lim +h +x +h +h +h +→ ++ + + + + + + + + + + + + (using formula for sin A – sin B) += +0 +0 +sin 2 +limcos +.lim +cos .1 +cos +2 +2 +h +h +h +h +x +x +x +h +→ +→ + + ++ += += + + + + +. +Example 17 Compute the derivative of tan x. +Solution Let f(x) = tan x. Then +( ) +df x +dx += +( +) +( ) +( +) +( ) +0 +0 +tan +tan +lim +lim +h +h +f x +h +f x +x +h +x +h +h +→ +→ ++ +− ++ +− += += +( +) +( +) +0 +sin +1 +sin +lim +cos +cos +h +x +h +x +h +x +h +x +→ + + ++ +− + + ++ + + + + +Reprint 2025-26 + +248 +MATHEMATICS += +( +) +( +) +( +) +0 +sin +cos +cos +sin +lim +cos +cos +h +x +h +x +x +h +x +h +x +h +x +→ + + ++ +− ++ + + ++ + + + + += +( +) +( +) +0 +sin +lim +cos +cos +h +x +h +x +h +x +h +x +→ ++ +− ++ + (using formula for sin (A + B)) += +( +) +0 +0 +sin +1 +lim +.lim cos +cos +h +h +h +h +x +h +x +→ +→ ++ += +2 +2 +1 +1. +sec +cos +x +x += +. +Example 18 Compute the derivative of f(x) = sin2 x. +Solution We use the Leibnitz product rule to evaluate this. +( +) +( ) +sin sin +df x +d +x +x +dx +dx += +( +) +( +) +sin +sin +sin +sin +x +x +x +x +′ +′ += ++ +( +) +( +) +cos +sin +sin +cos +x +x +x +x += ++ +2sin cos +sin 2 +x +x +x += += +. +EXERCISE 12.2 +1. +Find the derivative of x2 – 2 at x = 10. +2. +Find the derivative of x at x = 1. +3. +Find the derivative of 99x at x = l00. +4. +Find the derivative of the following functions from first principle. +(i) +3 +27 +x − + (ii) ( +)( +) +1 +2 +x +x +− +− +(iii) +2 +1 +x +(iv) +1 +1 +x +x ++ +− +5. +For the function +( ) +100 +99 +2 +. . . +1 +100 +99 +2 +x +x +x +f x +x += ++ ++ ++ ++ ++ . +Reprint 2025-26 + + LIMITS AND DERIVATIVES 249 +Prove that +( ) +( ) +1 +100 +0 +f +f +′ +′ += +. +6. +Find the derivative of +1 +2 +2 +1 +. . . +n +n +n +n +n +x +ax +a x +a +x +a +− +− +− ++ ++ ++ ++ ++ + for some fixed real +number a. +7. +For some constants a and b, find the derivative of + (i) +( +) ( +) +x +a +x +b +− +− + (ii) ( +) +2 +2 +ax +b ++ + (iii) x +a +x +b +− +− +8. +Find the derivative of +n +n +x +a +x +a +− +− + for some constant a. +9. +Find the derivative of +(i) +3 +2 +4 +x − +(ii) ( +) ( +) +3 +5 +3 +1 +1 +x +x +x ++ +− +− +(iii) +( +) +3 5 +3 +x +x +− ++ +(iv) +( +) +5 +9 +3 +6 +x +x− +− +(v) +( +) +4 +5 +3 +4 +x +x +− +− +− +(vi) +2 +2 +1 +3 +1 +x +x +x +− ++ +− +10. +Find the derivative of cos x from first principle. +11. +Find the derivative of the following functions: +(i) +sin cos +x +x +(ii) sec x +(iii) 5sec +4cos +x +x ++ +(iv) +cosec x +(v) 3cot +5cosec +x +x ++ +(vi) +5sin +6cos +7 +x +x +− ++ + (vii) 2tan +7sec +x +x +− +Miscellaneous Examples +Example 19 Find the derivative of f from the first principle, where f is given by +(i) +f (x) = 2 +3 +2 +x +x ++ +− +(ii) f (x) = +1 +x +x ++ +Solution (i) Note that function is not defined at x = 2. But, we have +( ) +( +) +( ) +( +) +0 +0 +2 +3 +2 +3 +2 +2 +lim +lim +h +h +x +h +x +f x +h +f x +x +h +x +f +x +h +h +→ +→ ++ ++ ++ +− ++ +− ++ +− +− +′ += += +Reprint 2025-26 + +250 +MATHEMATICS += +( +)( +) ( +)( +) +( +)( +) +0 +2 +2 +3 +2 +2 +3 +2 +lim +2 +2 +h +x +h +x +x +x +h +h x +x +h +→ ++ ++ +− +− ++ ++ +− +− ++ +− += +( +)( +) +( +) ( +)( +) +( +) +( +)( +) +0 +2 +3 +2 +2 +2 +2 +3 +2 +2 +3 +lim +2 +2 +h +x +x +h x +x +x +h +x +h x +x +h +→ ++ +− ++ +− +− ++ +− +− ++ +− ++ +− += +( +) ( +) +( +) +2 +0 +–7 +7 +lim +2 +2 +2 +h +x +x +h +x +→ += − +− ++ +− +− +Again, note that the function f ′ is also not defined at x = 2. +(ii) +The function is not defined at x = 0. But, we have +( ) +f +x +′ += +( +) +( ) +0 +0 +1 +1 +lim +lim +h +h +x +h +x +f x +h +f x +x +h +x +h +h +→ +→ + + + + ++ ++ +− ++ + + + + ++ +− ++ + + + + += += +0 +1 +1 +1 +lim +h +h +h +x +h +x +→ + + ++ +− + + ++ + + += +( +) +( +) +0 +0 +1 +1 +1 +lim +lim +1 +h +h +x +x +h +h +h +h +x x +h +h +x x +h +→ +→ + + + + + + +− +− ++ += +− + + + + + + + + ++ ++ + + + + + + + + + + += +( +) +2 +0 +1 +1 +lim 1 +1 +h +x x +h +x +→ + + +− += − + + ++ + + + + +Again, note that the function f ′ is not defined at x = 0. +Example 20 Find the derivative of f(x) from the first principle, where f(x) is +(i) sin +cos +x +x ++ +(ii) sin +x +x +Solution (i) we have +( ) +'f +x = +( +) +( ) +f +x +h +f x +h ++ +− += +( +) +( +) +0 +sin +cos +sin +cos +lim +h +x +h +x +h +x +x +h +→ ++ ++ ++ +− +− += +0 +sin +cos +cos +sin +cos cos +sin +sin +sin +cos +lim +h +x +h +x +h +x +h +x +h +x +x +h +→ ++ ++ +− +− +− +Reprint 2025-26 + + LIMITS AND DERIVATIVES 251 += +( +) +( +) +( +) +0 +sin +cos +sin +sin +cos +1 +cos +cos +1 +lim +h +h +x +x +x +h +x +h +h +→ +− ++ +− ++ +− += +( +) +( +) +0 +0 +cos +1 +sin +lim +cos +sin +limsin +h +h +h +h +x +x +x +h +h +→ +→ +− +− ++ + +( +) +0 +cos +1 +limcos +h +h +x +h +→ +− ++ += cos +sin +x +x +− +(ii) +( ) +'f +x += +( +) +( ) +( +) +( +) +0 +0 +sin +sin +lim +lim +h +h +f x +h +f x +x +h +x +h +x +x +h +h +→ +→ ++ +− ++ ++ +− += += +( +)( +) +0 +sin +cos +sin +cos +sin +lim +h +x +h +x +h +h +x +x +x +h +→ ++ ++ +− += +( +) +( +) +0 +sin +cos +1 +cos sin +sin cos +sin +cos +lim +h +x +x +h +x +x +h +h +x +h +h +x +h +→ +− ++ ++ ++ + = +( +) +0 +0 +sin +cos +1 +sin +lim +lim +cos +h +h +x +x +h +h +x +x +h +h +→ +→ +− ++ +( +) +0 +lim sin cos +sin cos +h +x +h +h +x +→ ++ ++ + = +cos +sin +x +x +x ++ +Example 21 Compute derivative of +(i) f(x) = sin 2x +(ii) g(x) = cot x +Solution (i) Recall the trigonometric formula sin 2x = 2 sin x cos x. Thus +( ) +df x +dx += +( +) +( +) +2sin cos +2 +sin cos +d +d +x +x +x +x +dx +dx += +( +) +( +) +2 +sin +cos +sin +cos +x +x +x +x + + +′ +′ += ++ + + + + +( +) +( +) +2 +cos +cos +sin +sin +x +x +x +x + + += ++ +− + + +( +) +2 +2 +2 cos +sin +x +x += +− +(ii) By definition, g(x) = +cos +cot +sin +x +x +x += +. We use the quotient rule on this function +wherever it is defined. +cos +(cot ) +sin +dg +d +d +x +x +dx +dx +dx +x + + += += + + + + +Reprint 2025-26 + +252 +MATHEMATICS += +2 +(cos ) (sin ) (cos )(sin ) +(sin ) +x +x +x +x +x +′ +′ +− += +2 +( sin )(sin ) (cos )(cos ) +(sin ) +x +x +x +x +x +− +− += +2 +2 +2 +2 +sin +cos +cosec +sin +x +x +x +x ++ +− +=− +Alternatively, this may be computed by noting that +1 +cot +tan +x +x += +. Here, we use the fact +that the derivative of tan x is sec2 x which we saw in Example 17 and also that the +derivative of the constant function is 0. +dg +dx = +1 +(cot ) +tan +d +d +x +dx +dx +x + + += + + + + += +2 +(1) (tan ) (1)(tan ) +(tan ) +x +x +x +′ +′ +− += +2 +2 +(0)(tan ) (sec ) +(tan ) +x +x +x +− += +2 +2 +2 +sec +cosec +tan +x +x +x +− += − +Example 22 Find the derivative of +(i) +5 +cos +sin +x +x +x +− +(ii) +cos +tan +x +x +x ++ +Solution (i) Let +5 +cos +( ) +sin +x +x +h x +x +− += +. We use the quotient rule on this function wherever +it is defined. +5 +5 +2 +( +cos ) sin +( +cos )(sin ) +( ) +(sin ) +x +x +x +x +x +x +h x +x +′ +′ +− +− +− +′ += +Reprint 2025-26 + + LIMITS AND DERIVATIVES 253 += +4 +5 +2 +(5 +sin )sin +( +cos )cos +sin +x +x +x +x +x +x +x ++ +− +− += +5 +4 +2 +cos +5 +sin +1 +(sin ) +x +x +x +x +x +− ++ ++ + (ii) +We use quotient rule on the function +cos +tan +x +x +x ++ + wherever it is defined. +( ) +h x +′ += +2 +( +cos ) tan +( +cos )(tan ) +(tan ) +x +x +x +x +x +x +x +′ +′ ++ +− ++ += +2 +2 +(1 +sin )tan +( +cos )sec +(tan ) +x +x +x +x +x +x +− +− ++ +Miscellaneous Exercise on Chapter 12 +1. Find the derivative of the following functions from first principle: + (i) +x +− +(ii) +1 +( +)x − +− +(iii) sin (x + 1) +(iv) cos (x – π +8 ) +Find the derivative of the following functions (it is to be understood that a, b, c, d, +p, q, r and s are fixed non-zero constants and m and n are integers): +2. (x + a) +3. (px + q) +r +s +x + + ++ + + + + +4. ( +)( +) +2 +ax +b +cx +d ++ ++ +5. +ax +b +cx +d ++ ++ +6. +1 +1 +1 +1 +x +x ++ +− +7. +2 +1 +ax +bx +c ++ ++ +8. +2 +ax +b +px +qx +r ++ ++ ++ +9. +2 +px +qx +r +ax +b ++ ++ ++ +10. +4 +2 +cos +a +b +x +x +x +− ++ +11. 4 +2 +x − +12. ( +)n +ax +b ++ +13. ( +) ( +) +n +m +ax +b +cx +d ++ ++ +14. sin (x + a) +15. cosec x cot x +16. +cos +1 +sin +x +x ++ +Reprint 2025-26 + +254 +MATHEMATICS +17. sin +cos +sin +cos +x +x +x +x ++ +− +18. sec +1 +sec +1 +x +x +− ++ +19. sinn x +20. +sin +cos +a +b +x +c +d +x ++ ++ +21. sin( +) +cos +x +a +x ++ +22. +4(5sin +3cos ) +x +x +x +− +23. ( +) +2 +1 cos +x +x ++ +24. ( +)( +) +2 +sin +cos +ax +x +p +q +x ++ ++ +25. ( +) ( +) +cos +tan +x +x +x +x ++ +− +26. +4 +5sin +3 +7cos +x +x +x +x ++ ++ +27. +2 cos 4 +sin +x +x +π + + + + + + +28. 1 +tan +x +x ++ +29. ( +) ( +) +sec +tan +x +x +x +x ++ +− +30. +sinn +x +x +Summary +®The expected value of the function as dictated by the points to the left of a +point defines the left hand limit of the function at that point. Similarly the right +hand limit. +®Limit of a function at a point is the common value of the left and right hand +limits, if they coincide. +®For a function f and a real number a, lim +x +a +→ f(x) and f (a) may not be same (In +fact, one may be defined and not the other one). +®For functions f and g the following holds: +[ +] +lim +( ) +( ) +lim +( ) +lim ( ) +x +a +x +a +x +a +f x +g x +f x +g x +→ +→ +→ +± += +± +[ +] +lim +( ). ( ) +lim +( ).lim +( ) +x +a +x +a +x +a +f x g x +f x +g x +→ +→ +→ += +lim +( ) +( ) +lim +( ) +lim ( ) +x +a +x +a +x +a +f x +f x +g x +g x +→ +→ +→ + += + + + + +®Following are some of the standard limits +1 +lim +n +n +n +x +a +x +a +na +x +a +− +→ +− += +− +Reprint 2025-26 + + LIMITS AND DERIVATIVES 255 +0 +sin +lim +1 +x +x +x +→ += +0 +1 +cos +lim +0 +x +x +x +→ +− += +®The derivative of a function f at a is defined by +0 +( +) +( ) +( ) lim +h +f a +h +f a +f +a +h +→ ++ +− +′ += +®Derivative of a function f at any point x is defined by +0 +( ) +( +) +( ) +( ) +lim +h +df x +f x +h +f x +f +x +dx +h +→ ++ +− +′ += += +®For functions u and v the following holds: +( +) +u +v +u +v +′ +′ +′ +± += +± +( +) +uv +u v uv +′ +′ +′ += ++ +2 +u +u v uv +v +v +′ +′ +′ +− + += + + + + + provided all are defined. +®Following are some of the standard derivatives. +1 +( +) +n +n +d +x +nx +dx +− += +(sin ) cos +d +x +x +dx += +(cos ) +sin +d +x +x +dx +=− +Historical Note +In the history of mathematics two names are prominent to share the credit for +inventing calculus, Issac Newton (1642 – 1727) and G.W. Leibnitz (1646 – 1717). +Both of them independently invented calculus around the seventeenth century. +After the advent of calculus many mathematicians contributed for further +development of calculus. The rigorous concept is mainly attributed to the great +Reprint 2025-26 + +256 +MATHEMATICS +mathematicians, A.L. Cauchy, J.L.Lagrange and Karl Weierstrass. Cauchy gave +the foundation of calculus as we have now generally accepted in our textbooks. +Cauchy used D’ Alembert’s limit concept to define the derivative of a function. +Starting with definition of a limit, Cauchy gave examples such as the limit of +sinα +α + for α = 0. He wrote +( +) +( ) , +y +f x i +f x +x +i +∆ ++ +− += +∆ + and called the limit for +0, +i → +the “function derive’e, y′ for f ′ (x)”. +Before 1900, it was thought that calculus is quite difficult to teach. So calculus +became beyond the reach of youngsters. But just in 1900, John Perry and others +in England started propagating the view that essential ideas and methods of calculus +were simple and could be taught even in schools. F.L. Griffin, pioneered the +teaching of calculus to first year students. This was regarded as one of the most +daring act in those days. +Today not only the mathematics but many other subjects such as Physics, +Chemistry, Economics and Biological Sciences are enjoying the fruits of calculus. +— v — +Reprint 2025-26" +class_11,13,Statistics,ncert_books/class_11/kemh1dd/kemh113.pdf,"v“Statistics may be rightly called the science of averages and their +estimates.” – A.L.BOWLEY & A.L. BODDINGTON v +13.1 Introduction +We know that statistics deals with data collected for specific +purposes. We can make decisions about the data by +analysing and interpreting it. In earlier classes, we have +studied methods of representing data graphically and in +tabular form. This representation reveals certain salient +features or characteristics of the data. We have also studied +the methods of finding a representative value for the given +data. This value is called the measure of central tendency. +Recall mean (arithmetic mean), median and mode are three +measures of central tendency. A measure of central +tendency gives us a rough idea where data points are +centred. But, in order to make better interpretation from the +data, we should also have an idea how the data are scattered or how much they are +bunched around a measure of central tendency. +Consider now the runs scored by two batsmen in their last ten matches as follows: +Batsman A : 30, 91, 0, 64, 42, 80, 30, 5, 117, 71 +Batsman B : 53, 46, 48, 50, 53, 53, 58, 60, 57, 52 +Clearly, the mean and median of the data are +Batsman A +Batsman B +Mean +53 +53 +Median +53 +53 +Recall that, we calculate the mean of a data (denoted by x ) by dividing the sum +of the observations by the number of observations, i.e., +13 +Chapter +STATISTICS +Karl Pearson + (1857-1936) +Reprint 2025-26 + +258 +MATHEMATICS +1 +1 +n +i +i +x +x +n += += ∑ +Also, the median is obtained by first arranging the data in ascending or descending +order and applying the following rule. +If the number of observations is odd, then the median is +th +1 +2 +n + + + + + + + observation. +If the number of observations is even, then median is the mean of +th +2 +n + + + + + + and +th +1 +2 +n + + ++ + + + + observations. +We find that the mean and median of the runs scored by both the batsmen A and +B are same i.e., 53. Can we say that the performance of two players is same? Clearly +No, because the variability in the scores of batsman A is from 0 (minimum) to 117 +(maximum). Whereas, the range of the runs scored by batsman B is from 46 to 60. +Let us now plot the above scores as dots on a number line. We find the following +diagrams: +For batsman A +For batsman B +We can see that the dots corresponding to batsman B are close to each other and +are clustering around the measure of central tendency (mean and median), while those +corresponding to batsman A are scattered or more spread out. +Thus, the measures of central tendency are not sufficient to give complete +information about a given data. Variability is another factor which is required to be +studied under statistics. Like ‘measures of central tendency’ we want to have a +single number to describe variability. This single number is called a ‘measure of +dispersion’. In this Chapter, we shall learn some of the important measures of dispersion +and their methods of calculation for ungrouped and grouped data. +Fig 13.1 +Fig 13.2 +Reprint 2025-26 + + STATISTICS 259 +13.2 Measures of Dispersion +The dispersion or scatter in a data is measured on the basis of the observations and the +types of the measure of central tendency, used there. There are following measures of +dispersion: +(i) Range, (ii) Quartile deviation, (iii) Mean deviation, (iv) Standard deviation. +In this Chapter, we shall study all of these measures of dispersion except the +quartile deviation. +13.3 +Range +Recall that, in the example of runs scored by two batsmen A and B, we had some idea +of variability in the scores on the basis of minimum and maximum runs in each series. +To obtain a single number for this, we find the difference of maximum and minimum +values of each series. This difference is called the ‘Range’ of the data. +In case of batsman A, Range = 117 – 0 = 117 and for batsman B, Range = 60 – 46 = 14. +Clearly, Range of A > Range of B. Therefore, the scores are scattered or dispersed in +case of A while for B these are close to each other. +Thus, Range of a series = Maximum value – Minimum value. +The range of data gives us a rough idea of variability or scatter but does not tell +about the dispersion of the data from a measure of central tendency. For this purpose, +we need some other measure of variability. Clearly, such measure must depend upon +the difference (or deviation) of the values from the central tendency. +The important measures of dispersion, which depend upon the deviations of the +observations from a central tendency are mean deviation and standard deviation. Let +us discuss them in detail. +13.4 +Mean Deviation +Recall that the deviation of an observation x from a fixed value ‘a’ is the difference +x – a. In order to find the dispersion of values of x from a central value ‘a’ , we find the +deviations about a. An absolute measure of dispersion is the mean of these deviations. +To find the mean, we must obtain the sum of the deviations. But, we know that a +measure of central tendency lies between the maximum and the minimum values of +the set of observations. Therefore, some of the deviations will be negative and some +positive. Thus, the sum of deviations may vanish. Moreover, the sum of the deviations +from mean ( x ) is zero. +Also +Mean of deviations +Sum of deviations +0 +0 +Number of observations +n += += += +Thus, finding the mean of deviations about mean is not of any use for us, as far +as the measure of dispersion is concerned. +Reprint 2025-26 + +260 +MATHEMATICS +Remember that, in finding a suitable measure of dispersion, we require the distance +of each value from a central tendency or a fixed number ‘a’. Recall, that the absolute +value of the difference of two numbers gives the distance between the numbers when +represented on a number line. Thus, to find the measure of dispersion from a fixed +number ‘a’ we may take the mean of the absolute values of the deviations from the +central value. This mean is called the ‘mean deviation’. Thus mean deviation about a +central value ‘a’ is the mean of the absolute values of the deviations of the observations +from ‘a’. The mean deviation from ‘a’ is denoted as M.D. (a). Therefore, +M.D.(a) = Sum of absolute values of deviations from ' ' +Number of observations +a . +Remark Mean deviation may be obtained from any measure of central tendency. +However, mean deviation from mean and median are commonly used in statistical +studies. +Let us now learn how to calculate mean deviation about mean and mean deviation +about median for various types of data +13.4.1 Mean deviation for ungrouped data Let n observations be x1, x2, x3, ...., xn. +The following steps are involved in the calculation of mean deviation about mean or +median: +Step 1 Calculate the measure of central tendency about which we are to find the mean +deviation. Let it be ‘a’. +Step 2 Find the deviation of each xi from a, i.e., x1 – a, x2 – a, x3 – a,. . . , xn– a +Step 3 Find the absolute values of the deviations, i.e., drop the minus sign (–), if it is +there, i.e., +a +x +a +x +a +x +a +x +n − +− +− +− +...., +, +, +, +3 +2 +1 +Step 4 Find the mean of the absolute values of the deviations. This mean is the mean +deviation about a, i.e., +1 +( ) +M.D. +n +i +i +x +a +a +n += +− += ∑ +Thus +M.D. ( x ) = +1 +1 +n +i +i +x +x +n += +− +∑ +, where x = Mean +and +M.D. (M) = +1 +1 +M +n +i +i +x +n += +− +∑ +, where M = Median +Reprint 2025-26 + + STATISTICS 261 +ANote In this Chapter, we shall use the symbol M to denote median unless stated +otherwise.Let us now illustrate the steps of the above method in following examples. +Example 1 Find the mean deviation about the mean for the following data: +6, 7, 10, 12, 13, 4, 8, 12 +Solution We proceed step-wise and get the following: +Step 1 Mean of the given data is +6 +7 +10 +12 +13 +4 +8 12 +72 +9 +8 +8 +x ++ ++ ++ ++ ++ ++ ++ += += += +Step 2 The deviations of the respective observations from the mean ,x i.e., xi– x are +6 – 9, 7 – 9, 10 – 9, 12 – 9, 13 – 9, 4 – 9, 8 – 9, 12 – 9, +or + –3, –2, 1, 3, 4, –5, –1, 3 +Step 3 The absolute values of the deviations, i.e., +ix +x +− + are + 3, 2, 1, 3, 4, 5, 1, 3 +Step 4 The required mean deviation about the mean is +M.D. ( ) +x = +8 +1 +8 +i +i +x +x += +− +∑ += 3 +2 +1 3 +4 +5 1 3 +22 +2 75 +8 +8 +. ++ ++ + ++ ++ ++ + += += +ANote Instead of carrying out the steps every time, we can carry on calculation, +step-wise without referring to steps. +Example 2 Find the mean deviation about the mean for the following data : +12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5 +Solution We have to first find the mean ( x ) of the given data +20 +1 +1 +20 +i +i +x +x += += +∑ + = 20 +200 + = 10 +Reprint 2025-26 + +262 +MATHEMATICS +The respective absolute values of the deviations from mean, i.e., +x +xi − + are +2, 7, 8, 7, 6, 1, 7, 9, 10, 5, 2, 7, 8, 7, 6, 1, 7, 9, 10, 5 +Therefore +20 +1 +124 +i +i +x +x += +− += +∑ +and +M.D. ( x ) = 124 +20 = 6.2 +Example 3 Find the mean deviation about the median for the following data: +3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21. +Solution Here the number of observations is 11 which is odd. Arranging the data into +ascending order, we have 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21 +Now +Median = +th +11 +1 +2 ++ + + + + + + + or 6th observation = 9 +The absolute values of the respective deviations from the median, i.e., +M +ix − + are +6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12 +Therefore +11 +1 +M +58 +i +i +x += +− += +∑ +and +( +) +11 +1 +1 +1 +M.D. M +M +58 +5.27 +11 +11 +i +i +x += += +− += +× += +∑ +13.4.2 Mean deviation for grouped data We know that data can be grouped into +two ways : +(a) Discrete frequency distribution, +(b) Continuous frequency distribution. +Let us discuss the method of finding mean deviation for both types of the data. +(a) Discrete frequency distribution Let the given data consist of n distinct values +x1, x2, ..., xn occurring with frequencies f1, f2 , ..., fn respectively. This data can be +represented in the tabular form as given below, and is called discrete frequency +distribution: +x : x1 + x2 +x3 ... xn +f : f1 + f2 +f3 ... fn +Reprint 2025-26 + + STATISTICS 263 +(i) Mean deviation about mean +First of all we find the mean x of the given data by using the formula +1 +1 +1 +1 +N +n +i +i +n +i +i +i +n +i +i +i +x f +x +x f +f += += += += += +∑ +∑ +∑ +, +where ∑ += +n +i +i +i f +x +1 + denotes the sum of the products of observations xi with their respective +frequencies fi and +∑ += += +n +i +if +1 +N + is the sum of the frequencies. +Then, we find the deviations of observations xi from the mean x and take their +absolute values, i.e., +x +xi − +for all i =1, 2,..., n. +After this, find the mean of the absolute values of the deviations, which is the +required mean deviation about the mean. Thus +1 +1 +M.D. ( ) +n +i +i +i +n +i +i +f x +x +x +f += += +− += ∑ +∑ + = +x +x +f +i +n +i +i +− +∑ +=1 +N +1 +(ii) Mean deviation about median To find mean deviation about median, we find the +median of the given discrete frequency distribution. For this the observations are arranged +in ascending order. After this the cumulative frequencies are obtained. Then, we identify +the observation whose cumulative frequency is equal to or just greater than +N +2 , where +N is the sum of frequencies. This value of the observation lies in the middle of the data, +therefore, it is the required median. After finding median, we obtain the mean of the +absolute values of the deviations from median.Thus, +1 +1 +M.D.(M) +M +N +n +i +i +i +f x += += +− +∑ +Example 4 Find mean deviation about the mean for the following data : +xi +2 +5 +6 +8 +10 +12 +f i +2 +8 10 +7 + 8 + 5 +Reprint 2025-26 + +264 +MATHEMATICS +Solution Let us make a Table 13.1 of the given data and append other columns after +calculations. +Table 13.1 +xi +f i +fixi +x +xi − +fi +x +xi − +2 +2 +4 +5.5 +11 +5 +8 +40 +2.5 +20 +6 +10 +60 +1.5 +15 +8 +7 +56 +0.5 +3.5 +10 +8 +80 +2.5 +20 +12 +5 +60 +4.5 +22.5 +40 +300 + 92 +40 +N +6 +1 += += ∑ += +i +if +, +300 +6 +1 += +∑ += +i +i +ix +f +, +92 +6 +1 += +− +∑ += +x +x +f +i +i +i +Therefore +6 +1 +1 +1 +300 +7.5 +N +40 +i +i +i +x +f x += += += +× += +∑ +and +6 +1 +1 +1 +M. D. ( ) +92 +2.3 +N +40 +i +i +i +x +f +x +x += += +− += +× += +∑ +Example 5 Find the mean deviation about the median for the following data: +xi +3 +6 +9 +12 +13 +15 +21 +22 +f i +3 +4 +5 +2 +4 +5 +4 +3 +Solution The given observations are already in ascending order. Adding a row +corresponding to cumulative frequencies to the given data, we get (Table 13.2). +Table 13.2 +xi +3 +6 +9 +12 +13 +15 +21 +22 +f i +3 +4 +5 +2 +4 +5 +4 +3 +c.f. +3 +7 +12 +14 +18 +23 +27 +30 +Now, N=30 which is even. +Reprint 2025-26 + + STATISTICS 265 +Median is the mean of the 15th and 16th observations. Both of these observations +lie in the cumulative frequency 18, for which the corresponding observation is 13. +th +th +15 +observation +16 +observation +13 13 +Therefore, Median M +13 +2 +2 ++ ++ += += += +Now, absolute values of the deviations from median, i.e., +M +ix − + are shown in +Table 13.3. +Table 13.3 +M +ix − +10 +7 +4 +1 +0 +2 +8 +9 +f i +3 +4 +5 +2 +4 +5 +4 +3 +fi +M +ix − +30 +28 +20 +2 +0 +10 +32 +27 +We have +8 +8 +1 +1 +30 and +M +149 +i +i +i +i +i +f +f x += += += +− += +∑ +∑ +Therefore +8 +1 +1 +M. D. (M) +M +N +i +i +i +f x += += +− +∑ + = 1 +149 +30 × + = 4.97. +(b) Continuous frequency distribution A continuous frequency distribution is a series +in which the data are classified into different class-intervals without gaps alongwith +their respective frequencies. +For example, marks obtained by 100 students are presented in a continuous +frequency distribution as follows : +Marks obtained +0-10 +10-20 20-30 30-40 +40-50 +50-60 +Number of Students +12 +18 +27 +20 +17 +6 +(i) Mean deviation about mean While calculating the mean of a continuous frequency +distribution, we had made the assumption that the frequency in each class is centred at +its mid-point. Here also, we write the mid-point of each given class and proceed further +as for a discrete frequency distribution to find the mean deviation. +Let us take the following example. +Reprint 2025-26 + +266 +MATHEMATICS +Example 6 Find the mean deviation about the mean for the following data. +Marks obtained +10-20 20-30 30-40 40-50 50-60 60-70 70-80 +Number of students +2 +3 +8 +14 +8 +3 +2 +Solution We make the following Table 13.4 from the given data : +Table 13.4 +Marks +Number of +Mid-points +fixi +x +xi − + fi +x +xi − +obtained +students +f i +xi +10-20 +2 +15 +30 +30 +60 +20-30 +3 +25 +75 +20 +60 +30-40 +8 +35 +280 +10 +80 +40-50 +14 +45 +630 +0 +0 +50-60 +8 +55 +440 +10 +80 +60-70 +3 +65 +195 +20 +60 +70-80 +2 +75 +150 +30 +60 +40 +1800 +400 +Here +7 +7 +7 +1 +1 +1 +N +40, +1800, +400 +i +i +i +i +i +i +i +i +f +f x +f x +x += += += += += += +− += +∑ +∑ +∑ +Therefore +7 +1 +1 +1800 +45 +N +40 +i +i +i +x +f x += += += += +∑ +and +( ) +7 +1 +1 +1 +M.D. +400 +10 +N +40 +i +i +i +x +f x +x += += +− += +× += +∑ +Shortcut method for calculating mean deviation about mean We can avoid the +tedious calculations of computing x by following step-deviation method. Recall that in +this method, we take an assumed mean which is in the middle or just close to it in the +data. Then deviations of the observations (or mid-points of classes) are taken from the +Reprint 2025-26 + + STATISTICS 267 +assumed mean. This is nothing but the shifting of origin from zero to the assumed mean +on the number line, as shown in Fig 13.3 +If there is a common factor of all the deviations, we divide them by this common +factor to further simplify the deviations. These are known as step-deviations. The +process of taking step-deviations is the change of scale on the number line as shown in +Fig 13.4 +The deviations and step-deviations reduce the size of the observations, so that the +computations viz. multiplication, etc., become simpler. Let, the new variable be denoted +by +h +a +x +d +i +i +− += +, where ‘a’ is the assumed mean and h is the common factor. Then, the +mean x by step-deviation method is given by + +1 +N +n +f d +i +i +i +x +a +h +∑ += += ++ +× +Let us take the data of Example 6 and find the mean deviation by using step- +deviation method. +Fig 13.3 +Fig 13.4 +Reprint 2025-26 + +268 +MATHEMATICS +Number of +students +Marks +obtained +Take the assumed mean a = 45 and h = 10, and form the following Table 13.5. +Table 13.5 +Mid-points +45 +10 +i +i +x +d +− += +i +i +f d +x +xi − +fi +x +xi − +f i +xi +10-20 +2 +15 +– 3 +– 6 +30 +60 +20-30 +3 +25 +– 2 +– 6 +20 +60 +30-40 +8 +35 +– 1 +– 8 +10 +80 +40-50 +14 +45 +0 +0 +0 +0 +50-60 +8 +55 +1 +8 +10 +80 +60-70 +3 +65 +2 +6 +20 +60 +70-80 +2 +75 +3 +6 +30 +60 +40 +0 +400 +Therefore +7 + +1 + +N + +f d +i +i +i +x +a +h +∑ += += ++ +× + = +0 +45 +10 +45 +40 ++ +× += +and + +7 +1 +1 +400 +M D +( ) +10 +N +40 +i +i +i +x +f x +x += += +− += += +∑ +. . +ANote The step deviation method is applied to compute x . Rest of the procedure +is same. +(ii) Mean deviation about median The process of finding the mean deviation about +median for a continuous frequency distribution is similar as we did for mean deviation +about the mean. The only difference lies in the replacement of the mean by median +while taking deviations. +Let us recall the process of finding median for a continuous frequency distribution. +The data is first arranged in ascending order. Then, the median of continuous +frequency distribution is obtained by first identifying the class in which median lies +(median class) and then applying the formula +Reprint 2025-26 + + STATISTICS 269 +frequency +N +C +2 +Median +l +h +f +− += + +× +where median class is the class interval whose cumulative frequency is just greater +than or equal to N +2 , N is the sum of frequencies, l, f, h and C are, respectively the lower +limit , the frequency, the width of the median class and C the cumulative frequency of +the class just preceding the median class. After finding the median, the absolute values +of the deviations of mid-point xi of each class from the median i.e., +M +ix − + are obtained. +Then +1 +M.D. (M) +M +1 +N +n +f x +i +i +i += +− +∑ += +The process is illustrated in the following example: +Example 7 Calculate the mean deviation about median for the following data : +Class +0-10 +10-20 +20-30 +30-40 +40-50 +50-60 +Frequency +6 +7 +15 +16 +4 +2 +Solution Form the following Table 13.6 from the given data : +Table 13.6 +Class +Frequency +Cumulative +Mid-points +Med. +xi − +fi +Med. +xi − +f i +(c.f.) +xi +0-10 +6 +6 +5 +23 +138 +10-20 +7 +13 +15 +13 +91 +20-30 +15 +28 +25 +3 +45 +30-40 +16 +44 +35 +7 +112 +40-50 +4 +48 +45 +17 +68 +50-60 +2 +50 +55 +27 +54 +50 +508 +Reprint 2025-26 + +270 +MATHEMATICS +The class interval containing +th +N +2 +or 25th item is 20-30. Therefore, 20–30 is the median +class. We know that +Median = +N +C +2 +l +h +f +− ++ +× +Here l = 20, C = 13, f = 15, h = 10 and N = 50 +Therefore, +Median +25 +13 +20 +10 +15 +− += ++ +× + = 20 + 8 = 28 +Thus, Mean deviation about median is given by +M.D. (M) = +6 +1 +1 +M +N +i +i +i +f x += +− +∑ + = 1 +508 +50 × + = 10.16 +EXERCISE 13.1 +Find the mean deviation about the mean for the data in Exercises 1 and 2. +1. +4, 7, 8, 9, 10, 12, 13, 17 +2. +38, 70, 48, 40, 42, 55, 63, 46, 54, 44 +Find the mean deviation about the median for the data in Exercises 3 and 4. +3. +13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 +4. +36, 72, 46, 42, 60, 45, 53, 46, 51, 49 +Find the mean deviation about the mean for the data in Exercises 5 and 6. +5. +xi 5 +10 +15 +20 +25 +f i 7 +4 +6 +3 +5 +6. +xi 10 +30 +50 +70 +90 +f i 4 +24 +28 +16 +8 +Find the mean deviation about the median for the data in Exercises 7 and 8. +7. +xi 5 +7 +9 +10 +12 +15 +f i 8 +6 +2 +2 +2 +6 +8. +xi 15 +21 +27 +30 +35 +f i 3 +5 +6 +7 +8 +Reprint 2025-26 + + STATISTICS 271 +Find the mean deviation about the mean for the data in Exercises 9 and 10. +9. +Income per 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 + day in ` + Number +4 +8 +9 +10 +7 +5 +4 +3 +of persons +10. + Height +95-105 +105-115 +115-125 +125-135 +135-145 +145-155 + in cms +Number of +9 +13 +26 +30 +12 +10 + boys +11. +Find the mean deviation about median for the following data : + Marks +0-10 +10-20 +20-30 +30-40 +40-50 +50-60 +Number of +6 +8 +14 +16 +4 +2 + Girls +12. +Calculate the mean deviation about median age for the age distribution of 100 +persons given below: + Age +16-20 +21-25 +26-30 +31-35 +36-40 +41-45 +46-50 +51-55 +(in years) + Number +5 +6 +12 +14 +26 +12 +16 +9 +[Hint Convert the given data into continuous frequency distribution by subtracting 0.5 +from the lower limit and adding 0.5 to the upper limit of each class interval] +13.4.3 Limitations of mean deviation In a series, where the degree of variability is +very high, the median is not a representative central tendency. Thus, the mean deviation +about median calculated for such series can not be fully relied. +The sum of the deviations from the mean (minus signs ignored) is more than the sum +of the deviations from median. Therefore, the mean deviation about the mean is not very +scientific.Thus, in many cases, mean deviation may give unsatisfactory results. Also mean +deviation is calculated on the basis of absolute values of the deviations and therefore, +cannot be subjected to further algebraic treatment. This implies that we must have some +other measure of dispersion. Standard deviation is such a measure of dispersion. +13.5 +Variance and Standard Deviation +Recall that while calculating mean deviation about mean or median, the absolute values +of the deviations were taken. The absolute values were taken to give meaning to the +mean deviation, otherwise the deviations may cancel among themselves. +Another way to overcome this difficulty which arose due to the signs of deviations, +is to take squares of all the deviations. Obviously all these squares of deviations are +Reprint 2025-26 + +272 +MATHEMATICS +non-negative. Let x1, x2, x3, ..., xn be n observations and x be their mean. Then +2 +2 +2 +2 +2 +1 +1 +( +) +( +) +....... +( +) +( +) +n +n +i +i +x +x +x +x +x +x +x +x += +− ++ +− ++ ++ +− += +− + +. +If this sum is zero, then each +) +( +x +xi − +has to be zero. This implies that there is no +dispersion at all as all observations are equal to the mean x . +If ∑ += +− +n +i +i +x +x +1 +2) +( +is small , this indicates that the observations x1, x2, x3,...,xn are +close to the mean x and therefore, there is a lower degree of dispersion. On the +contrary, if this sum is large, there is a higher degree of dispersion of the observations +from the mean x . Can we thus say that the sum ∑ += +− +n +i +i +x +x +1 +2) +( + is a reasonable indicator +of the degree of dispersion or scatter? +Let us take the set A of six observations 5, 15, 25, 35, 45, 55. The mean of the +observations is x = 30. The sum of squares of deviations from x for this set is +∑ += +− +6 +1 +2) +( +i +i +x +x += (5–30)2 + (15–30)2 + (25–30)2 + (35–30)2 + (45–30)2 +(55–30)2 + = 625 + 225 + 25 + 25 + 225 + 625 = 1750 +Let us now take another set B of 31 observations 15, 16, 17, 18, 19, 20, 21, 22, 23, +24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45. The +mean of these observations is y = 30 +Note that both the sets A and B of observations have a mean of 30. +Now, the sum of squares of deviations of observations for set B from the mean y is +given by +∑ += +− +31 +1 +2) +( +i +i +y +y += (15–30)2 +(16–30)2 + (17–30)2 + ...+ (44–30)2 +(45–30)2 += (–15)2 +(–14)2 + ...+ (–1)2 + 02 + 12 + 22 + 32 + ...+ 142 + 152 += 2 [152 + 142 + ... + 12] += +15 (15 +1) (30 +1) +2 +6 +× ++ ++ +× += 5 × 16 × 31 = 2480 +(Because sum of squares of first n natural numbers = +( +1) (2 +1) +6 +n n +n ++ ++ +. Here n = 15) +Reprint 2025-26 + + STATISTICS 273 +If ∑ += +− +n +i +i +x +x +1 +2) +( + is simply our measure of dispersion or scatter about mean, we +will tend to say that the set A of six observations has a lesser dispersion about the mean +than the set B of 31 observations, even though the observations in set A are more +scattered from the mean (the range of deviations being from –25 to 25) than in the set +B (where the range of deviations is from –15 to 15). +This is also clear from the following diagrams. +For the set A, we have +For the set B, we have +Thus, we can say that the sum of squares of deviations from the mean is not a proper +measure of dispersion. To overcome this difficulty we take the mean of the squares of +the deviations, i.e., we take ∑ += +− +n +i +i +x +x +n +1 +2) +( +1 +. In case of the set A, we have +1 +Mean +6 += +× 1750 = 291.67 and in case of the set B, it is 1 +31× 2480 = 80. +This indicates that the scatter or dispersion is more in set A than the scatter or dispersion +in set B, which confirms with the geometrical representation of the two sets. +Thus, we can take ∑ +− +2) +( +1 +x +x +n +i +as a quantity which leads to a proper measure +of dispersion. This number, i.e., mean of the squares of the deviations from mean is +called the variance and is denoted by +2 +σ +(read as sigma square). Therefore, the +variance of n observations x1, x2,..., xn is given by +Fig 13.5 +Fig 13.6 +Reprint 2025-26 + +274 +MATHEMATICS +Deviations from mean +(xi– x ) +∑ += +− += +n +i +i +x +x +n +1 +2 +2 +) +( +1 +σ +13.5.1 Standard Deviation In the calculation of variance, we find that the units of +individual observations xi and the unit of their mean x are different from that of variance, +since variance involves the sum of squares of (xi– x ). For this reason, the proper +measure of dispersion about the mean of a set of observations is expressed as positive +square-root of the variance and is called standard deviation. Therefore, the standard +deviation, usually denoted by σ , is given by +∑ += +− += +n +i +i +x +x +n +1 +2) +( +1 +σ +... (1) +Let us take the following example to illustrate the calculation of variance and +hence, standard deviation of ungrouped data. +Example 8 Find the variance of the following data: +6, 8, 10, 12, 14, 16, 18, 20, 22, 24 +Solution From the given data we can form the following Table 13.7. The mean is +calculated by step-deviation method taking 14 as assumed mean. The number of +observations is n = 10 +Table 13.7 +xi +14 +2 +i +i +x +d +− += +(xi– x ) +6 +–4 +–9 +81 +8 +–3 +–7 +49 +10 +–2 +–5 +25 +12 +–1 +–3 +9 +14 +0 +–1 +1 +16 +1 +1 +1 +18 +2 +3 +9 +20 +3 +5 +25 +22 +4 +7 +49 +24 +5 +9 +81 +5 +330 +Reprint 2025-26 + + STATISTICS 275 +Therefore +Mean x = assumed mean + +h +n +d +n +i +i +× +∑ +=1 + = +5 +14 +2 +15 +10 ++ +× += +and +Variance ( +2 +σ ) = +10 +2 +1 +1 +) +i +i +( x +x +n += +− +∑ += 1 +330 +10 × + = 33 +Thus Standard deviation (σ ) = +33 +5 74 +. += +13.5.2 Standard deviation of a discrete frequency distribution Let the given discrete +frequency distribution be +x : +x1, x2, x3 ,. . . , xn +f : +f1, f2, f3 ,. . . , fn +In this case standard deviation ( ) +2 +1 +1 +( +) +N +n +i +i +i +f x +x +σ += += +− +∑ +... (2) +where +1 +N +n +i +i +f += +=∑ +. +Let us take up following example. +Example 9 Find the variance and standard deviation for the following data: +xi +4 +8 +11 +17 +20 +24 +32 +f i +3 +5 + 9 +5 +4 +3 +1 +Solution Presenting the data in tabular form (Table 13.8), we get +Table 13.8 +xi +f i +fi xi +xi – x +2) +( +x +xi − +fi +2) +( +x +xi − +4 +3 +12 +–10 +100 +300 +8 +5 +40 +–6 +36 +180 +11 +9 +99 +–3 +9 +81 +17 +5 +85 +3 +9 +45 +20 +4 +80 +6 +36 +144 +24 +3 +72 +10 +100 +300 +32 +1 +32 +18 +324 +324 +30 +420 +1374 +Reprint 2025-26 + +276 +MATHEMATICS +N = 30, +( +) +7 +7 +2 +1 +1 +420, +1374 +i +i +i +i +i +i +f x +f +x +x += += += +− += +∑ +∑ +Therefore +7 +1 +1 +420 +14 +N +30 +i +i +i +f x +x += += += +× += +∑ +Hence +variance +2 +( +) +σ + = +7 +2 +1 +1 +( +) +N +i +i +i +f x +x += +− +∑ += 1 +30 × 1374 = 45.8 +and + Standard deviation +8. +45 +) +( += +σ + = 6.77 +13.5.3 Standard deviation of a continuous frequency distribution The given +continuous frequency distribution can be represented as a discrete frequency distribution +by replacing each class by its mid-point. Then, the standard deviation is calculated by +the technique adopted in the case of a discrete frequency distribution. +If there is a frequency distribution of n classes each class defined by its mid-point +xi with frequency fi, the standard deviation will be obtained by the formula +2 +1 +1 +( +) +N +n +i +i +i +f x +x +σ += += +− +∑ +, +where x is the mean of the distribution and +1 +N +n +i +i +f += +=∑ +. +Another formula for standard deviation We know that +Variance +2 +( +) +σ + = +2 +1 +1 +( +) +N +n +i +i +i +f x +x += +− +∑ += +2 +2 +1 +1 +( +2 +) +N +n +i +i +i +i +f x +x +x x += ++ +− +∑ += +2 +2 +1 +1 +1 +1 +2 +N +n +n +n +i +i +i +i +i +i +i +i +f x +x f +x f x += += += + + ++ +− + + + + +∑ +∑ +∑ + = +2 +2 +1 +1 +1 +1 +2 +N +n +n +n +i +i +i +i +i +i +i +i +f x +x +f +x +x f += += += + + ++ +− + + + + +∑ +∑ +∑ +Reprint 2025-26 + + STATISTICS 277 += +2 +2 +1 +1 +N +2 . N +N += + + ++ +− + + + + + +n +i +i +i +f x +x +x +x +1 +1 +1 +Here +or +N +N +n +n +i +i +i +i +i +i +x f +x +x f +x += += + + += += + + + + +∑ +∑ += +2 +2 +2 +1 +2 +1 +N +n +i +i +i +x +x +f x += ++ +− +∑ + +2 +2 +1 +1 +N +n +i +i +i +x +f x += += +− +∑ +or +2 +σ += +2 +2 +2 +2 +=1 +2 +1 +1 +=1 +1 +1 +N +N +N +N +n +i +i +n +n +n +i +i +i +i +i +i +i +i +i +i +f x +f x +f x +f x +− += + + + + + + + + + + + + +− += +− + + + + + + + + + + + + + +∑ +∑ +∑ +∑ +Thus, standard deviation ( ) +2 +2 +1 +=1 +1 +N +N +n +n +i +i +i +i +i +i +f x +f x +σ += += +− + + + + + +∑ +∑ + ... (3) +Example 10 Calculate the mean, variance and standard deviation for the following +distribution : +Class +30-40 +40-50 +50-60 +60-70 +70-80 +80-90 +90-100 +Frequency +3 +7 +12 +15 +8 +3 +2 +Solution From the given data, we construct the following Table 13.9. +Table 13.9 +Class +Frequency +Mid-point +fixi +(xi– x )2 +fi(xi– x )2 +(fi) +(xi) +30-40 +3 +35 +105 +729 +2187 +40-50 +7 +45 +315 +289 +2023 +50-60 +12 +55 +660 +49 +588 +60-70 +15 +65 +975 +9 +135 +70-80 +8 +75 +600 +169 +1352 +80-90 +3 +85 +255 +529 +1587 +90-100 +2 +95 +190 +1089 +2178 +50 +3100 +10050 +Reprint 2025-26 + +278 +MATHEMATICS +Thus +7 +1 +1 +3100 +Mean +62 +N +50 +i +i +i +x +f x += += += += +∑ +Variance ( +) +2 +σ + = +7 +2 +1 +1 +( +) +N +i +i +i +f x +x += +− +∑ += 1 +10050 +50 × + = 201 +and +Standard deviation ( ) +201 +14 18 +. +σ += += +Example 11 Find the standard deviation for the following data : +xi +3 +8 +13 +18 +23 +f i +7 +10 +15 +10 +6 +Solution Let us form the following Table 13.10: +Table 13.10 +xi +f i +fixi +xi +2 +fixi +2 +3 +7 +21 +9 +63 +8 +10 +80 +64 +640 +13 +15 +195 +169 +2535 +18 +10 +180 +324 +3240 +23 +6 +138 +529 +3174 +48 +614 +9652 +Now, by formula (3), we have +σ = +( +) +2 +2 +1 +N +N +i +i +i +i +f x +f x +− +∑ +∑ += +2 +1 +48 +9652 +(614) +48 +× +− += 1 +463296 +376996 +48 +− +Reprint 2025-26 + + STATISTICS 279 += 1 +293 77 +48 +. +× + = 6.12 +Therefore, +Standard deviation (σ ) = 6.12 +13.5.4. Shortcut method to find variance and standard deviation Sometimes the +values of xi in a discrete distribution or the mid points xi of different classes in a +continuous distribution are large and so the calculation of mean and variance becomes +tedious and time consuming. By using step-deviation method, it is possible to simplify +the procedure. +Let the assumed mean be ‘A’ and the scale be reduced to h +1 + times (h being the +width of class-intervals). Let the step-deviations or the new values be yi. +i.e. +A +i +i +x +y +h +− += + or xi = A + hyi +... (1) +We know that +1 +N +n +i +i +i +f x +x += += +∑ +... (2) +Replacing xi from (1) in (2), we get +x = +1 +A +) +N +n +i +i +i +f ( +hy += ++ +∑ += +1 +1 +1 +A +N +n +n +i +i +i +i +i +f +h f y += += ++ + + + + + + +∑ +∑ += +1 +1 +1 +A +N +i +n +n +i +i +i +i +f +h +f y += += + + ++ + + + + +∑ +∑ += +1 +N +A N +N +n +i +i +i +f y +. +h += ++ ∑ + +1 +because +N +n +i +i +f += + + += + + + + +∑ +Thus +x = A + h y +... (3) +Now +Variance of the variable x, +2 +2 +1 +1 +) +N +n +x +i +i +i +f ( x +x +σ += += +− +∑ += +2 +1 +1 +(A +A +) +N +n +i +i +i +f +hy +h y += ++ +− +− +∑ +(Using (1) and (3)) +Reprint 2025-26 + +280 +MATHEMATICS += +2 +2 +1 +1 +( +) +N +n +i +i +i +f h +y +y += +− +∑ += +2 +2 +1 +( +) +N +n +i +i +i +h +f +y +y += +− +∑ += h2 × variance of the variable yi +i.e. +2 +x +σ += +2 +2 +y +h σ +or +x +σ = +y +hσ +... (4) +From (3) and (4), we have +x +σ = +2 +2 +1 +1 +N +N +n +n +i +i +i +i +i +i +h +f y +f y += += + + +− + + + +∑ +∑ +... (5) +Let us solve Example 11 by the short-cut method and using formula (5) +Examples 12 Calculate mean, variance and standard deviation for the following +distribution. +Classes +30-40 40-50 50-60 60-70 70-80 80-90 90-100 +Frequency +3 +7 +12 +15 +8 +3 +2 +Solution Let the assumed mean A = 65. Here h = 10 +We obtain the following Table 13.11 from the given data : +Table 13.11 +Class +Frequency +Mid-point +yi= +65 +10 +ix − +yi +2 +fi yi +fi yi +2 +f i +xi +30-40 +3 +35 +– 3 +9 +– 9 +27 +40-50 +7 +45 +– 2 +4 +– 14 +28 +50-60 +12 +55 +– 1 +1 +– 12 +12 +60-70 +15 +65 +0 +0 +0 +0 +70-80 +8 +75 +1 +1 +8 +8 +80-90 +3 +85 +2 +4 +6 +12 +9 0-100 +2 +95 +3 +9 +6 +18 +N=50 +– 15 +105 +Reprint 2025-26 + + STATISTICS 281 +Therefore +x = +15 +A +65 +10 +62 +50 +50 +i +i +f y +h ++ +× += +− +× += +∑ +Variance +2 +σ + = +( +) +2 +2 +2 +N +2 +N +i +i +h +f y +f y +i i + + +− +∑ +∑ + + + + += ( +)2 +10 +2 +50 105 +(–15) +2 +(50) + + +× +− + + + + += 1 [5250 +225] +201 +25 +− += +and standard deviation ( ) +201 +σ = + = 14.18 + EXERCISE 13.2 +Find the mean and variance for each of the data in Exercies 1 to 5. + 1. +6, 7, 10, 12, 13, 4, 8, 12 + 2. +First n natural numbers + 3. +First 10 multiples of 3 + 4. +xi +6 +10 +14 +18 +24 +28 +30 +f i +2 +4 +7 +12 +8 +4 +3 + 5. +xi +92 +93 +97 +98 +102 +104 +109 +f i +3 +2 +3 +2 +6 +3 +3 + 6. +Find the mean and standard deviation using short-cut method. +xi +60 +61 +62 +63 +64 +65 +66 +67 +68 +f i +2 +1 +12 +29 +25 +12 +10 +4 +5 +Find the mean and variance for the following frequency distributions in Exercises +7 and 8. + 7. +Classes +0-30 30-60 60-90 90-120 120-150 150-180 180-210 +Frequencies +2 +3 +5 +10 +3 +5 +2 +Reprint 2025-26 + +282 +MATHEMATICS + 8. +Classes +0-10 +10-20 +20-30 +30-40 +40-50 +Frequencies +5 +8 +15 +16 +6 + 9. +Find the mean, variance and standard deviation using short-cut method +Height +70-75 75-80 80-85 85-90 90-95 95-100 100-105105-110 110-115 +in cms +No. of +3 +4 +7 +7 +15 +9 +6 +6 +3 +children +10. +The diameters of circles (in mm) drawn in a design are given below: + Diameters + 33-36 37-40 + 41-44 + 45-48 +49-52 + No. of circles 15 17 21 + 22 + 25 +Calculate the standard deviation and mean diameter of the circles. +[ Hint First make the data continuous by making the classes as 32.5-36.5, 36.5-40.5, +40.5-44.5, 44.5 - 48.5, 48.5 - 52.5 and then proceed.] +Miscellaneous Examples +Example 13 The variance of 20 observations is 5. If each observation is multiplied by +2, find the new variance of the resulting observations. +Solution Let the observations be x1, x2, ..., x20 and x be their mean. Given that +variance = 5 and n = 20. We know that +Variance ( +) +2 +20 +2 +1 +1 +( +) +i +i +x +x +n +σ += += +− +∑ +, i.e., +20 +2 +1 +1 +5 +( +) +20 +i +i +x +x += += +− +∑ +or +20 +2 +1 +( +) +i +i +x +x += +− +∑ += 100 +... (1) +If each observation is multiplied by 2, and the new resulting observations are yi , then +yi = 2xi i.e., xi = +iy +2 +1 +Reprint 2025-26 + + STATISTICS 283 +Therefore +20 +20 +1 +1 +1 +1 +2 +20 +i +i +i +i +y +y +x +n += += += += +∑ +∑ += +20 +1 +1 +2 20 +i +i +. +x +=∑ +i.e. +y = 2 x or x = +y +2 +1 +Substituting the values of xi and x in (1), we get +2 +20 +1 +1 +1 +100 +2 +2 +i +i +y +y += + + +− += + + + + +∑ +, i.e., ∑ += += +− +20 +1 +2 +400 +) +( +i +i +y +y +Thus the variance of new observations = +2 +1 +400 +20 +2 +5 +20 × += += +× +ANote The reader may note that if each observation is multiplied by a constant +k, the variance of the resulting observations becomes k2 times the original variance. +Example14 The mean of 5 observations is 4.4 and their variance is 8.24. If three of +the observations are 1, 2 and 6, find the other two observations. +Solution Let the other two observations be x and y. +Therefore, the series is 1, 2, 6, x, y. +Now +Mean x = 4.4 = 1 +2 +6 +5 +x +y ++ ++ ++ ++ +or +22 = 9 + x + y +Therefore +x + y = 13 +... (1) +Also +variance = 8.24 = +2 +5 +1 +) +( +1 +x +x +n i +i +∑ += +− +i.e. 8.24 = +( +) +( +) +( +) +( +) +2 +2 +2 +2 +2 +2 +1 +3 4 +2 4 +1 6 +2 +4 4 ( +) +2 +4 4 +5 +. +. +. +x +y +. +x +y +. + + ++ ++ ++ ++ +−× ++ ++ +× + + +or 41.20 = 11.56 + 5.76 + 2.56 + x2 + y2 –8.8 × 13 + 38.72 +Therefore +x2 + y2 = 97 +... (2) +But from (1), we have +x2 + y2 + 2xy = 169 +... (3) +From (2) and (3), we have +2xy = 72 +... (4) +Subtracting (4) from (2), we get +Reprint 2025-26 + +284 +MATHEMATICS +x2 + y2 – 2xy = 97 – 72 i.e. (x – y)2 = 25 +or +x – y = ± 5 +... (5) +So, from (1) and (5), we get +x = 9, y = 4 when x – y = 5 +or +x = 4, y = 9 when x – y = – 5 +Thus, the remaining observations are 4 and 9. +Example 15 If each of the observation x1, x2, ...,xn is increased by ‘a’, where a is a +negative or positive number, show that the variance remains unchanged. +Solution Let x be the mean of x1, x2, ...,xn . Then the variance is given by +2 +1 +σ + = +2 +1 +1 +( +) +n +i +i +x +x +n += +− +∑ +If ‘a is added to each observation, the new observations will be +yi = xi + a +... (1) +Let the mean of the new observations be y . Then +y = +1 +1 +1 +1 +( +) +n +n +i +i +i +i +y +x +a +n +n += += += ++ +∑ +∑ += +1 +1 +1 +n +n +i +i +i +x +a +n += += + + ++ + + + + +∑ +∑ + = +a +x +n +na +x +n +n +i +i ++ += ++ +∑ +=1 +1 +i.e. +y = x + a +... (2) +Thus, the variance of the new observations +2 +2 +σ += +2 +1 +1 +( +) +n +i +i +y +y +n += +− +∑ += +2) +( +1 +1 +a +x +a +x +n +n +i +i +− +− ++ +∑ += +[Using (1) and (2)] += +2 +1 +1 +( +) +n +i +i +x +x +n += +− +∑ += +2 +1 +σ +Thus, the variance of the new observations is same as that of the original observations. +ANote We may note that adding (or subtracting) a positive number to (or from) +each observation of a group does not affect the variance. +Example 16 The mean and standard deviation of 100 observations were calculated as +40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one +observation. What are the correct mean and standard deviation? +Reprint 2025-26 + + STATISTICS 285 +Solution Given that number of observations (n) = 100 +Incorrect mean ( x ) = 40, +Incorrect standard deviation (σ) = 5.1 +We know that + +∑ += += +n +i +ix +n +x +1 +1 +i.e. +100 +1 +1 +40 +100 +i +i +x += += +∑ + or + +100 +1 +i +i +x +=∑ += 4000 +i.e. +Incorrect sum of observations = 4000 +Thus +the correct sum of observations = Incorrect sum – 50 + 40 += 4000 – 50 + 40 = 3990 +Hence +Correct mean = correct sum +3990 +100 +100 += += 39.9 +Also +Standard deviation σ = +2 +2 +2 +1 +1 +1 +1 +n +n +i +i +i +i +x +x +n +n += += + + +− + + + + +∑ +∑ += +( ) +2 +1 +2 +1 +x +x +n +n +i +i − +∑ += +i.e. +5.1 = +2 +2 +1 +1 +Incorrect +(40) +100 +n +i +i +x += +× +− +∑ +or +26.01 = +2 +1 +1 +Incorrect +100 +n +i +i +x += +× +∑ +– 1600 +Therefore +Incorrect +2 +1 +n +i +i +x +=∑ += 100 (26.01 + 1600) = 162601 +Now +Correct +2 +1 +n +i +i +x +=∑ += Incorrect ∑ += +n +i +ix +1 +2 – (50)2 + (40)2 += 162601 – 2500 + 1600 = 161701 +Therefore +Correct standard deviation +Reprint 2025-26 + +286 +MATHEMATICS += +2 +2 +Correct +(Correct mean) +ix +n +− +∑ += +2 +161701 +(39 9) +100 +. +− += +1617 01 +1592 01 +. +. +− + = +25 = 5 +Miscellaneous Exercise On Chapter 13 +1. +The mean and variance of eight observations are 9 and 9.25, respectively. If six +of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations. +2. +The mean and variance of 7 observations are 8 and 16, respectively. If five of the +observations are 2, 4, 10, 12, 14. Find the remaining two observations. +3. +The mean and standard deviation of six observations are 8 and 4, respectively. If +each observation is multiplied by 3, find the new mean and new standard deviation +of the resulting observations. +4. +Given that x is the mean and σ2 is the variance of n observations x1, x2, ...,xn. +Prove that the mean and variance of the observations ax1, ax2, ax3, ...., axn are +a x and a2 σ2, respectively, (a ≠ 0). +5. +The mean and standard deviation of 20 observations are found to be 10 and 2, +respectively. On rechecking, it was found that an observation 8 was incorrect. +Calculate the correct mean and standard deviation in each of the following cases: +(i) If wrong item is omitted. +(ii) If it is replaced by 12. +6. +The mean and standard deviation of a group of 100 observations were found to +be 20 and 3, respectively. Later on it was found that three observations were +incorrect, which were recorded as 21, 21 and 18. Find the mean and standard +deviation if the incorrect observations are omitted. +Summary +®Measures of dispersion Range, Quartile deviation, mean deviation, variance, +standard deviation are measures of dispersion. +Range = Maximum Value – Minimum Value +®Mean deviation for ungrouped data +M +M.D. ( ) +M.D. (M) +i +i +x – x +x – +x +, +n +n += += +∑ +∑ +Reprint 2025-26 + + STATISTICS 287 +®Mean deviation for grouped data +M.D. +N +M.D. M +M +N +where N +( ) +– +, +( +) +– +, +x +f x +x +f x +f +i +i +i +i +i += += += +∑ +∑ +∑ +®Variance and standard deviation for ungrouped data +2 +2 +1 +( +) +ix – x +n +σ += +∑ +, +2 +1 +( +– +) +ix +x +n +σ = +∑ +®Variance and standard deviation of a discrete frequency distribution +( +) +( +) +2 +2 +2 +1 +1 +, +N +N +i +i +i +i +f +x +x +f +x +x +σ +σ += +− += +− +∑ +∑ +®Variance and standard deviation of a continuous frequency distribution +( +) +( +) +2 +2 +2 +2 +1 +1 +, +N +N +N +i +i +i +i +i +i +f +x +x +f x +f x +σ +σ += +− += +− +∑ +∑ +∑ +®Shortcut method to find variance and standard deviation. +( +) +2 +2 +2 +2 +2 +N +N +i +i +i +i +h +f y +f y +σ + + += +− + + + + +∑ +∑ +, +( +) +2 +2 +N +N +i +i +i +i +h +f y +f y +σ += +− +∑ +∑ +, +where +A +i +i +x +y +h +− += +Historical Note +‘Statistics’ is derived from the Latin word ‘status’ which means a political +state. This suggests that statistics is as old as human civilisation. In the year 3050 +B.C., perhaps the first census was held in Egypt. In India also, about 2000 years +ago, we had an efficient system of collecting administrative statistics, particularly, +during the regime of Chandra Gupta Maurya (324-300 B.C.). The system of +collecting data related to births and deaths is mentioned in Kautilya’s Arthshastra +(around 300 B.C.) A detailed account of administrative surveys conducted during +Akbar’s regime is given in Ain-I-Akbari written by Abul Fazl. +Captain John Graunt of London (1620-1674) is known as father of vital +statistics due to his studies on statistics of births and deaths. Jacob Bernoulli +(1654-1705) stated the Law of Large numbers in his book “Ars Conjectandi’, +published in 1713. +Reprint 2025-26 + +288 +MATHEMATICS +— v — +The theoretical development of statistics came during the mid seventeenth +century and continued after that with the introduction of theory of games and +chance (i.e., probability). Francis Galton (1822-1921), an Englishman, pioneered +the use of statistical methods, in the field of Biometry. Karl Pearson (1857-1936) +contributed a lot to the development of statistical studies with his discovery +of Chi square test and foundation of statistical laboratory in England (1911). +Sir Ronald A. Fisher (1890-1962), known as the Father of modern statistics, +applied it to various diversified fields such as Genetics, Biometry, Education, +Agriculture, etc. +Reprint 2025-26" +class_11,14,Probability,ncert_books/class_11/kemh1dd/kemh114.pdf,"vWhere a mathematical reasoning can be had, it is as great a folly to +make use of any other, as to grope for a thing in the dark, when +you have a candle in your hand. – JOHN ARBUTHNOT v +14.1 Event +We have studied about random experiment and sample space associated with an +experiment. The sample space serves as an universal set for all questions concerned +with the experiment. +Consider the experiment of tossing a coin two times. An associated sample space +is S = {HH, HT, TH, TT}. +Now suppose that we are interested in those outcomes which correspond to the +occurrence of exactly one head. We find that HT and TH are the only elements of S +corresponding to the occurrence of this happening (event). These two elements form +the set E = { HT, TH} +We know that the set E is a subset of the sample space S . Similarly, we find the +following correspondence between events and subsets of S. +Description of events +Corresponding subset of ‘S’ +Number of tails is exactly 2 +A = {TT} +Number of tails is atleast one +B = {HT, TH, TT} +Number of heads is atmost one +C = {HT, TH, TT} +Second toss is not head +D = { HT, TT} +Number of tails is atmost two +S = {HH, HT, TH, TT} +Number of tails is more than two +φ +The above discussion suggests that a subset of sample space is associated with +an event and an event is associated with a subset of sample space. In the light of this +we define an event as follows. +Definition Any subset E of a sample space S is called an event. +14 +Chapter +PROBABILITY +Reprint 2025-26 + +290 +MATHEMATICS +14.1.1 Occurrence of an event Consider the experiment of throwing a die. Let E +denotes the event “ a number less than 4 appears”. If actually ‘1’ had appeared on the +die then we say that event E has occurred. As a matter of fact if outcomes are 2 or 3, +we say that event E has occurred +Thus, the event E of a sample space S is said to have occurred if the outcome +ωof the experiment is such that ω∈ E. If the outcome ω is such that ω ∉ E, we say +that the event E has not occurred. +14.1.2 Types of events Events can be classified into various types on the basis of the +elements they have. +1. Impossible and Sure Events The empty set φ and the sample space S describe +events. In fact φ is called an impossible event and S, i.e., the whole sample space is +called the sure event. +To understand these let us consider the experiment of rolling a die. The associated +sample space is +S = {1, 2, 3, 4, 5, 6} +Let E be the event “ the number appears on the die is a multiple of 7”. Can you +write the subset associated with the event E? +Clearly no outcome satisfies the condition given in the event, i.e., no element of +the sample space ensures the occurrence of the event E. Thus, we say that the empty +set only correspond to the event E. In other words we can say that it is impossible to +have a multiple of 7 on the upper face of the die. Thus, the event E = φ is an impossible +event. +Now let us take up another event F “the number turns up is odd or even”. Clearly +F = {1, 2, 3, 4, 5, 6,} = S, i.e., all outcomes of the experiment ensure the occurrence of +the event F. Thus, the event F = S is a sure event. +2. Simple Event If an event E has only one sample point of a sample space, it is +called a simple (or elementary) event. +In a sample space containing n distinct elements, there are exactly n simple +events. +For example in the experiment of tossing two coins, a sample space is +S={HH, HT, TH, TT} +There are four simple events corresponding to this sample space. These are +E1= {HH}, E2={HT}, E3= { TH} and E4={TT}. +Reprint 2025-26 + + PROBABILITY 291 +3. Compound Event If an event has more than one sample point, it is called a +Compound event. +For example, in the experiment of “tossing a coin thrice” the events +E: ‘Exactly one head appeared’ +F: ‘Atleast one head appeared’ +G: ‘Atmost one head appeared’ etc. +are all compound events. The subsets of S associated with these events are +E={HTT,THT,TTH} +F={HTT,THT, TTH, HHT, HTH, THH, HHH} +G= {TTT, THT, HTT, TTH} +Each of the above subsets contain more than one sample point, hence they are all +compound events. +14.1.3 Algebra of events In the Chapter on Sets, we have studied about different +ways of combining two or more sets, viz, union, intersection, difference, complement +of a set etc. Like-wise we can combine two or more events by using the analogous set +notations. +Let A, B, C be events associated with an experiment whose sample space is S. +1. Complementary Event For every event A, there corresponds another event +A′called the complementary event to A. It is also called the event ‘not A’. +For example, take the experiment ‘of tossing three coins’. An associated sample +space is +S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} +Let A={HTH, HHT, THH} be the event ‘only one tail appears’ +Clearly for the outcome HTT, the event A has not occurred. But we may say that +the event ‘not A’ has occurred. Thus, with every outcome which is not in A, we say +that ‘not A’ occurs. +Thus the complementary event ‘not A’ to the event A is +A′ = {HHH, HTT, THT, TTH, TTT} +or +A′ = {ω : ω ∈ S and ω ∉A} = S – A. +2. The Event ‘A or B’ Recall that union of two sets A and B denoted by A ∪ B +contains all those elements which are either in A or in B or in both. +When the sets A and B are two events associated with a sample space, then +‘A ∪ B’ is the event ‘either A or B or both’. This event ‘A ∪ B’ is also called ‘A or B’. +Therefore +Event ‘A or B’ = A ∪ B += {ω : ω ∈ A or ω ∈ B} +Reprint 2025-26 + +292 +MATHEMATICS +3. The Event ‘A and B’ We know that intersection of two sets A ∩ B is the set of +those elements which are common to both A and B. i.e., which belong to both +‘A and B’. +If A and B are two events, then the set A ∩ B denotes the event ‘A and B’. +Thus, A ∩ B = {ω : ω ∈ A and ω ∈ B} +For example, in the experiment of ‘throwing a die twice’ Let A be the event +‘score on the first throw is six’ and B is the event ‘sum of two scores is atleast 11’ then +A = {(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}, and B = {(5,6), (6,5), (6,6)} +so +A ∩ B = {(6,5), (6,6)} +Note that the set A ∩ B = {(6,5), (6,6)} may represent the event ‘the score on the first +throw is six and the sum of the scores is atleast 11’. +4. The Event ‘A but not B’ We know that A–B is the set of all those elements +which are in A but not in B. Therefore, the set A–B may denote the event ‘A but not +B’.We know that + A – B = A ∩ B´ +Example 1 Consider the experiment of rolling a die. Let A be the event ‘getting a +prime number’, B be the event ‘getting an odd number’. Write the sets representing +the events (i) Aor B (ii) A and B (iii) A but not B (iv) ‘not A’. +Solution Here +S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5} +Obviously +(i) +‘A or B’ = A ∪ B = {1, 2, 3, 5} +(ii) +‘A and B’ = A ∩ B = {3,5} +(iii) +‘A but not B’ = A – B = {2} +(iv) +‘not A’ = A′ = {1,4,6} +14.1.4 Mutually exclusive events In the experiment of rolling a die, a sample space +is S = {1, 2, 3, 4, 5, 6}. Consider events, A ‘an odd number appears’ and B ‘an even +number appears’ +Clearly the event A excludes the event B and vice versa. In other words, there is +no outcome which ensures the occurrence of events A and B simultaneously. Here +A = {1, 3, 5} and B = {2, 4, 6} +Clearly A ∩ B = φ, i.e., A and B are disjoint sets. +In general, two events A and B are called mutually exclusive events if the +occurrence of any one of them excludes the occurrence of the other event, i.e., if they +can not occur simultaneously. In this case the sets A and B are disjoint. +Reprint 2025-26 + + PROBABILITY 293 +Again in the experiment of rolling a die, consider the events A ‘an odd number +appears’ and event B ‘a number less than 4 appears’ +Obviously A = {1, 3, 5} and B = {1, 2, 3} +Now 3 ∈ A as well as 3 ∈ B +Therefore, A and B are not mutually exclusive events. +Remark Simple events of a sample space are always mutually exclusive. +14.1.5 Exhaustive events Consider the experiment of throwing a die. We have +S = {1, 2, 3, 4, 5, 6}. Let us define the following events +A: ‘a number less than 4 appears’, +B: ‘a number greater than 2 but less than 5 appears’ +and +C: ‘a number greater than 4 appears’. +Then A = {1, 2, 3}, B = {3,4} and C = {5, 6}. We observe that +A ∪ B ∪ C = {1, 2, 3} ∪ {3, 4} ∪ {5, 6} = S. +Such events A, B and C are called exhaustive events. In general, if E1, E2, ..., En are n +events of a sample space S and if +1 +2 +3 +1 +E +E +E +E +E +S +n +n +i +i +... += +∪ +∪ +∪ +∪ += ∪ += +then E1, E2, ...., En are called exhaustive events.In other words, events E1, E2, ..., En +are said to be exhaustive if atleast one of them necessarily occurs whenever the +experiment is performed. +Further, if Ei ∩ Ej = φ for i ≠ j i.e., events Ei and Ej are pairwise disjoint and +S +E +1 += +∪ += +i +n +i +, then events E1, E2, ..., En are called mutually exclusive and exhaustive +events. +We now consider some examples. +Example 2 Two dice are thrown and the sum of the numbers which come up on the +dice is noted. Let us consider the following events associated with this experiment +A: ‘the sum is even’. +B: ‘the sum is a multiple of 3’. +C: ‘the sum is less than 4’. +D: ‘the sum is greater than 11’. +Which pairs of these events are mutually exclusive? +Reprint 2025-26 + +294 +MATHEMATICS +Solution There are 36 elements in the sample space S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}. +Then +A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), + (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)} +B = {(1, 2), (2, 1), (1, 5), (5, 1), (3, 3), (2, 4), (4, 2), (3, 6), (6, 3), (4, 5), (5, 4), + (6, 6)} +C = {(1, 1), (2, 1), (1, 2)} and D = {(6, 6)} +We find that +A ∩ B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)} ≠ φ +Therefore, A and B are not mutually exclusive events. +Similarly A ∩ C ≠ φ, A ∩ D ≠ φ, B ∩ C ≠ φ and B ∩ D ≠ φ. +Thus, the pairs of events, (A, C), (A, D), (B, C), (B, D) are not mutually exclusive +events. +Also C ∩ D = φ and so C and D are mutually exclusive events. +Example 3 A coin is tossed three times, consider the following events. +A: ‘No head appears’, B: ‘Exactly one head appears’ and C: ‘Atleast two heads +appear’. +Do they form a set of mutually exclusive and exhaustive events? +Solution The sample space of the experiment is +S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} +and A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH} +Now +A ∪ B ∪ C = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH} = S +Therefore, A, B and C are exhaustive events. +Also, +A ∩ B = φ, A ∩ C = φ and B ∩ C = φ +Therefore, the events are pair-wise disjoint, i.e., they are mutually exclusive. +Hence, A, B and C form a set of mutually exclusive and exhaustive events. +EXERCISE 14.1 +1. +A die is rolled. Let E be the event “die shows 4” and F be the event “die shows +even number”. Are E and F mutually exclusive? +2. +A die is thrown. Describe the following events: +(i) +A: a number less than 7 +(ii) +B: a number greater than 7 +(iii) +C: a multiple of 3 +(iv) +D: a number less than 4 +(v) +E: an even number greater than 4 +(vi) +F: a number not less than 3 +Also find A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E, E ∩ F′, F′ +Reprint 2025-26 + + PROBABILITY 295 +3. +An experiment involves rolling a pair of dice and recording the numbers that +come up. Describe the following events: +A: the sum is greater than 8, B: 2 occurs on either die +C: the sum is at least 7 and a multiple of 3. +Which pairs of these events are mutually exclusive? +4. +Three coins are tossed once. Let A denote the event ‘three heads show”, B +denote the event “two heads and one tail show”, C denote the event” three tails +show and D denote the event ‘a head shows on the first coin”. Which events are +(i) mutually exclusive? +(ii) simple? +(iii) Compound? +5. +Three coins are tossed. Describe +(i) +Two events which are mutually exclusive. +(ii) +Three events which are mutually exclusive and exhaustive. +(iii) +Two events, which are not mutually exclusive. +(iv) +Two events which are mutually exclusive but not exhaustive. +(v) +Three events which are mutually exclusive but not exhaustive. +6. +Two dice are thrown. The events A, B and C are as follows: +A: getting an even number on the first die. +B: getting an odd number on the first die. +C: getting the sum of the numbers on the dice ≤ 5. +Describe the events +(i) +A′ +(ii) +not B +(iii) +A or B +(iv) +A and B +(v) +A but not C +(vi) +B or C +(vii) +B and C +(viii) +A ∩ B′ ∩ C′ +7. +Refer to question 6 above, state true or false: (give reason for your answer) +(i) +A and B are mutually exclusive +(ii) +A and B are mutually exclusive and exhaustive +(iii) +A = B′ +(iv) +A and C are mutually exclusive +(v) +A and B′ are mutually exclusive. +(vi) +A′, B′, C are mutually exclusive and exhaustive. +14.2 Axiomatic Approach to Probability +In earlier sections, we have considered random experiments, sample space and +events associated with these experiments. In our day to day life we use many words +about the chances of occurrence of events. Probability theory attempts to quantify +these chances of occurrence or non occurrence of events. +Reprint 2025-26 + +296 +MATHEMATICS +In earlier classes, we have studied some methods of assigning probability to an +event associated with an experiment having known the number of total outcomes. +Axiomatic approach is another way of describing probability of an event. In this +approach some axioms or rules are depicted to assign probabilities. +Let S be the sample space of a random experiment. The probability P is a real +valued function whose domain is the power set of S and range is the interval [0,1] +satisfying the following axioms +(i) +For any event E, P (E) ≥0 +(ii) +P (S) = 1 +(iii) +If E and F are mutually exclusive events, then P(E ∪ F) = P(E) + P(F). +It follows from (iii) that P(φ) = 0. To prove this, we take F = φ and note that E and φ +are disjoint events. Therefore, from axiom (iii), we get +P (E ∪ φ) = P (E) + P (φ) or +P(E) = P(E) + P (φ) i.e. P (φ) = 0. +Let S be a sample space containing outcomes +1 +2 +, +,..., +n +ω ω +ω , i.e., +S = {ω1, ω2, ..., ωn} +It follows from the axiomatic definition of probability that +(i) +0 ≤ P (ωi) ≤ 1 for each ωi ∈ S +(ii) +P (ω1) + P (ω2) + ... + P (ωn) = 1 +(iii) +For any event A, P(A) = ∑ P(ωi ), ωi ∈ A. +ANote It may be noted that the singleton {ωi} is called elementary event and +for notational convenience, we write P(ωi ) for P({ωi }). +For example, in ‘a coin tossing’ experiment we can assign the number +1 +2 to each +of the outcomes H and T. +i.e. +P(H) = +1 +2 and P(T) = +1 +2 +(1) +Clearly this assignment satisfies both the conditions i.e., each number is neither +less than zero nor greater than 1 and +P(H) + P(T) = 2 +1 + 2 +1 = 1 +Therefore, in this case we can say that probability of H = 2 +1 , and probability of T = 2 +1 +If we take P(H) = 4 +1 and P(T) = 4 +3 +... (2) +Reprint 2025-26 + + PROBABILITY 297 +Does this assignment satisfy the conditions of axiomatic approach? +Yes, in this case, probability of H = +1 +4 and probability of T = 4 +3 . +We find that both the assignments (1) and (2) are valid for probability of +H and T. +In fact, we can assign the numbers p and (1 – p) to both the outcomes such that +0 ≤ p ≤ 1 and P(H) + P(T) = p + (1 – p) = 1 +This assignment, too, satisfies both conditions of the axiomatic approach of +probability. Hence, we can say that there are many ways (rather infinite) to assign +probabilities to outcomes of an experiment. We now consider some examples. +Example 4 Let a sample space be S = {ω1, ω2,..., ω6}.Which of the following +assignments of probabilities to each outcome are valid? +Outcomes +ω 1 +ω 2 +ω 3 +ω 4 +ω 5 +ω 6 +(a) +6 +1 +6 +1 +6 +1 +6 +1 +6 +1 +6 +1 +(b) + 1 +0 +0 +0 +0 +0 +(c) +8 +1 +3 +2 +3 +1 +3 +1 +4 +1 +− +3 +1 +− +(d) +12 +1 +12 +1 +6 +1 +6 +1 +6 +1 +2 +3 +(e) +0.1 +0.2 +0.3 +0.4 +0.5 +0.6 +Solution (a) Condition (i): Each of the number p(ωi) is positive and less than one. +Condition (ii): Sum of probabilities += +1 +6 +1 +6 +1 +6 +1 +6 +1 +6 +1 +6 +1 += ++ ++ ++ ++ ++ +Therefore, the assignment is valid +(b) +Condition (i): Each of the number p(ωi) is either 0 or 1. +Condition (ii) Sum of the probabilities = 1 + 0 + 0 + 0 + 0 + 0 = 1 +Therefore, the assignment is valid +(c) +Condition (i) Two of the probabilities p(ω5) and p(ω6) are negative, the assignment +is not valid +(d) +Since p(ω6) = 3 +2 > 1, the assignment is not valid +Reprint 2025-26 + +298 +MATHEMATICS +(e) +Since, sum of probabilities = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1, the assignment +is not valid. +14.2.1 Probability of an event Let S be a sample space associated with the experiment +‘examining three consecutive pens produced by a machine and classified as Good +(non-defective) and bad (defective)’. We may get 0, 1, 2 or 3 defective pens as result +of this examination. +A sample space associated with this experiment is +S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG}, +where B stands for a defective or bad pen and G for a non – defective or good pen. +Let the probabilities assigned to the outcomes be as follows +Sample point: +BBB +BBG +BGB +GBB +BGG +GBG +GGB +GGG +Probability: +8 +1 +8 +1 +8 +1 +8 +1 +8 +1 +8 +1 +8 +1 +8 +1 +Let event A: there is exactly one defective pen and event B: there are atleast two +defective pens. +Hence A = {BGG, GBG, GGB} and B = {BBG, BGB, GBB, BBB} +Now +P(A) = +P(ω ), ω +A +i +i +∑ +∀ +∈ += P(BGG) + P(GBG) + P(GGB) = +8 +3 +8 +1 +8 +1 +8 +1 += ++ ++ +and +P(B) = +P(ω ), ω +B +i +i +∑ +∀ +∈ += P(BBG) + P(BGB) + P(GBB) + P(BBB) = +2 +1 +8 +4 +8 +1 +8 +1 +8 +1 +8 +1 += += ++ ++ ++ +Let us consider another experiment of ‘tossing a coin “twice” +The sample space of this experiment is S = {HH, HT, TH, TT} +Let the following probabilities be assigned to the outcomes +P(HH) = 4 +1 , P(HT) = 7 +1 , P(TH) = 7 +2 , P(TT) = 28 +9 +Clearly this assignment satisfies the conditions of axiomatic approach. Now, let +us find the probability of the event E: ‘Both the tosses yield the same result’. +Here +E = {HH, TT} +Now +P(E) = Σ P(wi), for all wi ∈ E +Reprint 2025-26 + + PROBABILITY 299 + = P(HH) + P(TT) = +7 +4 +28 +9 +4 +1 += ++ +For the event F: ‘exactly two heads’, we have F = {HH} +and +P(F) = P(HH) = +1 +4 +14.2.2 Probabilities of equally likely outcomes Let a sample space of an +experiment be +S = {ω1, ω2,..., ωn}. +Let all the outcomes are equally likely to occur, i.e., the chance of occurrence of each +simple event must be same. +i.e. +P(ωi) = p, for all ωi ∈ S where 0 ≤ p ≤ 1 +Since +1 +P(ω ) 1 +n +i +i= += +∑ + i.e., p + p + ... + p (n times) = 1 +or +np = 1 i.e., p = 1 +n +Let S be a sample space and E be an event, such that n(S) = n and n(E) = m. If +each out come is equally likely, then it follows that +P(E) +m +n += += +Number of outcomesfavourable toE +Totalpossibleoutcomes +14.2.3 Probability of the event ‘A or B’ Let us now find the probability of event +‘A or B’, i.e., P (A ∪ B) +Let +A = {HHT, HTH, THH} and B = {HTH, THH, HHH} be two events associated +with ‘tossing of a coin thrice’ +Clearly A ∪ B = {HHT, HTH, THH, HHH} +Now +P (A ∪ B) = P(HHT) + P(HTH) + P(THH) + P(HHH) +If all the outcomes are equally likely, then +( +) +1 +1 +1 +1 +4 +1 +P A +B +8 +8 +8 +8 +8 +2 +∪ += ++ ++ ++ += += +Also +P(A) = P(HHT) + P(HTH) + P(THH) = 3 +8 +Reprint 2025-26 + +300 +MATHEMATICS +and +P(B) = P(HTH) + P(THH) + P(HHH) = 3 +8 +Therefore +P(A) + P(B) = 3 +3 +6 +8 +8 +8 ++ += +It is clear that +P(A ∪B) ≠ P(A) + P(B) +The points HTH and THH are common to both A and B. In the computation of +P(A) + P(B) the probabilities of points HTH and THH, i.e., the elements of A ∩B are +included twice. Thus to get the probability P(A∪B) we have to subtract the probabilities +of the sample points in A ∩ B from P(A) + P(B) +i.e. +P(A +B) +∪ += P(A) +P(B) +P(ω ) +ω +A +B +i +i +, ++ +−∑ +∀ +∈ +∩ += P A +P B +P A +B +( +) +( ) +( +) ++ +− +∩ +Thus we observe that, +) +B +A +( +P +) +B +( +P +) +A +( +P +) +B +A +( +P +∩ +− ++ += +∪ +In general, if A and B are any two events associated with a random experiment, +then by the definition of probability of an event, we have +( +) +( +) +P A +B +ω +ω +A +B +i +i +p +, +∪ += ∑ +∀ +∈ +∪ +. +Since +A +B = (A–B) +(A +B) +(B–A) +∪ +∪ +∩ +∪ +, +we have +P(A ∪ B) = [ +] [ +] +P(ω ) ω +(A–B) +P(ω ) +ω +A +B + +i +i +i +i +∑ +∀ +∈ ++ ∑ +∀ +∈ +∩ +[ +] +P(ω ) ω +B – A +i +i +∑ +∀ +∈ +(because A–B, A ∩ B and B – A are mutually exclusive) + ... (1) +Also +[ +] [ +] +P(A) +P(B) +(ω ) +A + +(ω ) +ω +B +i +i +i +i +p +p +ω ++ += ∑ +∀ +∈ +∑ +∀ +∈ += [ +] +P(ω ) ω +(A–B) +(A +B) + +i +i +∑ +∀ +∈ +∪ +∩ +[ +] +P(ω ) ω +(B – A) +(A +B) +i +i +∑ +∀ +∈ +∪ +∩ += [ +] [ +] +P(ω ) ω +(A – B) + +P(ω ) ω +(A +B) +i +i +i +i +∑ +∀ +∈ +∑ +∀ +∈ +∩ ++ [ +] +P(ω ) ω +(B–A) +i +i +∑ +∀ +∈ ++ + [ +] +P(ω ) ω +(A +B) +i +i +∑ +∀ +∈ +∩ += +[ +] +P(A +B) +P(ω ) +ω +A +B +i +i +∪ ++ ∑ +∀ +∈ +∩ + [using (1)] += P(A +B)+P(A +B) +∪ +∩ +. +Hence P(A +B) +P (A)+P(B) – P(A +B) +∪ += +∩ +. +Alternatively, it can also be proved as follows: +A ∪ B = A ∪ (B – A), where A and B – A are mutually exclusive, +and B = (A ∩ B) ∪ (B – A), where A ∩ B and B – A are mutually exclusive. +Using Axiom (iii) of probability, we get +Reprint 2025-26 + + PROBABILITY 301 +P (A ∪B) = P (A) + P (B – A) +... (2) +and +P(B) = P ( A ∩ B) + P (B – A) +... (3) +Subtracting (3) from (2) gives +P (A ∪ B) – P(B) = P(A) – P (A ∩ B) +or +P(A ∪ B) = P(A) + P (B) – P (A ∩ B) +The above result can further be verified by observing the Venn Diagram (Fig 14.1) +If A and B are disjoint sets, i.e., they are mutually exclusive events, then A ∩ B = φ +Therefore +P(A +B) = P ( ) = 0 +∩ +φ +Thus, for mutually exclusive events A and B, we have +) +B +( +P +) +A +( +P +) +B +A +( +P ++ += +∪ +, +which is Axiom (iii) of probability. +14.2.4 Probability of event ‘not A’ Consider the event A = {2, 4, 6, 8} associated +with the experiment of drawing a card from a deck of ten cards numbered from +1 to 10. Clearly the sample space is S = {1, 2, 3, ...,10} +If all the outcomes 1, 2, ...,10 are considered to be equally likely, then the probability +of each outcome is 10 +1 +Now +P(A) = P(2) + P(4) + P(6) + P(8) += 1 +1 +1 +1 +4 +2 +10 +10 +10 +10 +10 +5 ++ ++ ++ += += +Also event ‘not A’ = A′ = {1, 3, 5, 7, 9, 10} +Now +P(A′) = P(1) + P(3) + P(5) + P(7) + P(9) + P(10) +Fig 14.1 +Reprint 2025-26 + +302 +MATHEMATICS += 6 +3 +10 +5 += +Thus, +P(A′) = 3 +5 = +) +A +( +P +1 +5 +2 +1 +− += +− +Also, we know that A′ and A are mutually exclusive and exhaustive events i.e., +A ∩ A′ = φ and A ∪ A′ = S +or +P(A ∪ A′) = P(S) +Now +P(A) + P(A′) = 1, +by using axioms (ii) and (iii). +or +P( A′ ) = P(not A) = 1 – P(A) +We now consider some examples and exercises having equally likely outcomes +unless stated otherwise. +Example 5 One card is drawn from a well shuffled deck of 52 cards. If each outcome +is equally likely, calculate the probability that the card will be +(i) +a diamond +(ii) +not an ace +(iii) +a black card (i.e., a club or, a spade) +(iv) +not a diamond +(v) +not a black card. +Solution When a card is drawn from a well shuffled deck of 52 cards, the number of +possible outcomes is 52. +(i) +Let A be the event 'the card drawn is a diamond' +Clearly the number of elements in set A is 13. +Therefore, +P(A) = 13 +1 +52 +4 += +i.e. probability of a diamond card = +1 +4 +(ii) +We assume that the event ‘Card drawn is an ace’ is B +Therefore ‘Card drawn is not an ace’ should be B′. +We know that P(B′) = 1 – P(B) = +13 +12 +13 +1 +1 +52 +4 +1 += +− += +− +(iii) +Let C denote the event ‘card drawn is black card’ +Therefore, number of elements in the set C = 26 +i.e. +P(C) = +2 +1 +52 +26 = +Reprint 2025-26 + + PROBABILITY 303 +Thus, probability of a black card = 2 +1 +. +(iv) +We assumed in (i) above that A is the event ‘card drawn is a diamond’, +so the event ‘card drawn is not a diamond’ may be denoted as A' or ‘not A’ +Now P(not A) = 1 – P(A) = +4 +3 +4 +1 +1 += +− +(v) +The event ‘card drawn is not a black card’ may be denoted as C′ or ‘not C’. +We know that P(not C) = 1 – P(C) = +2 +1 +2 +1 +1 += +− +Therefore, probability of not a black card = 2 +1 +Example 6 A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The +discs are similar in shape and size. A disc is drawn at random from the bag. Calculate +the probability that it will be (i) red, (ii) yellow, (iii) blue, (iv) not blue, +(v) either red or blue. +Solution There are 9 discs in all so the total number of possible outcomes is 9. +Let the events A, B, C be defined as +A: ‘the disc drawn is red’ +B: ‘the disc drawn is yellow’ +C: ‘the disc drawn is blue’. +(i) +The number of red discs = 4, i.e., n (A) = 4 +Hence +P(A) = 9 +4 +(ii) +The number of yellow discs = 2, i.e., n (B) = 2 +Therefore, +P(B) = 9 +2 +(iii) The number of blue discs = 3, i.e., n(C) = 3 +Therefore, +P(C) = +3 +1 +9 +3 = +(iv) +Clearly the event ‘not blue’ is ‘not C’. We know that P(not C) = 1 – P(C) +Reprint 2025-26 + +304 +MATHEMATICS +Therefore +P(not C) = +3 +2 +3 +1 +1 += +− +(v) +The event ‘either red or blue’ may be described by the set ‘A or C’ +Since, +A and C are mutually exclusive events, we have +P(A or C) = P (A ∪ C) = P(A) + P(C) = +9 +7 +3 +1 +9 +4 += ++ +Example 7 Two students Anil and Ashima appeared in an examination. The probability +that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination +is 0.10. The probability that both will qualify the examination is 0.02. Find the +probability that +(a) +Both Anil and Ashima will not qualify the examination. +(b) + Atleast one of them will not qualify the examination and +(c) +Only one of them will qualify the examination. +Solution Let E and F denote the events that Anil and Ashima will qualify the examination, +respectively. Given that +P(E) = 0.05, P(F) = 0.10 and P(E ∩ F) = 0.02. +Then +(a) +The event ‘both Anil and Ashima will not qualify the examination’ may be +expressed as E´ ∩ F´. +Since, +E´ is ‘not E’, i.e., Anil will not qualify the examination and F´ is ‘not F’, i.e., +Ashima will not qualify the examination. +Also +E´ ∩ F´ = (E ∪ F)´ (by Demorgan's Law) +Now +P(E ∪ F) = P(E) + P(F) – P(E ∩ F) +or +P(E ∪ F) = 0.05 + 0.10 – 0.02 = 0.13 +Therefore P(E´ ∩ F´) = P(E ∪ F)´ = 1 – P(E ∪ F) = 1 – 0.13 = 0.87 +(b) +P (atleast one of them will not qualify) += 1 – P(both of them will qualify) += 1 – 0.02 = 0.98 +(c) +The event only one of them will qualify the examination is same as the event +either (Anil will qualify, and Ashima will not qualify) or (Anil will not qualify and Ashima +Reprint 2025-26 + + PROBABILITY 305 +will qualify) i.e., E ∩ F´ or E´ ∩ F, where E ∩ F´ and E´ ∩ F are mutually exclusive. +Therefore, P(only one of them will qualify) += P(E ∩ F´ or E´ ∩ F) += P(E ∩ F´) + P(E´ ∩ F) = P (E) – P(E ∩ F) + P(F) – P (E ∩ F) += 0.05 – 0.02 + 0.10 – 0.02 = 0.11 +Example 8 A committee of two persons is selected from two men and two women. +What is the probability that the committee will have (a) no man? (b) one man? (c) two +men? +Solution The total number of persons = 2 + 2 = 4. Out of these four person, two can +be selected in 4 +2 +C ways. +(a) +No men in the committee of two means there will be two women in the committee. +Out of two women, two can be selected in 2 +2 +C +1 += way. +Therefore +( +) +2 +2 +4 +2 +C +1 2 1 +1 +P no man +4 +3 +6 +C +× × += += += +× +(b) +One man in the committee means that there is one woman. One man out of 2 +can be selected in 2 +1 +C ways and one woman out of 2 can be selected in 2 +1 +C ways. +Together they can be selected in 2 +2 +1 +1 +C +C +× + ways. +Therefore +( +) +2 +2 +1 +1 +4 +2 +C +C +2 +2 +2 +P One man +2 3 +3 +C +× +× += += += +× +(c) +Two men can be selected in 2 +2 +C way. +Hence +( +) +2 +2 +4 +4 +2 +2 +C +1 +1 +P Two men +6 +C +C += += += +EXERCISE 14.2 +1. +Which of the following can not be valid assignment of probabilities for outcomes +of sample Space S = { +} +1 +2 +3 +4 +5 +6 +7 +, +, +, +, +, +, +ω ω ω ω ω ω ω +Reprint 2025-26 + +306 +MATHEMATICS +Assignment +ω 1 +ω 2 +ω 3 +ω 4 +ω 5 +ω 6 +ω 7 +(a) +0.1 +0.01 +0.05 +0.03 +0.01 +0.2 +0.6 +(b) +7 +1 +7 +1 +7 +1 +7 +1 +7 +1 +7 +1 +7 +1 +(c) +0.1 +0.2 +0.3 +0.4 +0.5 +0.6 +0.7 +(d) + – 0.1 +0.2 +0.3 +0.4 + – 0.2 +0.1 +0.3 +(e) +14 +1 +14 +2 +14 +3 +14 +4 +14 +5 +14 +6 +14 +15 +2. +A coin is tossed twice, what is the probability that atleast one tail occurs? +3. +A die is thrown, find the probability of following events: +(i) +A prime number will appear, +(ii) +A number greater than or equal to 3 will appear, +(iii) +A number less than or equal to one will appear, +(iv) +A number more than 6 will appear, +(v) +A number less than 6 will appear. +4. +A card is selected from a pack of 52 cards. +(a) +How many points are there in the sample space? +(b) +Calculate the probability that the card is an ace of spades. +(c) +Calculate the probability that the card is (i) an ace (ii) black card. +5. +A fair coin with 1 marked on one face and 6 on the other and a fair die are both +tossed. find the probability that the sum of numbers that turn up is (i) 3 (ii) 12 +6. +There are four men and six women on the city council. If one council member is +selected for a committee at random, how likely is it that it is a woman? +7. +A fair coin is tossed four times, and a person win Re 1 for each head and lose +Rs 1.50 for each tail that turns up. +From the sample space calculate how many different amounts of money you can +have after four tosses and the probability of having each of these amounts. +8. +Three coins are tossed once. Find the probability of getting +(i) +3 heads +(ii) +2 heads +(iii) +atleast 2 heads +(iv) +atmost 2 heads +(v) +no head +(vi) +3 tails +(vii) +exactly two tails +(viii) +no tail +(ix) +atmost two tails +9. +If 11 +2 is the probability of an event, what is the probability of the event ‘not A’. +10. +A letter is chosen at random from the word ‘ASSASSINATION’. Find the +probability that letter is (i) a vowel (ii) a consonant +Reprint 2025-26 + + PROBABILITY 307 +11. +In a lottery, a person choses six different natural numbers at random from 1 to 20, +and if these six numbers match with the six numbers already fixed by the lottery +committee, he wins the prize. What is the probability of winning the prize in the +game? [Hint order of the numbers is not important.] +12. +Check whether the following probabilities P(A) and P(B) are consistently defined +(i) +P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6 +(ii) +P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8 +13. +Fill in the blanks in following table: +P(A) +P(B) +P(A ∩B) +P(A ∪B) +(i) +1 +3 +1 +5 +1 +15 +. . . +(ii) +0.35 +. . . +0.25 +0.6 +(iii) +0.5 +0.35 +. . . +0.7 +14. +Given P(A) = 5 +3 and P(B) = 5 +1 . Find P(A or B), if A and B are mutually exclusive +events. +15. +If E and F are events such that P(E) = 4 +1 , P(F) = 2 +1 and P(E and F) = 8 +1 , find +(i) P(E or F), (ii) P(not E and not F). +16. +Events E and F are such that P(not E or not F) = 0.25, State whether E and F are +mutually exclusive. +17. +A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. +Determine (i) P(not A), (ii) P(not B) and +(iii) P(A or B) +18. +In Class XI of a school 40% of the students study Mathematics and 30% study +Biology. 10% of the class study both Mathematics and Biology. If a student is +selected at random from the class, find the probability that he will be studying +Mathematics or Biology. +19. +In an entrance test that is graded on the basis of two examinations, the probability +of a randomly chosen student passing the first examination is 0.8 and the probability +of passing the second examination is 0.7. The probability of passing atleast one of +them is 0.95. What is the probability of passing both? +20. +The probability that a student will pass the final examination in both English and +Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of +passing the English examination is 0.75, what is the probability of passing the +Hindi examination? +Reprint 2025-26 + +308 +MATHEMATICS +21. +In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for +both NCC and NSS. If one of these students is selected at random, find the +probability that +(i) +The student opted for NCC or NSS. +(ii) +The student has opted neither NCC nor NSS. +(iii) +The student has opted NSS but not NCC. +Miscellaneous Examples +Example 9 On her vacations Veena visits four cities (A, B, C and D) in a random +order. What is the probability that she visits +(i) +A before B? +(ii) +A before B and B before C? +(iii) +A first and B last? +(iv) +A either first or second? +(v) +A just before B? +Solution The number of arrangements (orders) in which Veena can visit four cities A, +B, C, or D is 4! i.e., 24.Therefore, n (S) = 24. +Since the number of elements in the sample space of the experiment is 24 all of these +outcomes are considered to be equally likely. A sample space for the +experiment is +S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB +BACD, BADC, BDAC, BDCA, BCAD, BCDA +CABD, CADB, CBDA, CBAD, CDAB, CDBA +DABC, DACB, DBCA, DBAC, DCAB, DCBA} +(i) +Let the event ‘she visits A before B’ be denoted by E +Therefore,E = {ABCD, CABD, DABC, ABDC, CADB, DACB + ACBD, ACDB, ADBC, CDAB, DCAB, ADCB} +Thus +( ) +( ) +( ) +E +12 +1 +P E +S +24 +2 +n +n += += += +(ii) +Let the event ‘Veena visits A before B and B before C’ be denoted by F. +Here +F = {ABCD, DABC, ABDC, ADBC} +Therefore, +( ) +( ) +( ) +F +4 +1 +P F +S +24 +6 +n +n += += += +Students are advised to find the probability in case of (iii), (iv) and (v). +Reprint 2025-26 + + PROBABILITY 309 +Example 10 Find the probability that when a hand of 7 cards is drawn from a well +shuffled deck of 52 cards, it contains (i) all Kings (ii) 3 Kings (iii) atleast 3 Kings. +Solution Total number of possible hands = 52 +7 +C + (i) Number of hands with 4 Kings = 4 +48 +4 +3 +C +C +× + (other 3 cards must be chosen from +the rest 48 cards) +Hence +P (a hand will have 4 Kings) = +4 +48 +4 +3 +52 +7 +C +C +1 +7735 +C +× += + (ii) +Number of hands with 3 Kings and 4 non-King cards = 4 +48 +3 +4 +C +C +× +Therefore +P (3 Kings) = +4 +48 +3 +4 +52 +7 +C +C +9 +1547 +C +× += +(iii) +P(atleast 3 King) += P(3 Kings or 4 Kings) += P(3 Kings) + P(4 Kings) += +9 +1 +46 +1547 +7735 +7735 ++ += +Example 11 If A, B, C are three events associated with a random experiment, +prove that +( +) +P A +B +C +∪ +∪ += ( ) +( ) +( ) +( +) +( +) +P A +P B +P C +P A +B +P A +C ++ +− +∩ +− +∩ +– P ( B ∩ C) + P ( A ∩ B ∩ C) +Solution Consider E = B ∪ C so that +P (A ∪ B ∪ C ) = P (A ∪ E ) + = ( ) +( ) +( +) +P A +P E +P A +E ++ +− +∩ +... (1) +Now +( ) +( +) +P E +P B +C += +∪ + +( ) +( ) +( +) +P B +P C +P B +C += ++ +− +∩ +... (2) +Also +( +) +A +E +A +B +C +∩ += +∩ +∪ + = ( +) +( +) +A +B +A +C +∩ +∪ +∩ +[using distribution property of +intersection of sets over the union]. Thus +( +) +( +) +( +) +P A +E +P A +B +P A +C +∩ += +∩ ++ +∩ +– +( +) +( +) +P +A +B +A +C + + +∩ +∩ +∩ + + +Reprint 2025-26 + +310 +MATHEMATICS + = ( +) +( +) +P A +B +P A +C +∩ ++ +∩ + – [ +] +P A +B +C +∩ +∩ +... (3) +Using (2) and (3) in (1), we get +[ +] +( ) +( ) +( ) +( +) +P A +B +C +P A +P B +P C +P B +C +∪ +∪ += ++ ++ +− +∩ + – ( +) +( +) +( +) +P A +B +P A +C +P A +B +C +∩ +− +∩ ++ +∩ +∩ +Example 12 In a relay race there are five teams A, B, C, D and E. +(a) +What is the probability that A, B and C finish first, second and third, +respectively. +(b) +What is the probability that A, B and C are first three to finish (in any order) +(Assume that all finishing orders are equally likely) +Solution If we consider the sample space consisting of all finishing orders in the first +three places, we will have 5 +3P , i.e., ( +) +5! +5 +3 ! +− + = 5 × 4 × 3 = 60 sample points, each with +a probability of 1 +60 . +(a) +A, B and C finish first, second and third, respectively. There is only one finishing +order for this, i.e., ABC. +Thus P(A, B and C finish first, second and third respectively) = 1 +60 . +(b) +A, B and C are the first three finishers. There will be 3! arrangements for A, B +and C. +Therefore, the sample points corresponding to this event will be 3! in +number. +So +P (A, B and C are first three to finish) +3! +6 +1 +60 +60 +10 += += += +Miscellaneous Exercise on Chapter 14 +1. +A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles +are drawn from the box, what is the probability that +(i) all will be blue? (ii) atleast one will be green? +2. +4 cards are drawn from a well – shuffled deck of 52 cards. What is the probability +of obtaining 3 diamonds and one spade? +Reprint 2025-26 + + PROBABILITY 311 +3. +A die has two faces each with number ‘1’, three faces each with number ‘2’ and +one face with number ‘3’. If die is rolled once, determine +(i) P(2) +(ii) P(1 or 3) +(iii) P(not 3) +4. +In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded. +What is the probability of not getting a prize if you buy (a) one ticket (b) two +tickets (c) 10 tickets. +5. +Out of 100 students, two sections of 40 and 60 are formed. If you and your friend +are among the 100 students, what is the probability that +(a) you both enter the same section? +(b) you both enter the different sections? +6. +Three letters are dictated to three persons and an envelope is addressed to each +of them, the letters are inserted into the envelopes at random so that each envelope +contains exactly one letter. Find the probability that at least one letter is in its +proper envelope. +7. +A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. +Find (i) P(A ∪ B) (ii) P(A´ ∩ B´) (iii) P(A ∩ B´) (iv) P(B ∩ A´) +8. +From the employees of a company, 5 persons are selected to represent them in +the managing committee of the company. Particulars of five persons are as follows: +S. No. +Name +Sex +Age in years +1. +Harish +M +30 +2. +Rohan +M +33 +3. +Sheetal +F +46 +4. +Alis +F +28 +5. +Salim +M +41 +A person is selected at random from this group to act as a spokesperson. What is +the probability that the spokesperson will be either male or over 35 years? +9. +If 4-digit numbers greater than 5,000 are randomly formed from the digits +0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when, +(i) the digits are repeated? (ii) the repetition of digits is not allowed? +10. +The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., +from 0 to 9. The lock opens with a sequence of four digits with no repeats. What +is the probability of a person getting the right sequence to open the suitcase? +Reprint 2025-26 + +312 +MATHEMATICS +Summary +In this Chapter, we studied about the axiomatic approach of probability. The main +features of this Chapter are as follows: +®Event: A subset of the sample space +®Impossible event : The empty set +®Sure event: The whole sample space +®Complementary event or ‘not event’ : The set A′ or S – A +®Event A or B: The set A ∪ B +®Event A and B: The set A ∩ B +®Event A and not B: The set A – B +®Mutually exclusive event: A and B are mutually exclusive if A ∩ B = φ +®Exhaustive and mutually exclusive events: Events E1, E2,..., En are mutually +exclusive and exhaustive if E1 ∪ E2 ∪ ...∪ En = S and Ei ∩ Ej = φ V i ≠ j +®Probability: Number P (ωi) associated with sample point ω i such that +(i) +0 ≤ P (ωi) ≤ 1 +(ii) +( +) +P ωi +∑ + for all ωi ∈ S = 1 +(iii) +P(A) = +( +) +P ωi +∑ +for all ωi ∈A. The number P (ωi) is called probability +of the outcome ωi. +®Equally likely outcomes: All outcomes with equal probability +®Probability of an event: For a finite sample space with equally likely outcomes +Probability of an event +(A) +P(A) +(S) +n +n += +, where n(A) = number of elements in +the set A, n(S) = number of elements in the set S. +®If A and B are any two events, then +P(A or B) = P(A) + P(B) – P(A and B) +equivalently, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) +®If A and B are mutually exclusive, then P(A or B) = P(A) + P(B) +®If A is any event, then +P(not A) = 1 – P(A) +Reprint 2025-26 + + PROBABILITY 313 +— v — +Historical Note +Probability theory like many other branches of mathematics, evolved out of +practical consideration. It had its origin in the 16th century when an Italian +physician and mathematician Jerome Cardan (1501–1576) wrote the first book +on the subject “Book on Games of Chance” (Biber de Ludo Aleae). It was +published in 1663 after his death. +In 1654, a gambler Chevalier de Metre approached the well known French +Philosopher and Mathematician Blaise Pascal (1623–1662) for certain dice +problem. Pascal became interested in these problems and discussed with famous +French Mathematician Pierre de Fermat (1601–1665). Both Pascal and Fermat +solved the problem independently. Besides, Pascal and Fermat, outstanding +contributions to probability theory were also made by Christian Huygenes (1629– +1665), a Dutchman, J. Bernoulli (1654–1705), De Moivre (1667–1754), a +Frenchman Pierre Laplace (1749–1827), the Russian P.L Chebyshev (1821– +1897), A. A Markov (1856–1922) and A. N Kolmogorove (1903–1987). +Kolmogorov is credited with the axiomatic theory of probability. His book +‘Foundations of Probability’ published in 1933, introduces probability as a set +function and is considered a classic. +Reprint 2025-26" +class_12,1,Relations and Functions,ncert_books/class_12/lemh1dd/lemh101.pdf,"vThere is no permanent place in the world for ugly mathematics ... . It may +be very hard to define mathematical beauty but that is just as true of +beauty of any kind, we may not know quite what we mean by a +beautiful poem, but that does not prevent us from recognising +one when we read it. — G. H. HARDY v +1.1 Introduction +Recall that the notion of relations and functions, domain, +co-domain and range have been introduced in Class XI +along with different types of specific real valued functions +and their graphs. The concept of the term ‘relation’ in +mathematics has been drawn from the meaning of relation +in English language, according to which two objects or +quantities are related if there is a recognisable connection +or link between the two objects or quantities. Let A be +the set of students of Class XII of a school and B be the +set of students of Class XI of the same school. Then some +of the examples of relations from A to B are +(i) +{(a, b) ∈A × B: a is brother of b}, +(ii) +{(a, b) ∈A × B: a is sister of b}, +(iii) +{(a, b) ∈A × B: age of a is greater than age of b}, +(iv) +{(a, b) ∈A × B: total marks obtained by a in the final examination is less than +the total marks obtained by b in the final examination}, +(v) +{(a, b) ∈ A × B: a lives in the same locality as b}. However, abstracting from +this, we define mathematically a relation R from A to B as an arbitrary subset +of A × B. +If (a, b) ∈ R, we say that a is related to b under the relation R and we write as +a R b. In general, (a, b) ∈ R, we do not bother whether there is a recognisable +connection or link between a and b. As seen in Class XI, functions are special kind of +relations. +In this chapter, we will study different types of relations and functions, composition +of functions, invertible functions and binary operations. +Chapter 1 +RELATIONS AND FUNCTIONS +Lejeune Dirichlet + (1805-1859) +Reprint 2025-26 + +MATHEMATICS +2 +1.2 Types of Relations +In this section, we would like to study different types of relations. We know that a +relation in a set A is a subset of A × A. Thus, the empty set φ and A × A are two +extreme relations. For illustration, consider a relation R in the set A = {1, 2, 3, 4} given by +R = {(a, b): a – b = 10}. This is the empty set, as no pair (a, b) satisfies the condition +a – b = 10. Similarly, R′ = {(a, b) : | a – b | ≥ 0} is the whole set A × A, as all pairs +(a, b) in A × A satisfy | a – b | ≥ 0. These two extreme examples lead us to the +following definitions. +Definition 1 A relation R in a set A is called empty relation, if no element of A is +related to any element of A, i.e., R = φ ⊂ A × A. +Definition 2 A relation R in a set A is called universal relation, if each element of A +is related to every element of A, i.e., R = A × A. +Both the empty relation and the universal relation are some times called trivial +relations. +Example 1 Let A be the set of all students of a boys school. Show that the relation R +in A given by R = {(a, b) : a is sister of b} is the empty relation and R′ = {(a, b) : the +difference between heights of a and b is less than 3 meters} is the universal relation. +Solution Since the school is boys school, no student of the school can be sister of any +student of the school. Hence, R = φ, showing that R is the empty relation. It is also +obvious that the difference between heights of any two students of the school has to be +less than 3 meters. This shows that R′ = A × A is the universal relation. +Remark In Class XI, we have seen two ways of representing a relation, namely raster +method and set builder method. However, a relation R in the set {1, 2, 3, 4} defined by R += {(a, b) : b = a + 1} is also expressed as a R b if and only if +b = a + 1 by many authors. We may also use this notation, as and when convenient. +If (a, b) ∈ R, we say that a is related to b and we denote it as a R b. +One of the most important relation, which plays a significant role in Mathematics, +is an equivalence relation. To study equivalence relation, we first consider three +types of relations, namely reflexive, symmetric and transitive. +Definition 3 A relation R in a set A is called +(i) +reflexive, if (a, a) ∈ R, for every a ∈ A, +(ii) +symmetric, if (a1, a2) ∈ R implies that (a2, a1) ∈ R, for all a1, a2 ∈ A. +(iii) +transitive, if (a1, a2) ∈ R and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, +a3 ∈ A. +Reprint 2025-26 + +RELATIONS AND FUNCTIONS +3 +Definition 4 A relation R in a set A is said to be an equivalence relation if R is +reflexive, symmetric and transitive. +Example 2 Let T be the set of all triangles in a plane with R a relation in T given by +R = {(T1, T2) : T1 is congruent to T2}. Show that R is an equivalence relation. +Solution R is reflexive, since every triangle is congruent to itself. Further, +(T1, T2) ∈ R ⇒ T1 is congruent to T2 ⇒ T2 is congruent to T1 ⇒ (T2, T1) ∈ R. Hence, +R is symmetric. Moreover, (T1, T2), (T2, T3) ∈ R ⇒ T1 is congruent to T2 and T2 is +congruent to T3 ⇒ T1 is congruent to T3 ⇒ (T1, T3) ∈ R. Therefore, R is an equivalence +relation. +Example 3 Let L be the set of all lines in a plane and R be the relation in L defined as +R = {(L1, L2) : L1 is perpendicular to L2}. Show that R is symmetric but neither +reflexive nor transitive. +Solution R is not reflexive, as a line L1 can not be perpendicular to itself, i.e., (L1, L1) +∉ R. R is symmetric as (L1, L2) ∈ R +⇒ +L1 is perpendicular to L2 +⇒ +L2 is perpendicular to L1 +⇒ +(L2, L1) ∈ R. +R is not transitive. Indeed, if L1 is perpendicular to L2 and +L2 is perpendicular to L3, then L1 can never be perpendicular to +L3. In fact, L1 is parallel to L3, i.e., (L1, L2) ∈ R, (L2, L3) ∈ R but (L1, L3) ∉ R. +Example 4 Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2), +(3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive. +Solution R is reflexive, since (1, 1), (2, 2) and (3, 3) lie in R. Also, R is not symmetric, +as (1, 2) ∈ R but (2, 1) ∉ R. Similarly, R is not transitive, as (1, 2) ∈ R and (2, 3) ∈ R +but (1, 3) ∉ R. +Example 5 Show that the relation R in the set Z of integers given by +R = {(a, b) : 2 divides a – b} +is an equivalence relation. +Solution R is reflexive, as 2 divides (a – a) for all a ∈ Z. Further, if (a, b) ∈ R, then +2 divides a – b. Therefore, 2 divides b – a. Hence, (b, a) ∈ R, which shows that R is +symmetric. Similarly, if (a, b) ∈ R and (b, c) ∈ R, then a – b and b – c are divisible by +2. Now, a – c = (a – b) + (b – c) is even (Why?). So, (a – c) is divisible by 2. This +shows that R is transitive. Thus, R is an equivalence relation in Z. +Fig 1.1 +Reprint 2025-26 + +MATHEMATICS +4 +In Example 5, note that all even integers are related to zero, as (0, ± 2), (0, ± 4) +etc., lie in R and no odd integer is related to 0, as (0, ± 1), (0, ± 3) etc., do not lie in R. +Similarly, all odd integers are related to one and no even integer is related to one. +Therefore, the set E of all even integers and the set O of all odd integers are subsets of +Z satisfying following conditions: +(i) +All elements of E are related to each other and all elements of O are related to +each other. +(ii) +No element of E is related to any element of O and vice-versa. +(iii) +E and O are disjoint and Z = E ∪ O. +The subset E is called the equivalence class containing zero and is denoted by +[0]. Similarly, O is the equivalence class containing 1 and is denoted by [1]. Note that +[0] ≠ [1], [0] = [2r] and [1] = [2r + 1], r ∈ Z. Infact, what we have seen above is true +for an arbitrary equivalence relation R in a set X. Given an arbitrary equivalence +relation R in an arbitrary set X, R divides X into mutually disjoint subsets Ai called +partitions or subdivisions of X satisfying: +(i) +all elements of Ai are related to each other, for all i. +(ii) +no element of Ai is related to any element of Aj , i ≠ j. +(iii) +∪ Aj = X and Ai ∩ Aj = φ, i ≠ j. +The subsets Ai are called equivalence classes. The interesting part of the situation +is that we can go reverse also. For example, consider a subdivision of the set Z given +by three mutually disjoint subsets A1, A2 and A3 whose union is Z with +A1 = {x ∈ Z : x is a multiple of 3} = {..., – 6, – 3, 0, 3, 6, ...} +A2 = {x ∈ Z : x – 1 is a multiple of 3} = {..., – 5, – 2, 1, 4, 7, ...} +A3 = {x ∈ Z : x – 2 is a multiple of 3} = {..., – 4, – 1, 2, 5, 8, ...} +Define a relation R in Z given by R = {(a, b) : 3 divides a – b}. Following the +arguments similar to those used in Example 5, we can show that R is an equivalence +relation. Also, A1 coincides with the set of all integers in Z which are related to zero, A2 +coincides with the set of all integers which are related to 1 and A3 coincides with the +set of all integers in Z which are related to 2. Thus, A1 = [0], A2 = [1] and A3 = [2]. +In fact, A1 = [3r], A2 = [3r + 1] and A3 = [3r + 2], for all r ∈ Z. +Example 6 Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by +R = {(a, b) : both a and b are either odd or even}. Show that R is an equivalence +relation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to each +other and all the elements of the subset {2, 4, 6} are related to each other, but no +element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}. +Reprint 2025-26 + +RELATIONS AND FUNCTIONS +5 +Solution Given any element a in A, both a and a must be either odd or even, so +that (a, a) ∈ R. Further, (a, b) ∈ R ⇒ both a and b must be either odd or even +⇒ (b, a) ∈ R. Similarly, (a, b) ∈ R and (b, c) ∈ R ⇒ all elements a, b, c, must be +either even or odd simultaneously ⇒ (a, c) ∈ R. Hence, R is an equivalence relation. +Further, all the elements of {1, 3, 5, 7} are related to each other, as all the elements +of this subset are odd. Similarly, all the elements of the subset {2, 4, 6} are related to +each other, as all of them are even. Also, no element of the subset {1, 3, 5, 7} can be +related to any element of {2, 4, 6}, as elements of {1, 3, 5, 7} are odd, while elements +of {2, 4, 6} are even. +EXERCISE 1.1 +1. +Determine whether each of the following relations are reflexive, symmetric and +transitive: +(i) Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as +R = {(x, y) : 3x – y = 0} +(ii) Relation R in the set N of natural numbers defined as +R = {(x, y) : y = x + 5 and x < 4} +(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as +R = {(x, y) : y is divisible by x} +(iv) Relation R in the set Z of all integers defined as +R = {(x, y) : x – y is an integer} +(v) Relation R in the set A of human beings in a town at a particular time given by +(a) R = {(x, y) : x and y work at the same place} +(b) R = {(x, y) : x and y live in the same locality} +(c) R = {(x, y) : x is exactly 7 cm taller than y} +(d) R = {(x, y) : x is wife of y} +(e) R = {(x, y) : x is father of y} +2. +Show that the relation R in the set R of real numbers, defined as +R = {(a, b) : a ≤ b2} is neither reflexive nor symmetric nor transitive. +3. +Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as +R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive. +4. +Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and +transitive but not symmetric. +5. +Check whether the relation R in R defined by R = {(a, b) : a ≤ b3} is reflexive, +symmetric or transitive. +Reprint 2025-26 + +MATHEMATICS +6 +6. +Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is +symmetric but neither reflexive nor transitive. +7. +Show that the relation R in the set A of all the books in a library of a college, +given by R = {(x, y) : x and y have same number of pages} is an equivalence +relation. +8. +Show that the relation R in the set A = {1, 2, 3, 4, 5} given by +R = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the +elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are +related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}. +9. +Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by +(i) R = {(a, b) : |a – b| is a multiple of 4} +(ii) R = {(a, b) : a = b} +is an equivalence relation. Find the set of all elements related to 1 in each case. +10. +Give an example of a relation. Which is +(i) Symmetric but neither reflexive nor transitive. +(ii) Transitive but neither reflexive nor symmetric. +(iii) Reflexive and symmetric but not transitive. +(iv) Reflexive and transitive but not symmetric. +(v) Symmetric and transitive but not reflexive. +11. +Show that the relation R in the set A of points in a plane given by +R = {(P, Q) : distance of the point P from the origin is same as the distance of the +point Q from the origin}, is an equivalence relation. Further, show that the set of +all points related to a point P ≠ (0, 0) is the circle passing through P with origin as +centre. +12. +Show that the relation R defined in the set A of all triangles as R = {(T1, T2) : T1 +is similar to T2}, is equivalence relation. Consider three right angle triangles T1 +with sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Which +triangles among T1, T2 and T3 are related? +13. +Show that the relation R defined in the set A of all polygons as R = {(P1, P2) : +P1 and P2 have same number of sides}, is an equivalence relation. What is the +set of all elements in A related to the right angle triangle T with sides 3, 4 and 5? +14. +Let L be the set of all lines in XY plane and R be the relation in L defined as +R = {(L1, L2) : L1 is parallel to L2}. Show that R is an equivalence relation. Find +the set of all lines related to the line y = 2x + 4. +Reprint 2025-26 + +RELATIONS AND FUNCTIONS +7 +15. +Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4,4), +(1, 3), (3, 3), (3, 2)}. Choose the correct answer. +(A) R is reflexive and symmetric but not transitive. +(B) R is reflexive and transitive but not symmetric. +(C) R is symmetric and transitive but not reflexive. +(D) R is an equivalence relation. +16. +Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose +the correct answer. +(A) (2, 4) ∈ R +(B) (3, 8) ∈ R +(C) (6, 8) ∈ R +(D) (8, 7) ∈ R +1.3 Types of Functions +The notion of a function along with some special functions like identity function, constant +function, polynomial function, rational function, modulus function, signum function etc. +along with their graphs have been given in Class XI. +Addition, subtraction, multiplication and division of two functions have also been +studied. As the concept of function is of paramount importance in mathematics and +among other disciplines as well, we would like to extend our study about function from +where we finished earlier. In this section, we would like to study different types of +functions. +Consider the functions f1, f2, f3 and f4 given by the following diagrams. +In Fig 1.2, we observe that the images of distinct elements of X1 under the function +f1 are distinct, but the image of two distinct elements 1 and 2 of X1 under f2 is same, +namely b. Further, there are some elements like e and f in X2 which are not images of +any element of X1 under f1, while all elements of X3 are images of some elements of X1 +under f3. The above observations lead to the following definitions: +Definition 5 A function f : X → Y is defined to be one-one (or injective), if the images +of distinct elements of X under f are distinct, i.e., for every x1, x2 ∈ X, f(x1) = f(x2) +implies x1 = x2. Otherwise, f is called many-one. +The function f1 and f4 in Fig 1.2 (i) and (iv) are one-one and the function f2 and f3 +in Fig 1.2 (ii) and (iii) are many-one. +Definition 6 A function f : X → Y is said to be onto (or surjective), if every element +of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an +element x in X such that f (x) = y. +The function f3 and f4 in Fig 1.2 (iii), (iv) are onto and the function f1 in Fig 1.2 (i) is +not onto as elements e, f in X2 are not the image of any element in X1 under f1. +Reprint 2025-26 + +MATHEMATICS +8 +Remark f : X → Y is onto if and only if Range of f = Y. +Definition 7 A function f : X → Y is said to be one-one and onto (or bijective), if f is +both one-one and onto. +The function f4 in Fig 1.2 (iv) is one-one and onto. +Example 7 Let A be the set of all 50 students of Class X in a school. Let f : A → N be +function defined by f (x) = roll number of the student x. Show that f is one-one +but not onto. +Solution No two different students of the class can have same roll number. Therefore, +f must be one-one. We can assume without any loss of generality that roll numbers of +students are from 1 to 50. This implies that 51 in N is not roll number of any student of +the class, so that 51 can not be image of any element of X under f. Hence, f is not onto. +Example 8 Show that the function f : N → N, given by f(x) = 2x, is one-one but not +onto. +Solution The function f is one-one, for f (x1) = f(x2) ⇒ 2x1 = 2x2 ⇒ x1 = x2. Further, +f is not onto, as for 1 ∈ N, there does not exist any x in N such that f(x) = 2x = 1. +Fig 1.2 (i) to (iv) +Reprint 2025-26 + +RELATIONS AND FUNCTIONS +9 +Example 9 Prove that the function f : R → R, given by f (x) = 2x, is one-one and onto. +Solution f is one-one, as f (x1) = f (x2) ⇒ 2x1 = 2x2 ⇒ x1 = x2. Also, given any real +number y in R, there exists 2 +y in R such that f ( 2 +y ) = 2 . ( 2 +y ) = y. Hence, f is onto. +Fig 1.3 +Example 10 Show that the function f : N → N, given by f (1) = f (2) = 1 and f(x) = x – 1, +for every x > 2, is onto but not one-one. +Solution f is not one-one, as f (1) = f (2) = 1. But f is onto, as given any y ∈ N, y ≠ 1, +we can choose x as y + 1 such that f (y + 1) = y + 1 – 1 = y. Also for 1 ∈ N, we +have f (1) = 1. +Example 11 Show that the function f : R → R, +defined as f (x) = x2, is neither one-one nor onto. +Solution Since f (– 1) = 1 = f (1), f is not one- +one. Also, the element – 2 in the co-domain R is +not image of any element x in the domain R +(Why?). Therefore f is not onto. +Example 12 Show that f : N → N, given by +1,if +is odd, +( ) +1,if +is even +x +x +f x +x +x ++ + += +− + +is both one-one and onto. +Fig 1.4 +Reprint 2025-26 + +MATHEMATICS +10 +Solution Suppose f (x1) = f (x2). Note that if x1 is odd and x2 is even, then we will have +x1 + 1 = x2 – 1, i.e., x2 – x1 = 2 which is impossible. Similarly, the possibility of x1 being +even and x2 being odd can also be ruled out, using the similar argument. Therefore, +both x1 and x2 must be either odd or even. Suppose both x1 and x2 are odd. Then +f (x1) = f (x2) ⇒ x1 + 1 = x2 + 1 ⇒ x1 = x2. Similarly, if both x1 and x2 are even, then also +f (x1) = f (x2) ⇒ x1 – 1 = x2 – 1 ⇒ x1 = x2. Thus, f is one-one. Also, any odd number +2r + 1 in the co-domain N is the image of 2r + 2 in the domain N and any even number +2r in the co-domain N is the image of 2r – 1 in the domain N. Thus, f is onto. +Example 13 Show that an onto function f : {1, 2, 3} → {1, 2, 3} is always one-one. +Solution Suppose f is not one-one. Then there exists two elements, say 1 and 2 in the +domain whose image in the co-domain is same. Also, the image of 3 under f can be +only one element. Therefore, the range set can have at the most two elements of the +co-domain {1, 2, 3}, showing that f is not onto, a contradiction. Hence, f must be one-one. +Example 14 Show that a one-one function f : {1, 2, 3} → {1, 2, 3} must be onto. +Solution Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different +elements of the co-domain {1, 2, 3} under f. Hence, f has to be onto. +Remark The results mentioned in Examples 13 and 14 are also true for an arbitrary +finite set X, i.e., a one-one function f : X → X is necessarily onto and an onto map +f : X → X is necessarily one-one, for every finite set X. In contrast to this, Examples 8 +and 10 show that for an infinite set, this may not be true. In fact, this is a characteristic +difference between a finite and an infinite set. +EXERCISE 1.2 +1. +Show that the function f : R∗ → R∗ defined by f (x) = 1 +x is one-one and onto, +where R∗ is the set of all non-zero real numbers. Is the result true, if the domain +R∗ is replaced by N with co-domain being same as R∗? +2. +Check the injectivity and surjectivity of the following functions: +(i) f : N → N given by f(x) = x2 +(ii) f : Z → Z given by f(x) = x2 +(iii) f : R → R given by f(x) = x2 +(iv) f : N → N given by f(x) = x3 +(v) f : Z → Z given by f(x) = x3 +3. +Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither +one-one nor onto, where [x] denotes the greatest integer less than or equal to x. +Reprint 2025-26 + +RELATIONS AND FUNCTIONS +11 +4. +Show that the Modulus Function f : R → R, given by f (x) = | x|, is neither one- +one nor onto, where | x | is x, if x is positive or 0 and |x | is – x, if x is negative. +5. +Show that the Signum Function f : R → R, given by +f x +x +x +x +( ) +, +, + , += +> += +< + + + +1 +0 +0 +0 +1 +0 +if +if +if +is neither one-one nor onto. +6. +Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function +from A to B. Show that f is one-one. +7. +In each of the following cases, state whether the function is one-one, onto or +bijective. Justify your answer. +(i) f : R → R defined by f (x) = 3 – 4x +(ii) f : R → R defined by f (x) = 1 + x2 +8. +Let A and B be sets. Show that f : A × B → B × A such that f (a, b) = (b, a) is +bijective function. +9. +Let f : N → N be defined by f (n) = +n +n +n +n ++ + + + + + + +1 +2 +2 +, +, +if +is odd +if +is even + for all n ∈ N. +State whether the function f is bijective. Justify your answer. +10. +Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by +f (x) = +2 +3 +x +x +− + + + + +− + +. Is f one-one and onto? Justify your answer. +11. +Let f : R → R be defined as f(x) = x4. Choose the correct answer. +(A) f is one-one onto +(B) f is many-one onto +(C) f is one-one but not onto +(D) f is neither one-one nor onto. +12. +Let f : R → R be defined as f (x) = 3x. Choose the correct answer. +(A) f is one-one onto +(B) f is many-one onto +(C) f is one-one but not onto +(D) f is neither one-one nor onto. +Reprint 2025-26 + +MATHEMATICS +12 +1.4 Composition of Functions and Invertible Function +Definition 8 Let f : A → B and g : B → C be two functions. Then the composition of +f and g, denoted by gof, is defined as the function gof : A → C given by +gof (x) = g(f (x)), ∀ x ∈ A. +Fig 1.5 +Example 15 Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be +functions defined as f (2) = 3, f(3) = 4, f(4) = f(5) = 5 and g (3) = g (4) = 7 and +g (5) = g(9) = 11. Find gof. +Solution We have gof (2) = g (f (2)) = g (3) = 7, gof (3) = g (f (3)) = g (4) = 7, +gof(4) = g (f(4)) = g (5) = 11 and gof(5) = g (5) = 11. +Example 16 Find gof and fog, if f : R → R and g : R → R are given by f(x) = cos x +and g(x) = 3x2. Show that gof ≠ fog. +Solution We have gof(x) = g (f(x)) = g(cos x) = 3 (cos x)2 = 3 cos2 x. Similarly, +fog(x) = f(g(x)) = f (3x2) = cos (3x2). Note that 3cos2 x ≠ cos 3x2, for x = 0. Hence, +gof ≠ fog. +Definition 9 A function f : X → Y is defined to be invertible, if there exists a function +g : Y → X such that gof = IX and fog = IY. The function g is called the inverse of f and +is denoted by f –1. +Thus, if f is invertible, then f must be one-one and onto and conversely, if f is +one-one and onto, then f must be invertible. This fact significantly helps for proving a +function f to be invertible by showing that f is one-one and onto, specially when the +actual inverse of f is not to be determined. +Example 17 Let f : N → Y be a function defined as f (x) = 4x + 3, where, +Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse. +Solution Consider an arbitrary element y of Y. By the definition of Y, y = 4x + 3, +for some x in the domain N. This shows that +( +3) +4 +y +x +− += +. Define g : Y → N by +Reprint 2025-26 + +RELATIONS AND FUNCTIONS +13 +( +3) +( ) +4 +y +g y +− += +. Now, gof(x) = g (f(x)) = g(4x + 3) = (4 +3 +3) +4 +x +x ++ +− += + and +fog(y) = f (g (y)) = f +( +3) +4 ( +3) +3 +4 +4 +y +y +− +− + += ++ + + + + + = y – 3 + 3 = y. This shows that gof = IN +and fog = IY, which implies that f is invertible and g is the inverse of f. +Miscellaneous Examples +Example 18 If R1 + and R2 are equivalence relations in a set A, show that R1 ∩ R2 is +also an equivalence relation. +Solution Since R1 + and R2 are equivalence relations, (a, a) ∈ R1, and (a, a) ∈ R2 ∀a ∈ A. +This implies that (a, a) ∈ R1 ∩ R2, ∀a, showing R1 ∩ R2 is reflexive. Further, +(a, b) ∈ R1 ∩ R2 ⇒ (a, b) ∈ R1 and (a, b) ∈ R2 ⇒ (b, a) ∈ R1 and (b, a) ∈ R2 ⇒ +(b, a) ∈ R1 ∩ R2, hence, R1 ∩ R2 is symmetric. Similarly, (a, b) ∈ R1 ∩ R2 and +(b, c) ∈ R1 ∩ R2 ⇒ (a, c) ∈ R1 and (a, c) ∈ R2 ⇒ (a, c) ∈ R1 ∩ R2. This shows that +R1 ∩ R2 is transitive. Thus, R1 ∩ R2 is an equivalence relation. +Example 19 Let R be a relation on the set A of ordered pairs of positive integers +defined by (x, y) R (u, v) if and only if xv = yu. Show that R is an equivalence relation. +Solution Clearly, (x, y) R (x, y), ∀(x, y) ∈ A, since xy = yx. This shows that R is +reflexive. Further, (x, y) R (u, v) ⇒ xv = yu ⇒ uy = vx and hence (u, v) R (x, y). This +shows that R is symmetric. Similarly, (x, y) R (u, v) and (u, v) R (a, b) ⇒ xv = yu and +ub = va ⇒ +a +a +xv +yu +u +u += +⇒ +b +a +xv +yu +v +u += + ⇒ xb = ya and hence (x, y) R (a, b). Thus, R +is transitive. Thus, R is an equivalence relation. +Example 20 Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R1 be a relation in X given +by R1 = {(x, y) : x – y is divisible by 3} and R2 be another relation on X given by +R2 = {(x, y): {x, y} ⊂ {1, 4, 7}} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9}}. Show that +R1 = R2. +Solution Note that the characteristic of sets {1, 4, 7}, {2, 5, 8} and {3, 6, 9} is +that difference between any two elements of these sets is a multiple of 3. Therefore, +(x, y) ∈ R1 ⇒ x – y is a multiple of 3 ⇒ {x, y} ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} +or {x, y} ⊂ {3, 6, 9} ⇒ (x, y) ∈ R2. Hence, R1 ⊂ R2. Similarly, {x, y} ∈ R2 ⇒ {x, y} +Reprint 2025-26 + +MATHEMATICS +14 +⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} ⇒ x – y is divisible by +3 ⇒ {x, y} ∈ R1. This shows that R2 ⊂ R1. Hence, R1 = R2. +Example 21 Let f : X → Y be a function. Define a relation R in X given by +R = {(a, b): f(a) = f(b)}. Examine whether R is an equivalence relation or not. +Solution For every a ∈ X, (a, a) ∈ R, since f(a) = f (a), showing that R is reflexive. +Similarly, (a, b) ∈ R ⇒ f (a) = f (b) ⇒ f (b) = f (a) ⇒ (b, a) ∈ R. Therefore, R is +symmetric. Further, (a, b) ∈ R and (b, c) ∈ R ⇒ f (a) = f (b) and f(b) = f (c) ⇒ f (a) += f(c) ⇒ (a, c) ∈ R, which implies that R is transitive. Hence, R is an equivalence +relation. +Example 22 Find the number of all one-one functions from set A = {1, 2, 3} to itself. +Solution One-one function from {1, 2, 3} to itself is simply a permutation on three +symbols 1, 2, 3. Therefore, total number of one-one maps from {1, 2, 3} to itself is +same as total number of permutations on three symbols 1, 2, 3 which is 3! = 6. +Example 23 Let A = {1, 2, 3}. Then show that the number of relations containing (1, 2) +and (2, 3) which are reflexive and transitive but not symmetric is three. +Solution The smallest relation R1 containing (1, 2) and (2, 3) which is reflexive and +transitive but not symmetric is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Now, if we add +the pair (2, 1) to R1 to get R2, then the relation R2 will be reflexive, transitive but not +symmetric. Similarly, we can obtain R3 by adding (3, 2) to R1 to get the desired relation. +However, we can not add two pairs (2, 1), (3, 2) or single pair (3, 1) to R1 at a time, as +by doing so, we will be forced to add the remaining pair in order to maintain transitivity +and in the process, the relation will become symmetric also which is not required. Thus, +the total number of desired relations is three. +Example 24 Show that the number of equivalence relation in the set {1, 2, 3} containing +(1, 2) and (2, 1) is two. +Solution The smallest equivalence relation R1 containing (1, 2) and (2, 1) is {(1, 1), +(2, 2), (3, 3), (1, 2), (2, 1)}. Now we are left with only 4 pairs namely (2, 3), (3, 2), +(1, 3) and (3, 1). If we add any one, say (2, 3) to R1, then for symmetry we must add +(3, 2) also and now for transitivity we are forced to add (1, 3) and (3, 1). Thus, the only +equivalence relation bigger than R1 is the universal relation. This shows that the total +number of equivalence relations containing (1, 2) and (2, 1) is two. +Example 25 Consider the identity function IN : N → N defined as IN (x) = x ∀x ∈ N. +Show that although IN is onto but IN + IN : N → N defined as +(IN + IN) (x) = IN (x) + IN (x) = x + x = 2x is not onto. +Reprint 2025-26 + +RELATIONS AND FUNCTIONS +15 +Solution Clearly IN is onto. But IN + IN is not onto, as we can find an element 3 +in the co-domain N such that there does not exist any x in the domain N with +(IN + IN) (x) = 2x = 3. +Example 26 Consider a function f : 0, 2 +π + +→ + + + + +R given by f (x) = sin x and +g : 0, 2 +π + +→ + + + + +R given by g(x) = cos x. Show that f and g are one-one, but f + g is not +one-one. +Solution Since for any two distinct elements x1 and x2 in 0, 2 +π + + + + + + +, sin x1 ≠ sin x2 and +cos x1 ≠ cos x2, both f and g must be one-one. But (f + g) (0) = sin 0 + cos 0 = 1 and +(f + g) +2 +π + + + + + + + = sin +cos +1 +2 +2 +π +π ++ += . Therefore, f + g is not one-one. +Miscellaneous Exercise on Chapter 1 +1. +Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by +( ) +1 | +| +x +f x +x += + +, +x ∈ R is one one and onto function. +2. +Show that the function f : R → R given by f (x) = x3 is injective. +3. +Given a non empty set X, consider P(X) which is the set of all subsets of X. +Define the relation R in P(X) as follows: +For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation +on P(X)? Justify your answer. +4. +Find the number of all onto functions from the set {1, 2, 3,......, n} to itself. +5. +Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be functions defined +by f (x) = x2 – x, x ∈ A and +1 +( ) +2 +1, +2 +g x +x += +− +− + x ∈ A. Are f and g equal? +Justify your answer. (Hint: One may note that two functions f : A → B and +g : A → B such that f (a) = g(a) ∀a ∈ A, are called equal functions). +Reprint 2025-26 + +MATHEMATICS +16 +6. +Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are +reflexive and symmetric but not transitive is +(A) 1 +(B) 2 +(C) 3 +(D) 4 +7. +Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is +(A) 1 +(B) 2 +(C) 3 +(D) 4 +Summary +In this chapter, we studied different types of relations and equivalence relation, +composition of functions, invertible functions and binary operations. The main features +of this chapter are as follows: +® +Empty relation is the relation R in X given by R = φ ⊂ X × X. +® +Universal relation is the relation R in X given by R = X × X. +® +Reflexive relation R in X is a relation with (a, a) ∈ R ∀a ∈ X. +® +Symmetric relation R in X is a relation satisfying (a, b) ∈ R implies (b, a) ∈ R. +® +Transitive relation R in X is a relation satisfying (a, b) ∈ R and (b, c) ∈ R +implies that (a, c) ∈ R. +® +Equivalence relation R in X is a relation which is reflexive, symmetric and +transitive. +® +Equivalence class [a] containing a ∈ X for an equivalence relation R in X is +the subset of X containing all elements b related to a. +® +A function f : X → Y is one-one (or injective) if +f (x1) = f(x2) ⇒ x1 = x2 ∀ x1, x2 ∈ X. +® +A function f : X → Y is onto (or surjective) if given any y ∈ Y, ∃ x ∈ X such +that f (x) = y. +® +A function f : X → Y is one-one and onto (or bijective), if f is both one-one +and onto. +® +Given a finite set X, a function f : X → X is one-one (respectively onto) if and +only if f is onto (respectively one-one). This is the characteristic property of a +finite set. This is not true for infinite set +Reprint 2025-26 + +RELATIONS AND FUNCTIONS +17 +—v— +Historical Note +The concept of function has evolved over a long period of time starting from +R. Descartes (1596-1650), who used the word ‘function’ in his manuscript +“Geometrie” in 1637 to mean some positive integral power xn of a variable x +while studying geometrical curves like hyperbola, parabola and ellipse. James +Gregory (1636-1675) in his work “ Vera Circuli et Hyperbolae Quadratura” +(1667) considered function as a quantity obtained from other quantities by +successive use of algebraic operations or by any other operations. Later G. W. +Leibnitz (1646-1716) in his manuscript “Methodus tangentium inversa, seu de +functionibus” written in 1673 used the word ‘function’ to mean a quantity varying +from point to point on a curve such as the coordinates of a point on the curve, the +slope of the curve, the tangent and the normal to the curve at a point. However, +in his manuscript “Historia” (1714), Leibnitz used the word ‘function’ to mean +quantities that depend on a variable. He was the first to use the phrase ‘function +of x’. John Bernoulli (1667-1748) used the notation φx for the first time in 1718 to +indicate a function of x. But the general adoption of symbols like f, F, φ, ψ ... to +represent functions was made by Leonhard Euler (1707-1783) in 1734 in the first +part of his manuscript “Analysis Infinitorium”. Later on, Joeph Louis Lagrange +(1736-1813) published his manuscripts “Theorie des functions analytiques” in +1793, where he discussed about analytic function and used the notion f (x), F(x), +φ(x) etc. for different function of x. Subsequently, Lejeunne Dirichlet +(1805-1859) gave the definition of function which was being used till the set +theoretic definition of function presently used, was given after set theory was +developed by Georg Cantor (1845-1918). The set theoretic definition of function +known to us presently is simply an abstraction of the definition given by Dirichlet +in a rigorous manner. +Reprint 2025-26" +class_12,2,Inverse Trigonometric Functions,ncert_books/class_12/lemh1dd/lemh102.pdf,"18 +MATHEMATICS +vMathematics, in general, is fundamentally the science of +self-evident things. — FELIX KLEIN v +2.1 Introduction +In Chapter 1, we have studied that the inverse of a function +f, denoted by f –1, exists if f is one-one and onto. There are +many functions which are not one-one, onto or both and +hence we can not talk of their inverses. In Class XI, we +studied that trigonometric functions are not one-one and +onto over their natural domains and ranges and hence their +inverses do not exist. In this chapter, we shall study about +the restrictions on domains and ranges of trigonometric +functions which ensure the existence of their inverses and +observe their behaviour through graphical representations. +Besides, some elementary properties will also be discussed. +The inverse trigonometric functions play an important +role in calculus for they serve to define many integrals. +The concepts of inverse trigonometric functions is also used in science and engineering. +2.2 Basic Concepts +In Class XI, we have studied trigonometric functions, which are defined as follows: +sine function, i.e., sine : R → [– 1, 1] +cosine function, i.e., cos : R → [– 1, 1] +tangent function, i.e., tan : R – { x : x = (2n + 1) 2 +π , n ∈ Z} → R +cotangent function, i.e., cot : R – { x : x = nπ, n ∈ Z} → R +secant function, i.e., sec : R – { x : x = (2n + 1) 2 +π , n ∈ Z} → R – (– 1, 1) +cosecant function, i.e., cosec : R – { x : x = nπ, n ∈ Z} → R – (– 1, 1) +Chapter 2 +INVERSE TRIGONOMETRIC +FUNCTIONS +Aryabhata + (476-550 A. D.) +Reprint 2025-26 + +INVERSE TRIGONOMETRIC FUNCTIONS 19 +We have also learnt in Chapter 1 that if f : X→Y such that f (x) = y is one-one and +onto, then we can define a unique function g : Y→X such that g(y) = x, where x ∈ X +and y = f (x), y ∈ Y. Here, the domain of g = range of f and the range of g = domain +of f. The function g is called the inverse of f and is denoted by f –1. Further, g is also +one-one and onto and inverse of g is f. Thus, g–1 = (f –1)–1 = f. We also have +(f –1 o f ) (x) = f –1 (f (x)) = f –1(y) = x +and +(f o f –1) (y) = f (f –1(y)) = f (x) = y +Since the domain of sine function is the set of all real numbers and range is the +closed interval [–1, 1]. If we restrict its domain to +, +2 +2 +−π π + + + + + + +, then it becomes one-one +and onto with range [– 1, 1]. Actually, sine function restricted to any of the intervals +− + + + + +3 +2 +2 +π +π +,  +, +, +2 +2 +−π π + + + + + + +, +3 +, +2 +2 +π +π + + + + + + +etc., is one-one and its range is [–1, 1]. We can, +therefore, define the inverse of sine function in each of these intervals. We denote the +inverse of sine function by sin–1 (arc sine function). Thus, sin–1 is a function whose +domain is [– 1, 1] and range could be any of the intervals +3 , +2 +2 +−π −π + + + + + + +, +, +2 +2 +−π π + + + + + + + or +3 +, +2 +2 +π +π + + + + + + +, and so on. Corresponding to each such interval, we get a branch of the +function sin–1. The branch with range +, +2 +2 +−π π + + + + + + is called the principal value branch, +whereas other intervals as range give different branches of sin–1. When we refer +to the function sin–1, we take it as the function whose domain is [–1, 1] and range is +, +2 +2 +−π π + + + + + + +. We write sin–1 : [–1, 1] → +, +2 +2 +−π π + + + + + + +From the definition of the inverse functions, it follows that sin (sin–1 x) = x +if – 1 ≤ x ≤ 1 and sin–1 (sin x) = x if +2 +2 +x +π +π +− +≤ +≤ +. In other words, if y = sin–1 x, then +sin y = x. +Remarks +(i) +We know from Chapter 1, that if y = f(x) is an invertible function, then x = f –1 (y). +Thus, the graph of sin–1 function can be obtained from the graph of original +function by interchanging x and y axes, i.e., if (a, b) is a point on the graph of +sine function, then (b, a) becomes the corresponding point on the graph of inverse +Reprint 2025-26 + + 20 +MATHEMATICS +of sine function. Thus, the graph of the function y = sin–1 x can be obtained from +the graph of y = sin x by interchanging x and y axes. The graphs of y = sin x and +y = sin–1 x are as given in Fig 2.1 (i), (ii), (iii). The dark portion of the graph of +y = sin–1 x represent the principal value branch. +(ii) +It can be shown that the graph of an inverse function can be obtained from the +corresponding graph of original function as a mirror image (i.e., reflection) along +the line y = x. This can be visualised by looking the graphs of y = sin x and +y = sin–1 x as given in the same axes (Fig 2.1 (iii)). +Like sine function, the cosine function is a function whose domain is the set of all +real numbers and range is the set [–1, 1]. If we restrict the domain of cosine function +to [0, π], then it becomes one-one and onto with range [–1, 1]. Actually, cosine function +Fig 2.1 (ii) +Fig 2.1 (iii) +Fig 2.1 (i) +Reprint 2025-26 + +INVERSE TRIGONOMETRIC FUNCTIONS 21 +restricted to any of the intervals [– π, 0], [0,π], [π, 2π] etc., is bijective with range as +[–1, 1]. We can, therefore, define the inverse of cosine function in each of these +intervals. We denote the inverse of the cosine function by cos–1 (arc cosine function). +Thus, cos–1 is a function whose domain is [–1, 1] and range +could be any of the intervals [–π, 0], [0, π], [π, 2π] etc. +Corresponding to each such interval, we get a branch of the +function cos–1. The branch with range [0, π] is called the principal +value branch of the function cos–1. We write +cos–1 : [–1, 1] → [0, π]. +The graph of the function given by y = cos–1 x can be drawn +in the same way as discussed about the graph of y = sin–1 x. The +graphs of y = cos x and y = cos–1x are given in Fig 2.2 (i) and (ii). +Fig 2.2 (ii) +Let us now discuss cosec–1x and sec–1x as follows: +Since, cosec x = +1 +sin x , the domain of the cosec function is the set {x : x ∈ R and +x ≠ nπ, n ∈ Z} and the range is the set {y : y ∈ R, y ≥ 1 or y ≤ –1} i.e., the set +R – (–1, 1). It means that y = cosec x assumes all real values except –1 < y < 1 and is +not defined for integral multiple of π. If we restrict the domain of cosec function to +, +2 2 +π π + + +− + + + +– {0}, then it is one to one and onto with its range as the set R – (– 1, 1). Actually, +cosec function restricted to any of the intervals +3 , +{ +} +2 +2 +−π −π + +−−π + + + + +, +, +2 +2 +−π π + + + + + + – {0}, +3 +, +{ } +2 +2 +π +π + +−π + + + + + etc., is bijective and its range is the set of all real numbers R – (–1, 1). +Fig 2.2 (i) +Reprint 2025-26 + + 22 +MATHEMATICS +Thus cosec–1 can be defined as a function whose domain is R – (–1, 1) and range could +be any of the intervals +− +− + + + +−− +3 +2 +2 +π +π +π +, +{ +}, +− + + + +− +π π +2 +2 +0 +, +{ }, +3 +, +{ } +2 +2 +π +π + +−π + + + + +etc. The +function corresponding to the range +, +{0} +2 +2 +−π π + +− + + + + +is called the principal value branch +of cosec–1. We thus have principal branch as +cosec–1 : R – (–1, 1) → +, +{0} +2 +2 +−π π + +− + + + + +The graphs of y = cosec x and y = cosec–1 x are given in Fig 2.3 (i), (ii). +Also, since sec x = +1 +cos x , the domain of y = sec x is the set R – {x : x = (2n + 1) 2 +π , +n ∈ Z} and range is the set R – (–1, 1). It means that sec (secant function) assumes +all real values except –1 < y < 1 and is not defined for odd multiples of 2 +π . If we +restrict the domain of secant function to [0, π] – { 2 +π }, then it is one-one and onto with +Fig 2.3 (i) +Fig 2.3 (ii) +Reprint 2025-26 + +INVERSE TRIGONOMETRIC FUNCTIONS 23 +its range as the set R – (–1, 1). Actually, secant function restricted to any of the +intervals [–π, 0] – { 2 +−π }, [0, ] – +2 +π + +π + + +, [π, 2π] – { 3 +2 +π } etc., is bijective and its range +is R – {–1, 1}. Thus sec–1 can be defined as a function whose domain is R– (–1, 1) and +range could be any of the intervals [– π, 0] – { 2 +−π }, [0, π] – { 2 +π }, [π, 2π] – { 3 +2 +π } etc. +Corresponding to each of these intervals, we get different branches of the function sec–1. +The branch with range [0, π] – { 2 +π } is called the principal value branch of the +function sec–1. We thus have +sec–1 : R – (–1,1) → [0, π] – { 2 +π } +The graphs of the functions y = sec x and y = sec-1 x are given in Fig 2.4 (i), (ii). +Finally, we now discuss tan–1 and cot–1 +We know that the domain of the tan function (tangent function) is the set +{x : x ∈ R and x ≠ (2n +1) 2 +π , n ∈ Z} and the range is R. It means that tan function +is not defined for odd multiples of 2 +π . If we restrict the domain of tangent function to +Fig 2.4 (i) +Fig 2.4 (ii) +Reprint 2025-26 + + 24 +MATHEMATICS +, +2 +2 +−π π + + + + + + +, then it is one-one and onto with its range as R. Actually, tangent function +restricted to any of the intervals +3 , +2 +2 +−π −π + + + + + + +, +, +2 +2 +−π π + + + + + + +, +3 +, +2 +2 +π +π + + + + + + + etc., is bijective +and its range is R. Thus tan–1 can be defined as a function whose domain is R and +range could be any of the intervals +3 , +2 +2 +−π −π + + + + + +, +, +2 +2 +−π π + + + + + +, +3 +, +2 +2 +π +π + + + + + + and so on. These +intervals give different branches of the function tan–1. The branch with range +, +2 +2 +−π π + + + + + + +is called the principal value branch of the function tan–1. +We thus have +tan–1 : R → +, +2 +2 +−π π + + + + + + +The graphs of the function y = tan x and y = tan–1x are given in Fig 2.5 (i), (ii). +Fig 2.5 (i) +Fig 2.5 (ii) +We know that domain of the cot function (cotangent function) is the set +{x : x ∈ R and x ≠ nπ, n ∈ Z} and range is R. It means that cotangent function is not +defined for integral multiples of π. If we restrict the domain of cotangent function to +(0, π), then it is bijective with and its range as R. In fact, cotangent function restricted +to any of the intervals (–π, 0), (0, π), (π, 2π) etc., is bijective and its range is R. Thus +cot –1 can be defined as a function whose domain is the R and range as any of the +Reprint 2025-26 + +INVERSE TRIGONOMETRIC FUNCTIONS 25 +intervals (–π, 0), (0, π), (π, 2π) etc. These intervals give different branches of the +function cot –1. The function with range (0, π) is called the principal value branch of +the function cot –1. We thus have +cot–1 : R → (0, π) +The graphs of y = cot x and y = cot–1x are given in Fig 2.6 (i), (ii). +Fig 2.6 (i) +Fig 2.6 (ii) +The following table gives the inverse trigonometric function (principal value +branches) along with their domains and ranges. +sin–1 +: +[–1, 1] +→ +, +2 2 +π π + + +− + + + + +cos–1 +: +[–1, 1] +→ +[0, π] +cosec–1 +: +R – (–1,1) +→ +, +2 2 +π π + + +− + + + +– {0} +sec–1 +: +R – (–1, 1) +→ +[0, π] – { } +2 +π +tan–1 +: +R +→ +, +2 +2 +−π π + + + + + + +cot–1 +: +R +→ +(0, π) +Reprint 2025-26 + + 26 +MATHEMATICS +ANote +1. sin–1x should not be confused with (sin x)–1. In fact (sin x)–1 = +1 +sin x and +similarly for other trigonometric functions. +2. Whenever no branch of an inverse trigonometric functions is mentioned, we +mean the principal value branch of that function. +3. The value of an inverse trigonometric functions which lies in the range of +principal branch is called the principal value of that inverse trigonometric +functions. +We now consider some examples: +Example 1 Find the principal value of sin–1 +1 +2 + + + + + +. +Solution Let sin–1 +1 +2 + + + + + + += y. Then, sin y = 1 +2 . +We know that the range of the principal value branch of sin–1 is +− + + + + +π π +2 +2 +, + and +sin 4 +π + + + + + + += +1 +2 . Therefore, principal value of sin–1 +1 +2 + + + + + + is 4 +π +Example 2 Find the principal value of cot–1 +1 +3 +− + + + + + + +Solution Let cot–1 +1 +3 +− + + + + + + + = y. Then, +1 +cot +cot 3 +3 +y +− +π + + += += − + + + + = cot +3 +π + + +π − + + + + = +2 +cot +3 +π + + + + + + +We know that the range of principal value branch of cot–1 is (0, π) and +cot 2 +3 +π + + + + + + += +1 +3 +−. Hence, principal value of cot–1 +1 +3 +− + + + + + + is 2 +3 +π +EXERCISE 2.1 +Find the principal values of the following: +1. +sin–1 +1 +2 + + +− + + + + +2. cos–1 +3 +2 + + + + + + + + +3. cosec–1 (2) +4. +tan–1 ( +3) +− +5. cos–1 +1 +2 + + +− + + + + +6. tan–1 (–1) +Reprint 2025-26 + +INVERSE TRIGONOMETRIC FUNCTIONS 27 +7. +sec–1 +2 +3 + + + + + + +8. cot–1 ( 3) +9. cos–1 +1 +2 + + +− + + + + +10. +cosec–1 ( +2 +− +) +Find the values of the following: +11. +tan–1(1) + cos–1 +1 +2 + +− + + sin–1 +1 +2 + +− + +12. cos–1 1 +2 + + + 2 sin–1 1 +2 + + +13. +If sin–1 x = y, then +(A) 0 ≤ y ≤ π +(B) +2 +2 +y +π +π +− +≤ +≤ +(C) 0 < y < π +(D) +2 +2 +y +π +π +− +< +< +14. +tan–1 +( +) +1 +3 +sec +2 +− +− +− + is equal to +(A) π +(B) +3 +π +− +(C) +3 +π +(D) 2 +3 +π +2.3 Properties of Inverse Trigonometric Functions +In this section, we shall prove some important properties of inverse trigonometric +functions. It may be mentioned here that these results are valid within the principal +value branches of the corresponding inverse trigonometric functions and wherever +they are defined. Some results may not be valid for all values of the domains of inverse +trigonometric functions. In fact, they will be valid only for some values of x for which +inverse trigonometric functions are defined. We will not go into the details of these +values of x in the domain as this discussion goes beyond the scope of this textbook. +Let us recall that if y = sin–1x, then x = sin y and if x = sin y, then y = sin–1x. This +is equivalent to +sin (sin–1 x) = x, x ∈ [– 1, 1] and sin–1 (sin x) = x, x ∈ +, +2 +2 +−π π + + + + + + +For suitable values of domain similar results follow for remaining trigonometric +functions. +Reprint 2025-26 + + 28 +MATHEMATICS +We now consider some examples. +Example 3 Show that +(i) +sin–1 ( +) +2 +2 +1 +x +x +− + = 2 sin��1 x, +1 +1 +2 +2 +x +− +≤ +≤ +(ii) +sin–1 ( +) +2 +2 +1 +x +x +− + = 2 cos–1 x, +1 +1 +2 +x +≤ +≤ +Solution +(i) +Let x = sin θ. Then sin–1 x = θ. We have +sin–1 ( +) +2 +2 +1 +x +x +− + = sin–1 ( +) +2 +2sin +1 +sin +θ +− +θ += sin–1 (2sinθ cosθ) = sin–1 (sin2θ) = 2θ += 2 sin–1 x +(ii) +Take x = cos θ, then proceeding as above, we get, sin–1 ( +) +2 +2 +1 +x +x +− += 2 cos–1 x +Example 4 Express +1 +cos +tan +1 +sin +x +x +− + + +− +, +3 +2 +2 +π +π +− +< +< +x + in the simplest form. +Solution We write +2 +2 +1 +–1 +2 +2 +cos +sin +cos +2 +2 +tan +tan +1 +sin +cos +sin +2sin +cos +2 +2 +2 +2 +x +x +x +x +x +x +x +x +− + + +− + + + += + + + + +− + + + + ++ +− + + += +–1 +2 +cos +sin +cos +sin +2 +2 +2 +2 +tan +cos +sin +2 +2 +x +x +x +x +x +x + + + + + ++ +− + + + + + + + + + + + + + + +− + + + + + + + + += +–1 cos +sin +2 +2 +tan +cos +sin +2 +2 +x +x +x +x + + ++ + + + + + + +− + + + +–1 1 +tan 2 +tan +1 +tan 2 +x +x + + ++ + + += + + + + +− + + += +–1 +tan +tan 4 +2 +4 +2 +x +x + + +π +π + + ++ += ++ + + + + + + + + +Reprint 2025-26 + +INVERSE TRIGONOMETRIC FUNCTIONS 29 +Example 5 Write +–1 +2 +1 +cot +1 +x + + + + +− + + +, x > 1 in the simplest form. +Solution Let x = sec θ, then +2 +1 +x −= +2 +sec +1 +tan +θ −= +θ +Therefore, +–1 +2 +1 +cot +1 +x − += cot–1 (cot θ) = θ = sec–1 x, which is the simplest form. +EXERCISE 2.2 +Prove the following: +1. +3sin–1 x = sin–1 (3x – 4x3), +1 +1 +– +, +2 +2 +x + + +∈ + + + +2. +3cos–1 x = cos–1 (4x3 – 3x), +1 , 1 +2 +x + + +∈ + + + +Write the following functions in the simplest form: +3. +2 +1 +1 +1 +tan +x +x +− ++ +−, x ≠ 0 +4. +1 +1 +cos +tan +1 +cos +x +x +− + +− + + + + ++ + + +, 0 < x < π +5. +1 cos +sin +tan +cos +sin +x +x +x +x +− + +− + + ++ + + +, 4 +−π < x < 3 +4 +π +6. +1 +2 +2 +tan +x +a +x +− +− +, |x| < a +7. +2 +3 +1 +3 +2 +3 +tan +3 +a x +x +a +ax +− + +− + + +− + + +, a > 0; +3 +3 +− +< +< +a +a +x +Find the values of each of the following: +8. +–1 +–1 1 +tan +2cos 2sin +2 + + + + + + + + + + + + +9. +2 +–1 +–1 +2 +2 +1 +2 +1 +tan +sin +cos +2 +1 +1 +x +y +x +y + + +− ++ + + ++ ++ + + +, |x | < 1, y > 0 and xy < 1 +Reprint 2025-26 + + 30 +MATHEMATICS +Find the values of each of the expressions in Exercises 16 to 18. +10. +–1 +2 +sin +sin 3 +π + + + + + + +11. +–1 +3 +tan +tan 4 +π + + + + + + +12. +–1 +–1 +3 +3 +tan sin +cot +5 +2 + + ++ + + + + +13. +1 +7 +cos +cos +is equal to +6 +− +π + + + + + + +(A) 7 +6 +π +(B) 5 +6 +π +(C) +3 +π +(D) +6 +π +14. +1 +1 +sin +sin +( +) +3 +2 +− +π + + +− +− + + + + is equal to +(A) 1 +2 +(B) 1 +3 +(C) 1 +4 +(D) 1 +15. +1 +1 +tan +3 +cot +( +3) +− +− +− +− + is equal to +(A) π +(B) +2 +π +− +(C) 0 +(D) 2 3 +Miscellaneous Examples +Example 6 Find the value of +1 +3 +sin +(sin +) +5 +− +π +Solution We know that +1 +sin +(sin ) +x +x +− += +. Therefore, +1 +3 +3 +sin +(sin +) +5 +5 +− +π +π += +But +3 +, +5 +2 2 +π +π π + + +∉− + + + +, which is the principal branch of sin–1 x +However +3 +3 +2 +sin ( +) +sin( +) +sin +5 +5 +5 +π +π +π += +π − += + and 2 +, +5 +2 2 +π +π π + + +∈− + + + + +Therefore +1 +1 +3 +2 +2 +sin +(sin +) +sin +(sin +) +5 +5 +5 +− +− +π +π +π += += +Reprint 2025-26 + +INVERSE TRIGONOMETRIC FUNCTIONS 31 +Miscellaneous Exercise on Chapter 2 +Find the value of the following: +1. +–1 +13 +cos +cos 6 +π + + + + + + +2. +–1 +7 +tan +tan 6 +π + + + + + + +Prove that +3. +–1 +–1 +3 +24 +2sin +tan +5 +7 += +4. +–1 +–1 +–1 +8 +3 +77 +sin +sin +tan +17 +5 +36 ++ += +5. +–1 +–1 +–1 +4 +12 +33 +cos +cos +cos +5 +13 +65 ++ += +6. +–1 +–1 +–1 +12 +3 +56 +cos +sin +sin +13 +5 +65 ++ += +7. +–1 +–1 +–1 +63 +5 +3 +tan +sin +cos +16 +13 +5 += ++ +Prove that +8. +–1 +–1 +1 +1 +tan +cos +2 +1 +x +x +x +− + += + ++ +, x ∈ [0, 1] +9. +–1 +1 +sin +1 +sin +cot +2 +1 +sin +1 +sin +x +x +x +x +x + + ++ ++ +− += + + + + ++ +− +− + + +, +0, 4 +x +π + + +∈ + + + +10. +–1 +–1 +1 +1 +1 +tan +cos +4 +2 +1 +1 +x +x +x +x +x + + ++ +− +− +π += +− + + + + ++ ++ +− + + +, +1 +1 +2 +x +− +≤ +≤ [Hint: Put x = cos 2θ] +Solve the following equations: +11. +2tan–1 (cos x) = tan–1 (2 cosec x) +12. +–1 +–1 +1 +1 +tan +tan +,( +0) +1 +2 +x +x x +x +− += +> ++ +13. +sin (tan–1 x), |x| < 1 is equal to +(A) +2 +1 +x +x +− +(B) +2 +1 +1 +x +− +(C) +2 +1 +1 +x ++ +(D) +2 +1 +x +x ++ +14. +sin–1 (1 – x) – 2 sin–1 x = 2 +π , then x is equal to +(A) 0, 1 +2 +(B) 1, 1 +2 +(C) 0 +(D) 1 +2 +Reprint 2025-26 + + 32 +MATHEMATICS +Summary +® The domains and ranges (principal value branches) of inverse trigonometric +functions are given in the following table: +Functions +Domain +Range +(Principal Value Branches) +y = sin–1 x +[–1, 1] +, +2 +2 +−π π + + + + + + +y = cos–1 x +[–1, 1] + [0, π] +y = cosec–1 x +R – (–1,1) +, +2 +2 +−π π + + + + + + +– {0} +y = sec–1 x +R – (–1, 1) +[0, π] – { } +2 +π +y = tan–1 x +R +, +2 2 +π π + + +− + + + + +y = cot–1 x +R +(0, π) +® sin–1x should not be confused with (sin x)–1. In fact (sin x)–1 = +1 +sin x and +similarly for other trigonometric functions. +® The value of an inverse trigonometric functions which lies in its principal +value branch is called the principal value of that inverse trigonometric +functions. +For suitable values of domain, we have +® y = sin–1 x ⇒ x = sin y +® x = sin y ⇒ y = sin–1 x +® sin (sin–1 x) = x +® sin–1 (sin x) = x +Reprint 2025-26 + +INVERSE TRIGONOMETRIC FUNCTIONS 33 +Historical Note +The study of trigonometry was first started in India. The ancient Indian +Mathematicians, Aryabhata (476A.D.), Brahmagupta (598 A.D.), Bhaskara I +(600 A.D.) and Bhaskara II (1114 A.D.) got important results of trigonometry. All +this knowledge went from India to Arabia and then from there to Europe. The +Greeks had also started the study of trigonometry but their approach was so +clumsy that when the Indian approach became known, it was immediately adopted +throughout the world. +In India, the predecessor of the modern trigonometric functions, known as +the sine of an angle, and the introduction of the sine function represents one of +the main contribution of the siddhantas (Sanskrit astronomical works) to +mathematics. +Bhaskara I (about 600 A.D.) gave formulae to find the values of sine functions +for angles more than 90°. A sixteenth century Malayalam work Yuktibhasa +contains a proof for the expansion of sin (A + B). Exact expression for sines or +cosines of 18°, 36°, 54°, 72°, etc., were given by Bhaskara II. +The symbols sin–1 x, cos–1 x, etc., for arc sin x, arc cos x, etc., were suggested +by the astronomer Sir John F.W. Hersehel (1813) The name of Thales +(about 600 B.C.) is invariably associated with height and distance problems. He +is credited with the determination of the height of a great pyramid in Egypt by +measuring shadows of the pyramid and an auxiliary staff (or gnomon) of known +height, and comparing the ratios: +H +S +h +s += + = tan (sun’s altitude) +Thales is also said to have calculated the distance of a ship at sea through +the proportionality of sides of similar triangles. Problems on height and distance +using the similarity property are also found in ancient Indian works. +—v— +Reprint 2025-26" +class_12,3,Matrices,ncert_books/class_12/lemh1dd/lemh103.pdf,"34 +MATHEMATICS +vThe essence of Mathematics lies in its freedom. — CANTOR v +3.1 Introduction +The knowledge of matrices is necessary in various branches of mathematics. Matrices +are one of the most powerful tools in mathematics. This mathematical tool simplifies +our work to a great extent when compared with other straight forward methods. The +evolution of concept of matrices is the result of an attempt to obtain compact and +simple methods of solving system of linear equations. Matrices are not only used as a +representation of the coefficients in system of linear equations, but utility of matrices +far exceeds that use. Matrix notation and operations are used in electronic spreadsheet +programs for personal computer, which in turn is used in different areas of business +and science like budgeting, sales projection, cost estimation, analysing the results of an +experiment etc. Also, many physical operations such as magnification, rotation and +reflection through a plane can be represented mathematically by matrices. Matrices +are also used in cryptography. This mathematical tool is not only used in certain branches +of sciences, but also in genetics, economics, sociology, modern psychology and industrial +management. +In this chapter, we shall find it interesting to become acquainted with the +fundamentals of matrix and matrix algebra. +3.2 Matrix +Suppose we wish to express the information that Radha has 15 notebooks. We may +express it as [15] with the understanding that the number inside [ ] is the number of +notebooks that Radha has. Now, if we have to express that Radha has 15 notebooks +and 6 pens. We may express it as [15 6] with the understanding that first number +inside [ ] is the number of notebooks while the other one is the number of pens possessed +by Radha. Let us now suppose that we wish to express the information of possession +Chapter 3 +MATRICES +Reprint 2025-26 + +MATRICES 35 +of notebooks and pens by Radha and her two friends Fauzia and Simran which +is as follows: +Radha +has +15 +notebooks +and +6 pens, +Fauzia +has +10 +notebooks +and +2 pens, +Simran +has +13 +notebooks +and +5 pens. +Now this could be arranged in the tabular form as follows: +Notebooks +Pens +Radha +15 +6 +Fauzia +10 +2 +Simran +13 +5 +and this can be expressed as +or +Radha +Fauzia +Simran +Notebooks +15 +10 +13 +Pens +6 +2 +5 +which can be expressed as: +In the first arrangement the entries in the first column represent the number of +note books possessed by Radha, Fauzia and Simran, respectively and the entries in the +second column represent the number of pens possessed by Radha, Fauzia and Simran, +Reprint 2025-26 + + 36 +MATHEMATICS +respectively. Similarly, in the second arrangement, the entries in the first row represent +the number of notebooks possessed by Radha, Fauzia and Simran, respectively. The +entries in the second row represent the number of pens possessed by Radha, Fauzia +and Simran, respectively. An arrangement or display of the above kind is called a +matrix. Formally, we define matrix as: +Definition 1 A matrix is an ordered rectangular array of numbers or functions. The +numbers or functions are called the elements or the entries of the matrix. +We denote matrices by capital letters. The following are some examples of matrices: +5 +– 2 +A +0 +5 +3 +6 + + + + +=  + + + + + +, +1 +2 +3 +2 +B +3.5 +–1 +2 +5 +3 +5 +7 +i + + ++ +− + + + + +=  + + + + + + + +, +3 +1 +3 +C +cos +tan +sin +2 +x +x +x +x +x + + ++ +=  + ++ + + +In the above examples, the horizontal lines of elements are said to constitute, rows +of the matrix and the vertical lines of elements are said to constitute, columns of the +matrix. Thus A has 3 rows and 2 columns, B has 3 rows and 3 columns while C has 2 +rows and 3 columns. +3.2.1 Order of a matrix +A matrix having m rows and n columns is called a matrix of order m × n or simply m × n +matrix (read as an m by n matrix). So referring to the above examples of matrices, we +have A as 3 × 2 matrix, B as 3 × 3 matrix and C as 2 × 3 matrix. We observe that A has +3 × 2 = 6 elements, B and C have 9 and 6 elements, respectively. +In general, an m × n matrix has the following rectangular array: +or +A = [aij]m × n, 1≤ i ≤ m, 1≤ j ≤ n i, j ∈ N +Thus the ith row consists of the elements ai1, ai2, ai3,..., ain, while the jth column +consists of the elements a1j, a2j, a3j,..., amj, +In general aij, is an element lying in the ith row and jth column. We can also call +it as the (i, j)th element of A. The number of elements in an m × n matrix will be +equal to mn. +Reprint 2025-26 + +MATRICES 37 +ANote In this chapter +1. +We shall follow the notation, namely A = [aij]m × n to indicate that A is a matrix +of order m × n. +2. +We shall consider only those matrices whose elements are real numbers or +functions taking real values. +We can also represent any point (x, y) in a plane by a matrix (column or row) as +x +y + + + (or [x, y]). For example point P(0, 1) as a matrix representation may be given as +0 +P +1 + +=  + + or [0 1]. +Observe that in this way we can also express the vertices of a closed rectilinear +figure in the form of a matrix. For example, consider a quadrilateral ABCD with vertices +A (1, 0), B (3, 2), C (1, 3), D (–1, 2). +Now, quadrilateral ABCD in the matrix form, can be represented as +2 +4 +A +B +C D +1 +3 +1 +1 +X +0 2 +3 +2 +× +− + + +=  + + + +or +4 2 +A 1 +0 +B +3 +2 +Y +C +1 +3 +D +1 +2 +× + + + + + + += + + + + +− + +Thus, matrices can be used as representation of vertices of geometrical figures in +a plane. +Now, let us consider some examples. +Example 1 Consider the following information regarding the number of men and women +workers in three factories I, II and III +Men workers +Women workers +I +30 +25 +II +25 +31 +III +27 +26 +Represent the above information in the form of a 3 × 2 matrix. What does the entry +in the third row and second column represent? +Reprint 2025-26 + + 38 +MATHEMATICS +Solution The information is represented in the form of a 3 × 2 matrix as follows: +30 +25 +A +25 +31 +27 +26 + + + + +=  + + + + + +The entry in the third row and second column represents the number of women +workers in factory III. +Example 2 If a matrix has 8 elements, what are the possible orders it can have? +Solution We know that if a matrix is of order m × n, it has mn elements. Thus, to find +all possible orders of a matrix with 8 elements, we will find all ordered pairs of natural +numbers, whose product is 8. +Thus, all possible ordered pairs are (1, 8), (8, 1), (4, 2), (2, 4) +Hence, possible orders are 1 × 8, 8 ×1, 4 × 2, 2 × 4 +Example 3 Construct a 3 × 2 matrix whose elements are given by +1 | +3 | +2 +ij +a +i +j += +− +. +Solution In general a 3 × 2 matrix is given by +11 +12 +21 +22 +31 +32 +A +a +a +a +a +a +a + + + + +=  + + + + + +. +Now +1 | +3 | +2 +ij +a +i +j += +− +, i = 1, 2, 3 and j = 1, 2. +Therefore +11 +1 |1 3 1| +1 +2 +a += +−× += +12 +1 +5 +|1 3 2| +2 +2 +a += +−× += +21 +1 +1 +| 2 +3 1| +2 +2 +a += +−× += +22 +1 | 2 +3 2| +2 +2 +a += +−× += +31 +1 |3 +3 1| +0 +2 +a += +−× += +32 +1 +3 +|3 +3 2 | +2 +2 +a += +−× += +Hence the required matrix is given by +5 +1 +2 +1 +A +2 +2 +3 +0 +2 + + + + + + +=  + + + + + + + +. +Reprint 2025-26 + +MATRICES 39 +3.3 Types of Matrices + In this section, we shall discuss different types of matrices. +(i) +Column matrix +A matrix is said to be a column matrix if it has only one column. +For example, +0 +3 +A +1 +1/ 2 + + + + + + + + += +− + + + + + + + is a column matrix of order 4 × 1. +In general, A = [aij] m × 1 is a column matrix of order m × 1. +(ii) +Row matrix +A matrix is said to be a row matrix if it has only one row. +For example, +1 4 +1 +B +5 2 3 +2 +× + + += − + + + + + is a row matrix. +In general, B = [bij] 1 × n is a row matrix of order 1 × n. +(iii) +Square matrix +A matrix in which the number of rows are equal to the number of columns, is +said to be a square matrix. Thus an m × n matrix is said to be a square matrix if +m = n and is known as a square matrix of order ‘n’. +For example +3 +1 +0 +3 +A +3 2 +1 +2 +4 +3 +1 +− + + + + + + +=  + + + +− + + + is a square matrix of order 3. +In general, A = [aij] m × m is a square matrix of order m. +ANote If A = [aij] is a square matrix of order n, then elements (entries) a11, a22, ..., ann +are said to constitute the diagonal, of the matrix A. Thus, if +1 +3 +1 +A +2 +4 +1 +3 +5 +6 +− + + + + += +− + + + + + + +. +Then the elements of the diagonal of A are 1, 4, 6. +Reprint 2025-26 + + 40 +MATHEMATICS +(iv) +Diagonal matrix +A square matrix B = [bij] m × m is said to be a diagonal matrix if all its non +diagonal elements are zero, that is a matrix B = [bij] m × m is said to be a diagonal +matrix if bij = 0, when i ≠ j. +For example, A = [4], +1 +0 +B +0 +2 +− + + +=  + + + +, +1.1 +0 +0 +C +0 +2 +0 +0 +0 +3 +− + + + + +=  + + + + + +, are diagonal matrices +of order 1, 2, 3, respectively. +(v) +Scalar matrix +A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, +that is, a square matrix B = [bij] n × n is said to be a scalar matrix if +bij = 0, when i ≠ j +bij = k, when i = j, for some constant k. +For example +A = [3], +1 +0 +B +0 +1 +− + + +=  + +− + + +, +3 +0 +0 +C +0 +3 +0 +0 +0 +3 + + + + +=  + + + + + +are scalar matrices of order 1, 2 and 3, respectively. +(vi) +Identity matrix +A square matrix in which elements in the diagonal are all 1 and rest are all zero +is called an identity matrix. In other words, the square matrix A = [aij] n × n is an +identity matrix, if +1 +if +0 +if +ij +i +j +a +i +j += + +=  +≠ + +. +We denote the identity matrix of order n by In. When order is clear from the +context, we simply write it as I. +For example [1], +1 +0 +0 +1 + + + + + +, +1 +0 +0 +0 +1 +0 +0 +0 +1 + + + + + + + + + + + are identity matrices of order 1, 2 and 3, +respectively. +Observe that a scalar matrix is an identity matrix when k = 1. But every identity +matrix is clearly a scalar matrix. +Reprint 2025-26 + +MATRICES 41 +(vii) +Zero matrix +A matrix is said to be zero matrix or null matrix if all its elements are zero. +For example, [0], +0 +0 +0 +0 + + + + + +, +0 +0 +0 +0 +0 +0 + + + + + +, [0, 0] are all zero matrices. We denote +zero matrix by O. Its order will be clear from the context. +3.3.1 Equality of matrices +Definition 2 Two matrices A = [aij] and B = [bij] are said to be equal if +(i) +they are of the same order +(ii) +each element of A is equal to the corresponding element of B, that is aij = bij for +all i and j. +For example, 2 +3 +2 +3 +and +0 +1 +0 +1 + + + + + + + + + + + + + are equal matrices but 3 +2 +2 +3 +and +0 +1 +0 +1 + + + + + + + + + + + + + are +not equal matrices. Symbolically, if two matrices A and B are equal, we write A = B. +If +1.5 +0 +2 +6 +3 +2 +x +y +z +a +b +c +− + + + + + + + +=  + + + + + + + + + + + +, then x = – 1.5, y = 0, z = 2, a = +6 , b = 3, c = 2 +Example 4 If +3 +4 +2 +7 +0 +6 +3 +2 +6 +1 +0 +6 +3 +2 +2 +3 +21 +0 +2 +4 +21 +0 +x +z +y +y +a +c +b +b ++ ++ +− +− + + + + + + + + +− +− += − +− ++ + + + + + + + + +− +− ++ +− + + + + +Find the values of a, b, c, x, y and z. +Solution As the given matrices are equal, therefore, their corresponding elements +must be equal. Comparing the corresponding elements, we get +x + 3 = 0, +z + 4 = 6, +2y – 7 = 3y – 2 +a – 1 = – 3, +0 = 2c + 2 +b – 3 = 2b + 4, +Simplifying, we get +a = – 2, b = – 7, c = – 1, x = – 3, y = –5, z = 2 +Example 5 Find the values of a, b, c, and d from the following equation: +2 +2 +4 +3 +5 +4 +3 +11 +24 +a +b +a +b +c +d +c +d ++ +− +− + + + + += + + + + +− ++ + + + + +Reprint 2025-26 + + 42 +MATHEMATICS +Solution By equality of two matrices, equating the corresponding elements, we get +2a + b = 4 +5c – d = 11 +a – 2b = – 3 +4c + 3d = 24 +Solving these equations, we get +a = 1, b = 2, c = 3 and d = 4 +EXERCISE 3.1 +1. +In the matrix +2 +5 +19 +7 +5 +A +35 +2 +12 +2 +17 +3 +1 +5 + + +− + + + + += +− + + + + +− + + +, write: +(i) The order of the matrix, +(ii) The number of elements, +(iii) Write the elements a13, a21, a33, a24, a23. +2. +If a matrix has 24 elements, what are the possible orders it can have? What, if it +has 13 elements? +3. +If a matrix has 18 elements, what are the possible orders it can have? What, if it +has 5 elements? +4. +Construct a 2 × 2 matrix, A = [aij], whose elements are given by: +(i) +2 +( +) +2 +ij +i +j +a ++ += +(ii) +ij +i +a +j += +(iii) +2 +( +2 ) +2 +ij +i +j +a ++ += +5. +Construct a 3 × 4 matrix, whose elements are given by: +(i) +1 | 3 +| +2 +ij +a +i +j += +− ++ +(ii) +2 +ij +a +i +j += +− +6. +Find the values of x, y and z from the following equations: +(i) +4 +3 +5 +1 +5 +y +z +x + + + + += + + + + + + + + +(ii) +2 +6 +2 +5 +5 +8 +x +y +z +xy ++ + + + + += + + + + ++ + + + + +(iii) +9 +5 +7 +x +y +z +x +z +y +z ++ ++ + + + + + + ++ += + + + + + + ++ + + + +7. +Find the value of a, b, c and d from the equation: +2 +1 +5 +2 +3 +0 +13 +a +b +a +c +a +b +c +d +− ++ +− + + + + += + + + + +− ++ + + + + +Reprint 2025-26 + +MATRICES 43 +8. +A = [aij]m × n\ is a square matrix, if +(A) m < n +(B) m > n +(C) m = n +(D) None of these +9. +Which of the given values of x and y make the following pair of matrices equal +3 +7 +5 +1 +2 +3 +x +y +x ++ + + + + ++ +− + + , +0 +2 +8 +4 +y − + + + + + + +(A) +1, +7 +3 +x +y +− += += +(B) Not possible to find +(C) y = 7, +2 +3 +x +− += +(D) +1 +2 +, +3 +3 +x +y +− +− += += +10. +The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is: +(A) 27 +(B) 18 +(C) 81 +(D) 512 +3.4 Operations on Matrices +In this section, we shall introduce certain operations on matrices, namely, addition of +matrices, multiplication of a matrix by a scalar, difference and multiplication of matrices. +3.4.1 Addition of matrices +Suppose Fatima has two factories at places A and B. Each factory produces sport +shoes for boys and girls in three different price categories labelled 1, 2 and 3. The +quantities produced by each factory are represented as matrices given below: +Suppose Fatima wants to know the total production of sport shoes in each price +category. Then the total production +In category 1 : for boys (80 + 90), for girls (60 + 50) +In category 2 : for boys (75 + 70), for girls (65 + 55) +In category 3 : for boys (90 + 75), for girls (85 + 75) +This can be represented in the matrix form as +80 +90 +60 +50 +75 +70 +65 +55 +90 +75 +85 +75 ++ ++ + + + + ++ ++ + + + + ++ ++ + + +. +Reprint 2025-26 + + 44 +MATHEMATICS +This new matrix is the sum of the above two matrices. We observe that the sum of +two matrices is a matrix obtained by adding the corresponding elements of the given +matrices. Furthermore, the two matrices have to be of the same order. +Thus, if +11 +12 +13 +21 +22 +23 +A +a +a +a +a +a +a + + +=  + + + + is a 2 × 3 matrix and +11 +12 +13 +21 +22 +23 +B +b +b +b +b +b +b + + +=  + + + + is another +2×3 matrix. Then, we define +11 +11 +12 +12 +13 +13 +21 +21 +22 +22 +23 +23 +A + B +a +b +a +b +a +b +a +b +a +b +a +b ++ ++ ++ + + +=  + ++ ++ ++ + + +. +In general, if A = [aij] and B = [bij] are two matrices of the same order, say m × n. +Then, the sum of the two matrices A and B is defined as a matrix C = [cij]m × n, where +cij = aij + bij, for all possible values of i and j. +Example 6 Given +3 +1 +1 +A +2 +3 +0 + + +− +=  + + + + and +2 +5 +1 +B +1 +2 +3 +2 + + + + +=  + +− + + + + +, find A + B +Since A, B are of the same order 2 × 3. Therefore, addition of A and B is defined +and is given by +2 +3 +1 +5 +1 1 +2 +3 +1 +5 +0 +A+B +1 +1 +2 +2 +3 +3 +0 +0 +6 +2 +2 + + + + ++ ++ +− ++ ++ + + + + += += + + + + +− ++ ++ + + + + + + + + +ANote +1. We emphasise that if A and B are not of the same order, then A + B is not +defined. For example if +2 +3 +A +1 +0 + + +=  + + + +, +1 +2 +3 +B +, +1 +0 +1 + + +=  + + + + then A + B is not defined. +2. We may observe that addition of matrices is an example of binary operation +on the set of matrices of the same order. +3.4.2 Multiplication of a matrix by a scalar +Now suppose that Fatima has doubled the production at a factory A in all categories +(refer to 3.4.1). +Reprint 2025-26 + +MATRICES 45 +Previously quantities (in standard units) produced by factory A were +Revised quantities produced by factory A are as given below: +Boys +Girls +2 80 +2 +60 +1 +2 2 +75 +2 +65 +3 2 90 +2 85 +× +× + + + + +× +× + + + + +× +× + + +This can be represented in the matrix form as +160 +120 +150 +130 +180 +170 + + + + + + + + + + +. We observe that +the new matrix is obtained by multiplying each element of the previous matrix by 2. +In general, we may define multiplication of a matrix by a scalar as follows: if +A = [aij] m × n is a matrix and k is a scalar, then kA is another matrix which is obtained +by multiplying each element of A by the scalar k. +In other words, kA = k[aij] m × n = [k (aij)] m × n, that is, (i, j)th element of kA is kaij +for all possible values of i and j. +For example, if +A = +3 +1 +1.5 +5 +7 +3 +2 +0 +5 + + + + +− + + + + + + +, then +3A = +3 +1 +1.5 +9 +3 +4.5 +3 +5 +7 +3 +3 5 +21 +9 +2 +0 +5 +6 +0 +15 + + + + + + + + +− += +− + + + + + + + + + + + + +Negative of a matrix The negative of a matrix is denoted by – A. We define +–A = (– 1) A. +Reprint 2025-26 + + 46 +MATHEMATICS +For example, let +A = +3 +1 +5 +x + + + + +− + + +, then – A is given by +– A = (– 1) +3 +1 +3 +1 +A +( 1) +5 +5 +x +x +− +− + + + + += − += + + + + +− +− + + + + +Difference of matrices If A = [aij], B = [bij] are two matrices of the same order, +say m × n, then difference A – B is defined as a matrix D = [dij], where dij = aij – bij, +for all value of i and j. In other words, D = A – B = A + (–1) B, that is sum of the matrix +A and the matrix – B. +Example 7 If +1 +2 +3 +3 +1 +3 +A +and B +2 +3 +1 +1 +0 +2 +− + + + + += += + + + + +− + + + + +, then find 2A – B. +Solution We have +2A – B = 2 1 +2 +3 +2 +3 1 +3 +1 +3 +1 +0 +2 + + + +− +− +− + + + + += +2 +4 +6 +3 1 +3 +4 +6 +2 +1 +0 +2 +− +− + + + + ++ + + + + +− + + + + += +2 +3 +4 +1 +6 +3 +1 +5 +3 +4 +1 +6 +0 +2 +2 +5 +6 +0 +− ++ +− +− + + + + += + + + + ++ ++ +− + + + + +3.4.3 Properties of matrix addition +The addition of matrices satisfy the following properties: +(i) +Commutative Law If A = [aij], B = [bij] are matrices of the same order, say +m × n, then A + B = B + A. +Now +A + B = [aij] + [bij] = [aij + bij] += [bij + aij] (addition of numbers is commutative) += ([bij] + [aij]) = B + A +(ii) +Associative Law For any three matrices A = [aij], B = [bij], C = [cij] of the +same order, say m × n, (A + B) + C = A + (B + C). +Now +(A + B) + C = ([aij] + [bij]) + [cij] += [aij + bij] + [cij] = [(aij + bij) + cij] += [aij + (bij + cij)] +(Why?) += [aij] + [(bij + cij)] = [aij] + ([bij] + [cij]) = A + (B + C) +Reprint 2025-26 + +MATRICES 47 +(iii) +Existence of additive identity Let A = [aij] be an m × n matrix and +O be an m × n zero matrix, then A + O = O + A = A. In other words, O is the +additive identity for matrix addition. +(iv) +The existence of additive inverse Let A = [aij]m × n be any matrix, then we +have another matrix as – A = [– aij]m × n such that A + (– A) = (– A) + A= O. So +– A is the additive inverse of A or negative of A. +3.4.4 Properties of scalar multiplication of a matrix +If A = [aij] and B = [bij] be two matrices of the same order, say m × n, and k and l are +scalars, then +(i) +k(A +B) = k A + kB, (ii) (k + l)A = k A + l A +(ii) +k (A + B) = k ([aij] + [bij]) += k [aij + bij] = [k (aij + bij)] = [(k aij) + (k bij)] += [k aij] + [k bij] = k [aij] + k [bij] = kA + kB +(iii) +( k + l) A = (k + l) [aij] += [(k + l) aij] + [k aij] + [l aij] = k [aij] + l [aij] = k A + l A +Example 8 If +8 +0 +2 +2 +A +4 +2 and B +4 +2 +3 +6 +5 1 +− + + + + + + + + += +− += + + + + + + + + +− + + + + +, then find the matrix X, such that +2A + 3X = 5B. +Solution We have 2A + 3X = 5B +or +2A + 3X – 2A = 5B – 2A +or +2A – 2A + 3X = 5B – 2A +(Matrix addition is commutative) +or +O + 3X = 5B – 2A +(– 2A is the additive inverse of 2A) +or +3X = 5B – 2A +(O is the additive identity) +or +X = 1 +3 (5B – 2A) +or +2 +2 +8 0 +1 +X +5 +4 +2 +2 4 +2 +3 +5 1 +3 6 + + +− + + + + + + + + + + += +− +− + + + + + + + + + + + + +− + + + + + + + = +10 +10 +16 +0 +1 +20 +10 +8 +4 +3 +25 +5 +6 +12 + + +− +− + + + + + + + + + + ++ +− + + + + + + + + + + + + +− +− +− + + + + + + +Reprint 2025-26 + + 48 +MATHEMATICS + = +10 +16 +10 +0 +1 +20 +8 +10 +4 +3 +25 +6 +5 12 +− +− ++ + + + + +− ++ + + + + +− +− +− + + + = +6 +10 +1 +12 +14 +3 +31 +7 +− +− + + + + + + + + +− +− + + + = +10 +2 +3 +14 +4 +3 +31 +7 +3 +3 +− + + +− + + + + + + + + + + +− +− + + + + + + +Example 9 Find X and Y, if +5 +2 +X +Y +0 +9 + + ++ +=  + + + + and +3 +6 +X +Y +0 +1 + + +− +=  + +− + + +. +Solution We have ( +) +( +) +5 +2 +3 +6 +X +Y +X +Y +0 +9 +0 +1 + + + + ++ ++ +− += ++ + + + + +− + + + + +. +or +(X + X) + (Y – Y) = +8 +8 +0 +8 + + + + + + + ⇒ +8 +8 +2X +0 +8 + + +=  + + + +or +X = +8 +8 +4 +4 +1 +0 +8 +0 +4 +2 + + + + += + + + + + + + + +Also +(X + Y) – (X – Y) = +5 +2 +3 +6 +0 +9 +0 +1 + + + + +− + + + + +− + + + + +or +(X – X) + (Y + Y) = +5 +3 +2 +6 +0 +9 +1 +− +− + + + + ++ + + + ⇒ +2 +4 +2Y +0 +10 +− + + +=  + + + +or +Y = +2 +4 +1 +2 +1 +0 +10 +0 +5 +2 +− +− + + + + += + + + + + + + + +Example 10 Find the values of x and y from the following equation: +5 +3 +4 +2 7 +3 +1 +2 +x +y +− + + + + ++ + + + + +− + + + + = +7 +6 +15 +14 + + + + + + +Solution We have + +5 +3 +4 +2 7 +3 +1 +2 +x +y +− + + + + ++ + + + + +− + + + + = +7 +6 +15 +14 + + + + + + + ⇒ 2 +10 +3 +4 +7 +6 +14 +2 +6 +1 +2 +15 +14 +x +y +− + + + + + + ++ += + + + + + + +− + + + + + + +Reprint 2025-26 + +MATRICES 49 +or +2 +3 +10 +4 +14 +1 +2 +6 +2 +x +y ++ +− + + + + ++ +− ++ + + = +7 +6 +15 14 + + + + + + + ⇒ 2 +3 +6 +7 +6 +15 +2 +4 +15 +14 +x +y ++ + + + + += + + + + +− + + + + +or +2x + 3 = 7 +and +2y – 4 = 14 +(Why?) +or +2x = 7 – 3 +and +2y = 18 +or +x = 4 +2 +and +y = 18 +2 +i.e. +x = 2 +and +y = 9. +Example 11 Two farmers Ramkishan and Gurcharan Singh cultivates only three +varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these +varieties of rice by both the farmers in the month of September and October are given +by the following matrices A and B. +(i) +Find the combined sales in September and October for each farmer in each +variety. +(ii) +Find the decrease in sales from September to October. +(iii) +If both farmers receive 2% profit on gross sales, compute the profit for each +farmer and for each variety sold in October. +Solution +(i) +Combined sales in September and October for each farmer in each variety is +given by +Reprint 2025-26 + + 50 +MATHEMATICS +(ii) +Change in sales from September to October is given by +(iii) +2% of B = 2 +B +100 × += 0.02 × B += 0.02 += +Thus, in October Ramkishan receives ` 100, ` 200 and ` 120 as profit in the +sale of each variety of rice, respectively, and Grucharan Singh receives profit of ` 400, +` 200 and ` 200 in the sale of each variety of rice, respectively. +3.4.5 Multiplication of matrices +Suppose Meera and Nadeem are two friends. Meera wants to buy 2 pens and 5 story +books, while Nadeem needs 8 pens and 10 story books. They both go to a shop to +enquire about the rates which are quoted as follows: +Pen – ` 5 each, story book – ` 50 each. +How much money does each need to spend? Clearly, Meera needs ` (5 × 2 + 50 × 5) +that is ` 260, while Nadeem needs (8 × 5 + 50 × 10) `, that is ` 540. In terms of matrix +representation, we can write the above information as follows: +Requirements +Prices per piece (in Rupees) Money needed (in Rupees) +2 +5 +8 +10 + + + + + + +5 +50 + + + + + + +5 +2 +5 50 +260 +8 5 +10 +50 +540 +× ++ +× + + + + += + + + + +× ++ +× + + + + +Suppose that they enquire about the rates from another shop, quoted as follows: +pen – ` 4 each, story book – ` 40 each. +Now, the money required by Meera and Nadeem to make purchases will be +respectively ` (4 × 2 + 40 × 5) = ` 208 and ` (8 × 4 + 10 × 40) = ` 432 +Reprint 2025-26 + +MATRICES 51 +Again, the above information can be represented as follows: +Requirements +Prices per piece (in Rupees) Money needed (in Rupees) +2 +5 +8 +10 + + + + + + +4 +40 + + + + + + +4 +2 +40 5 +208 +8 4 +10 4 0 +432 +× ++ +× + + + + += + + + + +× ++ +× + + + + +Now, the information in both the cases can be combined and expressed in terms of +matrices as follows: +Requirements +Prices per piece (in Rupees) +Money needed (in Rupees) +2 +5 +8 +10 + + + + + + +5 +4 +50 +40 + + + + + + +5 +2 +5 50 +4 +2 +40 5 +8 5 +10 5 0 8 4 +10 4 0 +× ++ × +× ++ +× + + + + +× ++ +× +× ++ +× + + += +260 +208 +540 +432 + + + + + + +The above is an example of multiplication of matrices. We observe that, for +multiplication of two matrices A and B, the number of columns in A should be equal to +the number of rows in B. Furthermore for getting the elements of the product matrix, +we take rows of A and columns of B, multiply them element-wise and take the sum. +Formally, we define multiplication of matrices as follows: +The product of two matrices A and B is defined if the number of columns of A is +equal to the number of rows of B. Let A = [aij] be an m × n matrix and B = [bjk] be an +n × p matrix. Then the product of the matrices A and B is the matrix C of order m × p. +To get the (i, k)th element cik of the matrix C, we take the ith row of A and kth column +of B, multiply them elementwise and take the sum of all these products. In other words, +if A = [aij]m × n, B = [bjk]n × p, then the ith row of A is [ai1 ai2 ... ain] and the kth column of +B is +1 +2 +. +. +. +k +k +nk +b +b +b + + + + + + + + + + + + + + + , then cik = ai1 b1k + ai2 b2k + ai3 b3k + ... + ain bnk = +1 +n +ij +jk +j +a b +=∑ +. +The matrix C = [cik]m × p is the product of A and B. +For example, if +1 +1 +2 +C +0 +3 4 +− + + +=  + + + + and +2 +7 +1 +D +1 +5 +4 + + + + += − + + + + +− + + + , then the product CD is defined +Reprint 2025-26 + + 52 +MATHEMATICS +and is given by +2 +7 +1 +1 +2 +CD +1 +1 +0 +3 +4 +5 +4 + + +− + + + += +− + + + + + + +− + + +. This is a 2 × 2 matrix in which each +entry is the sum of the products across some row of C with the corresponding entries +down some column of D. These four computations are +Thus +13 +2 +CD +17 +13 +− + + +=  + +− + + +Example 12 Find AB, if +6 +9 +2 +6 +0 +A +and B +2 +3 +7 +9 +8 + + + + += += + + + + + + + + +. +Solution The matrix A has 2 columns which is equal to the number of rows of B. +Hence AB is defined. Now +6(2) +9(7) +6(6) +9(9) +6(0) +9(8) +AB +2(2) +3(7) +2(6) +3(9) +2(0) +3(8) ++ ++ ++ + + +=  + ++ ++ ++ + + += +12 +63 +36 +81 +0 +72 +4 +21 +12 +27 +0 +24 ++ ++ ++ + + + + ++ ++ ++ + + + = +75 +117 +72 +25 +39 +24 + + + + + + +Reprint 2025-26 + +MATRICES 53 +Remark If AB is defined, then BA need not be defined. In the above example, AB is +defined but BA is not defined because B has 3 column while A has only 2 (and not 3) +rows. If A, B are, respectively m × n, k × l matrices, then both AB and BA are defined +if and only if n = k and l = m. In particular, if both A and B are square matrices of the +same order, then both AB and BA are defined. +Non-commutativity of multiplication of matrices +Now, we shall see by an example that even if AB and BA are both defined, it is not +necessary that AB = BA. +Example 13 If +2 3 +1 +2 +3 +A +and B +4 5 +4 +2 +5 +2 1 + + +− + + + + += += + + + + +− + + + + + + +, then find AB, BA. Show that +AB ≠ BA. +Solution Since A is a 2 × 3 matrix and B is 3 × 2 matrix. Hence AB and BA are both +defined and are matrices of order 2 × 2 and 3 × 3, respectively. Note that +2 3 +1 +2 +3 +AB +4 5 +4 +2 +5 +2 1 + + +− + + + +=  + + +− + + + + + = +2 +8 +6 +3 10 +3 +0 +4 +8 +8 +10 +12 +10 +5 +10 +3 +− ++ +− ++ +− + + + + += + + + + +−+ ++ +− ++ ++ + + + + +and +2 3 +2 12 +4 +6 +6 +15 +1 +2 +3 +BA +4 5 +4 +20 +8 10 +12 +25 +4 +2 +5 +2 1 +2 +4 +4 +2 +6 +5 +− +−+ ++ + + + + +− + + + + + + += += +− +−+ ++ + + + + + + +− + + + + + +− +−+ ++ + + + + + +10 +2 +21 +16 +2 +37 +2 +2 +11 +− + + + + += − + + + + +− +− + + +Clearly AB ≠ BA +In the above example both AB and BA are of different order and so AB ≠ BA. But +one may think that perhaps AB and BA could be the same if they were of the same +order. But it is not so, here we give an example to show that even if AB and BA are of +same order they may not be same. +Example 14 If +1 +0 +A +0 +1 + + +=  + +− + + + and +0 +1 +B +1 +0 + + +=  + + + +, then +0 +1 +AB +1 +0 + + +=  + +− + + +. +and +0 +1 +BA +1 +0 +− + + +=  + + + +. Clearly AB ≠ BA. +Thus matrix multiplication is not commutative. +Reprint 2025-26 + + 54 +MATHEMATICS +ANote This does not mean that AB ≠ BA for every pair of matrices A, B for +which AB and BA, are defined. For instance, +If +1 +0 +3 +0 +A +, B +0 +2 +0 +4 + + + + += += + + + + + + + + +, then AB = BA = +3 +0 +0 +8 + + + + + + +Observe that multiplication of diagonal matrices of same order will be commutative. +Zero matrix as the product of two non zero matrices +We know that, for real numbers a, b if ab = 0, then either a = 0 or b = 0. This need +not be true for matrices, we will observe this through an example. +Example 15 Find AB, if +0 +1 +A +0 +2 +− + + +=  + + + + and +3 +5 +B +0 +0 + + +=  + + + +. +Solution We have +0 +1 +3 +5 +0 +0 +AB +0 +2 +0 +0 +0 +0 +− + + + + + += += + + + + + + + + + + +. +Thus, if the product of two matrices is a zero matrix, it is not necessary that one of +the matrices is a zero matrix. +3.4.6 Properties of multiplication of matrices +The multiplication of matrices possesses the following properties, which we state without +proof. +1. +The associative law For any three matrices A, B and C. We have +(AB) C = A (BC), whenever both sides of the equality are defined. +2. +The distributive law For three matrices A, B and C. +(i) A (B+C) = AB + AC +(ii) (A+B) C = AC + BC, whenever both sides of equality are defined. +3. +The existence of multiplicative identity For every square matrix A, there +exist an identity matrix of same order such that IA = AI = A. +Now, we shall verify these properties by examples. +Example 16 If +1 +1 +1 +1 3 +1 +2 +3 +4 +A +2 +0 +3 , B +0 2 +and C +2 +0 +2 +1 +3 +1 +2 +1 4 +− + + + + +− + + + + + + += += +=  + + + + + +− + + + + + + +− +− + + + + +, find +A(BC), (AB)C and show that (AB)C = A(BC). +Reprint 2025-26 + +MATRICES 55 +Solution We have +1 +1 +1 +1 3 +1 +0 1 3 +2 +4 +2 +1 +AB +2 +0 +3 +0 2 +2 +0 +3 6 +0 12 +1 18 +3 +1 +2 +1 4 +3 +0 +2 9 +2 +8 +1 15 +− ++ ++ ++ +− + + + + + + + + + + + + + + += += ++ +− ++ ++ += − + + + + + + + + + + + + + + +− +− ++ +− +− ++ + + + + + + + +(AB) (C) +2 +2 +4 +0 +6 +2 +8 +1 +2 +1 +1 +2 +3 +4 +1 18 +1 +36 +2 +0 +3 +36 +4 +18 +2 +0 +2 +1 +1 15 +1 +30 +2 +0 +3 +30 +4 +15 ++ ++ +− +− ++ + + + + +− + + + + + + += − += −+ +−+ +−− ++ + + + + + + +− + + + + + + ++ ++ +− +− ++ + + + + += +4 +4 +4 +7 +35 +2 +39 +22 +31 +2 +27 +11 +− + + + + +− +− + + + + +− + + +Now +BC = +1 +6 +2 +0 +3 +6 +4 +3 +1 3 +1 +2 +3 +4 +0 2 +0 +4 +0 +0 +0 +4 +0 +2 +2 +0 +2 +1 +1 4 +1 +8 +2 +0 +3 +8 +4 +4 ++ ++ +− +−+ + + + + +− + + + + + + += ++ ++ +− ++ + + + + + + +− + + + + + + +− +−+ +−+ +−− ++ + + + + += +7 +2 +3 +1 +4 +0 +4 +2 +7 +2 +11 +8 +− +− + + + + +− + + + + +− +− + + +Therefore +A(BC) = +7 +2 +3 +1 +1 +1 +1 +2 +0 +3 +4 +0 +4 +2 +3 +1 +2 +7 +2 +11 +8 +− +− +− + + + + + + + + +− + + + + + + + + +− +− +− + + + += +7 +4 +7 +2 +0 +2 +3 +4 +11 +1 +2 +8 +14 +0 +21 +4 +0 +6 +6 +0 +33 +2 +0 +24 +21 +4 +14 +6 +0 +4 +9 +4 +22 +3 +2 +16 ++ +− ++ ++ +−− ++ +−+ +− + + + + ++ ++ ++ +− +−+ +− +−+ ++ + + + + +− ++ ++ +− +−+ +− +−− ++ + + += +4 +4 +4 +7 +35 +2 +39 +22 +31 +2 +27 +11 +− + + + + +− +− + + + + +− + + +. Clearly, (AB) C = A (BC) +Reprint 2025-26 + + 56 +MATHEMATICS +Example 17 If +0 +6 +7 +0 +1 +1 +2 +A +6 +0 +8 , B +1 +0 +2 , C +2 +7 +8 +0 +1 +2 +0 +3 + + + + + + + + + + + + += − += += − + + + + + + + + + + + + +− + + + + + + +Calculate AC, BC and (A + B)C. Also, verify that (A + B)C = AC + BC +Solution Now, +0 +7 +8 +A+B +5 +0 +10 +8 +6 +0 + + + + += − + + + + +− + + +So +(A + B) C = +0 +7 +8 +2 +0 +14 +24 +10 +5 +0 +10 +2 +10 +0 +30 +20 +8 +6 +0 +3 +16 +12 +0 +28 +− ++ + + + + + + + + + + + + + + +− +− += − ++ ++ += + + + + + + + + + + + + + + +− ++ ++ + + + + + + + +Further +AC = +0 +6 +7 +2 +0 12 +21 +9 +6 +0 +8 +2 +12 +0 +24 +12 +7 +8 +0 +3 +14 +16 +0 +30 +− ++ + + + + + + + + + + + + + + +− +− += − ++ ++ += + + + + + + + + + + + + + + +− ++ ++ + + + + + + + +and +BC = +0 +1 +1 +2 +0 +2 +3 +1 +1 +0 +2 +2 +2 +0 +6 +8 +1 +2 +0 +3 +2 +4 +0 +2 +− ++ + + + + + + + + + + + + + + +��� += ++ ++ += + + + + + + + + + + + + + + +− ++ +− + + + + + + + +So +AC + BC = +9 +1 +10 +12 +8 +20 +30 +2 +28 + + + + + + + + + + + + ++ += + + + + + + + + + + + + +− + + + + + + +Clearly, +(A + B) C = AC + BC +Example 18 If +1 +2 +3 +A +3 +2 +1 +4 +2 +1 + + + + += +− + + + + + + +, then show that A3 – 23A – 40 I = O +Solution We have +2 +1 +2 +3 +1 +2 +3 +19 +4 +8 +A +A.A +3 +2 +1 +3 +2 +1 +1 +12 +8 +4 +2 +1 +4 +2 +1 +14 +6 +15 + + + + + + + + + + += += +− +− += + + + + + + + + + + + + + + + +Reprint 2025-26 + +MATRICES 57 +So +A3 = A A2 = +1 +2 +3 +19 +4 +8 +63 +46 +69 +3 +2 +1 +1 +12 +8 +69 +6 +23 +4 +2 +1 +14 +6 +15 +92 +46 +63 + + + + + + + + + + +− += +− + + + + + + + + + + + + + + + +Now +A3 – 23A – 40I = +63 +46 +69 +1 +2 +3 +1 +0 +0 +69 +6 +23 – 23 3 +2 +1 – 40 0 +1 +0 +92 +46 +63 +4 +2 +1 +0 +0 1 + + + + + + + + + + + + +− +− + + + + + + + + + + + + + + + + + + += +63 +46 +69 +23 +46 +69 +40 +0 +0 +69 +6 +23 +69 +46 +23 +0 +40 +0 +92 +46 +63 +92 +46 +23 +0 +0 +40 +− +− +− +− + + + + + + + + + + + + +− ++ − +− ++ +− + + + + + + + + + + + + +− +− +− +− + + + + + + += +63 +23 +40 +46 +46 +0 +69 +69 +0 +69 +69 +0 +6 +46 +40 +23 +23 +0 +92 +92 +0 +46 +46 +0 +63 +23 +40 +− +− +− ++ +− ++ + + + + +− ++ +−+ +− +− ++ + + + + +− ++ +− ++ +− +− + + += +0 +0 +0 +0 +0 +0 +O +0 +0 +0 + + + += + + + + + + +Example 19 In a legislative assembly election, a political group hired a public relations +firm to promote its candidate in three ways: telephone, house calls, and letters. The +cost per contact (in paise) is given in matrix A as +A = +40 +Telephone +100 +Housecall +50 +Letter +Cost per contact + + + + + + + + + + +The number of contacts of each type made in two cities X and Y is given by +Telephone +Housecall +Letter +1000 +500 +5000 +X +B +Y +3000 +1000 10,000 +→ + + +=  +→ + + +. Find the total amount spent by the group in the two +cities X and Y. +Reprint 2025-26 + + 58 +MATHEMATICS +Solution We have +BA = +40,000 +50,000 +250,000 +X +Y +120,000 +100,000 +500,000 ++ ++ +→ + + + +→ + + += +340,000 +X +Y +720,000 +→ + + + +→ + + +So the total amount spent by the group in the two cities is 340,000 paise and +720,000 paise, i.e., ` 3400 and ` 7200, respectively. +EXERCISE 3.2 +1. +Let +2 +4 +1 +3 +2 +5 +A +, B +, C +3 +2 +2 +5 +3 +4 +− + + + + + + += += += + + + + + + +− + + + + + + +Find each of the following: +(i) A + B +(ii) A – B +(iii) 3A – C +(iv) AB +(v) BA +2. +Compute the following: +(i) +a +b +a +b +b +a +b +a + + + + ++ + + + + +− + + + +(ii) +2 +2 +2 +2 +2 +2 +2 +2 +2 +2 +2 +2 +a +b +b +c +ab +bc +ac +ab +a +c +a +b + + ++ ++ + + ++ + + + + +− +− ++ ++ + + + + + + +(iii) +1 +4 +6 +12 +7 +6 +8 +5 +16 +8 +0 +5 +2 +8 +5 +3 +2 +4 +− +− + + + + + + + + ++ + + + + + + + + + + + + +(iv) +2 +2 +2 +2 +2 +2 +2 +2 +cos +sin +sin +cos +sin +cos +cos +sin +x +x +x +x +x +x +x +x + + + + ++ + + + + + + + + + + + + +3. +Compute the indicated products. +(i) +a +b +a +b +b +a +b +a +− + + + + + + +− + + +(ii) +1 +2 +3 + + + + + +[2 3 4] +(iii) +1 +2 +1 +2 +3 +2 +3 +2 +3 +1 +− + + + + + + + + + +(iv) +2 +3 +4 +1 +3 +5 +3 +4 +5 +0 +2 +4 +4 +5 +6 +3 +0 +5 +− + + + + + + + + + + + + + + + +(v) +2 1 +1 +0 +1 +3 2 +1 +2 +1 +1 1 + + + + + + + +− + + + +− + +(vi) +2 +3 +3 +1 +3 +1 +0 +1 +0 +2 +3 +1 +− + + +− + + + + + + +− + + + + +Reprint 2025-26 + +MATRICES 59 +4. +If +1 +2 +3 +3 +1 +2 +4 +1 +2 +A +5 +0 +2 , B +4 +2 +5 +and C +0 +3 +2 +1 +1 +1 +2 +0 +3 +1 +2 +3 +− +− + + + + + + + + + + + + += += += + + + + + + + + + + + + +− +− + + + + + + +, then compute +(A+B) and (B – C). Also, verify that A + (B – C) = (A + B) – C. +5. +If +2 +5 +2 +3 +1 +1 +3 +3 +5 +5 +1 +2 +4 +1 +2 +4 +A +and B +3 +3 +3 +5 +5 +5 +7 +2 +7 +6 +2 +2 +3 +3 +5 +5 +5 + + + + + + + + + + + + + + + + += += + + + + + + + + + + + + + + + + + + + + +, then compute 3A – 5B. +6. +Simplify +cos +sin +sin +cos +cos ++ sin +sin +cos +cos +sin +θ +θ +θ +− +θ + + + + +θ +θ + + + + +− +θ +θ +θ +θ + + + + +7. +Find X and Y, if +(i) +7 +0 +3 +0 +X + Y +and X – Y +2 +5 +0 +3 + + + + += += + + + + + + + + +(ii) +2 +3 +2 +2 +2X + 3Y +and 3X +2Y +4 +0 +1 +5 +− + + + + += ++ += + + + + +− + + + + +8. +Find X, if Y = +3 +2 +1 +4 + + + + + + + and 2X + Y = +1 +0 +3 +2 + + + + +− + + +9. +Find x and y, if +1 +3 +0 +5 +6 +2 0 +1 +2 +1 +8 +y +x + + + + + + ++ += + + + + + + + + + + + + +10. +Solve the equation for x, y, z and t, if +1 +1 +3 +5 +2 +3 +3 +0 +2 +4 +6 +x +z +y +t +− + + + + + + ++ += + + + + + + + + + + + + +11. +If +2 +1 +10 +3 +1 +5 +x +y − + + + + + ++ += + + + + + + + + + + +, find the values of x and y. +12. +Given +6 +4 +3 +1 +2 +3 +x +y +x +x +y +z +w +w +z +w ++ + + + + + + += ++ + + + + + + +− ++ + + + + + + +, find the values of x, y, z and w. +Reprint 2025-26 + + 60 +MATHEMATICS +13. +If +cos +sin +0 +F ( ) +sin +cos +0 +0 +0 +1 +x +x +x +x +x +− + + + + +=  + + + + + +, show that F(x) F(y) = F(x + y). +14. +Show that +(i) +5 +1 +2 +1 +2 +1 +5 +1 +6 +7 +3 +4 +3 +4 +6 +7 +− +− + + + + + + +≠ + + + + + + + + + + + + +(ii) +1 +2 +3 +1 +1 +0 +1 +1 +0 +1 +2 +3 +0 +1 +0 +0 +1 +1 +0 +1 +1 +0 +1 +0 +1 +1 +0 +2 +3 +4 +2 +3 +4 +1 +1 +0 +− +− + + + + + + + + + + + + +− +≠ +− + + + + + + + + + + + + + + + + + + +15. +Find A2 – 5A + 6I, if +2 +0 +1 +A +2 +1 +3 +1 +1 +0 + + + + +=  + + + +− + + +16. +If +1 +0 +2 +A +0 +2 +1 +2 +0 +3 + + + + +=  + + + + + +, prove that A3 – 6A2 + 7A + 2I = 0 +17. +If +3 +2 +1 +0 +A +and I= +4 +2 +0 +1 +− + + + + +=  + + + +− + + + + +, find k so that A2 = kA – 2I +18. +If +0 +tan 2 +A +tan +0 +2 +α + + +− + + +=  + +α + + + + + + + and I is the identity matrix of order 2, show that +I + A = (I – A) +cos +sin +sin +cos +α +− +α + + + + +α +α + + +19. +A trust fund has ` 30,000 that must be invested in two different types of bonds. +The first bond pays 5% interest per year, and the second bond pays 7% interest +per year. Using matrix multiplication, determine how to divide ` 30,000 among +the two types of bonds. If the trust fund must obtain an annual total interest of: +(a) +` 1800 +(b) +` 2000 +Reprint 2025-26 + +MATRICES 61 +20. +The bookshop of a particular school has 10 dozen chemistry books, 8 dozen +physics books, 10 dozen economics books. Their selling prices are ` 80, ` 60 and +` 40 each respectively. Find the total amount the bookshop will receive from +selling all the books using matrix algebra. +Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, +respectively. Choose the correct answer in Exercises 21 and 22. +21. +The restriction on n, k and p so that PY + WY will be defined are: +(A) k = 3, p = n +(B) k is arbitrary, p = 2 +(C) p is arbitrary, k = 3 +(D) k = 2, p = 3 +22. +If n = p, then the order of the matrix 7X – 5Z is: +(A) p × 2 +(B) 2 × n +(C) n × 3 +(D) p × n +3.5. Transpose of a Matrix +In this section, we shall learn about transpose of a matrix and special types of matrices +such as symmetric and skew symmetric matrices. +Definition 3 If A = [aij] be an m × n matrix, then the matrix obtained by interchanging +the rows and columns of A is called the transpose of A. Transpose of the matrix A is +denoted by A′ or (AT). In other words, if A = [aij]m × n, then A′ = [aji]n × m. For example, +if +2 +3 +3 +2 +3 +5 +3 +3 +0 +A +3 +1 +, then A +1 +5 +1 +0 +1 +5 +5 +× +× + + + + + + + + += +′ = + + +− + + + + + + +− + + + + + + +3.5.1 Properties of transpose of the matrices +We now state the following properties of transpose of matrices without proof. These +may be verified by taking suitable examples. +For any matrices A and B of suitable orders, we have +(i) +(A′)′ = A, +(ii) +(kA)′ = kA′ (where k is any constant) +(iii) +(A + B)′ = A′ + B′ +(iv) +(A B)′ = B′ A′ +Example 20 If +2 +1 +2 +3 +3 +2 +A +and B +1 +2 +4 +4 +2 +0 + + +− + + += += + + + +�� + + + + +, verify that +(i) +(A′)′ = A, +(ii) +(A + B)′ = A′ + B′, +(iii) +(kB)′ = kB′, where k is any constant. +Reprint 2025-26 + + 62 +MATHEMATICS +Solution +(i) +We have +A = +( +) +3 +4 +3 +3 +2 +3 +3 +2 +A +3 2 +A +A +4 +2 +0 +4 +2 +0 +2 +0 + + + + + + + + +′ +′ +′ +⇒ += +⇒ += += + + + + + + + + + + + + + + +Thus +(A′)′ = A +(ii) +We have +A = 3 +3 +2 , +4 +2 +0 + + + + + + + B = +2 +1 +2 +5 +3 +1 +4 +A +B +1 +2 4 +5 +4 +4 + + +− + + +− +⇒ ++ +=  + + + + + + + +Therefore +(A + B)′ = +5 +5 +3 +1 +4 +4 +4 + + + + +− + + + + + + +Now +A′ = +3 +4 +2 1 +3 2 , B +1 2 , +2 +0 +2 4 + + + + + + + + +′ = − + + + + + + + + + + + + +So +A′ + B′ = +5 +5 +3 1 4 +4 +4 + + + + +− + + + + + + +Thus +(A + B)′ = A′ + B′ +(iii) +We have +kB = k 2 +1 +2 +2 +2 +1 +2 +4 +2 +4 +k +k +k +k +k +k +− +− + + + + += + + + + + + + + +Then +(kB)′ = +2 +2 1 +2 +1 2 +B +2 +4 +2 4 +k +k +k +k +k +k +k +k + + + + + + + + +′ +− += +− += + + + + + + + + + + + + +Thus +(kB)′ = kB′ +Reprint 2025-26 + +MATRICES 63 +Example 21 If +[ +] +2 +A +4 , B +1 +3 +6 +5 +− + + + + += += +− + + + + + + +, verify that (AB)′ = B′A′. +Solution We have +A = +[ +] +2 +4 , B +1 +3 +6 +5 +− + + + + += +− + + + + + + +then +AB = +[ +] +2 +4 +1 +3 +6 +5 +− + + + + +− + + + + + + + = +2 +6 +12 +4 +12 +24 +5 +15 +30 +− +− + + + + +− + + + + +− + + +Now +A′ = [–2 4 5] , +1 +B +3 +6 + + + + +′ =  + + + +− + + +B′A′ = +[ +] +1 +2 +4 +5 +3 +2 +4 +5 +6 +12 +15 +(AB) +6 +12 +24 +30 +− + + + + + + + + +′ +− += − += + + + + + + + + +− +− +− + + + + +Clearly +(AB)′ = B′A′ +3.6 Symmetric and Skew Symmetric Matrices +Definition 4 A square matrix A = [aij] is said to be symmetric if A′ = A, that is, +[aij] = [aji] for all possible values of i and j. +For example +3 +2 +3 +A +2 +1.5 +1 +3 +1 +1 + + + + += +− +− + + + + +− + + + is a symmetric matrix as A′ = A +Definition 5 A square matrix A = [aij] is said to be skew symmetric matrix if +A′ = – A, that is aji = – aij for all possible values of i and j. Now, if we put i = j, we +have aii = – aii. Therefore 2aii = 0 or aii = 0 for all i’s. +This means that all the diagonal elements of a skew symmetric matrix are zero. +Reprint 2025-26 + + 64 +MATHEMATICS +For example, the matrix +0 +B +0 +0 +e +f +e +g +f +g + + + + += − + + + + +− +− + + + is a skew symmetric matrix as B′= –B +Now, we are going to prove some results of symmetric and skew-symmetric +matrices. +Theorem 1 For any square matrix A with real number entries, A + A′ is a symmetric +matrix and A – A′ is a skew symmetric matrix. +Proof Let B = A + A′, then +B′ = (A + A′)′ += A′ + (A′)′ (as (A + B)′ = A′ + B′) += A′ + A (as (A′)′ = A) += A + A′ (as A + B = B + A) += B +Therefore +B = A + A′ is a symmetric matrix +Now let +C = A – A′ +C′ = (A – A′)′ = A′ – (A′)′ (Why?) += A′ – A (Why?) += – (A – A′) = – C +Therefore +C = A – A′ is a skew symmetric matrix. +Theorem 2 Any square matrix can be expressed as the sum of a symmetric and a +skew symmetric matrix. +Proof Let A be a square matrix, then we can write +1 +1 +A +(A +A ) +(A +A ) +2 +2 +′ +′ += ++ ++ +− +From the Theorem 1, we know that (A + A′) is a symmetric matrix and (A – A′) is +a skew symmetric matrix. Since for any matrix A, (kA)′ = kA′, it follows that 1 (A +A ) +2 +′ ++ +Reprint 2025-26 + +MATRICES 65 +is symmetric matrix and 1 (A +A ) +2 +′ +− + is skew symmetric matrix. Thus, any square +matrix can be expressed as the sum of a symmetric and a skew symmetric matrix. +Example 22 Express the matrix +2 +2 +4 +B +1 +3 +4 +1 +2 +3 +− +− + + + + += − + + + + +− +− + + + as the sum of a symmetric and a +skew symmetric matrix. +Solution Here +B′ = +2 +1 +1 +2 +3 +2 +4 +4 +3 +− + + + + +− +− + + + + +− +− + + +Let +P = +4 +3 +3 +1 +1 +(B + B ) +3 +6 +2 +2 +2 +3 +2 +6 +− +− + + + + +′ = +− + + + + +− +− + + + = +3 +3 +2 +2 +2 +3 +3 +1 +2 +3 +1 +3 +2 +− +− + + + + + + +− + + + + + + +− + + +− + + + + +, +Now +P′ = +3 +3 +2 +2 +2 +3 +3 +1 +2 +3 +1 +3 +2 +− +− + + + + + + +− + + + + + + +− + + +− + + + +�� += P +Thus +P = 1 (B + B ) +2 +′ is a symmetric matrix. +Also, let +Q = +1 +5 +0 +2 +2 +0 +1 +5 +1 +1 +1 +(B – B ) +1 +0 +6 +0 +3 +2 +2 +2 +5 +6 +0 +5 +3 +0 +2 +− +− + + + + +− +− + + + + + + + + +′ = += + + + + + + +− + + + + + + +− + + + + +Reprint 2025-26 + + 66 +MATHEMATICS +Then +Q′ = +1 +5 +0 +2 +3 +1 +0 +3 +Q +2 +5 +3 +0 +2 + + + + + + +− + + +− += − + + + + +− + + + + + + +Thus +Q = 1 (B – B ) +2 +′ is a skew symmetric matrix. +Now +3 +3 +1 +5 +2 +0 +2 +2 +2 +2 +2 +2 +4 +3 +1 +P + Q +3 +1 +0 +3 +1 +3 +4 +B +2 +2 +1 +2 +3 +3 +5 +1 +3 +3 +0 +2 +2 +− +− +− +− + + + + + + + + +− +− + + + + + + +− + + + + + + += ++ += − += + + + + + + + + +− +− + + + + + + +− + + + + +− +− + + + + + + + + +Thus, B is represented as the sum of a symmetric and a skew symmetric matrix. +EXERCISE 3.3 +1. +Find the transpose of each of the following matrices: +(i) +5 +1 +2 +1 + + + + + + + + + + +− + + +(ii) +1 +1 +2 +3 +− + + + + + + +(iii) +1 +5 +6 +3 +5 +6 +2 +3 +1 +− + + + + + + + + +− + + +2. +If +1 +2 +3 +4 +1 +5 +A +5 +7 +9 +and B +1 +2 +0 +2 +1 +1 +1 +3 +1 +− +− +− + + + + + + + + += += + + + + + + + + +− + + + + +, then verify that +(i) (A + B)′ = A′ + B′, +(ii) (A – B)′ = A′ – B′ +3. +If +3 4 +1 +2 +1 +A +1 2 +and B +1 +2 3 +0 1 + + +− + + + + +′ = − +=  + + + + + + + + + +, then verify that +(i) (A + B)′ = A′ + B′ +(ii) (A – B)′ = A′ – B′ +Reprint 2025-26 + +MATRICES 67 +4. +If +2 +3 +1 +0 +A +and B +1 +2 +1 +2 +− +− + + + + +′ = += + + + + + + + + +, then find (A + 2B)′ +5. +For the matrices A and B, verify that (AB)′ = B′A′, where +(i) +[ +] +1 +A +4 +, B +1 +2 +1 +3 + + + + += − += − + + + + + + +(ii) +[ +] +0 +A +1 +, B +1 +5 +7 +2 + + += += + + + +6. +If (i) +cos +sin +A +sin +cos +α +α + + +=  + +− +α +α + + +, then verify that A′ A = I +(ii) If +sin +cos +A +cos +sin +α +α + + +=  + +− +α +α + + +, then verify that A′ A = I +7. + (i) Show that the matrix +1 +1 +5 +A +1 +2 +1 +5 +1 +3 +− + + + + += − + + + + + + is a symmetric matrix. +(ii) Show that the matrix +0 +1 +1 +A +1 +0 +1 +1 +1 +0 +− + + + + += − + + + + +− + + + is a skew symmetric matrix. +8. +For the matrix +1 +5 +A +6 +7 + + +=  + + + +, verify that +(i) (A + A′) is a symmetric matrix +(ii) (A – A′) is a skew symmetric matrix +9. +Find +( +) +1 A +A +2 +′ ++ + and +( +) +1 A +A +2 +′ +− +, when +0 +A +0 +0 +a +b +a +c +b +c + + + + += − + + + + +− +− + + +10. +Express the following matrices as the sum of a symmetric and a skew symmetric +matrix: +Reprint 2025-26 + + 68 +MATHEMATICS +(i) +3 +5 +1 +1 + + + + +− + + +(ii) +6 +2 +2 +2 +3 +1 +2 +1 +3 +− + + + + +− +− + + + + +− + + +(iii) +3 +3 +1 +2 +2 +1 +4 +5 +2 +− + + + + +− +− + + + + +− +− + + +(iv) +1 +5 +1 +2 + + + + +− + + +Choose the correct answer in the Exercises 11 and 12. +11. +If A, B are symmetric matrices of same order, then AB – BA is a +(A) Skew symmetric matrix +(B) Symmetric matrix +(C) Zero matrix +(D) Identity matrix +12. +If +cos +sin +A +, +sin +cos +α +− +α + + +=  + +α +α + + +and A + A′ = I, then the value of α is +(A) 6 +π +(B) 3 +π +(C) π +(D) 3 +2 +π +3.7 Invertible Matrices +Definition 6 If A is a square matrix of order m, and if there exists another square +matrix B of the same order m, such that AB = BA = I, then B is called the inverse +matrix of A and it is denoted by A– 1. In that case A is said to be invertible. +For example, let +A = +2 +3 +1 +2 + + + + + + +and B = +2 +3 +1 +2 +− + + + + +− + + +be two matrices. +Now +AB = +2 +3 +2 +3 +1 +2 +1 +2 +− + + + + + + +− + + + += +4 +3 +6 +6 +1 +0 +I +2 +2 +3 +4 +0 +1 +− +−+ + + + + += += + + + + +− +−+ + + + + +Also +BA = +1 +0 +I +0 +1 + += + + + + +. Thus B is the inverse of A, in other +words B = A– 1 and A is inverse of B, i.e., A = B–1 +Reprint 2025-26 + +MATRICES 69 +ANote +1. A rectangular matrix does not possess inverse matrix, since for products BA +and AB to be defined and to be equal, it is necessary that matrices A and B +should be square matrices of the same order. +2. If B is the inverse of A, then A is also the inverse of B. +Theorem 3 (Uniqueness of inverse) Inverse of a square matrix, if it exists, is unique. +Proof Let A = [aij] be a square matrix of order m. If possible, let B and C be two +inverses of A. We shall show that B = C. +Since B is the inverse of A +AB = BA = I +... (1) +Since C is also the inverse of A +AC = CA = I +... (2) +Thus +B = BI = B (AC) = (BA) C = IC = C +Theorem 4 If A and B are invertible matrices of the same order, then (AB)–1 = B–1 A–1. +Proof From the definition of inverse of a matrix, we have +(AB) (AB)–1 = 1 +or +A–1 (AB) (AB)–1 = A–1I +(Pre multiplying both sides by A–1) +or +(A–1A) B (AB)–1 = A–1 +(Since A–1 I = A–1) +or +IB (AB)–1 = A–1 +or +B (AB)–1 = A–1 +or +B–1 B (AB)–1 = B–1 A–1 +or +I (AB)–1 = B–1 A–1 +Hence +(AB)–1 = B–1 A–1 +1. Matrices A and B will be inverse of each other only if + (A) AB = BA (B) AB = BA = 0 + (C) AB = 0, BA = I (D) AB = BA = I +EXERCISE 3.4 +Reprint 2025-26 + + 70 +MATHEMATICS +Miscellaneous Examples +Example 23 If +cos +sin +A +sin +cos +θ +θ + + +=  + +− +θ +θ + + +, then prove that +cos +sin +A +sin +cos +n +n +n +n +n +θ +θ + + +=  + +− +θ +θ + + +, n ∈ N. +Solution We shall prove the result by using principle of mathematical induction. +We have +P(n) : If +cos +sin +A +sin +cos +θ +θ + + +=  + +− +θ +θ + + +, then +cos +sin +A +sin +cos +n +n +n +n +n +θ +θ + + +=  + +− +θ +θ + + +, n ∈ N +P(1) : +cos +sin +A +sin +cos +θ +θ + + +=  + +− +θ +θ + + +, so +1 +cos +sin +A +sin +cos +θ +θ + + +=  + +− +θ +θ + + +Therefore, +the result is true for n = 1. +Let the result be true for n = k. So +P(k) : +cos +sin +A +sin +cos +θ +θ + + +=  + +− +θ +θ + + +, then +cos +sin +A +sin +cos +k +k +k +k +k +θ +θ + + +=  + +− +θ +θ + + +Now, we prove that the result holds for n = k +1 +Now +Ak + 1 = +cos +sin +cos +sin +A A +sin +cos +sin +cos +k +k +k +k +k +θ +θ +θ +θ + + + +⋅ +=  + + +− +θ +θ +− +θ +θ + + + += +cos cos +– sin sin +cos sin +sin cos +sin cos +cos sin +sin sin +cos cos +k +k +k +k +k +k +k +k +θ +θ +θ +θ +θ +θ + +θ +θ + + + + +− +θ +θ + +θ +θ +− +θ +θ + +θ +θ + + += +cos( +) +sin( +) +cos( +1) +sin ( +1) +sin( +) +cos( +) +sin ( +1) +cos( +1) +k +k +k +k +k +k +k +k +θ + θ +θ + θ ++ +θ ++ +θ + + + + += + + + + +− +θ + θ +θ + θ +− ++ +θ ++ +θ + + + + +Therefore, the result is true for n = k + 1. Thus by principle of mathematical induction, +we have +cos +sin +A +sin +cos +n +n +n +n +n +θ +θ + + +=  + +− +θ +θ + + +, holds for all natural numbers. +Example 24 If A and B are symmetric matrices of the same order, then show that AB +is symmetric if and only if A and B commute, that is AB = BA. +Solution Since A and B are both symmetric matrices, therefore A′ = A and B′ = B. +Reprint 2025-26 + +MATRICES 71 +Let +AB be symmetric, then (AB)′ = AB +But +(AB)′ = B′A′= BA (Why?) +Therefore +BA = AB +Conversely, if AB = BA, then we shall show that AB is symmetric. +Now +(AB)′ = B′A′ += B A (as A and B are symmetric) += AB +Hence AB is symmetric. +Example 25 Let +2 +1 +5 +2 +2 +5 +A +, B +, C +3 +4 +7 +4 +3 +8 +− + + + + + + += += += + + + + + + + + + + + + +. Find a matrix D such that +CD – AB = O. +Solution Since A, B, C are all square matrices of order 2, and CD – AB is well +defined, D must be a square matrix of order 2. +Let +D = a +b +c +d + + + + + + +. Then CD – AB = 0 gives +2 +5 +2 +1 +5 +2 +3 +8 +3 +4 +7 +4 +a +b +c +d +− + + + + + + +− + + + + + + + + + + + + + = O +or +2 +5 +2 +5 +3 +0 +3 +8 +3 +8 +43 +22 +a +c +b +d +a +c +b +d ++ ++ + + + + +− + + + + ++ ++ + + + + + = 0 +0 +0 +0 + + + + + + +or +2 +5 +3 +2 +5 +3 +8 +43 +3 +8 +22 +a +c +b +d +a +c +b +d ++ +− ++ + + + + ++ +− ++ +− + + + = 0 +0 +0 +0 + + + + + + +By equality of matrices, we get +2a + 5c – 3 = 0 +... (1) +3a + 8c – 43 = 0 +... (2) +2b + 5d = 0 +... (3) +and +3b + 8d – 22 = 0 +... (4) +Solving (1) and (2), we get a = –191, c = 77. Solving (3) and (4), we get b = – 110, +d = 44. +Reprint 2025-26 + + 72 +MATHEMATICS +Therefore +D = +191 +110 +77 +44 +a +b +c +d +− +− + + + + += + + + + + + + + +Miscellaneous Exercise on Chapter 3 +1. +If A and B are symmetric matrices, prove that AB – BA is a skew symmetric +matrix. +2. +Show that the matrix B′AB is symmetric or skew symmetric according as A is +symmetric or skew symmetric. +3. +Find the values of x, y, z if the matrix +0 +2 +A +y +z +x +y +z +x +y +z + + + + += +− + + + + +− + + + satisfy the equation +A′A = I. +4. +For what values of x : [ +] +1 +2 +0 +0 +1 +2 +1 +2 +0 +1 +2 +1 +0 +2 +x + + + + + + + + + + + = O? +5. +If +3 +1 +A +1 +2 + + +=  + +− + + +, show that A2 – 5A + 7I = 0. +6. +Find x, if [ +] +1 +0 +2 +5 +1 +0 +2 +1 +4 +O +2 +0 +3 +1 +x +x + + + + +− +− += + + + + + + +7. +A manufacturer produces three products x, y, z which he sells in two markets. +Annual sales are indicated below: +Market +Products +I +10,000 +2,000 +18,000 +II +6,000 +20,000 +8,000 +Reprint 2025-26 + +MATRICES 73 +(a) If unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively, +find the total revenue in each market with the help of matrix algebra. +(b) If the unit costs of the above three commodities are ` 2.00, ` 1.00 and +50 paise respectively. Find the gross profit. +8. +Find the matrix X so that +1 +2 +3 +7 +8 +9 +X 4 +5 +6 +2 +4 +6 +− +− +− + + + + += + + + + + + + + +Choose the correct answer in the following questions: +9. +If A = is such that A² = I, then +(A) 1 + α² + βγ = 0 +(B) 1 – α² + βγ = 0 +(C) 1 – α² – βγ = 0 +(D) 1 + α² – βγ = 0 +10. +If the matrix A is both symmetric and skew symmetric, then +(A) A is a diagonal matrix +(B) A is a zero matrix +(C) A is a square matrix +(D) None of these +11. +If A is square matrix such that A2 = A, then (I + A)³ – 7 A is equal to +(A) A +(B) I – A +(C) I +(D) 3A +Summary +® A matrix is an ordered rectangular array of numbers or functions. +® A matrix having m rows and n columns is called a matrix of order m × n. +® [aij]m × 1 is a column matrix. +® [aij]1 × n is a row matrix. +® An m × n matrix is a square matrix if m = n. +® A = [aij]m × m is a diagonal matrix if aij = 0, when i ≠ j. +® A = [aij]n × n is a scalar matrix if aij = 0, when i ≠ j, aij = k, (k is some +constant), when i = j. +® A = [aij]n × n is an identity matrix, if aij = 1, when i = j, aij = 0, when i ≠ j. +® A zero matrix has all its elements as zero. +® A = [aij] = [bij] = B if (i) A and B are of same order, (ii) aij = bij for all +possible values of i and j. +α +β +γ +α +− + + + + +Reprint 2025-26 + + 74 +MATHEMATICS +® kA = k[aij]m × n = [k(aij)]m × n +® – A = (–1)A +® A – B = A + (–1) B +® A + B = B + A +® (A + B) + C = A + (B + C), where A, B and C are of same order. +® k(A + B) = kA + kB, where A and B are of same order, k is constant. +® (k + l ) A = kA + lA, where k and l are constant. +® If A = [aij]m × n and B = [bjk]n × p, then AB = C = [cik]m × p, where = +® +(i) A(BC) = (AB)C, (ii) A(B + C) = AB + AC, (iii) (A + B)C = AC + BC +® If A = [aij]m × n, then A′ or AT = [aji]n × m +® +(i) (A′)′ = A, (ii) (kA)′ = kA′, (iii) (A + B)′ = A′ + B′, (iv) (AB)′ = B′A′ +® A is a symmetric matrix if A′ = A. +® A is a skew symmetric matrix if A′ = –A. +® Any square matrix can be represented as the sum of a symmetric and a +skew symmetric matrix. +® If A and B are two square matrices such that AB = BA = I, then B is the +inverse matrix of A and is denoted by A–1 and A is the inverse of B. +® Inverse of a square matrix, if it exists, is unique. +—v— +1 += +n +ik +ij +jk +j +c +a b +∑ +Reprint 2025-26 + +MATRICES 75 +NOTES +Reprint 2025-26" +class_12,4,Determinants,ncert_books/class_12/lemh1dd/lemh104.pdf,"76 +MATHEMATICS +v All Mathematical truths are relative and conditional. — C.P. STEINMETZ v +4.1 Introduction +In the previous chapter, we have studied about matrices +and algebra of matrices. We have also learnt that a system +of algebraic equations can be expressed in the form of +matrices. This means, a system of linear equations like +a1 x + b1 y = c1 +a2 x + b2 y = c2 +can be represented as +1 +1 +1 +2 +2 +2 +a +b +c +x +a +b +c +y + + + + += + + + + + + + + + + +. Now, this +system of equations has a unique solution or not, is +determined by the number a1 b2 – a2 b1. (Recall that if +1 +1 +2 +2 +a +b +a +b +≠ + or, a1 b2 – a2 b1 ≠ 0, then the system of linear +equations has a unique solution). The number a1 b2 – a2 b1 +which determines uniqueness of solution is associated with the matrix +1 +1 +2 +2 +A +a +b +a +b + + +=  + + + +and is called the determinant of A or det A. Determinants have wide applications in +Engineering, Science, Economics, Social Science, etc. +In this chapter, we shall study determinants up to order three only with real entries. +Also, we will study various properties of determinants, minors, cofactors and applications +of determinants in finding the area of a triangle, adjoint and inverse of a square matrix, +consistency and inconsistency of system of linear equations and solution of linear +equations in two or three variables using inverse of a matrix. +4.2 Determinant +To every square matrix A = [aij] of order n, we can associate a number (real or +complex) called determinant of the square matrix A, where aij = (i, j)th element of A. +Chapter 4 +DETERMINANTS +P.S. Laplace +(1749-1827) +Reprint 2025-26 + +DETERMINANTS 77 +This may be thought of as a function which associates each square matrix with a +unique number (real or complex). If M is the set of square matrices, K is the set of +numbers (real or complex) and f : M → K is defined by f (A) = k, where A ∈ M and +k ∈ K, then f (A) is called the determinant of A. It is also denoted by |A| or det A or ∆. +If A = +a +b +c +d + + + + + + +, then determinant of A is written as |A| = +a +b +c +d = det (A) +Remarks +(i) +For matrix A, |A| is read as determinant of A and not modulus of A. +(ii) +Only square matrices have determinants. +4.2.1 Determinant of a matrix of order one +Let A = [a] be the matrix of order 1, then determinant of A is defined to be equal to a +4.2.2 Determinant of a matrix of order two +Let +A = +11 +12 +21 +22 +a +a +a +a + + + + + + + be a matrix of order 2 × 2, +then the determinant of A is defined as: +det (A) = |A| = ∆ = + = a11a22 – a21a12 +Example 1 Evaluate +2 +4 +–1 +2 . +Solution We have +2 +4 +–1 +2 = 2(2) – 4(–1) = 4 + 4 = 8. +Example 2 Evaluate +1 +– 1 +x +x +x +x ++ +Solution We have +1 +– 1 +x +x +x +x ++ + = x (x) – (x + 1) (x – 1) = x2 – (x2 – 1) = x2 – x2 + 1 = 1 +4.2.3 Determinant of a matrix of order 3 × 3 +Determinant of a matrix of order three can be determined by expressing it in terms of +second order determinants. This is known as expansion of a determinant along +a row (or a column). There are six ways of expanding a determinant of order +Reprint 2025-26 + + 78 +MATHEMATICS +3 corresponding to each of three rows (R1, R2 and R3) and three columns (C1, C2 and +C3) giving the same value as shown below. +Consider the determinant of square matrix A = [aij]3 × 3 +i.e., +| A | = +21 +22 +23 +31 +32 +33 +a +a +a +a +a +a +11 +12 +13 +a +a +a +Expansion along first Row (R1) +Step 1 Multiply first element a11 of R1 by (–1)(1 + 1) [(–1)sum of suffixes in a11] and with the +second order determinant obtained by deleting the elements of first row (R1) and first +column (C1) of | A | as a11 lies in R1 and C1, +i.e., +(–1)1 + 1 a11 +22 +23 +32 +33 +a +a +a +a +Step 2 Multiply 2nd element a12 of R1 by (–1)1 + 2 [(–1)sum of suffixes in a12] and the second +order determinant obtained by deleting elements of first row (R1) and 2nd column (C2) +of | A | as a12 lies in R1 and C2, +i.e., +(–1)1 + 2 a12 +21 +23 +31 +33 +a +a +a +a +Step 3 Multiply third element a13 of R1 by (–1)1 + 3 [(–1)sum of suffixes in a13] and the second +order determinant obtained by deleting elements of first row (R1) and third column (C3) +of | A | as a13 lies in R1 and C3, +i.e., +(–1)1 + 3 a13 +21 +22 +31 +32 +a +a +a +a +Step 4 Now the expansion of determinant of A, that is, | A | written as sum of all three +terms obtained in steps 1, 2 and 3 above is given by +det A = |A| = (–1)1 + 1 a11 +22 +23 +21 +23 +1 +2 +12 +32 +33 +31 +33 +(–1) +a +a +a +a +a +a +a +a +a ++ ++ + + +21 +22 +1 +3 +13 +31 +32 +(–1) +a +a +a +a +a ++ +or +|A| = a11 (a22 a33 – a32 a23) – a12 (a21 a33 – a31 a23) ++ a13 (a21 a32 – a31 a22) +Reprint 2025-26 + +DETERMINANTS 79 += a11 a22 a33 – a11 a32 a23 – a12 a21 a33 + a12 a31 a23 + a13 a21 a32 +– a13 a31 a22 +... (1) +ANote We shall apply all four steps together. +Expansion along second row (R2) +| A | = +11 +12 +13 +31 +32 +33 +a +a +a +a +a +a +21 +22 +23 +a +a +a +Expanding along R2, we get +| A | = +12 +13 +11 +13 +2 +1 +2 +2 +21 +22 +32 +33 +31 +33 +(–1) +(–1) +a +a +a +a +a +a +a +a +a +a ++ ++ ++ +11 +12 +2 +3 +23 +31 +32 +(–1) +a +a +a +a +a ++ ++ += – a21 (a12 a33 – a32 a13) + a22 (a11 a33 – a31 a13) +– a23 (a11 a32 – a31 a12) +| A | = – a21 a12 a33 + a21 a32 a13 + a22 a11 a33 – a22 a31 a13 – a23 a11 a32 + ++ a23 a31 a12 += a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32 +– a13 a31 a22 +... (2) +Expansion along first Column (C1) +| A | = +12 +13 +22 +23 +32 +33 +11 +21 +31 +a +a +a +a +a +a +a +a +a +By expanding along C1, we get +| A | = +22 +23 +12 +13 +1 +1 +2 +1 +11 +21 +32 +33 +32 +33 +(–1) +( 1) +a +a +a +a +a +a +a +a +a +a ++ ++ ++ +− ++ +12 +13 +3 +1 +31 +22 +23 +(–1) +a +a +a +a +a ++ += a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22) +Reprint 2025-26 + + 80 +MATHEMATICS +| A | = a11 a22 a33 – a11 a23 a32 – a21 a12 a33 + a21 a13 a32 + a31 a12 a23 +– a31 a13 a22 += a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32 +– a13 a31 a22 +... (3) +Clearly, values of |A| in (1), (2) and (3) are equal. It is left as an exercise to the +reader to verify that the values of |A| by expanding along R3, C2 and C3 are equal to the +value of |A| obtained in (1), (2) or (3). +Hence, expanding a determinant along any row or column gives same value. +Remarks +(i) +For easier calculations, we shall expand the determinant along that row or column +which contains maximum number of zeros. +(ii) +While expanding, instead of multiplying by (–1)i + j, we can multiply by +1 or –1 +according as (i + j) is even or odd. +(iii) +Let A = 2 +2 +4 +0 + + + + + + + and B = +1 +1 +2 +0 + + + + + + + . Then, it is easy to verify that A = 2B. Also +|A| = 0 – 8 = – 8 and |B| = 0 – 2 = – 2. +Observe that, |A| = 4(– 2) = 22|B| or |A| = 2n|B|, where n = 2 is the order of +square matrices A and B. +In general, if A = kB where A and B are square matrices of order n, then | A| = kn +| B |, where n = 1, 2, 3 +Example 3 Evaluate the determinant ∆ = +1 +2 +4 +–1 +3 +0 +4 +1 +0 +. +Solution Note that in the third column, two entries are zero. So expanding along third +column (C3), we get +∆ = +–1 +3 +1 +2 +1 +2 +4 +– 0 +0 +4 +1 +4 +1 +–1 +3 ++ += 4 (–1 – 12) – 0 + 0 = – 52 +Example 4 Evaluate ∆ = +0 +sin +–cos +–sin +0 +sin +cos +–sin +0 +α +α +α +β +α +β +. +Reprint 2025-26 + +DETERMINANTS 81 +Solution Expanding along R1, we get +∆ = +0 +sin +–sin +sin +–sin +0 +0 +– sin +– cos +–sin +0 +cos +0 +cos +–sin +β +α +β +α +α +α +β +α +α +β += 0 – sin α (0 – sin β cos α) – cos α (sin α sin β – 0) += sin α sin β cos α – cos α sin α sin β = 0 +Example 5 Find values of x for which 3 +3 +2 +1 +4 +1 +x +x += +. +Solution We have 3 +3 +2 +1 +4 +1 +x +x += +i.e. +3 – x2 = 3 – 8 +i.e. +x2 = 8 +Hence +x = +2 2 +± +EXERCISE 4.1 +Evaluate the determinants in Exercises 1 and 2. +1. +2 +4 +–5 +–1 +2. +(i) +cos +–sin +sin +cos +θ +θ +θ +θ +(ii) +2 – +1 +– 1 +1 +1 +x +x +x +x +x ++ ++ ++ +3. +If +A = +1 +2 +4 +2 + + + + + + +, then show that | 2A | = 4 | A | +4. +If +A = +1 +0 +1 +0 +1 +2 +0 +0 +4 + + + + + + + + + + +, then show that | 3 A | = 27 | A | +5. +Evaluate the determinants +(i) +3 +–1 +–2 +0 +0 +–1 +3 +–5 +0 +(ii) +3 +– 4 +5 +1 +1 +–2 +2 +3 +1 +Reprint 2025-26 + + 82 +MATHEMATICS +(iii) +0 +1 +2 +–1 +0 +–3 +–2 +3 +0 +(iv) +2 +–1 +–2 +0 +2 +–1 +3 +–5 +0 +6. +If A = +1 +1 +–2 +2 +1 +–3 +5 +4 +–9 + + + + + + + + + + +, find | A | +7. +Find values of x, if +(i) +2 +4 +2 +4 +5 +1 +6 +x +x += +(ii) +2 +3 +3 +4 +5 +2 +5 +x +x += +8. +If +2 +6 +2 +18 +18 +6 +x +x = +, then x is equal to +(A) 6 +(B) ± 6 +(C) – 6 +(D) 0 +4.3 Area of a Triangle +In earlier classes, we have studied that the area of a triangle whose vertices are +(x1, y1), (x2, y2) and (x3, y3), is given by the expression 1 +2 [x1(y2–y3) + x2 (y3–y1) + +x3 (y1–y2)]. Now this expression can be written in the form of a determinant as +∆ = +1 +1 +2 +2 +3 +3 +1 +1 +1 +2 +1 +x +y +x +y +x +y +... (1) +Remarks +(i) +Since area is a positive quantity, we always take the absolute value of the +determinant in (1). +(ii) +If area is given, use both positive and negative values of the determinant for +calculation. +(iii) +The area of the triangle formed by three collinear points is zero. +Example 6 Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1). +Solution The area of triangle is given by +∆ = +3 +8 +1 +1 +4 +2 +1 +2 +5 +1 +1 +– +Reprint 2025-26 + +DETERMINANTS 83 += +( +) +( +) +( +) +1 3 2 –1 – 8 –4 – 5 +1 –4 –10 +2  ++ + + + + += +( +) +1 +61 +3 +72 +14 +2 +2 +– ++ += +Example 7 Find the equation of the line joining A(1, 3) and B (0, 0) using determinants +and find k if D(k, 0) is a point such that area of triangle ABD is 3sq units. +Solution Let P (x, y) be any point on AB. Then, area of triangle ABP is zero (Why?). So +0 +0 +1 +1 1 +3 +1 +2 +1 +x +y + = 0 +This gives +( +) +1 +3 +2 y – +x = 0 or y = 3x, +which is the equation of required line AB. +Also, since the area of the triangle ABD is 3 sq. units, we have +1 +3 +1 +1 0 +0 +1 +2 +0 +1 +k + = ± 3 +This gives, +3 +3 +2 +k +− += ± , i.e., k = ∓ 2. +EXERCISE 4.2 +1. +Find area of the triangle with vertices at the point given in each of the following : +(i) (1, 0), (6, 0), (4, 3) +(ii) (2, 7), (1, 1), (10, 8) +(iii) (–2, –3), (3, 2), (–1, –8) +2. +Show that points +A (a, b + c), B (b, c + a), C (c, a + b) are collinear. +3. +Find values of k if area of triangle is 4 sq. units and vertices are +(i) (k, 0), (4, 0), (0, 2) +(ii) (–2, 0), (0, 4), (0, k) +4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants. +(ii) Find equation of line joining (3, 1) and (9, 3) using determinants. +5. +If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is +(A) 12 +(B) –2 +(C) –12, –2 +(D) 12, –2 +Reprint 2025-26 + + 84 +MATHEMATICS +4.4 Minors and Cofactors +In this section, we will learn to write the expansion of a determinant in compact form +using minors and cofactors. +Definition 1 Minor of an element aij of a determinant is the determinant obtained by +deleting its ith row and jth column in which element aij lies. Minor of an element aij is +denoted by Mij. +Remark Minor of an element of a determinant of order n(n ≥ 2) is a determinant of +order n – 1. +Example 8 Find the minor of element 6 in the determinant +1 +2 +3 +4 +5 +6 +7 +8 +9 +∆= +Solution Since 6 lies in the second row and third column, its minor M23 is given by +M23 = +1 +2 +7 +8 = 8 – 14 = – 6 (obtained by deleting R2 and C3 in ∆). +Definition 2 Cofactor of an element aij, denoted by Aij is defined by +Aij = (–1)i + j Mij, where Mij is minor of aij. +Example 9 Find minors and cofactors of all the elements of the determinant +1 +–2 +4 +3 +Solution Minor of the element aij is Mij +Here a11 = 1. So M11 = Minor of a11= 3 +M12 + = Minor of the element a12 = 4 +M21 = Minor of the element a21 = –2 +M22 = Minor of the element a22 = 1 +Now, cofactor of aij is Aij. So +A11 = (–1)1 + 1 M11 = (–1)2 (3) = 3 +A12 = (–1)1 + 2 M12 = (–1)3 (4) = – 4 +A21 = (–1)2 + 1 M21 = (–1)3 (–2) = 2 +A22 = (–1)2 + 2 M22 = (–1)4 (1) = 1 +Reprint 2025-26 + +DETERMINANTS 85 +Example 10 Find minors and cofactors of the elements a11, a21 in the determinant +∆ = +11 +12 +13 +21 +22 +23 +31 +32 +33 +a +a +a +a +a +a +a +a +a +Solution By definition of minors and cofactors, we have +Minor of a11 = M11 = +22 +23 +32 +33 +a +a +a +a + = a22 a33– a23 a32 +Cofactor of a11 = A11 = (–1)1+1 M11 = a22 a33 – a23 a32 +Minor of a21 = M21 = +12 +13 +32 +33 +a +a +a +a + = a12 a33 – a13 a32 +Cofactor of a21 = A21 = (–1)2+1 M21 = (–1) (a12 a33 – a13 a32) = – a12 a33 + a13 a32 +Remark Expanding the determinant ∆, in Example 21, along R1, we have +∆ = (–1)1+1 a11 +22 +23 +32 +33 +a +a +a +a ++ (–1)1+2 a12 +21 +23 +31 +33 +a +a +a +a + + (–1)1+3 a13 +21 +22 +31 +32 +a +a +a +a + = a11 A11 + a12 A12 + a13 A13, where Aij is cofactor of aij + = sum of product of elements of R1 with their corresponding cofactors +Similarly, ∆ can be calculated by other five ways of expansion that is along R2, R3, +C1, C2 and C3. +Hence ∆ = sum of the product of elements of any row (or column) with their +corresponding cofactors. +ANote If elements of a row (or column) are multiplied with cofactors of any +other row (or column), then their sum is zero. For example, +∆ = a11 A21 + a12 A22 + a13 A23 += a11 (–1)1+1 +12 +13 +32 +33 +a +a +a +a ++ a12 (–1)1+2 +11 +13 +31 +33 +a +a +a +a ++ a13 (–1)1+3 +11 +12 +31 +32 +a +a +a +a += +11 +12 +13 +11 +12 +13 +31 +32 +33 +a +a +a +a +a +a +a +a +a + = 0 (since R1 and R2 are identical) +Similarly, we can try for other rows and columns. +Reprint 2025-26 + + 86 +MATHEMATICS +Example 11 Find minors and cofactors of the elements of the determinant +2 +3 +5 +6 +0 +4 +1 +5 +7 +– +– +and verify that a11 A31 + a12 A32 + a13 A33= 0 +Solution We have M11 = +0 +4 +5 +7 +– + = 0 –20 = –20; A11 = (–1)1+1 (–20) = –20 +M12 = +6 +4 +1 +7 +– + = – 42 – 4 = – 46; + A12 = (–1)1+2 (– 46) = 46 +M13 = +6 +0 +1 +5 = 30 – 0 = 30; + A13 = (–1)1+3 (30) = 30 +M21 = +3 +5 +5 +7 +– +– = 21 – 25 = – 4; + A21 = (–1)2+1 (– 4) = 4 +M22 = +2 +5 +1 +7 +– + = –14 – 5 = –19; + A22 = (–1)2+2 (–19) = –19 +M23 = +2 +3 +1 +5 +– + = 10 + 3 = 13; + A23 = (–1)2+3 (13) = –13 +M31 = +3 +5 +0 +4 +– + = –12 – 0 = –12; + A31 = (–1)3+1 (–12) = –12 +M32 = +2 +5 +6 +4 = 8 – 30 = –22; + A32 = (–1)3+2 (–22) = 22 +and +M33 = +2 +3 +6 +0 +– + = 0 + 18 = 18; + A33 = (–1)3+3 (18) = 18 +Now +a11 = 2, a12 = –3, a13 = 5; A31 = –12, A32 = 22, A33 = 18 +So +a11 A31 + a12 A32 + a13 A33 + = 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0 +Reprint 2025-26 + +DETERMINANTS 87 +EXERCISE 4.3 +Write Minors and Cofactors of the elements of following determinants: +1. +(i) 2 +4 +0 +3 +– +(ii) +a +c +b +d +2. +(i) +1 +0 +0 +0 +1 +0 +0 +0 +1 +(ii) +1 +0 +4 +3 +5 +1 +0 +1 +2 +– +3. +Using Cofactors of elements of second row, evaluate ∆ = +5 +3 +8 +2 +0 +1 +1 +2 +3 +. +4. +Using Cofactors of elements of third column, evaluate ∆ = +1 +1 +1 +x +yz +y +zx +z +xy +. +5. +If ∆ = +11 +12 +13 +21 +22 +23 +31 +32 +33 +a +a +a +a +a +a +a +a +a + and Aij is Cofactors of aij, then value of ∆ is given by +(A) +a11 A31+ a12 A32 + a13 A33 +(B) +a11 A11+ a12 A21 + a13 A31 +(C) +a21 A11+ a22 A12 + a23 A13 +(D) +a11 A11+ a21 A21 + a31 A31 +4.5 Adjoint and Inverse of a Matrix +In the previous chapter, we have studied inverse of a matrix. In this section, we shall +discuss the condition for existence of inverse of a matrix. +To find inverse of a matrix A, i.e., A–1 we shall first define adjoint of a matrix. +4.5.1 Adjoint of a matrix +Definition 3 The adjoint of a square matrix A = [aij]n × n is defined as the transpose of +the matrix [Aij]n × n, where Aij is the cofactor of the element aij. Adjoint of the matrix A +is denoted by adj A. +Let +11 +12 +13 +21 +22 +23 +31 +32 +33 +A = +a +a +a +a +a +a +a +a +a + + + + + + + + + + +Reprint 2025-26 + + 88 +MATHEMATICS +Then +11 +12 +13 +21 +22 +23 +31 +32 +33 +A +A +A +A =Transposeof A +A +A +A +A +A +adj + + + + + + + + + + +11 +21 +31 +12 +22 +32 +13 +23 +33 +A +A +A += A +A +A +A +A +A + + + + + + + + + + +Example 12 +2 +3 +Find + A for A = 1 +4 +adj + + + + + + +Solution We have A11 = 4, A12 = –1, A21 = –3, A22 = 2 +Hence +adj A = +11 +21 +12 +22 +A +A +4 +–3 + = +A +A +–1 +2 + + + + + + + + + + + + +Remark For a square matrix of order 2, given by +A = +11 +12 +21 +22 +a +a +a +a + + + + + + +The adj A can also be obtained by interchanging a11 and a22 and by changing signs +of a12 and a21, i.e., +We state the following theorem without proof. +Theorem 1 If A be any given square matrix of order n, then +A(adj A) = (adj A) A = A I , +where I is the identity matrix of order n +Verification +Let +A = +11 +12 +13 +21 +22 +23 +31 +32 +33 +a +a +a +a +a +a +a +a +a + + + + + + + + + + +, then adj A = +11 +21 +31 +12 +22 +32 +13 +23 +33 +A +A +A +A +A +A +A +A +A + + + + + + + + + + +Since sum of product of elements of a row (or a column) with corresponding +cofactors is equal to |A| and otherwise zero, we have +Reprint 2025-26 + +DETERMINANTS 89 +A (adj A) = +A +0 +0 +0 +A +0 +0 +0 +A + + + + + + + + + + + = A +1 +0 +0 +0 +1 +0 +0 +0 +1 + + + + + + + + + + + = A I +Similarly, we can show (adj A) A = A I +Hence A (adj A) = (adj A) A = A I +Definition 4 A square matrix A is said to be singular if A = 0. +For example, the determinant of matrix A = +1 +2 +4 +8 + + + + is zero +Hence A is a singular matrix. +Definition 5 A square matrix A is said to be non-singular if A ≠ 0 +Let +A = +1 +2 +3 +4 + + + + + + +. Then A = +1 +2 +3 +4 = 4 – 6 = – 2 ≠ 0. +Hence A is a nonsingular matrix +We state the following theorems without proof. +Theorem 2 If A and B are nonsingular matrices of the same order, then AB and BA +are also nonsingular matrices of the same order. +Theorem 3 The determinant of the product of matrices is equal to product of their +respective determinants, that is, AB = A B , where A and B are square matrices of +the same order +Remark We know that (adj A) A = A I = +A +A +A +A +0 +0 +0 +0 +0 +0 +0 + + + + + + + + + + +≠ +, +Writing determinants of matrices on both sides, we have +( +A)A +adj + = +A +0 +0 +0 +A +0 +0 +0 +A +Reprint 2025-26 + + 90 +MATHEMATICS +i.e. +|(adj A)| |A| = +3 +1 +0 +0 +A +0 +1 +0 +0 +0 +1 +(Why?) +i.e. +|(adj A)| |A| = |A|3 (1) +i.e. +|(adj A)| = | A |2 +In general, if A is a square matrix of order n, then |adj(A)| = |A|n – 1. +Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix. +Proof Let A be invertible matrix of order n and I be the identity matrix of order n. +Then, there exists a square matrix B of order n such that AB = BA = I +Now +AB = I. So AB = I or A B = 1 (since I +1, AB +A B ) += += +This gives +A ≠ 0. Hence A is nonsingular. +Conversely, let A be nonsingular. Then A ≠ 0 +Now +A (adj A) = (adj A) A = A I +(Theorem 1) +or +A +1 +1 +A +A A +I +| A | +| A | +adj +adj + + + + += += + + + + + + + + +or +AB = BA = I, where B = +1 +A +| A | adj +Thus +A is invertible and A–1 = +1 +A +| A | adj +Example 13 If A = +1 +3 +3 +1 +4 +3 +1 +3 +4 + + + + + + + + + + +, then verify that A adj A = |A| I. Also find A–1. +Solution We have A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0 +Now A11 = 7, A12 = –1, A13 = –1, A21 = –3, A22 = 1,A23 = 0, A31 = –3, A32 = 0, +A33 = 1 +Therefore +adj A = +7 +3 +3 +1 +1 +0 +1 +0 +1 +− +− + + + + +− + + + + +− + + +Reprint 2025-26 + +DETERMINANTS 91 +Now +A (adj A) = +1 +3 +3 +7 +3 +3 +1 +4 +3 +1 +1 +0 +1 +3 +4 +1 +0 +1 +− +− + + + + + + +− + + + + + + +− + + + += +7 +3 +3 +3 +3 +0 +3 +0 +3 +7 +4 +3 +3 +4 +0 +3 +0 +3 +7 +3 +4 +3 +3 +0 +3 +0 +4 +− +− +−+ ++ +−+ ++ + + + + +− +− +−+ ++ +−+ ++ + + + + +− +− +−+ ++ +−+ ++ + + += +1 +0 +0 +0 +1 +0 +0 +0 +1 + + + + + + + + + + + = (1) +1 +0 +0 +0 +1 +0 +0 +0 +1 + + + + + + + + + + + = A . I +Also +A–1 +1 +A +A +a d j += + = +7 +3 +3 +1 +1 +1 +0 +1 +1 +0 +1 +− +− + + + + +− + + + + +− + + + = +7 +3 +3 +1 +1 +0 +1 +0 +1 +− +− + + + + +− + + + + +− + + +Example 14 If A = +2 +3 +1 +2 +and B +1 +4 +1 +3 +− + + + + += + + + + +− +− + + + + +, then verify that (AB)–1 = B–1A–1. +Solution We have AB = +2 +3 +1 +2 +1 +5 +1 +4 +1 +3 +5 +14 +− +− + +��� + + + += + + + + + +− +− +− + + + + + +Since, +AB = –11 ≠ 0, (AB)–1 exists and is given by +(AB)–1 = +14 +5 +1 +1 +(AB) +5 +1 +AB +11 +adj +− +− + + +=− + + +− +− + + + +14 +5 +1 +5 +1 +11 + + += + + + + +Further, A = –11 ≠ 0 and B = 1 ≠ 0. Therefore, A–1 and B–1 both exist and are given by +A–1 = − +− +− +− + + + + +=  + + + +− +1 +11 +4 +3 +1 +2 +3 +2 +1 +1 +1 +,B +Therefore +B A +− +−= − + + + + +− +− +− + + + + +1 +1 +1 +11 +3 +2 +1 +1 +4 +3 +1 +2 + = − +− +− +− +− + + + + +1 +11 +14 +5 +5 +1 +14 +5 +1 +5 +1 +11 + + += + + + + +Hence (AB)–1 = B–1 A–1 +Reprint 2025-26 + + 92 +MATHEMATICS +Example 15 Show that the matrix A = +2 +3 +1 +2 + + + + satisfies the equation A2 – 4A + I = O, +where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix. Using this equation, find A–1. +Solution We have +2 +2 +3 +2 +3 +7 +12 +A +A.A +1 +2 +1 +2 +4 +7 + + + + + += += += + + + + + + + + + + +Hence +2 +7 +12 +8 +12 +1 +0 +A +4A +I +4 +7 +4 +8 +0 +1 + + + + + + +− ++ = +− ++ + + + + + + + + + + + + +0 +0 +O +0 +0 + + += += + + + + +Now +A2 – 4A + I = O +Therefore +A A – 4A = – I +or +A A (A–1) – 4 A A–1 = – I A–1 (Post multiplying by A–1 because |A| ≠ 0) +or +A (A A–1) – 4I = – A–1 +or +AI – 4I = – A–1 +or +A–1 = 4I – A = +4 +0 +2 +3 +2 +3 +0 +4 +1 +2 +1 +2 +− + + + + + + +− += + + + + + + +− + + + + + + +Hence +1 +2 +3 +A +1 +2 +− +− + + +=  + +− + + +EXERCISE 4.4 +Find adjoint of each of the matrices in Exercises 1 and 2. +1. +1 +2 +3 +4 + + + + +2. +1 +1 +2 +2 +3 +5 +2 +0 +1 +− +− + + + + + + + + + + +Verify A (adj A) = (adj A) A = |A| I in Exercises 3 and 4 +3. +2 +3 +4 +6 +− +− + + + + +4. +1 +1 +2 +3 +0 +2 +1 +0 +3 +− +− + + + + + + + + + + +Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. +5. +2 +2 +4 +3 +− + + + + +6. +− +− + + + + +1 +5 +3 +2 +7. +1 +2 +3 +0 +2 +4 +0 +0 +5 + + + + + + + + + + +Reprint 2025-26 + +DETERMINANTS 93 +8. +1 +0 +0 +3 +3 +0 +5 +2 +1 +− + + + + + + + + + + +9. +2 +1 +3 +4 +1 +0 +7 +2 +1 +− +− + + + + + + + + + + +10. +1 +1 +2 +0 +2 +3 +3 +2 +4 +− +− +− + + + + + + + + + + +11. +1 +0 +0 +0 +cos +sin +0 +sin +cos + + + + +α +α + + + + +α +− +α + + +12. +Let A = +3 +7 +2 +5 + + + + and B = 6 +8 +7 +9 + + + +. Verify that (AB)–1 = B–1 A–1. +13. +If A = +3 +1 +1 +2 +− + + + +, show that A2 – 5A + 7I = O. Hence find A–1. +14. +For the matrix A = +3 +2 +1 +1 + + + +, find the numbers a and b such that A2 + aA + bI = O. +15. +For the matrix A = +1 +1 +1 +1 +2 +3 +2 +1 +3 +− +− + + + + + + + + + + +Show that A3– 6A2 + 5A + 11 I = O. Hence, find A–1. +16. +If A = +2 +1 +1 +1 +2 +1 +1 +1 +2 +− +− +− +− + + + + + + + + + + +Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1 +17. +Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to +(A) | A | +(B) | A |2 +(C) | A |3 +(D) 3|A| +18. +If A is an invertible matrix of order 2, then det (A–1) is equal to +(A) det (A) +(B) +1 +det (A) +(C) 1 +(D) 0 +4.6 Applications of Determinants and Matrices +In this section, we shall discuss application of determinants and matrices for solving the +system of linear equations in two or three variables and for checking the consistency of +the system of linear equations. +Reprint 2025-26 + + 94 +MATHEMATICS +Consistent system A system of equations is said to be consistent if its solution (one +or more) exists. +Inconsistent system A system of equations is said to be inconsistent if its solution +does not exist. +ANote In this chapter, we restrict ourselves to the system of linear equations +having unique solutions only. +4.6.1 Solution of system of linear equations using inverse of a matrix +Let us express the system of linear equations as matrix equations and solve them using +inverse of the coefficient matrix. +Consider the system of equations +a1 x + b1 y + c1 z = d1 +a2 x + b2 y + c2 z = d 2 +a3 x + b3 y + c3 z = d 3 +Let +A = +1 +1 +1 +1 +2 +2 +2 +2 +3 +3 +3 +3 +, X +and B +a +b +c +x +d +a +b +c +y +d +a +b +c +z +d + + + + + + + + + + += += + + + + + + + + + + +�� + + + + +Then, the system of equations can be written as, AX = B, i.e., + +1 +1 +1 +2 +2 +2 +3 +3 +3 +a +b +c +x +a +b +c +y +a +b +c +z + + + + + + + + + + + + = +1 +2 +3 +d +d +d + + + + + + + + + + +Case I If A is a nonsingular matrix, then its inverse exists. Now +AX = B +or +A–1 (AX) = A–1 B +(premultiplying by A–1) +or +(A–1A) X = A–1 B +(by associative property) +or +I X = A–1 B +or +X = A–1 B +This matrix equation provides unique solution for the given system of equations as +inverse of a matrix is unique. This method of solving system of equations is known as +Matrix Method. +Case II If A is a singular matrix, then |A| = 0. +In this case, we calculate (adj A) B. +If (adj A) B ≠ O, (O being zero matrix), then solution does not exist and the +system of equations is called inconsistent. +Reprint 2025-26 + +DETERMINANTS 95 +If (adj A) B = O, then system may be either consistent or inconsistent according +as the system have either infinitely many solutions or no solution. +Example 16 Solve the system of equations +2x + 5y = 1 +3x + 2y = 7 +Solution The system of equations can be written in the form AX = B, where +A = +2 +5 +1 +,X +and B +3 +2 +7 +x +y + + + + += += + + + + + + + + +Now, A = –11 ≠ 0, Hence, A is nonsingular matrix and so has a unique solution. +Note that +A–1 = − +− +− + + + + +1 +11 +2 +5 +3 +2 +Therefore +X = A–1B = – 1 +11 +2 +5 +3 +2 +1 +7 +− +− + + + + + + + + +i.e. +x +y + + + = − +− + + + += − + + + + +1 +11 +33 +11 +3 +1 +Hence +x = 3, y = – 1 +Example 17 Solve the following system of equations by matrix method. +3x – 2y + 3z = 8 +2x + y – z = 1 +4x – 3y + 2z = 4 +Solution The system of equations can be written in the form AX = B, where +3 +2 +3 +8 +A +2 +1 +1 , X +and B +1 +4 +3 +2 +4 +x +y +z +− + + + + + + + + += +− += += + + + + + + + + +− + + + + +We see that +A = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0 +Hence, A is nonsingular and so its inverse exists. Now +A11 = –1, +A12 = – 8, +A13 = –10 +A21 = –5, +A22 = – 6, +A23 = 1 +A31 = –1, +A32 = 9, +A33 = 7 +Reprint 2025-26 + + 96 +MATHEMATICS +Therefore +A–1 = +1 +5 +1 +1 +8 +6 +9 +17 +10 +1 +7 +− +− +− + + + + +− +− +− + + + + +− + + +So +X = +–1 +1 +5 +1 +8 +1 +A B = +8 +6 +9 +1 +17 +10 +1 +7 +4 +− +− +− + + + + +− +− +− + + + + +− + + +i.e. +x +y +z + + + + + + = +17 +1 +1 +34 +2 +17 +51 +3 +− + + + + + + +− +− += + + + + + + +− + + + +Hence +x = 1, y = 2 and z = 3. +Example 18 The sum of three numbers is 6. If we multiply third number by 3 and add +second number to it, we get 11. By adding first and third numbers, we get double of the +second number. Represent it algebraically and find the numbers using matrix method. +Solution Let first, second and third numbers be denoted by x, y and z, respectively. +Then, according to given conditions, we have +x + y + z = 6 +y + 3z = 11 +x + z = 2y or x – 2y + z = 0 +This system can be written as A X = B, where +A = +1 +1 +1 +0 +1 +3 +1 +2 +1 + + + + + + + + + + + +, X = +x +y +z + + + + + + + + + + + and B = +6 +11 +0 + + + + + + + + + + +Here +( +) +( +) +A +1 1 +6 – (0 – 3) +0 –1 +9 +0 += ++ ++ += +≠ +. Now we find adj A +A11 = 1 (1 + 6) = 7, +A12 = – (0 – 3) = 3, +A13 = – 1 +A21 = – (1 + 2) = – 3, +A22 = 0, +A23 = – (– 2 – 1) = 3 +A31 = (3 – 1) = 2, +A32 = – (3 – 0) = – 3, +A33 = (1 – 0) = 1 +Hence +adj A = +7 +–3 +2 +3 +0 +–3 +–1 +3 +1 + + + + + + + + + + +Reprint 2025-26 + +DETERMINANTS 97 +Thus +A –1 = +1 +A adj (A) = +7 +3 +2 +1 +3 +0 +3 +9 +1 +3 +1 +– +– +– + + + + + + + + + + +Since +X = A–1 B +X = +7 +3 +2 +6 +1 +3 +0 +3 +11 +9 +1 +3 +1 +0 +– +– +– + + + + + + + + + + + + + + + +or +x +y +z + + + + + + + + + + + = 1 +9 +42 +33 +0 +18 +0 +0 +6 +33 +0 +− ++ + + + + ++ ++ + + + + +−+ ++ + + + = 1 +9 +9 +18 +27 + + + + + + + + + + + = +1 +2 +3 + + + + + + + + + + +Thus +x = 1, y = 2, z = 3 +EXERCISE 4.5 +Examine the consistency of the system of equations in Exercises 1 to 6. +1. +x + 2y = 2 +2. 2x – y = 5 +3. x + 3y = 5 +2x + 3y = 3 +x + y = 4 +2x + 6y = 8 +4. +x + y + z = 1 +5. 3x–y – 2z = 2 +6. 5x – y + 4z = 5 +2x + 3y + 2z = 2 +2y – z = –1 +2x + 3y + 5z = 2 +ax + ay + 2az = 4 +3x – 5y = 3 +5x – 2y + 6z = –1 +Solve system of linear equations, using matrix method, in Exercises 7 to 14. +7. +5x + 2y = 4 +8. 2x – y = –2 +9. 4x – 3y = 3 +7x + 3y = 5 +3x + 4y = 3 +3x – 5y = 7 +10. +5x + 2y = 3 +11. 2x + y + z = 1 +12. x – y + z = 4 +3x + 2y = 5 +x – 2y – z = 3 +2 +2x + y – 3z = 0 +3y – 5z = 9 +x + y + z = 2 +13. +2x + 3y +3 z = 5 +14. x – y + 2z = 7 +x – 2y + z = – 4 +3x + 4y – 5z = – 5 +3x – y – 2z = 3 +2x – y + 3z = 12 +Reprint 2025-26 + + 98 +MATHEMATICS +15. +If A = +2 +–3 +5 +3 +2 +– 4 +1 +1 +–2 + + + + + + + + + + +, find A–1. Using A–1 solve the system of equations +2x – 3y + 5z = 11 +3x + 2y – 4z = – 5 +x + y – 2z = – 3 +16. +The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ` 60. The cost of 2 kg onion, +4 kg wheat and 6 kg rice is ` 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice +is ` 70. Find cost of each item per kg by matrix method. +Miscellaneous Examples +Example 19 Use product +1 +1 +2 +0 +2 +3 +3 +2 +4 +2 +0 +1 +9 +2 +3 +6 +1 +2 + + + + + + + + + + + + + + + + + + + + + + + + + + + to solve the system of equations +x – y + 2z = 1 +2y – 3z = 1 +3x – 2y + 4z = 2 +Solution Consider the product +1 +1 +2 +2 +0 +1 +0 +2 +3 +9 +2 +3 +3 +2 +4 +6 +1 +2 +– +– +– +– +– +– + + + + + + + + + + + + + + + + = +2 +9 +12 +0 +2 +2 +1 +3 +4 +0 +18 18 +0 +4 +3 +0 +6 +6 +6 18 +24 +0 +4 +4 +3 +6 +8 +−− ++ +− ++ ++ +− + + + + ++ +− ++ +− +− ++ + + + + +−− ++ +− ++ ++ +− + + + = +1 +0 +0 +0 +1 +0 +0 +0 +1 + + + + + + + + + + +Hence + +1 +1 +2 +0 +2 +3 +3 +2 +4 +2 +0 +1 +9 +2 +3 +6 +1 +2 +1 +– +– +– +– +– +– +– + + + + + + + + + + += + + + + + + + + + + +Now, given system of equations can be written, in matrix form, as follows +1 +–1 +2 +0 +2 +–3 +3 +–2 +4 +x +y +z + + + + + + + + + + + + = +1 +1 +2 + + + + + +Reprint 2025-26 + +DETERMINANTS 99 +or + + + + +x +y +z + = +1 +1 +1 +2 +1 +0 +2 +3 +1 +3 +2 +4 +2 +− +− + + + + + + +− + + + + + + +− + + + + = + + + +2 +0 +1 +9 +2 +3 +6 +1 +2 +1 +1 +2 + + + + + + + + + + + + + + + + + + + + += +2 +0 +2 +0 +9 +2 +6 +5 +6 +1 +4 +3 +−+ ++ + + + + + + ++ +− += + + + + + + ++ − + + + +Hence +x = 0, y = 5 and z = 3 +Miscellaneous Exercises on Chapter 4 +1. +Prove that the determinant +sin +cos +–sin +– +1 +cos +1 +x +x +x +θ +θ +θ +θ + is independent of θ. +2. +Evaluate +cos +cos +cos +sin +–sin +–sin +cos +0 +sin +cos +sin +sin +cos +α +β +α +β +α +β +β +α +β +α +β +α +. +3. +If A–1 = +( +) +1 +3 +1 +1 +1 +2 +2 +15 +6 +5 and B +1 +3 +0 +, find AB +5 +2 +2 +0 +2 +1 +– +– +– +– +– +– +– +– + + + + + + + + += + + + + + + + + + + + + +4. +Let A = +1 +2 +1 +2 +3 +1 +1 +1 +5 + + + + + + + + + + + + +. Verify that +(i) [adj A]–1 = adj (A–1) +(ii) (A–1)–1 = A +5. +Evaluate +x +y +x +y +y +x +y +x +x +y +x +y ++ ++ ++ +6. +Evaluate +1 +1 +1 +x +y +x +y +y +x +x+ y ++ +Reprint 2025-26 + + 100 +MATHEMATICS +Using properties of determinants in Exercises 11 to 15, prove that: +7. +Solve the system of equations +2 +3 +10 +4 ++ ++ += +x +y +z +4 +6 +5 +1 ++ += +– +x +y +z +6 +9 +20 +2 ++ += +– +x +y +z +Choose the correct answer in Exercise 17 to 19. +8. +If x, y, z are nonzero real numbers, then the inverse of matrix +0 +0 +A +0 +0 +0 +0 +x +y +z + + + + +=  + + + + + +is +(A) +1 +1 +1 +0 +0 +0 +0 +0 +0 +x +y +z +− +− +− + + + + + + + + + + +(B) +1 +1 +1 +0 +0 +0 +0 +0 +0 +x +xyz +y +z +− +− +− + + + + + + + + + + +(C) +0 +0 +1 +0 +0 +0 +0 +x +y +xyz +z + + + + + + + + + + +(D) +1 +0 +0 +1 +0 +1 +0 +0 +0 +1 +xyz + + + + + + + + + + +9. +Let A = +1 +sin +1 +sin +1 +sin +1 +sin +1 +θ + + + + +− +θ +θ + + + + +− +− +θ + + +, where 0 ≤ θ ≤ 2π. Then +(A) Det (A) = 0 +(B) Det (A) ∈ (2, ∞) +(C) Det (A) ∈ (2, 4) +(D) Det (A) ∈ [2, 4] +Reprint 2025-26 + +DETERMINANTS 101 +Summary +® Determinant of a matrix A = [a11]1×1 is given by |a11| = a11 +® Determinant of a matrix A =  + + + +a +a +a +a +11 +12 +21 +22 + is given by +11 +12 +21 +22 +A +a +a +a +a += += a11 a22 – a12 a21 +® Determinant of a matrix A = + + + + + + + + + + +a +b +c +a +b +c +a +b +c +1 +1 +1 +2 +2 +2 +3 +3 +3 + is given by (expanding along R1) +1 +1 +1 +2 +2 +2 +2 +2 +2 +2 +2 +2 +1 +1 +1 +3 +3 +3 +3 +3 +3 +3 +3 +3 +A +a +b +c +b +c +a +c +a +b +a +b +c +a +b +c +b +c +a +c +a +b +a +b +c += += +− ++ +For any square matrix A, the |A| satisfy following properties. +® Area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by +1 +1 +2 +2 +3 +3 +1 +1 +1 +2 +1 +x +y +x +y +x +y +∆= +® Minor of an element aij of the determinant of matrix A is the determinant +obtained by deleting ith row and jth column and denoted by Mij. +® Cofactor of aij of given by Aij = (– 1)i+ j Mij +® Value of determinant of a matrix A is obtained by sum of product of elements +of a row (or a column) with corresponding cofactors. For example, +A = a11 A11 + a12 A12 + a13 A13. +® If elements of one row (or column) are multiplied with cofactors of elements +of any other row (or column), then their sum is zero. For example, a11 A21 + a12 +A22 + a13 A23 = 0 +Reprint 2025-26 + + 102 +MATHEMATICS +® If +11 +12 +13 +21 +22 +23 +31 +32 +33 +A +, +a +a +a +a +a +a +a +a +a + + + + +=  + + + + + + then +11 +21 +31 +12 +22 +32 +13 +23 +33 +A +A +A +A +A +A +A +A +A +A +adj + + + + +=  + + + + + +, where Aij is +cofactor of aij +® A (adj A) = (adj A) A = |A| I, where A is square matrix of order n. +® A square matrix A is said to be singular or non-singular according as +|A| = 0 or |A| ≠ 0. +® If AB = BA = I, where B is square matrix, then B is called inverse of A. +Also A–1 = B or B–1 = A and hence (A–1)–1 = A. +® A square matrix A has inverse if and only if A is non-singular. +® +–1 +1 +A +( +A) +A adj += +® If +a1 x + b1 y + c1 z = d1 +a2 x + b2 y + c2 z = d2 +a3 x + b3 y + c3 z = d3, +then these equations can be written as A X = B, where +1 +1 +1 +1 +2 +2 +2 +2 +3 +3 +3 +3 +A +,X= +and B= +a +b +c +x +d +a +b +c +y +d +a +b +c +z +d + + + + + + + + + + += + + + + + + + + + + + + + + +® Unique solution of equation AX = B is given by X = A–1 B, where A +0 +≠ +. +® A system of equation is consistent or inconsistent according as its solution +exists or not. +® For a square matrix A in matrix equation AX = B +(i) |A| ≠ 0, there exists unique solution +(ii) |A| = 0 and (adj A) B ≠ 0, then there exists no solution +(iii) |A| = 0 and (adj A) B = 0, then system may or may not be consistent. +Reprint 2025-26 + +DETERMINANTS 103 +Historical Note +The Chinese method of representing the coefficients of the unknowns of +several linear equations by using rods on a calculating board naturally led to the +discovery of simple method of elimination. The arrangement of rods was precisely +that of the numbers in a determinant. The Chinese, therefore, early developed the +idea of subtracting columns and rows as in simplification of a determinant +Mikami, China, pp 30, 93. +Seki Kowa, the greatest of the Japanese Mathematicians of seventeenth +century in his work ‘Kai Fukudai no Ho’ in 1683 showed that he had the idea of +determinants and of their expansion. But he used this device only in eliminating a +quantity from two equations and not directly in the solution of a set of simultaneous +linear equations. T. Hayashi, “The Fakudoi and Determinants in Japanese +Mathematics,” in the proc. of the Tokyo Math. Soc., V. +Vendermonde was the first to recognise determinants as independent functions. +He may be called the formal founder. Laplace (1772), gave general method of +expanding a determinant in terms of its complementary minors. In 1773 Lagrange +treated determinants of the second and third orders and used them for purpose +other than the solution of equations. In 1801, Gauss used determinants in his +theory of numbers. +The next great contributor was Jacques - Philippe - Marie Binet, (1812) who +stated the theorem relating to the product of two matrices of m-columns and n- +rows, which for the special case of m = n reduces to the multiplication theorem. +Also on the same day, Cauchy (1812) presented one on the same subject. He +used the word ‘determinant’ in its present sense. He gave the proof of multiplication +theorem more satisfactory than Binet’s. +The greatest contributor to the theory was Carl Gustav Jacob Jacobi, after +this the word determinant received its final acceptance. +Reprint 2025-26" +class_12,5,Continuity and Differentiability,ncert_books/class_12/lemh1dd/lemh105.pdf,"MATHEMATICS +104 +vThe whole of science is nothing more than a refinement +of everyday thinking.” — ALBERT EINSTEIN v +5.1 Introduction +This chapter is essentially a continuation of our study of +differentiation of functions in Class XI. We had learnt to +differentiate certain functions like polynomial functions and +trigonometric functions. In this chapter, we introduce the +very important concepts of continuity, differentiability and +relations between them. We will also learn differentiation +of inverse trigonometric functions. Further, we introduce a +new class of functions called exponential and logarithmic +functions. These functions lead to powerful techniques of +differentiation. We illustrate certain geometrically obvious +conditions through differential calculus. In the process, we +will learn some fundamental theorems in this area. +5.2 Continuity +We start the section with two informal examples to get a feel of continuity. Consider +the function +1, if +0 +( ) +2, if +0 +x +f x +x +≤ + +=  +> + +This function is of course defined at every +point of the real line. Graph of this function is +given in the Fig 5.1. One can deduce from the +graph that the value of the function at nearby +points on x-axis remain close to each other +except at x = 0. At the points near and to the +left of 0, i.e., at points like – 0.1, – 0.01, – 0.001, +the value of the function is 1. At the points near +and to the right of 0, i.e., at points like 0.1, 0.01, +Chapter 5 +CONTINUITY AND +DIFFERENTIABILITY +Sir Issac Newton +(1642-1727) +Fig 5.1 +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +105 +0.001, the value of the function is 2. Using the language of left and right hand limits, we +may say that the left (respectively right) hand limit of f at 0 is 1 (respectively 2). In +particular the left and right hand limits do not coincide. We also observe that the value +of the function at x = 0 concides with the left hand limit. Note that when we try to draw +the graph, we cannot draw it in one stroke, i.e., without lifting pen from the plane of the +paper, we can not draw the graph of this function. In fact, we need to lift the pen when +we come to 0 from left. This is one instance of function being not continuous at x = 0. +Now, consider the function defined as +f x +x +x +( ) +, +, += +≠ += + + + +1 +0 +2 +0 +if +if +This function is also defined at every point. Left and the right hand limits at x = 0 +are both equal to 1. But the value of the +function at x = 0 equals 2 which does not +coincide with the common value of the left +and right hand limits. Again, we note that we +cannot draw the graph of the function without +lifting the pen. This is yet another instance of +a function being not continuous at x = 0. +Naively, we may say that a function is +continuous at a fixed point if we can draw the +graph of the function around that point without +lifting the pen from the plane of the paper. +Mathematically, it may be phrased precisely as follows: +Definition 1 Suppose f is a real function on a subset of the real numbers and let c be +a point in the domain of f. Then f is continuous at c if +lim +( ) +( ) +x +c f x +f c +→ += +More elaborately, if the left hand limit, right hand limit and the value of the function +at x = c exist and equal to each other, then f is said to be continuous at x = c. Recall that +if the right hand and left hand limits at x = c coincide, then we say that the common +value is the limit of the function at x = c. Hence we may also rephrase the definition of +continuity as follows: a function is continuous at x = c if the function is defined at +x = c and if the value of the function at x = c equals the limit of the function at +x = c. If f is not continuous at c, we say f is discontinuous at c and c is called a point +of discontinuity of f. +Fig 5.2 +Reprint 2025-26 + + MATHEMATICS +106 +Example 1 Check the continuity of the function f given by f(x) = 2x + 3 at x = 1. +Solution First note that the function is defined at the given point x = 1 and its value is 5. +Then find the limit of the function at x = 1. Clearly +1 +1 +lim +( ) +lim(2 +3) +2(1) +3 +5 +x +x +f x +x +→ +→ += ++ += ++ += +Thus +1 +lim +( ) +5 +(1) +x +f x +f +→ += += +Hence, f is continuous at x = 1. +Example 2 Examine whether the function f given by f(x) = x2 is continuous at x = 0. +Solution First note that the function is defined at the given point x = 0 and its value is 0. +Then find the limit of the function at x = 0. Clearly +2 +2 +0 +0 +lim +( ) +lim +0 +0 +x +x +f x +x +→ +→ += += += +Thus +0 +lim +( ) +0 +(0) +x +f x +f +→ += += +Hence, f is continuous at x = 0. +Example 3 Discuss the continuity of the function f given by f(x) = | x | at x = 0. +Solution By definition +f(x) = +, if +0 +, if +0 +x +x +x +x +− +< + + +≥ + +Clearly the function is defined at 0 and f(0) = 0. Left hand limit of f at 0 is +0 +0 +lim +( ) +lim (– ) +0 +x +x +f x +x +− +− +→ +→ += += +Similarly, the right hand limit of f at 0 is +0 +0 +lim +( ) +lim +0 +x +x +f x +x ++ ++ +→ +→ += += +Thus, the left hand limit, right hand limit and the value of the function coincide at +x = 0. Hence, f is continuous at x = 0. +Example 4 Show that the function f given by +f(x) = +3 +3, if +0 +1, +if +0 +x +x +x + ++ +≠ + += + +is not continuous at x = 0. +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +107 +Solution The function is defined at x = 0 and its value at x = 0 is 1. When x ≠ 0, the +function is given by a polynomial. Hence, +0 +lim +( ) +x +f x +→ + = +3 +3 +0 +lim ( +3) +0 +3 +3 +x +x +��� ++ += ++ += +Since the limit of f at x = 0 does not coincide with f(0), the function is not continuous +at x = 0. It may be noted that x = 0 is the only point of discontinuity for this function. +Example 5 Check the points where the constant function f(x) = k is continuous. +Solution The function is defined at all real numbers and by definition, its value at any +real number equals k. Let c be any real number. Then +lim +( ) +x +c f x +→ + = lim +x +c k +k +→ += +Since f(c) = k = lim +x +c +→ f(x) for any real number c, the function f is continuous at +every real number. +Example 6 Prove that the identity function on real numbers given by f(x) = x is +continuous at every real number. +Solution The function is clearly defined at every point and f (c) = c for every real +number c. Also, +lim +( ) +x +c f x +→ + = lim +x +c x +c +→ += +Thus, lim +x +c +→f(x) = c = f(c) and hence the function is continuous at every real number. +Having defined continuity of a function at a given point, now we make a natural +extension of this definition to discuss continuity of a function. +Definition 2 A real function f is said to be continuous if it is continuous at every point +in the domain of f. +This definition requires a bit of elaboration. Suppose f is a function defined on a +closed interval [a, b], then for f to be continuous, it needs to be continuous at every +point in [a, b] including the end points a and b. Continuity of f at a means +lim +( ) +x +a f x ++ +→ += f (a) +and continuity of f at b means +– +lim +( ) +x +b f x +→ += f(b) +Observe that lim +( ) +x +a f x +− +→ + and lim +( ) +x +b f x ++ +→ +do not make sense. As a consequence +of this definition, if f is defined only at one point, it is continuous there, i.e., if the +domain of f is a singleton, f is a continuous function. +Reprint 2025-26 + + MATHEMATICS +108 +Example 7 Is the function defined by f(x) = | x |, a continuous function? +Solution We may rewrite f as +f (x) = +, if +0 +, if +0 +x +x +x +x +− +< + + +≥ + +By Example 3, we know that f is continuous at x = 0. +Let c be a real number such that c < 0. Then f(c) = – c. Also +lim +( ) +x +c f x +→ + = lim ( +) +– +x +c +x +c +→ +− += + (Why?) +Since lim +( ) +( ) +x +c f x +f c +→ += +, f is continuous at all negative real numbers. +Now, let c be a real number such that c > 0. Then f (c) = c. Also +lim +( ) +x +c f x +→ + = lim +x +c x +c +→ += (Why?) +Since lim +( ) +( ) +x +c f x +f c +→ += +, f is continuous at all positive real numbers. Hence, f +is continuous at all points. +Example 8 Discuss the continuity of the function f given by f (x) = x3 + x2 – 1. +Solution Clearly f is defined at every real number c and its value at c is c3 + c2 – 1. We +also know that +lim +( ) +x +c f x +→ + = +3 +2 +3 +2 +lim ( +1) +1 +x +c x +x +c +c +→ ++ +− += ++ +− +Thus lim +( ) +( ) +x +c f x +f c +→ += +, and hence f is continuous at every real number. This means +f is a continuous function. +Example 9 Discuss the continuity of the function f defined by f (x) = 1 +x , x ≠ 0. +Solution Fix any non zero real number c, we have +1 +1 +lim +( ) +lim +x +c +x +c +f x +x +c +→ +→ += += +Also, since for c ≠ 0, +1 +( ) +f c +c += +, we have lim +( ) +( ) +x +c f x +f c +→ += + and hence, f is continuous +at every point in the domain of f. Thus f is a continuous function. +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +109 +We take this opportunity to explain the concept of infinity. This we do by analysing +the function f (x) = 1 +x near x = 0. To carry out this analysis we follow the usual trick of +finding the value of the function at real numbers close to 0. Essentially we are trying to +find the right hand limit of f at 0. We tabulate this in the following (Table 5.1). +Table 5.1 +x +1 +0.3 +0.2 +0.1 = 10–1 +0.01 = 10–2 +0.001 = 10–3 +10–n +f (x) +1 +3.333... +5 +10 +100 = 102 +1000 = 103 +10n +We observe that as x gets closer to 0 from the right, the value of f (x) shoots up +higher. This may be rephrased as: the value of f (x) may be made larger than any given +number by choosing a positive real number very close to 0. In symbols, we write +0 +lim +( ) +x +f x ++ +→ += + ∞ +(to be read as: the right hand limit of f (x) at 0 is plus infinity). We wish to emphasise +that + ∞ is NOT a real number and hence the right hand limit of f at 0 does not exist (as +a real number). +Similarly, the left hand limit of f at 0 may be found. The following table is self +explanatory. +Table 5.2 +x +– 1 +– 0.3 +– 0.2 +– 10–1 +– 10–2 +– 10–3 +– 10–n +f (x) +– 1 +– 3.333... +– 5 +– 10 +– 102 +– 103 +– 10n +From the Table 5.2, we deduce that the +value of f(x) may be made smaller than any +given number by choosing a negative real +number very close to 0. In symbols, +we write +0 +lim +( ) +x +f x +− +→ += −∞ +(to be read as: the left hand limit of f (x) at 0 is +minus infinity). Again, we wish to emphasise +that – ∞ is NOT a real number and hence the +left hand limit of f at 0 does not exist (as a real +number). The graph of the reciprocal function +given in Fig 5.3 is a geometric representation +of the above mentioned facts. +Fig 5.3 +Reprint 2025-26 + + MATHEMATICS +110 +Example 10 Discuss the continuity of the function f defined by +f (x) = +2, if +1 +2, if +1 +x +x +x +x ++ +≤ + +− +> + +Solution The function f is defined at all points of the real line. +Case 1 If c < 1, then f(c) = c + 2. Therefore, lim +( ) +lim( +2) +2 +x +c +x +c +f x +x +c +→ +→ += ++ += ++ +Thus, f is continuous at all real numbers less than 1. +Case 2 If c > 1, then f (c) = c – 2. Therefore, +lim +( ) +lim +x +c +x +c +f x +→ +→ += +(x – 2) = c – 2 = f (c) +Thus, f is continuous at all points x > 1. +Case 3 If c = 1, then the left hand limit of f at +x = 1 is +– +– +1 +1 +lim +( ) +lim ( +2) +1 +2 +3 +x +x +f x +x +→ +→ += ++ += + += +The right hand limit of f at x = 1 is +1 +1 +lim +( ) +lim ( +2) +1 +2 +1 +x +x +f x +x ++ ++ +→ +→ += +− += − += − +Since the left and right hand limits of f at x = 1 +do not coincide, f is not continuous at x = 1. Hence +x = 1 is the only point of discontinuity of f. The graph of the function is given in Fig 5.4. +Example 11 Find all the points of discontinuity of the function f defined by +f (x) = +2, if +1 +0, +if +1 +2, if +1 +x +x +x +x +x ++ +< + + += + +− +> + +Solution As in the previous example we find that f +is continuous at all real numbers x ≠ 1. The left +hand limit of f at x = 1 is +– +1 +1 +lim +( ) +lim ( +2) +1 +2 +3 +x +x +f x +x +− +→ +→ += ++ += + += +The right hand limit of f at x = 1 is +1 +1 +lim +( ) +lim ( +2) +1 +2 +1 +x +x +f x +x ++ ++ +→ +→ += +− += − += − +Since, the left and right hand limits of f at x = 1 +do not coincide, f is not continuous at x = 1. Hence +x = 1 is the only point of discontinuity of f. The +graph of the function is given in the Fig 5.5. +Fig 5.4 +Fig 5.5 +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +111 +Example 12 Discuss the continuity of the function defined by +f(x) = +2, if +0 +2, if +0 +x +x +x +x ++ +< + +−+ +> + +Solution Observe that the function is defined at all real numbers except at 0. Domain +of definition of this function is +D1 ∪ D2 where D1 = {x ∈ R : x < 0} and + D2 = {x ∈ R : x > 0} +Case 1 If c ∈ D1, then lim +( ) +lim +x +c +x +c +f x +→ +→ += + (x + 2) += c + 2 = f (c) and hence f is continuous in D1. +Case 2 If c ∈ D2, then lim +( ) +lim +x +c +x +c +f x +→ +→ += + (– x + 2) += – c + 2 = f (c) and hence f is continuous in D2. +Since f is continuous at all points in the domain of f, +we deduce that f is continuous. Graph of this +function is given in the Fig 5.6. Note that to graph +this function we need to lift the pen from the plane +of the paper, but we need to do that only for those points where the function is not +defined. +Example 13 Discuss the continuity of the function f given by +f (x) = +2 +, +if +0 +, if +0 +x +x +x +x +≥ + +< + +Solution Clearly the function is defined at +every real number. Graph of the function is +given in Fig 5.7. By inspection, it seems prudent +to partition the domain of definition of f into +three disjoint subsets of the real line. +Let +D1 = {x ∈ R : x < 0}, D2 = {0} and +D3 = {x ∈ R : x > 0} +Case 1 At any point in D1, we have f(x) = x2 and it is easy to see that it is continuous +there (see Example 2). +Case 2 At any point in D3, we have f(x) = x and it is easy to see that it is continuous +there (see Example 6). +Fig 5.6 + Fig 5.7 +Reprint 2025-26 + + MATHEMATICS +112 +Case 3 Now we analyse the function at x = 0. The value of the function at 0 is f(0) = 0. +The left hand limit of f at 0 is +– +2 +2 +0 +0 +lim +( ) +lim +0 +0 +x +x +f x +x +− +→ +→ += += += +The right hand limit of f at 0 is +0 +0 +lim +( ) +lim +0 +x +x +f x +x ++ ++ +→ +→ += += +Thus +0 +lim +( ) +0 +x +f x +→ += += f(0) and hence f is continuous at 0. This means that f is +continuous at every point in its domain and hence, f is a continuous function. +Example 14 Show that every polynomial function is continuous. +Solution Recall that a function p is a polynomial function if it is defined by +p(x) = a0 + a1 x + ... + an xn for some natural number n, an ≠ 0 and ai ∈ R. Clearly this +function is defined for every real number. For a fixed real number c, we have +lim +( ) +( ) +x +c p x +p c +→ += +By definition, p is continuous at c. Since c is any real number, p is continuous at +every real number and hence p is a continuous function. +Example 15 Find all the points of discontinuity of the greatest integer function defined +by f (x) = [x], where [x] denotes the greatest integer less than or equal to x. +Solution First observe that f is defined for all real numbers. Graph of the function is +given in Fig 5.8. From the graph it looks like that f is discontinuous at every integral +point. Below we explore, if this is true. +Fig 5.8 +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +113 +Case 1 Let c be a real number which is not equal to any integer. It is evident from the +graph that for all real numbers close to c the value of the function is equal to [c]; i.e., +lim +( ) +lim [ ] +[ ] +x +c +x +c +f x +x +c +→ +→ += += +. Also f(c) = [c] and hence the function is continuous at all real +numbers not equal to integers. +Case 2 Let c be an integer. Then we can find a sufficiently small real number +r > 0 such that [c – r] = c – 1 whereas [c + r] = c. +This, in terms of limits mean that +lim +x +c− +→ +f (x) = c – 1, lim +x +c+ +→ +f (x) = c +Since these limits cannot be equal to each other for any c, the function is +discontinuous at every integral point. +5.2.1 Algebra of continuous functions +In the previous class, after having understood the concept of limits, we learnt some +algebra of limits. Analogously, now we will study some algebra of continuous functions. +Since continuity of a function at a point is entirely dictated by the limit of the function at +that point, it is reasonable to expect results analogous to the case of limits. +Theorem 1 Suppose f and g be two real functions continuous at a real number c. +Then +(1) +f + g is continuous at x = c. +(2) +f – g is continuous at x = c. +(3) +f . g is continuous at x = c. +(4) +f +g + + + + + + + is continuous at x = c, (provided g(c) ≠ 0). +Proof We are investigating continuity of (f + g) at x = c. Clearly it is defined at +x = c. We have +lim( +)( ) +x +c f +g +x +→ ++ + = lim[ ( ) +( )] +x +c f x +g x +→ ++ +(by definition of f + g) += lim +( ) +lim ( ) +x +c +x +c +f x +g x +→ +→ ++ +(by the theorem on limits) += f (c) + g(c) +(as f and g are continuous) += (f + g) (c) +(by definition of f + g) +Hence, f + g is continuous at x = c. +Proofs for the remaining parts are similar and left as an exercise to the reader. +Reprint 2025-26 + + MATHEMATICS +114 +Remarks +(i) +As a special case of (3) above, if f is a constant function, i.e., f (x) = λ for some +real number λ, then the function (λ . g) defined by (λ . g) (x) = λ . g(x) is also +continuous. In particular if λ = – 1, the continuity of f implies continuity of – f. +(ii) +As a special case of (4) above, if f is the constant function f (x) = λ, then the +function g +λ defined by +( ) +( ) +x +g +g x +λ +λ += +is also continuous wherever g(x) ≠ 0. In +particular, the continuity of g implies continuity of 1 +g . +The above theorem can be exploited to generate many continuous functions. They +also aid in deciding if certain functions are continuous or not. The following examples +illustrate this: +Example 16 Prove that every rational function is continuous. +Solution Recall that every rational function f is given by +( ) +( ) +, +( ) +0 +( ) +p x +f x +q x +q x += +≠ +where p and q are polynomial functions. The domain of f is all real numbers except +points at which q is zero. Since polynomial functions are continuous (Example 14), f is +continuous by (4) of Theorem 1. +Example 17 Discuss the continuity of sine function. +Solution To see this we use the following facts +0 +lim sin +0 +x +x +→ += +We have not proved it, but is intuitively clear from the graph of sin x near 0. +Now, observe that f (x) = sin x is defined for every real number. Let c be a real +number. Put x = c + h. If x → c we know that h → 0. Therefore +lim +( ) +x +c f x +→ + = lim sin +x +c +x +→ += +0 +lim sin( +) +h +c +h +→ ++ += +0 +lim [sin cos +cos sin ] +h +c +h +c +h +→ ++ += +0 +0 +lim [sin cos ] +lim [cos sin ] +h +h +c +h +c +h +→ +→ ++ += sin c + 0 = sin c = f (c) +Thus lim +x +c +→ f (x) = f(c) and hence f is a continuous function. +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +115 +Remark A similar proof may be given for the continuity of cosine function. +Example 18 Prove that the function defined by f(x) = tan x is a continuous function. +Solution The function f (x) = tan x = sin +cos +x +x . This is defined for all real numbers such +that cos x ≠ 0, i.e., x ≠ (2n +1) 2 +π . We have just proved that both sine and cosine +functions are continuous. Thus tan x being a quotient of two continuous functions is +continuous wherever it is defined. +An interesting fact is the behaviour of continuous functions with respect to +composition of functions. Recall that if f and g are two real functions, then +(f o g) (x) = f(g (x)) +is defined whenever the range of g is a subset of domain of f. The following theorem +(stated without proof) captures the continuity of composite functions. +Theorem 2 Suppose f and g are real valued functions such that (f o g) is defined at c. +If g is continuous at c and if f is continuous at g(c), then (f o g) is continuous at c. +The following examples illustrate this theorem. +Example 19 Show that the function defined by f(x) = sin (x2) is a continuous function. +Solution Observe that the function is defined for every real number. The function +f may be thought of as a composition g o h of the two functions g and h, where +g (x) = sin x and h(x) = x2. Since both g and h are continuous functions, by Theorem 2, +it can be deduced that f is a continuous function. +Example 20 Show that the function f defined by +f (x) = |1 – x + | x||, +where x is any real number, is a continuous function. +Solution Define g by g(x) = 1 – x + |x| and h by h(x) = |x| for all real x. Then +(h o g) (x) = h (g (x)) += h (1– x + |x |) += |1– x + | x|| = f (x) +In Example 7, we have seen that h is a continuous function. Hence g being a sum +of a polynomial function and the modulus function is continuous. But then f being a +composite of two continuous functions is continuous. +Reprint 2025-26 + + MATHEMATICS +116 +EXERCISE 5.1 +1. +Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5. +2. +Examine the continuity of the function f(x) = 2x2 – 1 at x = 3. +3. +Examine the following functions for continuity. +(a) +f(x) = x – 5 +(b) +f(x) = +1 +5 +x − +, x ≠ 5 +(c) +f(x) = +2 +25 +5 +x +x +− ++ +, x ≠ –5 +(d) +f(x) = |x – 5 | +4. +Prove that the function f(x) = xn is continuous at x = n, where n is a positive +integer. +5. +Is the function f defined by +, if +1 +( ) +5, if +>1 +x +x +f x +x +≤ + +=  + +continuous at x = 0? At x = 1? At x = 2? +Find all points of discontinuity of f, where f is defined by +6. +2 +3, if +2 +( ) +2 +3, if +> 2 +x +x +f x +x +x ++ +≤ + +=  +− + +7. +| +| 3, if +3 +( ) +2 , if +3 +< 3 +6 +2, if +3 +x +x +f x +x +x +x +x ++ +≤− + + += +− +− +< + + ++ +≥ + +8. +| +|, if +0 +( ) +0, +if +0 +x +x +f x +x +x + +≠ + +=  + += + +9. +, if +0 +| +| +( ) +1, +if +0 +x +x +x +f x +x + +< + +=  +− +≥ + +10. +2 +1, if +1 +( ) +1, if +1 +x +x +f x +x +x ++ +≥ + +=  ++ +< + +11. +3 +2 +3, if +2 +( ) +1, +if +2 +x +x +f x +x +x + +− +≤ + +=  ++ +> + +12. +10 +2 +1, if +1 +( ) +, +if +1 +x +x +f x +x +x + +− +≤ + +=  +> + +13. +Is the function defined by +5, if +1 +( ) +5, if +1 +x +x +f x +x +x ++ +≤ + += − +> + +a continuous function? +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +117 +Discuss the continuity of the function f, where f is defined by +14. +3, if 0 +1 +( ) +4, if 1 +3 +5, if 3 +10 +x +f x +x +x +≤ +≤ + + += +< +< + + +≤ +≤ + +15. +2 , if +0 +( ) +0, +if 0 +1 +4 , if +>1 +x +x +f x +x +x +x +< + + += +≤ +≤ + + +16. +2, if +1 +( ) +2 , if +1 +1 +2, +if +1 +x +f x +x +x +x +− +≤− + + += +−< +≤ + + +> + +17. +Find the relationship between a and b so that the function f defined by +1, if +3 +( ) +3, if +3 +ax +x +f x +bx +x ++ +≤ + +=  ++ +> + +is continuous at x = 3. +18. +For what value of λ is the function defined by +2 +( +2 ), if +0 +( ) +4 +1, +if +0 +x +x +x +f x +x +x +λ +− +≤ + +=  ++ +> + +continuous at x = 0? What about continuity at x = 1? +19. +Show that the function defined by g (x) = x – [x] is discontinuous at all integral +points. Here [x] denotes the greatest integer less than or equal to x. +20. +Is the function defined by f(x) = x2 – sin x + 5 continuous at x = π? +21. +Discuss the continuity of the following functions: +(a) +f(x) = sin x + cos x +(b) +f(x) = sin x – cos x +(c) +f(x) = sin x . cos x +22. +Discuss the continuity of the cosine, cosecant, secant and cotangent functions. +23. +Find all points of discontinuity of f, where +sin +, if +0 +( ) +1, +if +0 +x +x +f x +x +x +x + +< + +=  ++ +≥ + +24. +Determine if f defined by +2 +1 +sin +, if +0 +( ) +0, +if +0 +x +x +f x +x +x + +≠ + +=  + += + +is a continuous function? +Reprint 2025-26 + + MATHEMATICS +118 +25. +Examine the continuity of f, where f is defined by +sin +cos , if +0 +( ) +1, +if +0 +x +x +x +f x +x +− +≠ + += − += + +Find the values of k so that the function f is continuous at the indicated point in Exercises +26 to 29. +26. +cos , if +2 +2 +( ) +3, +if +2 +k +x +x +x +f x +x +π + +≠ +π − +=  +π + += + + at x = 2 +π +27. +2, if +2 +( ) +3, +if +2 +kx +x +f x +x + +≤ + +=  +> + + at x = 2 +28. +1, if +( ) +cos , +if +kx +x +f x +x +x ++ +≤π + +=  +> π + + at x = π +29. +1, if +5 +( ) +3 +5, if +5 +kx +x +f x +x +x ++ +≤ + +=  +− +> + + at x = 5 +30. +Find the values of a and b such that the function defined by +5, +if +2 +( ) +, if 2 +10 +21, +if +10 +x +f x +ax +b +x +x +≤ + + += ++ +< +< + + +≥ + +is a continuous function. +31. +Show that the function defined by f(x) = cos (x2) is a continuous function. +32. +Show that the function defined by f(x) = |cos x | is a continuous function. +33. +Examine that sin |x| is a continuous function. +34. +Find all the points of discontinuity of f defined by f(x) = |x | – |x + 1 |. +5.3. Differentiability +Recall the following facts from previous class. We had defined the derivative of a real +function as follows: +Suppose f is a real function and c is a point in its domain. The derivative of f at c is +defined by +0 +( +) +( ) +lim +h +f c +h +f c +h +→ ++ +− +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +119 +f (x) +xn +sin x +cos x +tan x +f ′(x) +nxn – 1 +cos x +– sin x +sec2 x +provided this limit exists. Derivative of f at c is denoted by f ′(c) or +( ( )) |c +d +f x +dx +. The +function defined by +0 +( +) +( ) +( ) +lim +h +f x +h +f x +f +x +h +→ ++ +− +′ += +wherever the limit exists is defined to be the derivative of f. The derivative of f is +denoted by f ′(x) or +( ( )) +d +f x +dx +or if y = f (x) by dy +dx or y′. The process of finding +derivative of a function is called differentiation. We also use the phrase differentiate +f(x) with respect to x to mean find f ′(x). +The following rules were established as a part of algebra of derivatives: +(1) +(u ± v)′ = u′ ± v′ +(2) +(uv)′ = u′v + uv′ (Leibnitz or product rule) +(3) +2 +u +u v +uv +v +v +′ +′ − +′ + += + + + + +, wherever v ≠ 0 (Quotient rule). +The following table gives a list of derivatives of certain standard functions: +Table 5.3 +Whenever we defined derivative, we had put a caution provided the limit exists. +Now the natural question is; what if it doesn’t? The question is quite pertinent and so is +its answer. If +0 +( +) +( ) +lim +h +f c +h +f c +h +→ ++ +− + does not exist, we say that f is not differentiable at c. +In other words, we say that a function f is differentiable at a point c in its domain if both +– +0 +( +) +( ) +lim +h +f c +h +f c +h +→ ++ +− + and +0 +( +) +( ) +lim +h +f c +h +f c +h ++ +→ ++ +− + are finite and equal. A function is said +to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b]. As +in case of continuity, at the end points a and b, we take the right hand limit and left hand +limit, which are nothing but left hand derivative and right hand derivative of the function +at a and b respectively. Similarly, a function is said to be differentiable in an interval +(a, b) if it is differentiable at every point of (a, b). +Reprint 2025-26 + + MATHEMATICS +120 +Theorem 3 If a function f is differentiable at a point c, then it is also continuous at that +point. +Proof Since f is differentiable at c, we have +( ) +( ) +lim +( ) +x +c +f x +f c +f c +x +c +→ +− += +′ +− +But for x ≠ c, we have +f (x) – f (c) = +( ) +( ) .( +) +f x +f c +x +c +x +c +− +− +− +Therefore +lim[ ( ) +( )] +x +c f x +f c +→ +− + = +( ) +( ) +lim +.( +) +x +c +f x +f c +x +c +x +c +→ +− + + +− + + +− + + +or +lim[ ( )] +lim[ ( )] +x +c +x +c +f x +f c +→ +→ +− + = +( ) +( ) +lim +.lim [( +)] +x +c +x +c +f x +f c +x +c +x +c +→ +→ +− + + +− + + +− + + += f′(c) . 0 = 0 +or +lim +( ) +x +c f x +→ + = f (c) +Hence f is continuous at x = c. +Corollary 1 Every differentiable function is continuous. +We remark that the converse of the above statement is not true. Indeed we have +seen that the function defined by f(x) = |x | is a continuous function. Consider the left +hand limit +– +0 +(0 +) +(0) +lim +1 +h +f +h +f +h +h +h +→ ++ +− +− += += − +The right hand limit +0 +(0 +) +(0) +lim +1 +h +f +h +f +h +h +h ++ +→ ++ +− += += +Since the above left and right hand limits at 0 are not equal, +0 +(0 +) +(0) +lim +h +f +h +f +h +→ ++ +− +does not exist and hence f is not differentiable at 0. Thus f is not a differentiable +function. +5.3.1 Derivatives of composite functions +To study derivative of composite functions, we start with an illustrative example. Say, +we want to find the derivative of f, where +f (x) = (2x + 1)3 +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +121 +One way is to expand (2x + 1)3 using binomial theorem and find the derivative as +a polynomial function as illustrated below. +( ) +d f x +dx + = +3 +(2 +1) +d +x +dx + + ++ + + += +3 +2 +(8 +12 +6 +1) +d +x +x +x +dx ++ ++ ++ += 24x2 + 24x + 6 += 6 (2x + 1)2 +Now, observe that +f (x) = (h o g) (x) +where g(x) = 2x + 1 and h(x) = x3. Put t = g(x) = 2x + 1. Then f(x) = h(t) = t3. Thus +df +dx = +6 (2x + 1)2 = 3(2x + 1)2 . 2 = 3t2 . 2 = dh dt +dt +dx +⋅ +The advantage with such observation is that it simplifies the calculation in finding +the derivative of, say, (2x + 1)100. We may formalise this observation in the following +theorem called the chain rule. +Theorem 4 (Chain Rule) Let f be a real valued function which is a composite of two +functions u and v; i.e., f = v o u. Suppose t = u(x) and if both dt +dx and dv +dt exist, we have +df +dv dt +dx +dt dx += +⋅ +We skip the proof of this theorem. Chain rule may be extended as follows. Suppose +f is a real valued function which is a composite of three functions u, v and w; i.e., +f = (w o u) o v. If t = v (x) and s = u (t), then +( o ) +df +d w u +dt +dw ds +dt +dx +dt +dx +ds +dt +dx += +⋅ += +⋅ +⋅ +provided all the derivatives in the statement exist. Reader is invited to formulate chain +rule for composite of more functions. +Example 21 Find the derivative of the function given by f(x) = sin (x2). +Solution Observe that the given function is a composite of two functions. Indeed, if +t = u(x) = x2 and v(t) = sin t, then +f (x) = (v o u) (x) = v(u(x)) = v(x2) = sin x2 +Reprint 2025-26 + + MATHEMATICS +122 +Put t = u(x) = x2. Observe that +cos +dv +t +dt = + and +2 +dt +x +dx = + exist. Hence, by chain rule +df +dx = +cos +2 +dv +dt +t +x +dt +dx +⋅ += +⋅ +It is normal practice to express the final result only in terms of x. Thus +df +dx = +2 +cos +2 +2 cos +t +x +x +x +⋅ += +EXERCISE 5.2 +Differentiate the functions with respect to x in Exercises 1 to 8. +1. +sin (x2 + 5) +2. cos (sin x) +3. sin (ax + b) +4. +sec (tan ( +x )) +5. +sin ( +) +cos ( +) +ax +b +cx +d ++ ++ +6. cos x3 . sin2 (x5) +7. +( +) +2 +2 cot x +8. +( +) +cos +x +9. +Prove that the function f given by +f (x) = |x – 1|, x ∈ R +is not differentiable at x = 1. +10. +Prove that the greatest integer function defined by +f (x) = [x], 0 < x < 3 +is not differentiable at x = 1 and x = 2. +5.3.2 Derivatives of implicit functions +Until now we have been differentiating various functions given in the form y = f(x). +But it is not necessary that functions are always expressed in this form. For example, +consider one of the following relationships between x and y: +x – y – π = 0 +x + sin xy – y = 0 +In the first case, we can solve for y and rewrite the relationship as y = x – π. In +the second case, it does not seem that there is an easy way to solve for y. Nevertheless, +there is no doubt about the dependence of y on x in either of the cases. When a +relationship between x and y is expressed in a way that it is easy to solve for y and +write y = f(x), we say that y is given as an explicit function of x. In the latter case it +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +123 +is implicit that y is a function of x and we say that the relationship of the second type, +above, gives function implicitly. In this subsection, we learn to differentiate implicit +functions. +Example 22 Find dy +dx if x – y = π. +Solution One way is to solve for y and rewrite the above as +y = x – π +But then +dy +dx = 1 +Alternatively, directly differentiating the relationship w.r.t., x, we have +( +) +d +x +y +dx +− + = d +dx +π +Recall that d +dx +π means to differentiate the constant function taking value π +everywhere w.r.t., x. Thus +( ) +( ) +d +d +x +y +dx +dx +− + = 0 +which implies that +dy +dx = +1 +dx +dx = +Example 23 Find dy +dx +, if y + sin y = cos x. +Solution We differentiate the relationship directly with respect to x, i.e., +(sin ) +dy +d +y +dx +dx ++ + = +(cos ) +d +x +dx +which implies using chain rule +cos +dy +dy +y +dx +dx ++ +⋅ + = – sin x +This gives +dy +dx = +sin +1 +cos +x +y +−+ +where +y ≠(2n + 1) π +Reprint 2025-26 + + MATHEMATICS +124 +5.3.3 Derivatives of inverse trigonometric functions +We remark that inverse trigonometric functions are continuous functions, but we will +not prove this. Now we use chain rule to find derivatives of these functions. +Example 24 Find the derivative of f given by f(x) = sin–1 x assuming it exists. +Solution Let y = sin–1 x. Then, x = sin y. +Differentiating both sides w.r.t. x, we get +1 = cos y dy +dx +which implies that +dy +dx = +1 +1 +1 +cos +cos(sin +) +y +x +− += +Observe that this is defined only for cos y ≠ 0, i.e., sin–1 x ≠ +, +2 2 +π π +− +, i.e., x ≠ – 1, 1, +i.e., x ∈ (– 1, 1). +To make this result a bit more attractive, we carry out the following manipulation. +Recall that for x ∈ (– 1, 1), sin (sin–1 x) = x and hence +cos2 y = 1 – (sin y)2 = 1 – (sin (sin–1 x))2 = 1 – x2 +Also, since y ∈ +, +2 2 +π π + + +− + + + +, cos y is positive and hence cos y = +2 +1 +x +− +Thus, for x ∈ (– 1, 1), +2 +1 +1 +cos +1 +dy +dx +y +x += += +− +2 +1 +1 +x +− +2 +1 +1 +x +− +− +2 +1 +1 +x ++ +f(x) sin–1 x cos-1 x tan-1x +Domain off (-1, 1) (-1, 1) R +f 1(x) +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +125 +EXERCISE 5.3 +Find dy +dx in the following: +1. +2x + 3y = sin x +2. 2x + 3y = sin y +3. ax + by2 = cos y +4. +xy + y2 = tan x + y +5. x2 + xy + y2 = 100 +6. x3 + x2y + xy2 + y3 = 81 +7. +sin2 y + cos xy = κ +8. sin2 x + cos2 y = 1 +9. y = sin–1 +2 +2 +1 +x +x + + + + ++ + + +10. +y = tan–1 +3 +2 +3 +, +1 +3 +x +x +x + + +− + + +− + + + +1 +1 +3 +3 +x +− +< +< +11. +2 +1 +2 +1 +, +cos +0 +1 +1 +x +y +x +x +− + +− += +< +< + + ++ + + +12. +2 +1 +2 +1 +, +sin +0 +1 +1 +x +y +x +x +− + +− += +< +< + + ++ + + +13. +1 +2 +2 +, +cos +1 +1 +1 +x +y +x +x +− + += +−< +< + + ++ + + +14. +( +) +1 +2 +1 +1 +, +sin +2 +1 +2 +2 +y +x +x +x +− += +− +− +< +< +15. +1 +2 +1 +1 +, +sec +0 +2 +1 +2 +y +x +x +− + += +< +< + + +− + + +5.4 Exponential and Logarithmic Functions +Till now we have learnt some aspects of different classes of functions like polynomial +functions, rational functions and trigonometric functions. In this section, we shall +learn about a new class of (related) functions called exponential functions and logarithmic +functions. It needs to be emphasized that many statements made in this section are +motivational and precise proofs of these are well beyond the scope of this text. +The Fig 5.9 gives a sketch of y = f1(x) = x, y = f2(x) = x2, y = f3(x) = x3 and y = f4(x) += x4. Observe that the curves get steeper as the power of x increases. Steeper the +curve, faster is the rate of growth. What this means is that for a fixed increment in the +value of x (> 1), the increment in the value of y = fn (x) increases as n increases for n += 1, 2, 3, 4. It is conceivable that such a statement is true for all positive values of n, +Reprint 2025-26 + + MATHEMATICS +126 +where fn (x) = xn. Essentially, this means +that the graph of y = fn (x) leans more +towards the y-axis as n increases. For +example, consider f10(x) = x10 and f15(x) += x15. If x increases from 1 to 2, f10 +increases from 1 to 210 whereas f15 +increases from 1 to 215. Thus, for the same +increment in x, f15 grow faster than f10. +Upshot of the above discussion is that +the growth of polynomial functions is +dependent on the degree of the polynomial +function – higher the degree, greater is +the growth. The next natural question is: +Is there a function which grows faster than any polynomial function. The answer is in +affirmative and an example of such a function is +y = f(x) = 10x. +Our claim is that this function f grows faster than fn (x) = xn for any positive integer n. +For example, we can prove that 10x grows faster than f100 (x) = x100. For large values +of x like x = 103, note that f100 (x) = (103)100 = 10300 whereas f(103) = +3 +10 +10 += 101000. +Clearly f (x) is much greater than f100 (x). It is not difficult to prove that for all +x > 103, f (x) > f100 (x). But we will not attempt to give a proof of this here. Similarly, by +choosing large values of x, one can verify that f(x) grows faster than fn (x) for any +positive integer n. +Definition 3 The exponential function with positive base b > 1 is the function +y = f (x) = bx +The graph of y = 10x is given in the Fig 5.9. +It is advised that the reader plots this graph for particular values of b like 2, 3 and 4. +Following are some of the salient features of the exponential functions: +(1) +Domain of the exponential function is R, the set of all real numbers. +(2) +Range of the exponential function is the set of all positive real numbers. +(3) +The point (0, 1) is always on the graph of the exponential function (this is a +restatement of the fact that b0 = 1 for any real b > 1). +(4) +Exponential function is ever increasing; i.e., as we move from left to right, the +graph rises above. +Fig 5.9 +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +127 +(5) +For very large negative values of x, the exponential function is very close to 0. In +other words, in the second quadrant, the graph approaches x-axis (but never +meets it). +Exponential function with base 10 is called the common exponential function. In +the Appendix A.1.4 of Class XI, it was observed that the sum of the series +1 +1 +1 +... +1! +2! ++ ++ ++ +is a number between 2 and 3 and is denoted by e. Using this e as the base we obtain an +extremely important exponential function y = ex. +This is called natural exponential function. +It would be interesting to know if the inverse of the exponential function exists and +has nice interpretation. This search motivates the following definition. +Definition 4 Let b > 1 be a real number. Then we say logarithm of a to base b is x if +bx = a. +Logarithm of a to base b is denoted by logb a. Thus logb a = x if bx = a. Let us +work with a few explicit examples to get a feel for this. We know 23 = 8. In terms of +logarithms, we may rewrite this as log2 8 = 3. Similarly, 104 = 10000 is equivalent to +saying log10 10000 = 4. Also, 625 = 54 = 252 is equivalent to saying log5 625 = 4 or +log25 625 = 2. +On a slightly more mature note, fixing a base b > 1, we may look at logarithm as +a function from positive real numbers to all real numbers. This function, called the +logarithmic function, is defined by +logb : R+ →R +x →logb x = y if by = x +As before if the base b = 10, we say it +is common logarithms and if b = e, then +we say it is natural logarithms. Often +natural logarithm is denoted by ln. In this +chapter, log x denotes the logarithm +function to base e, i.e., ln x will be written +as simply log x. The Fig 5.10 gives the plots +of logarithm function to base 2, e and 10. +Some of the important observations +about the logarithm function to any base +b > 1 are listed below: + Fig 5.10 +Reprint 2025-26 + + MATHEMATICS +128 + Fig 5.11 +(1) +We cannot make a meaningful definition of logarithm of non-positive numbers +and hence the domain of log function is R+. +(2) +The range of log function is the set of all real numbers. +(3) +The point (1, 0) is always on the graph of the log function. +(4) +The log function is ever increasing, +i.e., as we move from left to right +the graph rises above. +(5) +For x very near to zero, the value +of log x can be made lesser than +any given real number. In other +words in the fourth quadrant the +graph approaches y-axis (but +never meets it). +(6) +Fig 5.11 gives the plot of y = ex and +y = ln x. It is of interest to observe +that the two curves are the mirror +images of each other reflected in the line y = x. +Two properties of ‘log’ functions are proved below: +(1) +There is a standard change of base rule to obtain loga p in terms of logb p. Let +loga p = α, logb p = β and logb a = γ. This means aα = p, bβ = p and bγ = a. +Substituting the third equation in the first one, we have +(bγ)α = bγα = p +Using this in the second equation, we get +bβ = p = bγα +which implies +β = αγ or α = β +γ . But then +loga p = log +log +b +b +p +a +(2) +Another interesting property of the log function is its effect on products. Let +logb pq = α. Then bα = pq. If logb p = β and logb q = γ, then bβ = p and bγ = q. +But then bα = pq = bβbγ = bβ + γ +which implies α = β + γ, i.e., +logb pq = logb p + logb q +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +129 +A particularly interesting and important consequence of this is when p = q. In +this case the above may be rewritten as +logb p2 = logb p + logb p = 2 log p +An easy generalisation of this (left as an exercise!) is +logb pn = n log p +for any positive integer n. In fact this is true for any real number n, but we will +not attempt to prove this. On the similar lines the reader is invited to verify +logb +x +y = logb x – logb y +Example 25 Is it true that x = elog x for all real x? +Solution First, observe that the domain of log function is set of all positive real numbers. +So the above equation is not true for non-positive real numbers. Now, let y = elog x. If +y > 0, we may take logarithm which gives us log y = log (elog x) = log x . log e = log x. Thus +y = x. Hence x = elog x is true only for positive values of x. +One of the striking properties of the natural exponential function in differential +calculus is that it doesn’t change during the process of differentiation. This is captured +in the following theorem whose proof we skip. +Theorem 5* +(1) +The derivative of ex w.r.t., x is ex; i.e., d +dx (ex) = ex. +(2) +The derivative of log x w.r.t., x is 1 +x ; i.e., d +dx (log x) = 1 +x . +Example 26 Differentiate the following w.r.t. x: +(i) +e–x +(ii) sin (log x), x > 0 + (iii) cos–1 (ex) + (iv) ecos x +Solution +(i) +Let y = e– x. Using chain rule, we have +dy +dx = +x +d +e +dx +−⋅ + (– x) = – e– x +(ii) +Let y = sin (log x). Using chain rule, we have +dy +dx = +cos (log ) +cos (log ) +(log ) +d +x +x +x +dx +x +⋅ += +* Please see supplementary material on Page 222. +Reprint 2025-26 + + MATHEMATICS +130 +(iii) +Let y = cos–1 (ex). Using chain rule, we have +dy +dx = +2 +2 +1 +( +) +1 +( +) +1 +x +x +x +x +d +e +e +dx +e +e +− +− +⋅ += +− +− +(iv) +Let y = ecos x. Using chain rule, we have +dy +dx = +cos +cos +( sin ) +(sin ) +x +x +e +x +x e +⋅− += − +EXERCISE 5.4 +Differentiate the following w.r.t. x: +1. +sin +xe +x +2. +1 +sin +x +e +− +3. +3xe +4. +sin (tan–1 e–x) +5. log (cos ex) +6. +2 +5 +... +x +x +x +e +e +e ++ ++ ++ +7. +, +0 +x +e +x > +8. log (log x), x > 1 +9. +cos , +0 +log +x +x +x +> +10. +cos (log x + ex), x > 0 +5.5. Logarithmic Differentiation +In this section, we will learn to differentiate certain special class of functions given in +the form +y = f (x) = [u(x)]v (x) +By taking logarithm (to base e) the above may be rewritten as +log y = v(x) log [u(x)] +Using chain rule we may differentiate this to get +1 +1 +( ) +( ) +dy +v x +y dx +u x +⋅ += +⋅ + . u′(x) + v′(x) . log [u(x)] +which implies that +[ +] +( ) +( ) +( ) log +( ) +( ) +dy +v x +y +u x +v x +u x +dx +u x + + += +⋅′ ++ ′ +⋅ + + + + +The main point to be noted in this method is that f(x) and u(x) must always be +positive as otherwise their logarithms are not defined. This process of differentiation is +known as logarithms differentiation and is illustrated by the following examples: +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +131 +Example 27 Differentiate +2 +2 +( +3) ( +4) +3 +4 +5 +x +x +x +x +− ++ ++ ++ + w.r.t. x. +Solution Let +2 +2 +( +3) ( +4) +(3 +4 +5) +x +x +y +x +x +− ++ += ++ ++ +Taking logarithm on both sides, we have +log y = 1 +2 [log (x – 3) + log (x2 + 4) – log (3x2 + 4x + 5)] +Now, differentiating both sides w.r.t. x, we get +1 dy +y dx +⋅ + = +2 +2 +1 +1 +2 +6 +4 +2 ( +3) +4 +3 +4 +5 +x +x +x +x +x +x ++ + + ++ +− + + +− ++ ++ ++ + + +or +dy +dx = +2 +2 +1 +2 +6 +4 +2 +( +3) +4 +3 +4 +5 +y +x +x +x +x +x +x ++ + + ++ +− + + +− ++ ++ ++ + + += +2 +2 +2 +2 +1 +( +3)( +4) +1 +2 +6 +4 +2 +( +3) +3 +4 +5 +4 +3 +4 +5 +x +x +x +x +x +x +x +x +x +x +− ++ ++ + + ++ +− + + +− ++ ++ ++ ++ ++ + + +Example 28 Differentiate ax w.r.t. x, where a is a positive constant. +Solution Let y = ax. Then +log y = x log a +Differentiating both sides w.r.t. x, we have +1 dy +y dx = log a +or +dy +dx = y log a +Thus +( +) +x +d +a +dx + = ax log a +Alternatively +( +) +x +d +a +dx + = +log +log +( +) +( log ) +x +a +x +a +d +d +e +e +x +a +dx +dx += += ex log a . log a = ax log a. +Reprint 2025-26 + + MATHEMATICS +132 +Example 29 Differentiate xsin x, x > 0 w.r.t. x. +Solution Let y = xsin x. Taking logarithm on both sides, we have +log y = sin x log x +Therefore +1 . dy +y dx = sin +(log ) +log +(sin ) +d +d +x +x +x +x +dx +dx ++ +or +1 dy +y dx = +1 +(sin ) +log +cos +x +x +x +x + +or +dy +dx = +sin +cos +log +x +y +x +x +x + + ++ + + + + += +sin +sin +cos +log +x +x +x +x +x +x + + ++ + + + + += +sin +1 +sin +sin +cos +log +x +x +x +x +x +x +x +−⋅ ++ +⋅ +Example 30 Find dy +dx , if yx + xy + xx = ab. +Solution Given that yx + xy + xx = ab. +Putting u = yx, v = xy and w = xx, we get u + v + w = ab +Therefore +0 +du +dv +dw +dx +dx +dx ++ ++ += +... (1) +Now, u = yx. Taking logarithm on both sides, we have +log u = x log y +Differentiating both sides w.r.t. x, we have +1 du +u dx +⋅ + = +(log ) +log +( ) +d +d +x +y +y +x +dx +dx ++ += +1 +log +1 +dy +x +y +y dx +⋅ ++ +⋅ +So +du +dx = +log +log +x +x dy +x dy +u +y +y +y +y dx +y dx + + + + ++ += ++ + + + + + + + + ... (2) +Also v = xy +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +133 +Taking logarithm on both sides, we have +log v = y log x +Differentiating both sides w.r.t. x, we have +1 dv +v dx +⋅ + = +(log ) +log +d +dy +y +x +x +dx +dx ++ += +1 +log +dy +y +x +x +dx +⋅ ++ +⋅ +So +dv +dx = +log +y +dy +v +x +x +dx + + ++ + + + + += +log +y +y +dy +x +x +x +dx + + ++ + + + + +... (3) +Again +w = xx +Taking logarithm on both sides, we have +log w = x log x. +Differentiating both sides w.r.t. x, we have +1 dw +w dx +⋅ + = +(log ) +log +( ) +d +d +x +x +x +x +dx +dx ++ +⋅ += +1 +log +1 +x +x +x +⋅ ++ +⋅ +i.e. +dw +dx = w (1 + log x) += xx (1 + log x) +... (4) +From (1), (2), (3), (4), we have +log +log +x +y +x dy +y +dy +y +y +x +x +y dx +x +dx + + + + ++ ++ ++ + + + + + + + + + + xx (1 + log x) = 0 +or +(x . yx – 1 + xy . log x) dy +dx = – xx (1 + log x) – y . xy–1 – yx log y +Therefore +dy +dx = +1 +1 +[ +log +. +(1 +log )] +. +log +x +y +x +x +y +y +y +y x +x +x +x y +x +x +− +− +− ++ ++ ++ ++ +Reprint 2025-26 + + MATHEMATICS +134 +EXERCISE 5.5 +Differentiate the functions given in Exercises 1 to 11 w.r.t. x. +1. +cos x . cos 2x . cos 3x +2. +( +1) ( +2) +( +3) ( +4) ( +5) +x +x +x +x +x +− +− +− +− +− +3. +(log x)cos x +4. +xx – 2sin x +5. +(x + 3)2 . (x + 4)3 . (x + 5)4 +6. +1 +1 +1 +x +x +x +x +x + + ++ + + + + + + ++ ++ + + + + +7. +(log x)x + xlog x +8. +(sin x)x + sin–1 +x +9. +xsin x + (sin x)cos x +10. +2 +cos +2 +1 +1 +x +x +x +x +x ++ ++ +− +11. +(x cos x)x + +1 +( sin ) x +x +x +Find dy +dx of the functions given in Exercises 12 to 15. +12. +xy + yx = 1 +13. +yx = xy +14. +(cos x)y = (cos y)x +15. +xy = e(x – y) +16. +Find the derivative of the function given by f(x) = (1 + x) (1 + x2) (1 + x4) (1 + x8) +and hence find f′(1). +17. +Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned below: +(i) by using product rule +(ii) by expanding the product to obtain a single polynomial. +(iii) by logarithmic differentiation. +Do they all give the same answer? +18. +If u, v and w are functions of x, then show that +d +dx (u. v. w) = du +dx v. w + u . dv +dx . w + u . v dw +dx +in two ways - first by repeated application of product rule, second by logarithmic +differentiation. +5.6 Derivatives of Functions in Parametric Forms +Sometimes the relation between two variables is neither explicit nor implicit, but some +link of a third variable with each of the two variables, separately, establishes a relation +between the first two variables. In such a situation, we say that the relation between +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +135 +them is expressed via a third variable. The third variable is called the parameter. More +precisely, a relation expressed between two variables x and y in the form +x = f(t), y = g (t) is said to be parametric form with t as a parameter. +In order to find derivative of function in such form, we have by chain rule. +dy +dt = dy dx +dx dt +⋅ +or +dy +dx = +whenever +0 +dy +dx +dt +dx +dt +dt + + +≠ + + + + +Thus +dy +dx = +( ) +as +( ) and +( ) +( ) +g t +dy +dx +g t +f t +f t +dt +dt +′ + + += +′ += +′ + + +′ + + [provided f′(t) ≠ 0] +Example 31 Find dy +dx , if x = a cos θ, y = a sin θ. +Solution Given that +x = a cos θ, y = a sin θ +Therefore +dx +dθ = – a sin θ, dy +dθ = a cos θ +Hence +dy +dx = +cos +cot +sin +dy +a +d +dx +a +d +θ +θ = += − +θ +− +θ +θ +Example 32 Find dy +dx +, if x = at2, y = 2at. +Solution Given that x = at2, y = 2at +So +dx +dt = 2at and dy +dt = 2a +Therefore +dy +dx = +2 +1 +2 +dy +a +dt +dx +at +t +dt += += +Reprint 2025-26 + + MATHEMATICS +136 +Example 33 Find dy +dx +, if x = a (θ + sin θ), y = a (1 – cos θ). +Solution We have dx +dθ = a(1 + cos θ), dy +dθ = a (sin θ) +Therefore +dy +dx = +sin +tan +(1 +cos ) +2 +dy +a +d +dx +a +d +θ +θ +θ = += ++ +θ +θ +ANote It may be noted here that dy +dx is expressed in terms of parameter only +without directly involving the main variables x and y. +Example 34 Find +2 +2 +2 +3 +3 +3 +, if +dy +x +y +a +dx ++ += +. +Solution Let x = a cos3 θ, y = a sin3 θ. Then +2 +2 +3 +3 +x +y ++ + = +2 +2 +3 +3 +3 +3 +( cos +) +( sin +) +a +a +θ ++ +θ += +2 +2 +2 +2 +3 +3 +(cos +(sin +) +a +a +θ + +θ = +Hence, x = a cos3θ, y = a sin3θ is parametric equation of +2 +2 +2 +3 +3 +3 +x +y +a ++ += +Now +dx +dθ = – 3a cos2 θ sin θ and dy +dθ = 3a sin2 θ cos θ +Therefore +dy +dx = +2 +3 +2 +3 sin +cos +tan +3 cos +sin +dy +a +y +d +dx +x +a +d +θ +θ +θ = += − +θ = − +− +θ +θ +θ +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +137 +EXERCISE 5.6 +If x and y are connected parametrically by the equations given in Exercises 1 to 10, +without eliminating the parameter, Find dy +dx . +1. +x = 2at2, y = at4 +2. +x = a cos θ, y = b cos θ +3. +x = sin t, y = cos 2t +4. +x = 4t, y = 4 +t +5. +x = cos θ – cos 2θ, y = sin θ – sin 2θ +6. +x = a (θ – sin θ), y = a (1 + cos θ) +7. +x = +3 +sin +cos2 +t +t , +3 +cos +cos2 +t +y +t += +8. +cos +log tan 2 +t +x +a +t + + += ++ + + + + y = a sin t +9. +x = a sec θ, y = b tan θ +10. +x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ) +11. +If +1 +1 +sin +cos +, +, show that +t +t +dy +y +x +a +y +a +dx +x +− +− += += += − +5.7 Second Order Derivative +Let +y = f (x). Then +dy +dx = f ′(x) +... (1) +If f′(x) is differentiable, we may differentiate (1) again w.r.t. x. Then, the left hand +side becomes d +dy +dx +dx + + + + + + which is called the second order derivative of y w.r.t. x and +is denoted by +2 +2 +d y +dx +. The second order derivative of f(x) is denoted by f ″(x). It is also +denoted by D2 y or y″ or y2 if y = f(x). We remark that higher order derivatives may be +defined similarly. +Reprint 2025-26 + + MATHEMATICS +138 +Example 35 Find +2 +2 +d y +dx +, if y = x3 + tan x. +Solution Given that y = x3 + tan x. Then +dy +dx = 3x2 + sec2 x +Therefore +2 +2 +d y +dx + = +( +) +2 +2 +3 +sec +d +x +x +dx ++ += 6x + 2 sec x . sec x tan x = 6x + 2 sec2 x tan x +Example 36 If y = A sin x + B cos x, then prove that +2 +2 +0 +d y +y +dx ++ += +. +Solution We have +dy +dx = A cos x – B sin x +and +2 +2 +d y +dx + = d +dx (A cos x – B sin x) += – A sin x – B cos x = – y +Hence +2 +2 +d y +dx + + y = 0 +Example 37 If y = 3e2x + 2e3x, prove that +2 +2 +5 +6 +0 +d y +dy +y +dx +dx +− ++ += +. +Solution Given that y = 3e2x + 2e3x. Then +dy +dx = 6e2x + 6e3x = 6 (e2x + e3x) +Therefore +2 +2 +d y +dx + = 12e2x + 18e3x = 6 (2e2x + 3e3x) +Hence +2 +2 +5 +d y +dy +dx +dx +− + + 6y = 6 (2e2x + 3e3x) +– 30 (e2x + e3x) + 6 (3e2x + 2e3x) = 0 +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +139 +Example 38 If y = sin–1 x, show that (1 – x2) +2 +2 +0 +d y +dy +x dx +dx +− += +. +Solution We have y = sin–1x. Then +dy +dx = +2 +1 +(1 +) +x +− +or +2 +(1 +) +1 +dy +x +dx +− += +So +2 +(1 +) . +0 +d +dy +x +dx +dx + + +− += + + + + +or +( +) +2 +2 +2 +2 +(1 +) +(1 +) +0 +d y +dy +d +x +x +dx dx +dx +− +⋅ ++ +⋅ +− += +or +2 +2 +2 +2 +2 +(1 +) +0 +2 1 +d y +dy +x +x +dx +dx +x +− +⋅ +− +⋅ += +− +Hence +2 +2 +2 +(1 +) +0 +d y +dy +x +x dx +dx +− +− += +Alternatively, Given that y = sin–1 x, we have +1 +2 +1 +1 +y +x += +− +, i.e., ( +) +2 +2 +1 +1 +1 +x +y +− += +So +2 +2 +1 +2 +1 +(1 +) . 2 +(0 +2 ) +0 +x +y y +y +x +− ++ +− += +Hence +(1 – x2) y2 – xy1 = 0 +EXERCISE 5.7 +Find the second order derivatives of the functions given in Exercises 1 to 10. +1. +x2 + 3x + 2 +2. x20 +3. x . cos x +4. +log x +5. x3 log x +6. ex sin 5x +7. +e6x cos 3x +8. tan–1 x +9. log (log x) +10. +sin (log x) +11. +If y = 5 cos x – 3 sin x, prove that +2 +2 +0 +d y +y +dx ++ += +Reprint 2025-26 + + MATHEMATICS +140 +12. +If y = cos–1 x, Find +2 +2 +d y +dx +in terms of y alone. +13. +If y = 3 cos (log x) + 4 sin (log x), show that x2 y2 + xy1 + y = 0 +14. +If y = Aemx + Benx, show that +2 +2 +( +) +0 +d y +dy +m +n +mny +dx +dx +− ++ ++ += +15. +If y = 500e7x + 600e–7x, show that +2 +2 +49 +d y +y +dx += +16. +If ey + (x + 1) = 1, show that +2 +2 +2 +d y +dy +dx +dx + + +=  + + + +17. +If y = (tan–1 x)2, show that (x2 + 1)2 y2 + 2x (x2 + 1) y1 = 2 +Miscellaneous Examples +Example 39 Differentiate w.r.t. x, the following function: +(i) +2 +1 +3 +2 +2 +4 +x +x ++ ++ ++ +(ii) log7 (log x) +Solution +(i) +Let y = +2 +1 +3 +2 +2 +4 +x +x ++ ++ ++ += +1 +1 +2 +2 +2 +(3 +2) +(2 +4) +x +x +− ++ ++ ++ +Note that this function is defined at all real numbers +2 +3 +x > − +. Therefore +dy +dx = +1 +1 +1 +1 +2 +2 +2 +2 +1 +1 +(3 +2) +(3 +2) +(2 +4) +(2 +4) +2 +2 +d +d +x +x +x +x +dx +dx +− +−− + + ++ +⋅ ++ ++ − ++ +⋅ ++ + + + + += 1 +2 3 +2 +3 +1 +2 +2 +4 +4 +1 +2 +2 +3 +2 +( +) +( ) +( +) +x +x +x ++ +⋅ +− + + ++ +⋅ +− +− += +( +) +3 +2 +2 +3 +2 +2 3 +2 +2 +4 +x +x +x +− ++ ++ +This is defined for all real numbers +2 +3 +x > − +. +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +141 +(ii) +Let y = log7 (log x) = log (log ) +log7 +x (by change of base formula). +The function is defined for all real numbers x > 1. Therefore +dy +dx = +1 +(log (log )) +log7 +d +x +dx += +1 +1 +(log ) +log7 log +d +x +x dx +⋅ += +1 +log7 log +x +x +Example 40 Differentiate the following w.r.t. x. +(i) +cos –1 (sin x) +(ii) +1 +sin +tan +1 +cos +x +x +− + + + ++ + + + (iii) +1 +1 +2 +sin +1 +4 +x +x ++ +− + + + ++ + + +Solution +(i) +Let f (x) = cos –1 (sin x). Observe that this function is defined for all real numbers. +We may rewrite this function as +f(x) = cos –1 (sin x) += cos +cos +− +− + + + + + + + + +1 +2 +π +x += 2 +x +π − +Thus +f ′(x) = – 1. +(ii) +Let f(x) = tan –1 +sin +1 +cos +x +x + + + + ++ + + +. Observe that this function is defined for all real +numbers, where cos x ≠ – 1; i.e., at all odd multiplies of π. We may rewrite this +function as +f(x) = +1 +sin +tan +1 +cos +x +x +− + + + ++ + + += +1 +2 +2 sin +cos +2 +2 +tan +2cos 2 +x +x +x +− + + + + + + + + + + + + + + + + + + + + + + + + +Reprint 2025-26 + + MATHEMATICS +142 += +1 +tan +tan 2 +2 +x +x +− + + += + + + + + + + + +Observe that we could cancel cos +2 +x + + + + + +in both numerator and denominator as it +is not equal to zero. Thus f ′(x) = 1. +2 +(iii) +Let f(x) = sin–1 +1 +2 +1 +4 +x +x ++ + + + + ++ + + +. To find the domain of this function we need to find all +x such that +1 +2 +1 +1 +1 +4 +x +x ++ +−≤ +≤ ++ +. Since the quantity in the middle is always positive, +we need to find all x such that +1 +2 +1 +1 +4 +x +x ++ +≤ ++ +, i.e., all x such that 2x + 1 ≤ 1 + 4x. We +may rewrite this as 2 ≤ 1 +2x + 2x which is true for all x. Hence the function +is defined at every real number. By putting 2x = tan θ, this function may be +rewritten as +f(x) = +1 +1 +2 +sin +1 +4 +x +x ++ +− + + + ++ + + += sin− +⋅ ++ ( +) + + + + + + + + +1 +2 +2 +2 +1 +2 +x +x += +1 +2 +2tan +sin +1 +tan +− +θ + + + + ++ +θ + + += sin –1 [sin 2θ] += 2θ = 2 tan – 1 (2x) +Thus +f ′(x) = +( +) +2 +1 +2 +(2 ) +1 +2 +x +x +d +dx +⋅ +⋅ ++ += +2 +(2 )log2 +1 +4 +x +x ⋅ ++ += +1 +2 +log2 +1 +4 +x +x ++ ++ +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +143 +Example 41 Find f ′(x) if f(x) = (sin x)sin x for all 0 < x < π. +Solution The function y = (sin x)sin x is defined for all positive real numbers. Taking +logarithms, we have +log y = log (sin x)sin x = sin x log (sin x) +Then +1 dy +y dx = d +dx (sin x log (sin x)) += cos x log (sin x) + sin x . +1 +(sin ) +sin +d +x +x dx +⋅ += cos x log (sin x) + cos x += (1 + log (sin x)) cos x +Thus +dy +dx = y((1 + log (sin x)) cos x) = (1 + log (sin x)) ( sin x)sin x cos x +Example 42 For a positive constant a find dy +dx , where +1 +1 +, and +a +t t +y +a +x +t +t ++ + + += += ++ + + + + +Solution Observe that both y and x are defined for all real t ≠ 0. Clearly +dy +dt = +( +) +1 +t t +d +a +dt ++ + = +1 +1 +log +t t d +a +t +a +dt +t ++ + + ++ +⋅ + + + + += +1 +2 +1 +1 +log +t t +a +a +t ++  + +− + + + + +Similarly +dx +dt = +1 +1 +1 +a +d +a t +t +t +dt +t +− + + + + ++ +⋅ ++ + + + + + + + + += +1 +2 +1 +1 +1 +a +a t +t +t +− + + + + ++ +⋅ +− + + + + + + + + +dx +dt ≠ 0 only if t ≠ ± 1. Thus for t ≠ ± 1, +Reprint 2025-26 + + MATHEMATICS +144 +dy +dy +dt +dx +dx +dt += + = +a +t +a +a t +t +t +t t +a ++ +− +− + + + + ++ + + + + +⋅ +− + + + + +1 +2 +1 +2 +1 +1 +1 +1 +1 +log += +1 +1 +log +1 +t t +a +a +a +a t +t ++ +− + + ++ + + + + +Example 43 Differentiate sin2 x w.r.t. e cos x. +Solution Let u (x) = sin2 x and v (x) = e cos x. We want to find +/ +/ +du +du dx +dv +dv dx += +. Clearly +du +dx = 2 sin x cos x and dv +dx = e cos x (– sin x) = – (sin x) e cos x +Thus +du +dv = +cos +cos +2sin +cos +2cos +sin +x +x +x +x +x +x e +e += − +− +Miscellaneous Exercise on Chapter 5 +Differentiate w.r.t. x the function in Exercises 1 to 11. +1. +(3x2 – 9x + 5)9 +2. +sin3 x + cos6 x +3. +(5x)3 cos 2x +4. +sin–1(x +x ), 0 ≤ x ≤ 1 +5. +1 +cos +2 +2 +7 +x +x +− ++ +, – 2 < x < 2 +6. + +1 +1 +sin +1 sin +cot +1 +sin +1 sin +x +x +x +x +− + ++ ++ +− + + ++ +− +− + +, 0 < x < 2 +π +7. +(log x)log x, x > 1 +8. +cos (a cos x + b sin x), for some constant a and b. +9. +(sin x – cos x) (sin x – cos x), +3 +4 +4 +x +π +π +< +< +10. +xx + xa + ax + aa, for some fixed a > 0 and x > 0 +Reprint 2025-26 + +CONTINUITY AND DIFFERENTIABILITY +145 +11. +( +) +2 +2 3 +3 +x +xx +x +−+ +− +, for x > 3 +12. +Find dy +dx +, if y = 12 (1 – cos t), x = 10 (t – sin t), +2 +2 +t +π +π +− +< < +13. +Find dy +dx , if y = sin–1 x + sin–1 +2 +1 +x +− +, 0 < x < 1 +14. +If +1 +1 +0 +x +y +y +x ++ ++ ++ += +, for , – 1 < x < 1, prove that +( +)2 +1 +1 +dy +dx +x += − ++ +15. +If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that +3 +2 +2 +2 +2 +1 +dy +dx +d y +dx + + + + ++ + + + + + + + + +is a constant independent of a and b. +16. +If cos y = x cos (a + y), with cos a ≠ ± 1, prove that +2 +cos ( +) +sin +dy +a +y +dx +a ++ += +. +17. +If x = a (cos t + t sin t) and y = a (sin t – t cos t), find +2 +2 +d y +dx +. +18. +If f(x) = | x |3, show that f ″(x) exists for all real x and find it. +19. +Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, +obtain the sum formula for cosines. +20. +Does there exist a function which is continuous everywhere but not differentiable +at exactly two points? Justify your answer. +21. +If +( ) +( ) +( ) +f x +g x +h x +y +l +m +n +a +b +c += +, prove that +( ) +( ) +( ) +f +x +g x +h x +dy +l +m +n +dx +a +b +c +′ +′ +′ += +22. +If y = +1 +cos +a +x +e +− +, – 1 ≤ x ≤ 1, show that ( +) +2 +2 +2 +2 +1 +0 +d y +dy +x +x +a y +dx +dx +− +− +− += +. +Reprint 2025-26 + + MATHEMATICS +146 +Summary +® A real valued function is continuous at a point in its domain if the limit of the +function at that point equals the value of the function at that point. A function +is continuous if it is continuous on the whole of its domain. +® Sum, difference, product and quotient of continuous functions are continuous. +i.e., if f and g are continuous functions, then +(f ± g) (x) = f (x) ± g(x) is continuous. +(f . g) (x) = f (x) . g(x) is continuous. +( ) +( ) +( ) +f +f x +x +g +g x + + += + + + + + (wherever g(x) ≠ 0) is continuous. +® Every differentiable function is continuous, but the converse is not true. +® Chain rule is rule to differentiate composites of functions. If f = v o u, t = u (x) +and if both dt +dx and dv +dt exist then +df +dv dt +dx +dt dx += +⋅ +® Following are some of the standard derivatives (in appropriate domains): +( +) +1 +2 +1 +sin +1 +d +x +dx +x +− += +− +( +) +1 +2 +1 +cos +1 +d +x +dx +x +− += − +− +( +) +1 +2 +1 +tan +1 +d +x +dx +x +− += ++ +( +) +x +x +d +e +e +dx += +( +) +1 +log +d +x +dx +x += +® Logarithmic differentiation is a powerful technique to differentiate functions +of the form f (x) = [u (x)]v (x). Here both f (x) and u(x) need to be positive for +this technique to make sense. +—v— +Reprint 2025-26" +class_12,6,Application of Derivatives,ncert_books/class_12/lemh1dd/lemh106.pdf,"v With the Calculus as a key, Mathematics can be successfully applied +to the explanation of the course of Nature.” — WHITEHEAD v +6.1 Introduction +In Chapter 5, we have learnt how to find derivative of composite functions, inverse +trigonometric functions, implicit functions, exponential functions and logarithmic functions. +In this chapter, we will study applications of the derivative in various disciplines, e.g., in +engineering, science, social science, and many other fields. For instance, we will learn +how the derivative can be used (i) to determine rate of change of quantities, (ii) to find +the equations of tangent and normal to a curve at a point, (iii) to find turning points on +the graph of a function which in turn will help us to locate points at which largest or +smallest value (locally) of a function occurs. We will also use derivative to find intervals +on which a function is increasing or decreasing. Finally, we use the derivative to find +approximate value of certain quantities. +6.2 Rate of Change of Quantities +Recall that by the derivative ds +dt , we mean the rate of change of distance s with +respect to the time t. In a similar fashion, whenever one quantity y varies with another +quantity x, satisfying some rule +( ) +y +f x += +, then dy +dx (or f′(x)) represents the rate of +change of y with respect to x and +dy +dx +x x + + += 0 + (or f′(x0)) represents the rate of change +of y with respect to x at +0 +x +x += +. +Further, if two variables x and y are varying with respect to another variable t, i.e., +if +( ) +x +f t += +and +( ) +y +g t += +, then by Chain Rule +dy +dx = dy +dx +dt +dt , if +0 +dx +dt ≠ +Chapter 6 +APPLICATION OF +DERIVATIVES +Reprint 2025-26 + + MATHEMATICS +148 +Thus, the rate of change of y with respect to x can be calculated using the rate of +change of y and that of x both with respect to t. +Let us consider some examples. +Example 1 Find the rate of change of the area of a circle per second with respect to +its radius r when r = 5 cm. +Solution The area A of a circle with radius r is given by A = πr2. Therefore, the rate +of change of the area A with respect to its radius r is given by +2 +A +( +) +2 +d +d +r +r +dr +dr += +π += π . +When r = 5 cm, +A +10 +d +dr = +π . Thus, the area of the circle is changing at the rate of +10π cm2/s. +Example 2 The volume of a cube is increasing at a rate of 9 cubic centimetres per +second. How fast is the surface area increasing when the length of an edge is 10 +centimetres ? +Solution Let x be the length of a side, V be the volume and S be the surface area of +the cube. Then, V = x3 and S = 6x2, where x is a function of time t. +Now +V +d +dt = 9cm3/s (Given) +Therefore +9 = +3 +3 +V +( +) +( +) +d +d +d +dx +x +x +dt +dt +dx +dt += += +⋅ +(By Chain Rule) += +2 +3 +dx +x +dt +⋅ +or +dx +dt = +2 +3 +x +... (1) +Now +dS +dt = +2 +2 +(6 +) +(6 +) +d +d +dx +x +x +dt +dx +dt += +⋅ +(By Chain Rule) += +2 +3 +36 +12x +x +x + + +⋅ += + + + + +(Using (1)) +Hence, when +x = 10 cm, +2 +3.6 cm /s +dS +dt = +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +149 +Example 3 A stone is dropped into a quiet lake and waves move in circles at a speed +of 4cm per second. At the instant, when the radius of the circular wave is 10 cm, how +fast is the enclosed area increasing? +Solution The area A of a circle with radius r is given by A = πr2. Therefore, the rate +of change of area A with respect to time t is +A +d +dt = +2 +2 +( +) +( +) +d +d +dr +r +r +dt +dr +dt +π += +π +⋅ + = 2π r dr +dt +(By Chain Rule) +It is given that +dr +dt = 4cm/s +Therefore, when r = 10 cm, +A +d +dt = 2π(10) (4) = 80π +Thus, the enclosed area is increasing at the rate of 80π cm2/s, when r = 10 cm. +ANote dy +dx is positive if y increases as x increases and is negative if y decreases +as x increases. +Example 4 The length x of a rectangle is decreasing at the rate of 3 cm/minute and +the width y is increasing at the rate of 2cm/minute. When x =10cm and y = 6cm, find +the rates of change of (a) the perimeter and (b) the area of the rectangle. +Solution Since the length x is decreasing and the width y is increasing with respect to +time, we have +3 cm/min +dx +dt = − +and +2 cm/min +dy +dt = +(a) +The perimeter P of a rectangle is given by +P = 2 (x + y) +Therefore +P +d +dt = 2 +2 +3 +2 +2 +dx +dt +dy +dt ++ + + + += +−+ += − +( +) +cm/min +(b) +The area A of the rectangle is given by +A = x . y +Therefore +A +d +dt = dx +dy +y +x +dt +dt +⋅ ++ +⋅ += – 3(6) + 10(2) +(as x = 10 cm and y = 6 cm) += 2 cm2/min +Reprint 2025-26 + + MATHEMATICS +150 +Example 5 The total cost C(x) in Rupees, associated with the production of x units of +an item is given by +C(x) = 0.005 x3 – 0.02 x2 + 30x + 5000 +Find the marginal cost when 3 units are produced, where by marginal cost we +mean the instantaneous rate of change of total cost at any level of output. +Solution Since marginal cost is the rate of change of total cost with respect to the +output, we have +Marginal +cost (MC) = +2 +0.005(3 +) +0.02(2 ) +30 +dC +x +x +dx = +− ++ +When +x = 3, MC = +2 +0.015(3 ) +0.04(3) +30 +− ++ += 0.135 – 0.12 + 30 = 30.015 +Hence, the required marginal cost is ` 30.02 (nearly). +Example 6 The total revenue in Rupees received from the sale of x units of a product +is given by R(x) = 3x2 + 36x + 5. Find the marginal revenue, when x = 5, where by +marginal revenue we mean the rate of change of total revenue with respect to the +number of items sold at an instant. +Solution Since marginal revenue is the rate of change of total revenue with respect to +the number of units sold, we have +Marginal Revenue +(MR) = +R +6 +36 +d +x +dx = ++ +When +x = 5, MR = 6(5) + 36 = 66 +Hence, the required marginal revenue is ` 66. +EXERCISE 6.1 +1. +Find the rate of change of the area of a circle with respect to its radius r when +(a) +r = 3 cm +(b) +r = 4 cm +2. +The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the +surface area increasing when the length of an edge is 12 cm? +3. +The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate +at which the area of the circle is increasing when the radius is 10 cm. +4. +An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the +volume of the cube increasing when the edge is 10 cm long? +5. +A stone is dropped into a quiet lake and waves move in circles at the speed of +5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is +the enclosed area increasing? +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +151 +6. +The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of +increase of its circumference? +7. +The length x of a rectangle is decreasing at the rate of 5 cm/minute and the +width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find +the rates of change of (a) the perimeter, and (b) the area of the rectangle. +8. +A balloon, which always remains spherical on inflation, is being inflated by pumping +in 900 cubic centimetres of gas per second. Find the rate at which the radius of +the balloon increases when the radius is 15 cm. +9. +A balloon, which always remains spherical has a variable radius. Find the rate at +which its volume is increasing with the radius when the later is 10 cm. +10. +A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled +along the ground, away from the wall, at the rate of 2cm/s. How fast is its height +on the wall decreasing when the foot of the ladder is 4 m away from the wall ? +11. +A particle moves along the curve 6y = x3 +2. Find the points on the curve at +which the y-coordinate is changing 8 times as fast as the x-coordinate. +12. +The radius of an air bubble is increasing at the rate of 1 +2 cm/s. At what rate is the +volume of the bubble increasing when the radius is 1 cm? +13. +A balloon, which always remains spherical, has a variable diameter 3 (2 +1) +2 +x + +. +Find the rate of change of its volume with respect to x. +14. +Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone +on the ground in such a way that the height of the cone is always one-sixth of the +radius of the base. How fast is the height of the sand cone increasing when the +height is 4 cm? +15. +The total cost C (x) in Rupees associated with the production of x units of an +item is given by +C (x) = 0.007x3 – 0.003x2 + 15x + 4000. +Find the marginal cost when 17 units are produced. +16. +The total revenue in Rupees received from the sale of x units of a product is +given by +R (x) = 13x2 + 26x + 15. +Find the marginal revenue when x = 7. +Choose the correct answer for questions 17 and 18. +17. +The rate of change of the area of a circle with respect to its radius r at r = 6 cm is +(A) 10π +(B) 12π +(C) 8π +(D) 11π +Reprint 2025-26 + + MATHEMATICS +152 +18. +The total revenue in Rupees received from the sale of x units of a product is +given by +R(x) = 3x2 + 36x + 5. The marginal revenue, when x = 15 is +(A) 116 +(B) 96 +(C) 90 +(D) 126 +6.3 Increasing and Decreasing Functions +In this section, we will use differentiation to find out whether a function is increasing or +decreasing or none. +Consider the function f given by f (x) = x2, x Î R. The graph of this function is a +parabola as given in Fig 6.1. +Fig 6.1 +First consider the graph (Fig 6.1) to the right of the origin. Observe that as we +move from left to right along the graph, the height of the graph continuously increases. +For this reason, the function is said to be increasing for the real numbers x > 0. +Now consider the graph to the left of the origin and observe here that as we move +from left to right along the graph, the height of the graph continuously decreases. +Consequently, the function is said to be decreasing for the real numbers x < 0. +We shall now give the following analytical definitions for a function which is +increasing or decreasing on an interval. +Definition 1 Let I be an interval contained in the domain of a real valued function f. +Then f is said to be +(i) +increasing on I if x1 < x2 in I Þ f (x1) < f (x2) for all x1, x2 Î I. +(ii) +decreasing on I, if x1 < x2 in I Þ f (x1) > f (x2) for all x1, x2 Î I. +(iii) +constant on I, if f(x) = c for all x Î I, where c is a constant. +x +f (x) = x2 + –2 +4 +3 +2 + +9 +4 + –1 +1 +1 +2 + +1 +4 + 0 +0 +Values left to origin +as we move from left to right, the +height of the graph decreases +x +f (x) = x2 +0 +0 +1 +2 +1 +4 + 1 +1 +3 +2 +9 +4 + 2 +4 +Values right to origin +as we move from left to right, the +height of the graph increases +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +153 +(iv) +strictly increasing on I if x1 < x2 in I Þ f (x1) < f (x2) for all x1, x2 Î I. +(v) +strictly decreasing on I if x1 < x2 in I Þ f (x1) > f (x2) for all x1, x2 Î I. +For graphical representation of such functions see Fig 6.2. +Fig 6.2 +We shall now define when a function is increasing or decreasing at a point. +Definition 2 Let x0 be a point in the domain of definition of a real valued function f. +Then f is said to be increasing, decreasing at x0 if there exists an open interval I +containing x0 such that f is increasing, decreasing, respectively, in I. +Let us clarify this definition for the case of increasing function. +Example 7 Show that the function given by f(x) = 7x – 3 is increasing on R. +Solution Let x1 and x2 be any two numbers in R. Then +x1 < x2 Þ 7x1 < 7x2 Þ 7x1 – 3 < 7x2 – 3 Þ f (x1) < f (x2) +Thus, by Definition 1, it follows that f is strictly increasing on R. +We shall now give the first derivative test for increasing and decreasing functions. +The proof of this test requires the Mean Value Theorem studied in Chapter 5. +Theorem 1 Let f be continuous on [a, b] and differentiable on the open interval +(a,b). Then +(a) +f is increasing in [a,b] if f ¢(x) > 0 for each x Î (a, b) +(b) +f is decreasing in [a,b] if f ¢(x) < 0 for each x Î (a, b) +(c) +f is a constant function in [a,b] if f ¢(x) = 0 for each x Î (a, b) +Strictly Increasing function +(i) +Neither Increasing nor +Decreasing function +(iii) +Strictly Decreasing function +(ii) +Reprint 2025-26 + + MATHEMATICS +154 +Proof (a) Let x1, x2 Î [a, b] be such that x1 < x2. +Then, by Mean Value Theorem (Theorem 8 in Chapter 5), there exists a point c +between x1 and x2 such that +f(x2) – f(x1) = f ¢(c) (x2 – x1) +i.e. +f(x2) – f(x1) > 0 +(as f ¢(c) > 0 (given)) +i.e. +f(x2) > f (x1) +Thus, we have +1 +2 +1 +2 +1 +2 +( +) +( +), for all +, +[ , ] +x +x +f x +f x +x x +a b + + + + +Hence, f is an increasing function in [a,b]. +The proofs of part (b) and (c) are similar. It is left as an exercise to the reader. +Remarks +There is a more generalised theorem, which states that if f¢(x) > 0 for x in an interval +excluding the end points and f is continuous in the interval, then f is increasing. Similarly, +if f¢(x) < 0 for x in an interval excluding the end points and f is continuous in the +interval, then f is decreasing. +Example 8 Show that the function f given by +f (x) = x3 – 3x2 + 4x, x Î R +is increasing on R. +Solution Note that +f ¢(x) = 3x2 – 6x + 4 += 3(x2 – 2x + 1) + 1 += 3(x – 1)2 + 1 > 0, in every interval of R +Therefore, the function f is increasing on R. +Example 9 Prove that the function given by f (x) = cos x is +(a) +decreasing in (0, p) +(b) +increasing in (p, 2p), and +(c) +neither increasing nor decreasing in (0, 2p). +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +155 +Fig 6.4 +Solution Note that f ′(x) = – sin x +(a) +Since for each x ∈ (0, π), sin x > 0, we have f ′(x) < 0 and so f is decreasing in +(0, π). +(b) +Since for each x ∈ (π, 2π), sin x < 0, we have f ′(x) > 0 and so f is increasing in +(π, 2π). +(c) +Clearly by (a) and (b) above, f is neither increasing nor decreasing in (0, 2π). +Example 10 Find the intervals in which the function f given by f(x) = x2 – 4x + 6 is +(a) increasing +(b) decreasing +Solution We have +f (x) += x2 – 4x + 6 +or +f ′(x) = 2x – 4 +Therefore, f ′(x) = 0 gives x = 2. Now the point x = 2 divides the real line into two +disjoint intervals namely, (– ∞, 2) and (2, ∞) (Fig 6.3). In the interval (– ∞, 2), f ′(x) = 2x +– 4 < 0. +Therefore, f is decreasing in this interval. Also, in the interval (2, +) +∞, +( ) +0 +f +x > +′ +and so the function f is increasing in this interval. +Example 11 Find the intervals in which the function f given by f (x) = 4x3 – 6x2 – 72x ++ 30 is (a) increasing (b) decreasing. +Solution We have +f (x) += 4x3 – 6x2 – 72x + 30 +or +f ′(x) = 12x2 – 12x – 72 += 12(x2 – x – 6) += 12(x – 3) (x + 2) +Therefore, f ′(x) = 0 gives x = – 2, 3. The +points x = – 2 and x = 3 divides the real line into +three disjoint intervals, namely, (– ∞, – 2), (– 2, 3) +and (3, ∞). +Fig 6.3 +Reprint 2025-26 + + MATHEMATICS +156 +In the intervals (– ∞, – 2) and (3, ∞), f ′(x) is positive while in the interval (– 2, 3), +f ′(x) is negative. Consequently, the function f is increasing in the intervals +(– ∞, – 2) and (3, ∞) while the function is decreasing in the interval (– 2, 3). However, +f is neither increasing nor decreasing in R. +Interval +Sign of f ′(x) +Nature of function f +(– ∞, – 2) +(–) (–) > 0 +f is increasing +(– 2, 3) +(–) (+) < 0 +f is decreasing +(3, ∞) +(+) (+) > 0 +f is increasing +Example 12 Find intervals in which the function given by f (x) = sin 3x, x ∈ + + + +0 2 +, π is +(a) increasing (b) decreasing. +Solution We have +f (x) = sin 3x +or +f ′(x) = 3cos 3x +Therefore, f′(x) = 0 gives cos 3x = 0 which in turn gives +3 +3 +, +2 +2 +x +π +π += + (as x ∈ + + + +0 2 +, π +implies +3 +3 +0, 2 +x +π + + +∈ + + + +). So +6 +x +π += + and 2 +π . The point +6 +x +π += + divides the interval 0 2 +, π + + + + +into two disjoint intervals 0, 6 +π + + + + + and π π +6 2 +, + + + +. +Now, +( ) +0 +f +x > +′ + for all +0, 6 +x +π + + +∈ + + + as 0 +0 +3 +6 +2 +x +x +π +π +≤ +< +⇒ +≤ +< + and +( ) +0 +f +x < +′ + for +all +, +6 2 +x +π π + + +∈ + + + as +3 +3 +6 +2 +2 +2 +x +x +π +π +π +π +< +< +⇒ +< +< +. +Therefore, f is increasing in 0, 6 +π + + + + + and decreasing in +, +6 2 +π π + + + + + +. +Fig 6.5 +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +157 +Also, the given function is continuous at x = 0 and +6 +x +π += +. Therefore, by Theorem 1, +f is increasing on 0 6 +, π + + + + and decreasing on +π π +6 2 +, + + + +. +Example 13 Find the intervals in which the function f given by + f (x) = sin x + cos x, 0 ≤ x ≤ 2π +is increasing or decreasing. +Solution We have +f(x) = sin x + cos x, +or +f ′(x) = cos x – sin x +Now +( ) +0 +f +x = +′ + gives sin x = cos x which gives that +4 +x +π += +, 5 +4 +π as 0 +2 +x +≤ +≤π +The points +4 +x +π += + and +5 +4 +x +π += + divide the interval [0, 2π] into three disjoint intervals, +namely, 0, 4 +π + + + + +, π +π +4 +5 +4 +, + + + + and 5 ,2 +4 +π + + +π + + + + +. +Note that +5 +( ) +0 if +0, +,2 +4 +4 +f +x +x +π +π + + + + +′ +> +∈ +∪ +π + + + + + + + + +or +f is increasing in the intervals 0, 4 +π + + + + + and +5 ,2 +4 +π + + +π + + + + +Also +′ +< +∈ + + + +f +x +x +( ) +, +0 +4 +5 +4 +if +π +π +or + f is decreasing in +π +π +4 +5 +4 +, + + + + +Fig 6.6 +Reprint 2025-26 + + MATHEMATICS +158 +Interval +Sign of +( ) +f +x +′ +Nature of function +0, 4 +π + + + + + +> 0 +f is increasing +π +π +4 +5 +4 +, + + + + +< 0 +f is decreasing +5 ,2 +4 +π + + +π + + + + +> 0 +f is increasing +EXERCISE 6.2 +1. +Show that the function given by f (x) = 3x + 17 is increasing on R. +2. +Show that the function given by f (x) = e2x is increasing on R. +3. +Show that the function given by f (x) = sin x is +(a) +increasing in 0, 2 +π + + + + + + +(b) +decreasing in +, +2 +π + + +π + + + + +(c) neither increasing nor decreasing in (0, π) +4. +Find the intervals in which the function f given by f (x) = 2x2 – 3x is +(a) +increasing +(b) +decreasing +5. +Find the intervals in which the function f given by f(x) = 2x3 – 3x2 – 36x + 7 is +(a) +increasing +(b) +decreasing +6. +Find the intervals in which the following functions are strictly increasing or +decreasing: +(a) +x2 + 2x – 5 +(b) +10 – 6x – 2x2 +(c) +–2x3 – 9x2 – 12x + 1 +(d) +6 – 9x – x2 +(e) +(x + 1)3 (x – 3)3 +7. +Show that +2 +log(1 +) +2 +x +y +x +x += ++ +− ++ +, x > – 1, is an increasing function of x +throughout its domain. +8. +Find the values of x for which y = [x(x – 2)]2 is an increasing function. +9. +Prove that +4sin +(2 +cos ) +y +θ += +−θ ++ +θ + is an increasing function of θ in 0 2 +, π + + + +. +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +159 +10. +Prove that the logarithmic function is increasing on (0, ∞). +11. +Prove that the function f given by f(x) = x2 – x + 1 is neither strictly increasing +nor decreasing on (– 1, 1). +12. +Which of the following functions are decreasing on 0, 2 +π + + + + + + ? +(A) cos x +(B) cos 2x +(C) cos 3x +(D) tan x +13. +On which of the following intervals is the function f given by f (x) = x100 + sin x –1 +decreasing ? +(A) (0,1) +(B) +, +2 +π + + +π + + + + +(C) +0, 2 +π + + + + + + +(D) None of these +14. +For what values of a the function f given by f (x) = x2 + ax + 1 is increasing on +[1, 2]? +15. +Let I be any interval disjoint from [–1, 1]. Prove that the function f given by +1 +( ) +f x +x +x += ++ + is increasing on I. +16. +Prove that the function f given by f (x) = log sin x is increasing on 0 2 +, π + + + + and +decreasing on π π +2 , + + + +. +17. +Prove that the function f given by f (x) = log |cos x| is decreasing on 0, 2 +π + + + + + + + and +increasing on 3 , 2 +2 +π + + +π + + + + +. +18. +Prove that the function given by f (x) = x3 – 3x2 + 3x – 100 is increasing in R. +19. +The interval in which y = x2 e–x is increasing is +(A) (– ∞, ∞) +(B) (– 2, 0) +(C) (2, ∞) +(D) (0, 2) +6.4 Maxima and Minima +In this section, we will use the concept of derivatives to calculate the maximum or +minimum values of various functions. In fact, we will find the ‘turning points’ of the +graph of a function and thus find points at which the graph reaches its highest (or +Reprint 2025-26 + + MATHEMATICS +160 +lowest) locally. The knowledge of such points is very useful in sketching the graph of +a given function. Further, we will also find the absolute maximum and absolute minimum +of a function that are necessary for the solution of many applied problems. +Let us consider the following problems that arise in day to day life. +(i) +The profit from a grove of orange trees is given by P(x) = ax + bx2, where a,b +are constants and x is the number of orange trees per acre. How many trees per +acre will maximise the profit? +(ii) +A ball, thrown into the air from a building 60 metres high, travels along a path +given by +2 +( ) +60 +60 +x +h x +x += ++ +− +, where x is the horizontal distance from the building +and h(x) is the height of the ball . What is the maximum height the ball will +reach? +(iii) +An Apache helicopter of enemy is flying along the path given by the curve +f (x) = x2 + 7. A soldier, placed at the point (1, 2), wants to shoot the helicopter +when it is nearest to him. What is the nearest distance? +In each of the above problem, there is something common, i.e., we wish to find out +the maximum or minimum values of the given functions. In order to tackle such problems, +we first formally define maximum or minimum values of a function, points of local +maxima and minima and test for determining such points. +Definition 3 Let f be a function defined on an interval I. Then +(a) + f is said to have a maximum value in I, if there exists a point c in I such that +( ) +( ) +> +f c +f x , for all x ∈ I. +The number f (c) is called the maximum value of f in I and the point c is called a +point of maximum value of f in I. +(b) + f is said to have a minimum value in I, if there exists a point c in I such that +f (c) < f (x), for all x ∈ I. +The number f (c), in this case, is called the minimum value of f in I and the point +c, in this case, is called a point of minimum value of f in I. +(c) +f is said to have an extreme value in I if there exists a point c in I such that +f (c) is either a maximum value or a minimum value of f in I. +The number f (c), in this case, is called an extreme value of f in I and the point c +is called an extreme point. +Remark In Fig 6.7(a), (b) and (c), we have exhibited that graphs of certain particular +functions help us to find maximum value and minimum value at a point. Infact, through +graphs, we can even find maximum/minimum value of a function at a point at which it +is not even differentiable (Example 15). +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +161 +Fig 6.7 +Example 14 Find the maximum and the minimum values, if +any, of the function f given by +f (x) = x2, x ∈ R. +Solution From the graph of the given function (Fig 6.8), we +have f (x) = 0 if x = 0. Also +f (x) ≥0, for all x ∈ R. +Therefore, the minimum value of f is 0 and the point of +minimum value of f is x = 0. Further, it may be observed +from the graph of the function that f has no maximum value +and hence no point of maximum value of f in R. +ANote If we restrict the domain of f to [– 2, 1] only, +then f will have maximum value(– 2)2 = 4 at x = – 2. +Example 15 Find the maximum and minimum values +of f , if any, of the function given by f(x) = |x|, x ∈ R. +Solution From the graph of the given function +(Fig 6.9) , note that +f (x) ≥ 0, for all x ∈ R and f (x) = 0 if x = 0. +Therefore, the function f has a minimum value 0 +and the point of minimum value of f is x = 0. Also, the +graph clearly shows that f has no maximum value in R +and hence no point of maximum value in R. +ANote +(i) +If we restrict the domain of f to [– 2, 1] only, then f will have maximum value +|– 2| = 2. +Fig 6.8 +Fig 6.9 +Reprint 2025-26 + + MATHEMATICS +162 +Fig 6.10 +(ii) +One may note that the function f in Example 27 is not differentiable at +x = 0. +Example 16 Find the maximum and the minimum values, if any, of the function +given by +f (x) = x, x ∈ (0, 1). +Solution The given function is an increasing (strictly) function in the given interval +(0, 1). From the graph (Fig 6.10) of the function f , it +seems that, it should have the minimum value at a +point closest to 0 on its right and the maximum value +at a point closest to 1 on its left. Are such points +available? Of course, not. It is not possible to locate +such points. Infact, if a point x0 is closest to 0, then +we find +0 +0 +2 +x +x +< + for all +0 +(0,1) +x ∈ +. Also, if x1 is closest +to 1, then +1 +1 +1 +2 +x +x ++ +> + for all +1 +(0,1) +x ∈ +. +Therefore, the given function has neither the +maximum value nor the minimum value in the interval (0,1). +Remark The reader may observe that in Example 16, if we include the points 0 and 1 +in the domain of f , i.e., if we extend the domain of f to [0,1], then the function f has +minimum value 0 at x = 0 and maximum value 1 at x = 1. Infact, we have the following +results (The proof of these results are beyond the scope of the present text) +Every monotonic function assumes its maximum/minimum value at the end +points of the domain of definition of the function. +A more general result is +Every continuous function on a closed interval has a maximum and a minimum +value. +ANote By a monotonic function f in an interval I, we mean that f is either +increasing in I or decreasing in I. +Maximum and minimum values of a function defined on a closed interval will be +discussed later in this section. +Let us now examine the graph of a function as shown in Fig 6.11. Observe that at +points A, B, C and D on the graph, the function changes its nature from decreasing to +increasing or vice-versa. These points may be called turning points of the given +function. Further, observe that at turning points, the graph has either a little hill or a little +valley. Roughly speaking, the function has minimum value in some neighbourhood +(interval) of each of the points A and C which are at the bottom of their respective +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +163 +valleys. Similarly, the function has maximum value in some neighbourhood of points B +and D which are at the top of their respective hills. For this reason, the points A and C +may be regarded as points of local minimum value (or relative minimum value) and +points B and D may be regarded as points of local maximum value (or relative maximum +value) for the function. The local maximum value and local minimum value of the +function are referred to as local maxima and local minima, respectively, of the function. +We now formally give the following definition +Definition 4 Let f be a real valued function and let c be an interior point in the domain +of f. Then +(a) +c is called a point of local maxima if there is an h > 0 such that +f (c) ≥ f (x), for all x in (c – h, c + h), x ≠ c +The value f (c) is called the local maximum value of f. +(b) +c is called a point of local minima if there is an h > 0 such that +f (c) ≤ f (x), for all x in (c – h, c + h) +The value f (c) is called the local minimum value of f . +Geometrically, the above definition states that if x = c is a point of local maxima of f, +then the graph of f around c will be as shown in Fig 6.12(a). Note that the function f is +increasing (i.e., f ′(x) > 0) in the interval (c – h, c) and decreasing (i.e., f ′(x) < 0) in the +interval (c, c + h). +This suggests that f ′(c) must be zero. +Fig 6.11 +Fig 6.12 +Reprint 2025-26 + + MATHEMATICS +164 +Fig 6.13 +Similarly, if c is a point of local minima of f , then the graph of f around c will be as +shown in Fig 6.14(b). Here f is decreasing (i.e., f ′(x) < 0) in the interval (c – h, c) and +increasing (i.e., f ′(x) > 0) in the interval (c, c + h). This again suggest that f ′(c) must +be zero. +The above discussion lead us to the following theorem (without proof). +Theorem 2 Let f be a function defined on an open interval I. Suppose c ∈ I be any +point. If f has a local maxima or a local minima at x = c, then either f ′(c) = 0 or f is not +differentiable at c. +Remark The converse of above theorem need not +be true, that is, a point at which the derivative vanishes +need not be a point of local maxima or local minima. +For example, if f (x) = x3, then f ′(x) = 3x2 and so +f ′(0) = 0. But 0 is neither a point of local maxima nor +a point of local minima (Fig 6.13). +ANote A point c in the domain of a function f at +which either f ′(c) = 0 or f is not differentiable is +called a critical point of f. Note that if f is continuous +at c and f ′(c) = 0, then there exists an h > 0 such +that f is differentiable in the interval +(c – h, c + h). +We shall now give a working rule for finding points of local maxima or points of +local minima using only the first order derivatives. +Theorem 3 (First Derivative Test) Let f be a function defined on an open interval I. +Let f be continuous at a critical point c in I. Then +(i) +If f ′(x) changes sign from positive to negative as x increases through c, i.e., if +f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at +every point sufficiently close to and to the right of c, then c is a point of local +maxima. +(ii) +If f ′(x) changes sign from negative to positive as x increases through c, i.e., if +f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at +every point sufficiently close to and to the right of c, then c is a point of local +minima. +(iii) +If f ′(x) does not change sign as x increases through c, then c is neither a point of +local maxima nor a point of local minima. Infact, such a point is called point of +inflection (Fig 6.13). +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +165 +ANote If c is a point of local maxima of f , then f (c) is a local maximum value of +f. Similarly, if c is a point of local minima of f , then f(c) is a local minimum value of f. +Figures 6.13 and 6.14, geometrically explain Theorem 3. +Fig 6.14 +Example 17 Find all points of local maxima and local minima of the function f +given by +f (x) = x3 – 3x + 3. +Solution We have +f (x) = x3 – 3x + 3 +or +f ′(x) = 3x2 – 3 = 3(x – 1) (x + 1) +or +f ′(x) = 0 at x = 1 and x = – 1 +Thus, x = ± 1 are the only critical points which could possibly be the points of local +maxima and/or local minima of f . Let us first examine the point x = 1. +Note that for values close to 1 and to the right of 1, f ′(x) > 0 and for values close +to 1 and to the left of 1, f ′(x) < 0. Therefore, by first derivative test, x = 1 is a point +of local minima and local minimum value is f (1) = 1. In the case of x = –1, note that +f ′(x) > 0, for values close to and to the left of –1 and f ′(x) < 0, for values close to and +to the right of – 1. Therefore, by first derivative test, x = – 1 is a point of local maxima +and local maximum value is f (–1) = 5. +Values of x +Sign of f ′(x) = 3(x – 1) (x + 1) +Close to 1 +to the right (say 1.1 etc.) +>0 +to the left (say 0.9 etc.) +<0 +Close to –1 +to the right (say +0.9 etc.) +0 +to the left (say +1.1 etc.) +0 +− +< +− +> +Reprint 2025-26 + + MATHEMATICS +166 +Fig 6.15 +Example 18 Find all the points of local maxima and local minima of the function f +given by +f (x) = 2x3 – 6x2 + 6x +5. +Solution We have +f (x) = 2x3 – 6x2 + 6x + 5 +or +f ′(x) = 6x2 – 12x + 6 = 6(x – 1)2 +or +f ′(x) = 0 at x = 1 +Thus, x = 1 is the only critical point of f . We shall now examine this point for local +maxima and/or local minima of f. Observe that f ′(x) ≥ 0, for all x ∈ R and in particular +f ′(x) > 0, for values close to 1 and to the left and to the right of 1. Therefore, by first +derivative test, the point x = 1 is neither a point of local maxima nor a point of local +minima. Hence x = 1 is a point of inflexion. +Remark One may note that since f ′(x), in Example 30, never changes its sign on R, +graph of f has no turning points and hence no point of local maxima or local minima. +We shall now give another test to examine local maxima and local minima of a +given function. This test is often easier to apply than the first derivative test. +Theorem 4 (Second Derivative Test) Let f be a function defined on an interval I +and c ∈ I. Let f be twice differentiable at c. Then +(i) +x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0 +The value f (c) is local maximum value of f . +(ii) +x = c is a point of local minima if +( ) +0 +f +c = +′ + and f ″(c) > 0 +In this case, f (c) is local minimum value of f . +(iii) +The test fails if f ′(c) = 0 and f ″(c) = 0. +In this case, we go back to the first derivative test and find whether c is a point of +local maxima, local minima or a point of inflexion. +ANote As f is twice differentiable at c, we mean +second order derivative of f exists at c. +Example 19 Find local minimum value of the function f +given by f (x) = 3 + |x|, x ∈ R. +Solution Note that the given function is not differentiable +at x = 0. So, second derivative test fails. Let us try first +derivative test. Note that 0 is a critical point of f . Now to +the left of 0, f (x) = 3 – x and so f ′(x) = – 1 < 0. Also to +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +167 +the right of 0, f (x) = 3 + x and so f ′(x) = 1 > 0. Therefore, by first derivative test, x = +0 is a point of local minima of f and local minimum value of f is f (0) = 3. +Example 20 Find local maximum and local minimum values of the function f given by +f (x) = 3x4 + 4x3 – 12x2 + 12 +Solution We have +f (x) = 3x4 + 4x3 – 12x2 + 12 +or +f ′(x) = 12x3 + 12x2 – 24x = 12x (x – 1) (x + 2) +or +f ′(x) = 0 at x = 0, x = 1 and x = – 2. +Now +f ″(x) = 36x2 + 24x – 24 = 12 (3x2 + 2x – 2) +or +′′ += − +< +′′ += +> +′′ − += +> + + + +f +f +f +( ) +( ) +( +) +0 +24 +0 +1 +36 +0 +2 +72 +0 +Therefore, by second derivative test, x = 0 is a point of local maxima and local +maximum value of f at x = 0 is f (0) = 12 while x = 1 and x = – 2 are the points of local +minima and local minimum values of f at x = – 1 and – 2 are f (1) = 7 and f (–2) = –20, +respectively. +Example 21 Find all the points of local maxima and local minima of the function f +given by +f(x) = 2x3 – 6x2 + 6x +5. +Solution We have +f(x) = 2x3 – 6x2 + 6x +5 +or +2 +2 +( ) +6 +12 +6 +6( +1) +( ) +12( +1) +f +x +x +x +x +f +x +x +′ += +− ++ += +− +′′ += +− + +Now f ′(x) = 0 gives x =1. Also f ″(1) = 0. Therefore, the second derivative test +fails in this case. So, we shall go back to the first derivative test. +We have already seen (Example 18) that, using first derivative test, x =1 is neither +a point of local maxima nor a point of local minima and so it is a point of inflexion. +Example 22 Find two positive numbers whose sum is 15 and the sum of whose +squares is minimum. +Solution Let one of the numbers be x. Then the other number is (15 – x). Let S(x) +denote the sum of the squares of these numbers. Then +Reprint 2025-26 + + MATHEMATICS +168 +S(x) = x2 + (15 – x)2 = 2x2 – 30x + 225 +or +S ( ) +4 +30 +S ( ) +4 +x +x +x +′ += +− + +′′ += + +Now S′(x) = 0 gives +15 +2 +x = +. Also +15 +S +4 +0 +2 + + +′′ += +> + + + + +. Therefore, by second derivative +test, +15 +2 +x = + is the point of local minima of S. Hence the sum of squares of numbers is +minimum when the numbers are 15 +2 and +15 +15 +15 +2 +2 +− += +. +Remark Proceeding as in Example 34 one may prove that the two positive numbers, +whose sum is k and the sum of whose squares is minimum, are +and +2 +2 +k +k . +Example 23 Find the shortest distance of the point (0, c) from the parabola y = x2, +where 1 +2 ≤ c ≤ 5. +Solution Let (h, k) be any point on the parabola y = x2. Let D be the required distance +between (h, k) and (0, c). Then + +2 +2 +2 +2 +D +( +0) +( +) +( +) +h +k +c +h +k +c += +− ++ +− += ++ +− +... (1) +Since (h, k) lies on the parabola y = x2, we have k = h2. So (1) gives +D ≡D(k) = +2 +( +) +k +k +c ++ +− +or +D′(k) = +2 +1 +2( +) +2 +( +) +k +c +k +k +c ++ +− ++ +− +Now +D′(k) = 0 gives +2 +1 +2 +c +k +− += +Observe that when +2 +1 +2 +c +k +− +< +, then 2( +) +1 +0 +k +c +− ++ < +, i.e., D ( ) +0 +k +′ +< +. Also when +2 +1 +2 +c +k +− +> +, then D ( ) +0 +k +′ +> +. So, by first derivative test, D (k) is minimum at +2 +1 +2 +c +k +− += +. +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +169 +Hence, the required shortest distance is given by +2 +2 +1 +2 +1 +2 +1 +4 +1 +D +2 +2 +2 +2 +c +c +c +c +c +− +− +− +− + + + + += ++ +− += + + + + + + + + +ANote The reader may note that in Example 35, we have used first derivative +test instead of the second derivative test as the former is easy and short. +Example 24 Let AP and BQ be two vertical poles at +points A and B, respectively. If AP = 16 m, BQ = 22 m +and AB = 20 m, then find the distance of a point R on +AB from the point A such that RP2 + RQ2 is minimum. +Solution Let R be a point on AB such that AR = x m. +Then RB = (20 – x) m (as AB = 20 m). From Fig 6.16, +we have +RP2 = AR2 + AP2 +and +RQ2 = RB2 + BQ2 +Therefore +RP2 + RQ2 = AR2 + AP2 + RB2 + BQ2 += x2 + (16)2 + (20 – x)2 + (22)2 += 2x2 – 40x + 1140 +Let +S ≡ S(x) = RP2 + RQ2 = 2x2 – 40x + 1140. +Therefore +S′(x) = 4x – 40. +Now S′(x) = 0 gives x = 10. Also S″(x) = 4 > 0, for all x and so S″(10) > 0. +Therefore, by second derivative test, x = 10 is the point of local minima of S. Thus, the +distance of R from A on AB is AR = x =10 m. +Example 25 If length of three sides of a trapezium other than base are equal to 10cm, +then find the area of the trapezium when it is maximum. +Solution The required trapezium is as given in Fig 6.17. Draw perpendiculars DP and +Fig 6.16 +Fig 6.17 +Reprint 2025-26 + + MATHEMATICS +170 +CQ on AB. Let AP = x cm. Note that ∆APD ~ ∆BQC. Therefore, QB = x cm. Also, by +Pythagoras theorem, DP = QC = +2 +100 +x +− +. Let A be the area of the trapezium. Then +A ≡ A(x) = 1 +2 (sum of parallel sides) (height) += +( +) +2 +1 (2 +10 10) +100 +2 +x +x ++ ++ +− += +( +) +2 +( +10) +100 +x +x ++ +− +or +A′(x) = +( +) +2 +2 +( 2 ) +( +10) +100 +2 100 +x +x +x +x +− ++ ++ +− +− += +2 +2 +2 +10 +100 +100 +x +x +x +− +− ++ +− +Now +A′(x) = 0 gives 2x2 + 10x – 100 = 0, i.e., x = 5 and x = –10. +Since x represents distance, it can not be negative. +So, +x = 5. Now +A″(x) = +2 +2 +2 +2 +( 2 ) +100 +( 4 +10) +( 2 +10 +100) +2 100 +100 +x +x +x +x +x +x +x +− +− +− +− +−− +− ++ +− +− += +3 +3 +2 +2 +2 +300 +1000 +(100 +) +x +x +x +− +− +− + (on simplification) +or +A″(5) = +3 +3 +2 +2 +2(5) +300(5) +1000 +2250 +30 +0 +75 75 +75 +(100 +(5) ) +− +− +− +− += += +< +− +Thus, area of trapezium is maximum at x = 5 and the area is given by +A(5) = +2 +2 +(5 +10) 100 +(5) +15 75 +75 3 cm ++ +− += += +Example 26 Prove that the radius of the right circular cylinder of greatest curved +surface area which can be inscribed in a given cone is half of that of the cone. +Solution Let OC = r be the radius of the cone and OA = h be its height. Let a cylinder +with radius OE = x inscribed in the given cone (Fig 6.18). The height QE of the cylinder +is given by +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +171 +QE +OA = EC +OC (since ∆QEC ~ ∆AOC) +or +QE +h = r +x +r +− +or +QE = +( +) +h r +x +r +− +Let S be the curved surface area of the given +cylinder. Then +S ≡S(x) = 2 +( +) +xh r +x +r +π +− + = +2 +2 +( +) +h rx +x +r +π +− +or +2 +S ( ) +( +2 ) +4 +S ( ) +h +x +r +x +r +h +x +r +π +′ += +− + +−π +′′ += + +Now S′(x) = 0 gives +2 +r +x = +. Since S″(x) < 0 for all x, S +0 +2 +r + + +′′ +< + + + + +. So +2 +r +x = + is a +point of maxima of S. Hence, the radius of the cylinder of greatest curved surface area +which can be inscribed in a given cone is half of that of the cone. +6.4.1 Maximum and Minimum Values of a Function in a Closed Interval +Let us consider a function f given by +f (x) = x + 2, x ∈ (0, 1) +Observe that the function is continuous on (0, 1) and neither has a maximum value +nor has a minimum value. Further, we may note that the function even has neither a +local maximum value nor a local minimum value. +However, if we extend the domain of f to the closed interval [0, 1], then f still may +not have a local maximum (minimum) values but it certainly does have maximum value +3 = f (1) and minimum value 2 = f (0). The maximum value 3 of f at x = 1 is called +absolute maximum value (global maximum or greatest value) of f on the interval +[0, 1]. Similarly, the minimum value 2 of f at x = 0 is called the absolute minimum +value (global minimum or least value) of f on [0, 1]. +Consider the graph given in Fig 6.19 of a continuous function defined on a closed +interval [a, d]. Observe that the function f has a local minima at x = b and local +Fig 6.18 +Reprint 2025-26 + + MATHEMATICS +172 +minimum value is f(b). The function also has a local maxima at x = c and local maximum +value is f (c). +Also from the graph, it is evident that f has absolute maximum value f (a) and +absolute minimum value f (d). Further note that the absolute maximum (minimum) +value of f is different from local maximum (minimum) value of f. +We will now state two results (without proof) regarding absolute maximum and +absolute minimum values of a function on a closed interval I. +Theorem 5 Let f be a continuous function on an interval I = [a, b]. Then f has the +absolute maximum value and f attains it at least once in I. Also, f has the absolute +minimum value and attains it at least once in I. +Theorem 6 Let f be a differentiable function on a closed interval I and let c be any +interior point of I. Then +(i) f ′(c) = 0 if f attains its absolute maximum value at c. +(ii) f ′(c) = 0 if f attains its absolute minimum value at c. +In view of the above results, we have the following working rule for finding absolute +maximum and/or absolute minimum values of a function in a given closed interval +[a, b]. +Working Rule +Step 1: Find all critical points of f in the interval, i.e., find points x where either +( ) +0 +f +x = +′ + or f is not differentiable. +Step 2: Take the end points of the interval. +Step 3: At all these points (listed in Step 1 and 2), calculate the values of f . +Step 4: Identify the maximum and minimum values of f out of the values calculated in +Step 3. This maximum value will be the absolute maximum (greatest) value of +f and the minimum value will be the absolute minimum (least) value of f . +Fig 6.19 +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +173 +Example 27 Find the absolute maximum and minimum values of a function f given by +f (x) = 2x3 – 15x2 + 36x +1 on the interval [1, 5]. +Solution We have +f (x) = 2x3 – 15x2 + 36x + 1 +or +f ′(x) = 6x2 – 30x + 36 = 6 (x – 3) (x – 2) +Note that f ′(x) = 0 gives x = 2 and x = 3. +We shall now evaluate the value of f at these points and at the end points of the +interval [1, 5], i.e., at x = 1, x = 2, x = 3 and at x = 5. So +f (1) = 2(13) – 15(12) + 36 (1) + 1 = 24 +f (2) = 2(23) – 15(22) + 36 (2) + 1 = 29 +f (3) = 2(33) – 15(32) + 36 (3) + 1 = 28 +f (5) = 2(53) – 15(52) + 36 (5) + 1 = 56 +Thus, we conclude that absolute maximum value of f on [1, 5] is 56, occurring at +x =5, and absolute minimum value of f on [1, 5] is 24 which occurs at x = 1. +Example 28 Find absolute maximum and minimum values of a function f given by +4 +1 +3 +3 +( ) +12 +6 +, +[ 1, 1] +f x +x +x +x += +− +∈− +Solution We have +f (x) = +4 +1 +3 +3 +12 +6 +x +x +− +or +f ′(x) = +1 +3 +2 +2 +3 +3 +2 +2(8 +1) +16 +x +x +x +x +− +− += +Thus, f ′(x) = 0 gives +1 +8 +x = +. Further note that f ′(x) is not defined at x = 0. So the +critical points are x = 0 and +1 +8 +x = +. Now evaluating the value of f at critical points +x = 0, 1 +8 and at end points of the interval x = –1 and x = 1, we have +f (–1) = +4 +1 +3 +3 +12( 1) +6( 1) +18 +− +− +− += +f (0) = 12 (0) – 6(0) = 0 +Reprint 2025-26 + + MATHEMATICS +174 +1 +8 +f  + + + + + = +4 +1 +3 +3 +1 +1 +9 +12 +6 +8 +8 +4 +− + + + + +− += + + + + + + + + +f (1) = +4 +1 +3 +3 +12(1) +6(1) +6 +− += +Hence, we conclude that absolute maximum value of f is 18 that occurs at x = –1 +and absolute minimum value of f is +9 +4 +− that occurs at +1 +8 +x = +. +Example 29 An Apache helicopter of enemy is flying along the curve given by +y = x2 + 7. A soldier, placed at (3, 7), wants to shoot down the helicopter when it is +nearest to him. Find the nearest distance. +Solution For each value of x, the helicopter’s position is at point (x, x2 + 7). +Therefore, the distance between the helicopter and the soldier placed at (3,7) is +2 +2 +2 +( +3) +( +7 +7) +x +x +− ++ ++ +− +, i.e., +2 +4 +( +3) +x +x +− ++ +. +Let +f (x) = (x – 3)2 + x4 +or +f ′(x) = 2(x – 3) + 4x3 = 2(x – 1) (2x2 + 2x + 3) +Thus, f ′(x) = 0 gives x = 1 or 2x2 + 2x + 3 = 0 for which there are no real roots. +Also, there are no end points of the interval to be added to the set for which f ′ is zero, +i.e., there is only one point, namely, x = 1. The value of f at this point is given by +f (1) = (1 – 3)2 + (1)4 = 5. Thus, the distance between the solider and the helicopter is +(1) +5 +f += +. +Note that 5 is either a maximum value or a minimum value. Since +(0) +f + = +2 +4 +(0 +3) +(0) +3 +5 +− ++ += +> +, +it follows that +5 is the minimum value of +( ) +f x . Hence, +5 is the minimum +distance between the soldier and the helicopter. +EXERCISE 6.3 +1. +Find the maximum and minimum values, if any, of the following functions +given by +(i) f (x) = (2x – 1)2 + 3 +(ii) f (x) = 9x2 + 12x + 2 +(iii) f (x) = – (x – 1)2 + 10 +(iv) g(x) = x3 + 1 +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +175 +2. +Find the maximum and minimum values, if any, of the following functions +given by +(i) f (x) = |x + 2| – 1 +(ii) g(x) = – |x + 1| + 3 +(iii) h(x) = sin(2x) + 5 +(iv) f (x) = |sin 4x + 3| +(v) h(x) = x + 1, x ∈ (– 1, 1) +3. +Find the local maxima and local minima, if any, of the following functions. Find +also the local maximum and the local minimum values, as the case may be: +(i) f (x) = x2 +(ii) g(x) = x3 – 3x +(iii) h(x) = sin x + cos x, 0 +2 +x +π +< +< +(iv) f (x) = sin x – cos x, 0 +2 +x +< +< π +(v) f (x) = x3 – 6x2 + 9x + 15 +(vi) +2 +( ) +, +0 +2 +x +g x +x +x += ++ +> +(vii) +2 +1 +( ) +2 +g x +x += ++ +(viii) +( ) +1 +, 0 +1 += +− +< +< +f x +x +x +x +4. +Prove that the following functions do not have maxima or minima: +(i) f (x) = ex +(ii) g(x) = log x +(iii) h (x) = x3 + x2 + x +1 +5. +Find the absolute maximum value and the absolute minimum value of the following +functions in the given intervals: +(i) f (x) = x3, x ∈ [– 2, 2] +(ii) f (x) = sin x + cos x , x ∈ [0, π] +(iii) f (x) = +2 +1 +9 +4 +, +2, +2 +2 +x +x +x + + +− +∈− + + + + +(iv) +2 +( ) +( +1) +3, +[ 3,1] +f x +x +x += +− ++ +∈− +6. +Find the maximum profit that a company can make, if the profit function is +given by +p(x) = 41 – 72x – 18x2 +7. +Find both the maximum value and the minimum value of +3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3]. +8. +At what points in the interval [0, 2π], does the function sin 2x attain its maximum +value? +9. +What is the maximum value of the function sin x + cos x? +10. +Find the maximum value of 2x3 – 24x + 107 in the interval [1, 3]. Find the +maximum value of the same function in [–3, –1]. +Reprint 2025-26 + + MATHEMATICS +176 +11. +It is given that at x = 1, the function x4 – 62x2 + ax + 9 attains its maximum value, +on the interval [0, 2]. Find the value of a. +12. +Find the maximum and minimum values of x + sin 2x on [0, 2π]. +13. +Find two numbers whose sum is 24 and whose product is as large as possible. +14. +Find two positive numbers x and y such that x + y = 60 and xy3 is maximum. +15. +Find two positive numbers x and y such that their sum is 35 and the product x2 y5 +is a maximum. +16. +Find two positive numbers whose sum is 16 and the sum of whose cubes is +minimum. +17. +A square piece of tin of side 18 cm is to be made into a box without top, by +cutting a square from each corner and folding up the flaps to form the box. What +should be the side of the square to be cut off so that the volume of the box is the +maximum possible. +18. +A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, +by cutting off square from each corner and folding up the flaps. What should be +the side of the square to be cut off so that the volume of the box is maximum ? +19. +Show that of all the rectangles inscribed in a given fixed circle, the square has +the maximum area. +20. +Show that the right circular cylinder of given surface and maximum volume is +such that its height is equal to the diameter of the base. +21. +Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic +centimetres, find the dimensions of the can which has the minimum surface +area? +22. +A wire of length 28 m is to be cut into two pieces. One of the pieces is to be +made into a square and the other into a circle. What should be the length of the +two pieces so that the combined area of the square and the circle is minimum? +23. +Prove that the volume of the largest cone that can be inscribed in a sphere of +radius R is 8 +27 of the volume of the sphere. +24. +Show that the right circular cone of least curved surface and given volume has +an altitude equal to +2 time the radius of the base. +25. +Show that the semi-vertical angle of the cone of the maximum volume and of +given slant height is +1 +tan +2 +− +. +26. +Show that semi-vertical angle of right circular cone of given surface area and +maximum volume is sin− + + + +1 1 +3 . +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +177 +Choose the correct answer in Questions 27 and 29. +27. +The point on the curve x2 = 2y which is nearest to the point (0, 5) is +(A) (2 2,4) +(B) (2 2,0) +(C) (0, 0) +(D) (2, 2) +28. +For all real values of x, the minimum value of +2 +2 +1 +1 +x +x +x +x +− ++ ++ ++ + is +(A) 0 +(B) 1 +(C) 3 +(D) 1 +3 +29. +The maximum value of +1 +3 +[ ( +1) +1] +x x − ++ +, 0 +1 +x +≤ +≤ is +(A) +1 +3 +1 +3 + + + + +(B) 1 +2 +(C) 1 +(D) 0 +Miscellaneous Examples +Example 30 A car starts from a point P at time t = 0 seconds and stops at point Q. The +distance x, in metres, covered by it, in t seconds is given by +x +t +t += +− + + + + +2 2 +3 +Find the time taken by it to reach Q and also find distance between P and Q. +Solution Let v be the velocity of the car at t seconds. +Now +x = +2 2 +3 +t +t  + +− + + + + +Therefore +v = dx +dt = 4t – t2 = t(4 – t) +Thus, v = 0 gives t = 0 and/or t = 4. +Now v = 0 at P as well as at Q and at P, t = 0. So, at Q, t = 4. Thus, the car will +reach the point Q after 4 seconds. Also the distance travelled in 4 seconds is given by +x]t = 4 = +2 +4 +2 +32 +4 +2 +16 +m +3 +3 +3 + + + + +− += += + + + + + + + + +Reprint 2025-26 + + MATHEMATICS +178 +Example 31 A water tank has the shape of an inverted right circular cone with its axis +vertical and vertex lowermost. Its semi-vertical angle is tan–1(0.5). Water is poured +into it at a constant rate of 5 cubic metre per hour. Find the rate at which the level of +the water is rising at the instant when the depth of water in the tank is 4 m. +Solution Let r, h and α be as in Fig 6.20. Then +. +tan +r +h +α = +So +α = +1 +tan +r +h +− + + + + +. +But +α = tan–1(0.5) (given) +or +r +h = 0.5 +or +r = 2 +h +Let V be the volume of the cone. Then +V = +2 +3 +2 +1 +1 +3 +3 +2 +12 +h +h +r h +h +π + + +π += +π += + + + + +Therefore +V +d +dt = +3 +12 +d +h +dh +dh +dt + + +π +⋅ + + + + +(by Chain Rule) += +2 +4 +dh +h dt +π +Now rate of change of volume, i.e., V +5 +d +dt = +m3/h and h = 4 m. +Therefore +5 = +2 +(4) +4 +dh +dt +π +⋅ +or +dh +dt = 5 +35 +22 +m/h +4 +88 +7 + + += +π = + + +π + + +Thus, the rate of change of water level is 35 m/h +88 +. +Example 32 A man of height 2 metres walks at a uniform speed of 5 km/h away from +a lamp post which is 6 metres high. Find the rate at which the length of his shadow +increases. +Fig 6.20 +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +179 +Solution In Fig 6.21, Let AB be the lamp-post, the +lamp being at the position B and let MN be the man at +a particular time t and let AM = l metres. Then, MS is +the shadow of the man. Let MS = s metres. +Note that +∆MSN ~ ∆ASB +or +MS +AS = MN +AB +or +AS = 3s (as MN = +2 and AB = 6 (given)) +Thus +AM = 3s – s = 2s. But AM = l +So +l = 2s +Therefore +dl +dt = 2 ds +dt +Since +5 +dl +dt = +km/h. Hence, the length of the shadow increases at the rate 5 +2 km/h. +Example 33 Find intervals in which the function given by +f (x) = +4 +3 +2 +3 +4 +36 +3 +11 +10 +5 +5 +x +x +x +x +− +− ++ ++ +is (a) increasing (b) decreasing. +Solution We have +f (x) = +4 +3 +2 +3 +4 +36 +3 +11 +10 +5 +5 +x +x +x +x +− +− ++ ++ +Therefore +f ′(x) = +3 +2 +3 +4 +36 +(4 +) +(3 +) +3(2 ) +10 +5 +5 +x +x +x +− +− ++ += 6( +1)( +2)( +3) +5 x +x +x +− ++ +− +(on simplification) +Fig 6.21 +Reprint 2025-26 + + MATHEMATICS +180 +Now f ′(x) = 0 gives x = 1, x = – 2, or x = 3. The +points x = 1, – 2, and 3 divide the real line into four +disjoint intervals namely, (– ∞, – 2), (– 2, 1), (1, 3) +and (3, ∞) (Fig 6.22). +Consider the interval (– ∞, – 2), i.e., when – ∞ < x < – 2. +In this case, we have x – 1 < 0, x + 2 < 0 and x – 3 < 0. +(In particular, observe that for x = –3, f ′(x) = (x – 1) (x + 2) (x – 3) = (– 4) (– 1) +(– 6) < 0) +Therefore, +f ′(x) < 0 when – ∞ < x < – 2. +Thus, the function f is decreasing in (– ∞, – 2). +Consider the interval (– 2, 1), i.e., when – 2 < x < 1. +In this case, we have x – 1 < 0, x + 2 > 0 and x – 3 < 0 +(In particular, observe that for x = 0, f ′(x) = (x – 1) (x + 2) (x – 3) = (–1) (2) (–3) += 6 > 0) +So +f ′(x) > 0 when – 2 < x < 1. +Thus, +f is increasing in (– 2, 1). +Now consider the interval (1, 3), i.e., when 1 < x < 3. In this case, we have +x – 1 > 0, x + 2 > 0 and x – 3 < 0. +So, +f ′(x) < 0 when 1 < x < 3. +Thus, + f is decreasing in (1, 3). +Finally, consider the interval (3, ∞), i.e., when x > 3. In this case, we have x – 1 > 0, +x + 2 > 0 and x – 3 > 0. So f ′(x) > 0 when x > 3. +Thus, f is increasing in the interval (3, ∞). +Example 34 Show that the function f given by +f (x) = tan–1(sin x + cos x), x > 0 +is always an increasing function in 0 4 +, π + + + +. +Solution We have +f (x) = tan–1(sin x + cos x), x > 0 +Therefore +f ′(x) = +2 +1 +(cos +sin ) +1 +(sin +cos ) +x +x +x +x +− ++ ++ +Fig 6.22 +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +181 += cos +sin +2 +sin 2 +x +x +x +− ++ +(on simplification) +Note that 2 + sin 2x > 0 for all x in 0, 4 +π + + + + + + . +Therefore +f ′(x) > 0 if cos x – sin x > 0 +or +f ′(x) > 0 if cos x > sin x or cot x > 1 +Now +cot x > 1 if tan x < 1, i.e., if 0 +4 +x +π +< +< +Thus +f ′(x) > 0 in 0 4 +, π + + + + +Hence f is increasing function in 0, 4 +π + + + + + + +. +Example 35 A circular disc of radius 3 cm is being heated. Due to expansion, its +radius increases at the rate of 0.05 cm/s. Find the rate at which its area is increasing +when radius is 3.2 cm. +Solution Let r be the radius of the given disc and A be its area. Then +A = πr2 +or +A +d +dt = 2 +dr +r dt +π +(by Chain Rule) +Now approximate rate of increase of radius = dr = +0.05 +dr +t +dt ∆= +cm/s. +Therefore, the approximate rate of increase in area is given by +dA = +A ( +) +d +t +dt ∆ + = 2 +dr +r +t +dt + + +π +∆ + + + + += 2π (3.2) (0.05) = 0.320π cm2/s +(r = 3.2 cm) +Example 36 An open topped box is to be constructed by removing equal squares from +each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the +sides. Find the volume of the largest such box. +Reprint 2025-26 + + MATHEMATICS +182 +Solution Let x metre be the length of a side of the removed squares. Then, the height +of the box is x, length is 8 – 2x and breadth is 3 – 2x (Fig 6.23). If V(x) is the volume +of the box, then +Fig 6.23 +V(x) = x(3 – 2x) (8 – 2x) += 4x3 – 22x2 + 24x +Therefore +2 +V ( ) +12 +44 +24 +4( +3)(3 +2) +V ( ) +24 +44 +x +x +x +x +x +x +x +′ += +− ++ += +− +− +′′ += +− + +Now +V′(x) = 0 gives +2 +3, 3 +x = +. But x ≠ 3 (Why?) +Thus, we have +2 +3 +x = +. Now +2 +2 +V +24 +44 +28 +0 +3 +3 + + + + +′′ += +− += − +< + + + + + + + + +. +Therefore, +2 +3 +x = + is the point of maxima, i.e., if we remove a square of side 2 +3 +metre from each corner of the sheet and make a box from the remaining sheet, then +the volume of the box such obtained will be the largest and it is given by +2 +V 3 + + + + + + = +3 +2 +2 +2 +2 +4 +22 +24 +3 +3 +3 + + + + + + +− ++ + + + + + + + + + + + + += +3 +200 m +27 +Example 37 Manufacturer can sell x items at a price of rupees 5 +100 +− + + + + +x + each. The +cost price of x items is Rs +x +5 +500 ++ + + + +. Find the number of items he should sell to earn +maximum profit. +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +183 +Solution Let S(x) be the selling price of x items and let C(x) be the cost price of x +items. Then, we have +S(x) = +2 +5 +5 +100 +100 +x +x +x +x + + +− += +− + + + + +and +C(x) = +500 +5 +x + +Thus, the profit function P(x) is given by +P(x) = +2 +S( ) +C( ) +5 +500 +100 +5 +x +x +x +x +x +− += +− +− +− +i.e. +P(x) = +2 +24 +500 +5 +100 +x +x − +− +or +P′(x) = 24 +5 +50 +x +− +Now P′(x) = 0 gives x = 240. Also +1 +P ( ) +50 +x +− +′′ += +. So +1 +P (240) +0 +50 +− +′′ += +< +Thus, x = 240 is a point of maxima. Hence, the manufacturer can earn maximum +profit, if he sells 240 items. +Miscellaneous Exercise on Chapter 6 +1. +Show that the function given by +log +( ) +x +f x +x += + has maximum at x = e. +2. +The two equal sides of an isosceles triangle with fixed base b are decreasing at +the rate of 3 cm per second. How fast is the area decreasing when the two equal +sides are equal to the base ? +3. +Find the intervals in which the function f given by +4sin +2 +cos +( ) +2 +cos +x +x +x +x +f x +x +− +− += ++ +is (i) increasing (ii) decreasing. +4. +Find the intervals in which the function f given by +3 +3 +1 +( ) +, +0 +f x +x +x +x += ++ +≠ + is +(i) increasing +(ii) decreasing. +Reprint 2025-26 + + MATHEMATICS +184 +5. +Find the maximum area of an isosceles triangle inscribed in the ellipse +2 +2 +2 +2 +1 +x +y +a +b ++ += +with its vertex at one end of the major axis. +6. +A tank with rectangular base and rectangular sides, open at the top is to be +constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs +Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is +the cost of least expensive tank? +7. +The sum of the perimeter of a circle and square is k, where k is some constant. +Prove that the sum of their areas is least when the side of square is double the +radius of the circle. +8. +A window is in the form of a rectangle surmounted by a semicircular opening. +The total perimeter of the window is 10 m. Find the dimensions of the window to +admit maximum light through the whole opening. +9. +A point on the hypotenuse of a triangle is at distance a and b from the sides of +the triangle. +Show that the minimum length of the hypotenuse is +2 +2 +3 +3 +3 +2 +( +) +a +b ++ +. +10. +Find the points at which the function f given by f (x) = (x – 2)4 (x + 1)3 has +(i) local maxima +(ii) local minima +(iii) point of inflexion +11. +Find the absolute maximum and minimum values of the function f given by +f (x) = cos2 x + sin x, x ∈ [0, π] +12. +Show that the altitude of the right circular cone of maximum volume that can be +inscribed in a sphere of radius r is 4 +3 +r . +13. +Let f be a function defined on [a, b] such that f ′(x) > 0, for all x ∈ (a, b). Then +prove that f is an increasing function on (a, b). +14. +Show that the height of the cylinder of maximum volume that can be inscribed in +a sphere of radius R is 2R +3 +. Also find the maximum volume. +15. +Show that height of the cylinder of greatest volume which can be inscribed in a +right circular cone of height h and semi vertical angle α is one-third that of the +cone and the greatest volume of cylinder is +3 +2 +4 +tan +27 h +π +α . +Reprint 2025-26 + +APPLICATION OF DERIVATIVES +185 +16. +A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 +cubic metre per hour. Then the depth of the wheat is increasing at the rate of +(A) 1 m/h +(B) 0.1 m/h +(C) 1.1 m/h +(D) 0.5 m/h +Summary +® If a quantity y varies with another quantity x, satisfying some rule +( ) +y +f x += +, +then dy +dx (or +( ) +f +x +′ +) represents the rate of change of y with respect to x and +0 +x x +dy +dx += + + + (or +0 +( ) +f x +′ +) represents the rate of change of y with respect to x at +0 +x +x += +. +® If two variables x and y are varying with respect to another variable t, i.e., if +( ) +x +f t += +and +( ) +y +g t += +, then by Chain Rule +dy +dy +dx +dt +dt +dx = +, if +0 +dx +dt ≠ +. +® A function f is said to be +(a) increasing on an interval (a, b) if +x1 < x2 in (a, b) ⇒ f (x1) < f (x2) for all x1, x2 ∈ (a, b). +Alternatively, if f ′(x) ≥ 0 for each x in (a, b) +(b) decreasing on (a,b) if +x1 < x2 in (a, b) ⇒ f (x1) > f (x2) for all x1, x2 ∈ (a, b). +(c) constant in (a, b), if f (x) = c for all x ∈ (a, b), where c is a constant. +® A point c in the domain of a function f at which either f ′(c) = 0 or f is not +differentiable is called a critical point of f. +® First Derivative Test Let f be a function defined on an open interval I. Let +f be continuous at a critical point c in I. Then + (i) If f ′(x) changes sign from positive to negative as x increases through c, +i.e., if f ′(x) > 0 at every point sufficiently close to and to the left of c, +and f ′(x) < 0 at every point sufficiently close to and to the right of c, +then c is a point of local maxima. +Reprint 2025-26 + + MATHEMATICS +186 +(ii) +If f ′(x) changes sign from negative to positive as x increases through c, +i.e., if f ′(x) < 0 at every point sufficiently close to and to the left of c, +and f ′(x) > 0 at every point sufficiently close to and to the right of c, +then c is a point of local minima. +(iii) If f ′(x) does not change sign as x increases through c, then c is neither +a point of local maxima nor a point of local minima. Infact, such a point +is called point of inflexion. +® Second Derivative Test Let f be a function defined on an interval I and +c ∈ I. Let f be twice differentiable at c. Then + (i) x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0 +The values f (c) is local maximum value of f . +(ii) +x = c is a point of local minima if f ′(c) = 0 and f ″(c) > 0 +In this case, f (c) is local minimum value of f . +(iii) The test fails if f ′(c) = 0 and f ″(c) = 0. +In this case, we go back to the first derivative test and find whether c is +a point of maxima, minima or a point of inflexion. +® Working rule for finding absolute maxima and/or absolute minima +Step 1: Find all critical points of f in the interval, i.e., find points x where +either f ′(x) = 0 or f is not differentiable. +Step 2:Take the end points of the interval. +Step 3: At all these points (listed in Step 1 and 2), calculate the values of f . +Step 4: Identify the maximum and minimum values of f out of the values +calculated in Step 3. This maximum value will be the absolute maximum +value of f and the minimum value will be the absolute minimum value of f . +—v— +Reprint 2025-26" +class_12,7,Integrals,ncert_books/class_12/lemh2dd/lemh201.pdf,"INTEGRALS 225 +v Just as a mountaineer climbs a mountain – because it is there, so +a good mathematics student studies new material because +it is there. — JAMES B. BRISTOL v +7.1 Introduction +Differential Calculus is centred on the concept of the +derivative. The original motivation for the derivative was +the problem of defining tangent lines to the graphs of +functions and calculating the slope of such lines. Integral +Calculus is motivated by the problem of defining and +calculating the area of the region bounded by the graph of +the functions. +If a function f is differentiable in an interval I, i.e., its +derivative f ′exists at each point of I, then a natural question +arises that given f ′at each point of I, can we determine +the function? The functions that could possibly have given +function as a derivative are called anti derivatives (or +primitive) of the function. Further, the formula that gives +all these anti derivatives is called the indefinite integral of the function and such +process of finding anti derivatives is called integration. Such type of problems arise in +many practical situations. For instance, if we know the instantaneous velocity of an +object at any instant, then there arises a natural question, i.e., can we determine the +position of the object at any instant? There are several such practical and theoretical +situations where the process of integration is involved. The development of integral +calculus arises out of the efforts of solving the problems of the following types: +(a) +the problem of finding a function whenever its derivative is given, +(b) +the problem of finding the area bounded by the graph of a function under certain +conditions. +These two problems lead to the two forms of the integrals, e.g., indefinite and +definite integrals, which together constitute the Integral Calculus. +Chapter 7 +INTEGRALS +G .W. Leibnitz +(1646 -1716) +Reprint 2025-26 + +226 +MATHEMATICS +There is a connection, known as the Fundamental Theorem of Calculus, between +indefinite integral and definite integral which makes the definite integral as a practical +tool for science and engineering. The definite integral is also used to solve many interesting +problems from various disciplines like economics, finance and probability. +In this Chapter, we shall confine ourselves to the study of indefinite and definite +integrals and their elementary properties including some techniques of integration. +7.2 Integration as an Inverse Process of Differentiation +Integration is the inverse process of differentiation. Instead of differentiating a function, +we are given the derivative of a function and asked to find its primitive, i.e., the original +function. Such a process is called integration or anti differentiation. +Let us consider the following examples: +We know that +(sin ) +d +x +dx + = cos x +... (1) +3 +( +) +3 +d +x +dx + = x2 +... (2) +and +( +) +x +d +e +dx += ex +... (3) +We observe that in (1), the function cos x is the derived function of sin x. We say +that sin x is an anti derivative (or an integral) of cos x. Similarly, in (2) and (3), +3 +3 +x and +ex are the anti derivatives (or integrals) of x2 and ex, respectively. Again, we note that +for any real number C, treated as constant function, its derivative is zero and hence, we +can write (1), (2) and (3) as follows : +(sin ++ C) +cos += +d +x +x +dx +, +3 +2 +( ++ C) +3 += +d +x +x +dx +and +( ++ C) = +x +x +d +e +e +dx +Thus, anti derivatives (or integrals) of the above cited functions are not unique. +Actually, there exist infinitely many anti derivatives of each of these functions which +can be obtained by choosing C arbitrarily from the set of real numbers. For this reason +C is customarily referred to as arbitrary constant. In fact, C is the parameter by +varying which one gets different anti derivatives (or integrals) of the given function. +More generally, if there is a function F such that +F ( ) = +( ) +d +x +f x +dx +, ∀x ∈ I (interval), +then for any arbitrary real number C, (also called constant of integration) +[ +] +F ( ) + C +d +x +dx + = f (x), x ∈ I +Reprint 2025-26 + +INTEGRALS 227 +Thus, +{F + C, C ∈ R} denotes a family of anti derivatives of f. +Remark Functions with same derivatives differ by a constant. To show this, let g and h +be two functions having the same derivatives on an interval I. +Consider the function f = g – h defined by f (x) = g(x) – h(x), ∀x ∈ I +Then +df +dx = f′ = g′ – h′ giving f′ (x) = g′ (x) – h′ (x) ∀x ∈ I +or +f′ (x) = 0, ∀x ∈ I by hypothesis, +i.e., the rate of change of f with respect to x is zero on I and hence f is constant. +In view of the above remark, it is justified to infer that the family {F + C, C ∈ R} +provides all possible anti derivatives of f. +We introduce a new symbol, namely, +( ) +f x dx +∫ + which will represent the entire +class of anti derivatives read as the indefinite integral of f with respect to x. +Symbolically, we write +( ) += F ( ) + C +f x dx +x +∫ +. +Notation Given that +( ) +dy +f x +dx = +, we write y = +( ) +f x dx +∫ +. +For the sake of convenience, we mention below the following symbols/terms/phrases +with their meanings as given in the Table (7.1). +Table 7.1 +Symbols/Terms/Phrases +Meaning +( ) +f x dx +∫ +Integral of f with respect to x +f (x) in +( ) +f x dx +∫ +Integrand +x in +( ) +f x dx +∫ +Variable of integration +Integrate +Find the integral +An integral of f +A function F such that +F′(x) = f (x) +Integration +The process of finding the integral +Constant of Integration +Any real number C, considered as +constant function +Reprint 2025-26 + +228 +MATHEMATICS +We already know the formulae for the derivatives of many important functions. +From these formulae, we can write down immediately the corresponding formulae +(referred to as standard formulae) for the integrals of these functions, as listed below +which will be used to find integrals of other functions. +Derivatives +Integrals (Anti derivatives) +(i) +1 +1 +n +n +d +x +x +dx +n ++ + += + + ++ + + + ; +1 +C +1 +n +n +x +x dx +n ++ += ++ ++ +∫ +, n ≠ –1 +Particularly, we note that +( ) +1 +d +x +dx += ; + +C +dx +x += ++ +∫ +(ii) +( +) +sin +cos +d +x +x +dx += + ; +cos +sin +C +x dx +x += ++ +∫ +(iii) +( +) +– cos +sin +d +x +x +dx += + ; +sin +cos +C +x dx +– +x += ++ +∫ +(iv) +( +) +2 +tan +sec +d +x +x +dx += + ; +2 +sec +tan +C +x dx +x += ++ +∫ +(v) +( +) +2 +– cot +cosec +d +x +x +dx += + ; +2 +cosec +cot +C +x dx +– +x += ++ +∫ +(vi) +( +) +sec +sec +tan +d +x +x +x +dx += + ; +sec +tan +sec +C +x +x dx +x += ++ +∫ +(vii) +( +) +– cosec +cosec +cot +d +x +x +x +dx += + ; +cosec +cot +– cosec +C +x +x dx +x += ++ +∫ +(viii) +( +) +– 1 +2 +1 +sin +1 +d +x +dx +– x += + ; +– 1 +2 +sin +C +1 +dx +x +– x += ++ +∫ +(ix) +( +) +– 1 +2 +1 +– cos +1 +d +x +dx +– x += + ; +– 1 +2 +cos +C +1 +dx +– +x +– x += ++ +∫ +(x) +( +) +– 1 +2 +1 +tan +1 +d +x +dx +x += ++ + ; +– 1 +2 +tan +C +1 +dx +x +x += ++ ++ +∫ +(xi) +( +) +x +x +d +e +e +dx += + ; +C +x +x +e dx +e += ++ +∫ +Reprint 2025-26 + +INTEGRALS 229 +(xii) +( +) +1 +log | +| +d +x +dx +x += +; +1 +log | +| C +dx +x +x += ++ +∫ +(xiii) +x +x +d +a +a +dx +log a + += + + + + + ; +C +x +x +a +a dx +log a += ++ +∫ +ANote In practice, we normally do not mention the interval over which the various +functions are defined. However, in any specific problem one has to keep it in mind. +7.2.1 Some properties of indefinite integral +In this sub section, we shall derive some properties of indefinite integrals. +(I) +The process of differentiation and integration are inverses of each other in the +sense of the following results : +( ) +d +f x dx +dx ∫ + = f (x) +and +( ) +f x dx +′ +∫ + = f (x) + C, where C is any arbitrary constant. +Proof Let F be any anti derivative of f, i.e., +F( ) +d +x +dx + = f (x) +Then +( ) +f x dx +∫ + = F(x) + C +Therefore +( ) +d +f x dx +dx ∫ + = +( +) +F ( ) + C +d +x +dx += +F ( ) = +( ) +d +x +f x +dx +Similarly, we note that +f ′(x) = +( ) +d f x +dx +and hence +( ) +f x dx +′ +∫ + = f (x) + C +where C is arbitrary constant called constant of integration. +(II) +Two indefinite integrals with the same derivative lead to the same family of +curves and so they are equivalent. +Reprint 2025-26 + +230 +MATHEMATICS +Proof Let f and g be two functions such that +( ) +d +f x dx +dx ∫ + = +( ) +d +g x dx +dx ∫ +or +( ) +( ) +d +f x dx – +g x dx +dx + + + + +∫ +∫ + = 0 +Hence +( ) +( ) +f x dx – +g x dx +∫ +∫ += C, where C is any real number + (Why?) +or +( ) +f x dx +∫ + = +( ) +C +g x dx + +∫ +So the families of curves { +} +1 +1 +( ) +C , C +R +f x dx + +∈ +∫ +and +{ +} +2 +2 +( ) +C , C +R +g x dx + +∈ +∫ + are identical. +Hence, in this sense, +( ) +and +( ) +f x dx +g x dx +∫ +∫ + are equivalent. +A Note The equivalence of the families { +} +1 +1 +( ) ++ C ,C +f x dx +∈ +∫ +R and +{ +} +2 +2 +( ) ++ C ,C +g x dx +∈ +∫ +R is customarily expressed by writing +( ) += +( ) +f x dx +g x dx +∫ +∫ +, +without mentioning the parameter. +(III) +[ +] +( ) + +( ) +( ) ++ +( ) +f x +g x +dx +f x dx +g x dx += +∫ +∫ +∫ +Proof +By Property (I), we have +[ +( ) + +( )] +d +f x +g x +dx +dx + + + + +∫ + = f (x) + g (x) +... (1) + On the otherhand, we find that +( ) ++ +( ) +d +f x dx +g x dx +dx + + + + +∫ +∫ + = +( ) ++ +( ) +d +d +f x dx +g x dx +dx +dx +∫ +∫ += f (x) + g (x) +... (2) + Thus, in view of Property (II), it follows by (1) and (2) that +( +) +( ) +( ) +f x +g x +dx ++ +∫ += +( ) +( ) +f x dx +g x dx ++ +∫ +∫ +. +(IV) For any real number k, +( ) +( ) +k f x dx +k +f x dx += +∫ +∫ +Reprint 2025-26 + +INTEGRALS 231 +Proof By the Property (I), +( ) +( ) +d +k f x dx +k f x +dx += +∫ +. +Also +( ) +d +k +f x dx +dx + + + + +∫ + = +( ) += +( ) +d +k +f x dx +k f x +dx ∫ + Therefore, using the Property (II), we have +( ) +( ) +k f x dx +k +f x dx += +∫ +∫ +. +(V) +Properties (III) and (IV) can be generalised to a finite number of functions +f1, f2, ..., fn and the real numbers, k1, k2, ..., kn giving +[ +] +1 1 +2 +2 +( ) +( ) +( ) +n +n +k f x +k f +x +... +k f +x +dx ++ ++ ++ +∫ += +1 +1 +2 +2 +( ) +( ) +( ) +n +n +k +f x dx +k +f +x dx +... +k +f +x dx ++ ++ ++ +∫ +∫ +∫ +. +To find an anti derivative of a given function, we search intuitively for a function +whose derivative is the given function. The search for the requisite function for finding +an anti derivative is known as integration by the method of inspection. We illustrate it +through some examples. +Example 1 Write an anti derivative for each of the following functions using the +method of inspection: +(i) +cos 2x +(ii) +3x2 + 4x3 +(iii) +1 +x , x ≠ 0 +Solution +(i) +We look for a function whose derivative is cos 2x. Recall that +d +dx sin 2x = 2 cos 2x +or +cos 2x = 1 +2 +d +dx (sin 2x) = +1 sin 2 +2 +d +x +dx + + + + + + +Therefore, an anti derivative of cos 2x is 1 sin 2 +2 +x . +(ii) +We look for a function whose derivative is 3x2 + 4x3. Note that +( +) +3 +4 +d +x +x +dx ++ += 3x2 + 4x3. +Therefore, an anti derivative of 3x2 + 4x3 is x3 + x4. +Reprint 2025-26 + +232 +MATHEMATICS +(iii) +We know that +1 +1 +1 +(log ) +0 and +[log ( +)] +( +1) +0 +d +d +x +, x +– x +– +, x +dx +x +dx +– x +x += +> += += +< +Combining above, we get +( +) +1 +log +0 +d +x +, x +dx +x += +≠ +Therefore, 1 +log +dx +x +x += +∫ + is one of the anti derivatives of 1 +x +. +Example 2 Find the following integrals: +(i) +3 +2 +1 +x – dx +x +∫ +(ii) +2 +3 +( +1) +x +dx ++ +∫ +(iii) ∫ +3 +2 +1 +( +2 +– +) ++ +x +x +e +dx +x +Solution +(i) +We have +3 +2 +2 +1 +– +x – +dx +x dx – +x +dx +x += +∫ +∫ +∫ +(by Property V) += +1 +1 +2 +1 +1 +2 +C +C +1 +1 +2 +1 +– +x +x +– +– ++ ++ + + + + ++ ++ + + + + ++ ++ + + + + +; C1, C2 are constants of integration += +2 +1 +1 +2 +C +C +2 +1 +– +x +x +– +– +– ++ + = +2 +1 +2 +1 + C +C +2 +x +– +x ++ += +2 +1 + C +2 +x +x ++ +, where C = C1 – C2 is another constant of integration. +ANote From now onwards, we shall write only one constant of integration in the +final answer. +(ii) We have +2 +2 +3 +3 +( +1) +x +dx +x dx +dx ++ += ++ +∫ +∫ +∫ += +2 +1 +3 +C +2 +1 +3 +x +x ++ ++ ++ ++ + = +5 +3 +3 +C +5 x +x ++ ++ +Reprint 2025-26 + +INTEGRALS 233 +(iii) We have +3 +3 +2 +2 +1 +1 +( +2 +) +2 +x +x +x +e – +dx +x dx +e dx – +dx +x +x ++ += ++ +∫ +∫ +∫ +∫ += +3 +1 +2 +2 +– log ++ C +3 +1 +2 +x +x +e +x ++ ++ ++ += +5 +2 +2 +2 +– log ++ C +5 +x +x +e +x ++ +Example 3 Find the following integrals: +(i) +(sin +cos ) +x +x dx ++ +∫ +(ii) cosec +(cosec +cot ) +x +x +x dx ++ +∫ +(iii) +2 +1 +sin +cos +– +x dx +x +∫ +Solution +(i) +We have +(sin +cos ) +sin +cos +x +x dx +x dx +x dx ++ += ++ +∫ +∫ +∫ += – cos +sin +C +x +x ++ ++ +(ii) +We have +2 +(cosec +(cosec ++ cot ) +cosec +cosec +cot +x +x +x dx +x dx +x +x dx += ++ +∫ +∫ +∫ += – cot +cosec +C +x – +x + +(iii) +We have +2 +2 +2 +1 +sin +1 +sin +cos +cos +cos +– +x +x +dx +dx – +dx +x +x +x += +∫ +∫ +∫ += +2 +sec +tan +sec +x dx – +x +x dx +∫ +∫ += tan +sec +C +x – +x + +Example 4 Find the anti derivative F of f defined by f (x) = 4x3 – 6, where F (0) = 3 +Solution One anti derivative of f (x) is x4 – 6x since +4 +( +6 ) +d +x – x +dx + = 4x3 – 6 +Therefore, the anti derivative F is given by +F(x) = x4 – 6x + C, where C is constant. +Reprint 2025-26 + +234 +MATHEMATICS +Given that +F(0) = 3, which gives, +3 = 0 – 6 × 0 + C or C = 3 +Hence, the required anti derivative is the unique function F defined by +F(x) = x4 – 6x + 3. +Remarks +(i) +We see that if F is an anti derivative of f, then so is F + C, where C is any +constant. Thus, if we know one anti derivative F of a function f, we can write +down an infinite number of anti derivatives of f by adding any constant to F +expressed by F(x) + C, C ∈ R. In applications, it is often necessary to satisfy an +additional condition which then determines a specific value of C giving unique +anti derivative of the given function. +(ii) +Sometimes, F is not expressible in terms of elementary functions viz., polynomial, +logarithmic, exponential, trigonometric functions and their inverses etc. We are +therefore blocked for finding +( ) +f x dx +∫ +. For example, it is not possible to find +2 +– x +e +dx +∫ + by inspection since we can not find a function whose derivative is +2 +– x +e +(iii) +When the variable of integration is denoted by a variable other than x, the integral +formulae are modified accordingly. For instance +4 +1 +4 +5 +1 +C +C +4 +1 +5 +y +y dy +y ++ += ++ += ++ ++ +∫ +EXERCISE 7.1 +Find an anti derivative (or integral) of the following functions by the method of inspection. +1. +sin 2x +2. cos 3x +3. e2x +4. +(ax + b)2 +5. sin 2x – 4 e3x +Find the following integrals in Exercises 6 to 20: +6. +3 +(4 ++ 1) +x +e +dx +∫ +7. +2 +2 +1 +(1 – +) +x +dx +x +∫ +8. +2 +( +) +ax +bx +c dx ++ ++ +∫ +9. +2 +(2 +) +x +x +e +dx ++ +∫ +10. +2 +1 +x – +dx +x + + + + + + +∫ +11. +3 +2 +2 +5 +4 +x +x – +dx +x ++ +∫ +12. +3 +3 +4 +x +x +dx +x ++ ++ +∫ +13. +3 +2 +1 +1 +x +x +x – +dx +x – +− ++ +∫ +14. +(1 +) +– x +x dx +∫ +Reprint 2025-26 + +INTEGRALS 235 +15. +2 +( 3 +2 +3) +x +x +x +dx ++ ++ +∫ +16. +(2 +3cos +) +x +x – +x +e +dx ++ +∫ +17. +2 +(2 +3sin +5 +) +x – +x +x dx ++ +∫ +18. +sec +(sec +tan ) +x +x +x dx ++ +∫ +19. +2 +2 +sec +cosec +x dx +x +∫ +20. +2 +2 – 3sin +cos +x +x +∫ +dx. +Choose the correct answer in Exercises 21 and 22. +21. +The anti derivative of +1 +x +x + + ++ + + + + + equals +(A) +1 +1 +3 +2 +1 +2 +C +3 x +x ++ ++ +(B) +2 +2 +3 +2 +1 +C +3 +2 +x +x ++ ++ +(C) +3 +1 +2 +2 +2 +2 +C +3 x +x ++ ++ +(D) +3 +1 +2 +2 +3 +1 +C +2 +2 +x +x ++ ++ +22. +If +3 +4 +3 +( ) +4 +d f x +x +dx +x += +− + such that f (2) = 0. Then f (x) is +(A) +4 +3 +1 +129 +8 +x +x ++ +− +(B) +3 +4 +1 +129 +8 +x +x ++ ++ +(C) +4 +3 +1 +129 +8 +x +x ++ ++ +(D) +3 +4 +1 +129 +8 +x +x ++ +− +7.3 Methods of Integration +In previous section, we discussed integrals of those functions which were readily +obtainable from derivatives of some functions. It was based on inspection, i.e., on the +search of a function F whose derivative is f which led us to the integral of f. However, +this method, which depends on inspection, is not very suitable for many functions. +Hence, we need to develop additional techniques or methods for finding the integrals +by reducing them into standard forms. Prominent among them are methods based on: +1. +Integration by Substitution +2. +Integration using Partial Fractions +3. +Integration by Parts +7.3.1 Integration by substitution +In this section, we consider the method of integration by substitution. +The given integral +( ) +f x dx +∫ + can be transformed into another form by changing +the independent variable x to t by substituting x = g (t). +Reprint 2025-26 + +236 +MATHEMATICS +Consider +I = +( ) +f x dx +∫ +Put x = g(t) so that dx +dt = g′(t). +We write +dx = g′(t) dt +Thus +I = +( ) +( ( )) +( ) +f x dx +f g t +g t dt += +′ +∫ +∫ +This change of variable formula is one of the important tools available to us in the +name of integration by substitution. It is often important to guess what will be the useful +substitution. Usually, we make a substitution for a function whose derivative also occurs +in the integrand as illustrated in the following examples. +Example 5 Integrate the following functions w.r.t. x: +(i) +sin mx +(ii) +2x sin (x2 + 1) +(iii) +4 +2 +tan +sec +x +x +x +(iv) +1 +2 +sin (tan +) +1 +– x +x ++ +Solution +(i) +We know that derivative of mx is m. Thus, we make the substitution +mx = t so that mdx = dt. +Therefore, +1 +sin +sin +mx dx +t dt +m += +∫ +∫ + = – 1 +m cos t + C = – 1 +m cos mx + C +(ii) +Derivative of x2 + 1 is 2x. Thus, we use the substitution x2 + 1 = t so that +2x dx = dt. +Therefore, +2 +2 sin ( +1) +sin +x +x +dx +t dt ++ += +∫ +∫ + = – cos t + C = – cos (x2 + 1) + C +(iii) +Derivative of +x is +1 +2 +1 +1 +2 +2 +– +x +x += +. Thus, we use the substitution +1 +so that +giving +2 +x +t +dx +dt +x += += + dx = 2t dt. +Thus, +4 +2 +4 +2 +tan +sec +2 tan +sec +x +x +t +t +t dt +dx +t +x += +∫ +∫ + = +4 +2 +2 tan +sec +t +t dt +∫ +Again, we make another substitution tan t = u so that +sec2 t dt = du +Reprint 2025-26 + +INTEGRALS 237 +Therefore, +4 +2 +4 +2 tan +sec +2 +t +t dt +u du += +∫ +∫ + = +5 +2 +C +5 +u + += +5 +2 tan +C +5 +t + + (since u = tan t) += +5 +2 tan +C (since +) +5 +x +t +x ++ += +Hence, +4 +2 +tan +sec +x +x dx +x +∫ + = +5 +2 tan +C +5 +x + +Alternatively, make the substitution tan +x +t += +(iv) +Derivative of +1 +2 +1 +tan +1 +– x +x += ++ +. Thus, we use the substitution +tan–1 x = t so that +2 +1 +dx +x ++ + = dt. +Therefore , +1 +2 +sin (tan +) +sin +1 +– x dx +t dt +x += ++ +∫ +∫ + = – cos t + C = – cos(tan –1x) + C +Now, we discuss some important integrals involving trigonometric functions and +their standard integrals using substitution technique. These will be used later without +reference. +(i) ∫tan += log sec ++ C +x dx +x +We have +sin +tan +cos +x +x dx +dx +x += +∫ +∫ +Put cos x = t so that sin x dx = – dt +Then +tan +log +C +log cos +C +dt +x dx +– +– +t +– +x +t += += ++ += ++ +∫ +∫ +or +tan +log sec +C +x dx +x += ++ +∫ +(ii) ∫cot += log sin ++ C +x dx +x +We have +cos +cot +sin +x +x dx +dx +x += +∫ +∫ +Reprint 2025-26 + +238 +MATHEMATICS +Put sin x = t so that cos x dx = dt +Then +cot +dt +x dx +t += +∫ +∫ + = log +C +t + + = log sin +C +x + +(iii) ∫sec += log sec ++ tan ++ C +x dx +x +x +We have +sec +(sec +tan ) +sec +sec ++ tan +x +x +x +x dx +dx +x +x ++ += +∫ +∫ +Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt +Therefore, sec +log ++ C = log sec +tan +C +dt +x dx +t +x +x +t += += ++ ++ +∫ +∫ +(iv) ∫cosec += log cosec +– cot ++ C +x dx +x +x +We have +cosec +(cosec +cot ) +cosec +(cosec +cot ) +x +x +x +x dx +dx +x +x ++ += ++ +∫ +∫ +Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt +So +cosec +– +–log | | +– log |cosec +cot +| +C +dt +x dx +t +x +x +t += += += ++ ++ +∫ +∫ += +2 +2 +cosec +cot +– log +C +cosec +cot +x +x +x +x +− ++ +− += log cosec +cot +C +x – +x + +Example 6 Find the following integrals: +(i) +3 +2 +sin +cos +x +x dx +∫ +(ii) +sin +sin ( +) +x +dx +x +a ++ +∫ + (iii) +1 +1 +tan +dx +x ++ +∫ +Solution +(i) +We have +3 +2 +2 +2 +sin +cos +sin +cos +(sin ) +x +x dx +x +x +x dx += +∫ +∫ += +2 +2 +(1 – cos +) cos +(sin ) +x +x +x dx +∫ +Put t = cos x so that dt = – sin x dx +Reprint 2025-26 + +INTEGRALS 239 +Therefore, +2 +2 +sin +cos +(sin ) +x +x +x dx +∫ + = +2 +2 +(1 – +) +t +t dt +−∫ += +3 +5 +2 +4 +( +– +) +C +3 +5 +t +t +– +t +t +dt +– +– + + += ++ + + + + +∫ += +3 +5 +1 +1 +cos +cos +C +3 +5 +– +x +x ++ ++ +(ii) +Put x + a = t. Then dx = dt. Therefore +sin +sin ( +) +sin ( +) +sin +x +t – a +dx +dt +x +a +t += ++ +∫ +∫ += +sin cos +cos sin +sin +t +a – +t +a dt +t +∫ += cos +– sin +cot +a +dt +a +t dt +∫ +∫ += +1 +(cos ) +(sin ) log sin +C +a t – +a +t + + ++ + + += +1 +(cos ) ( +) +(sin ) log sin ( +) +C +a +x +a – +a +x +a + + ++ ++ ++ + + += +1 +cos +cos +(sin +) log sin ( +) +C sin +x +a +a +a – +a +x +a – +a ++ ++ +Hence, +sin +sin ( +) +x +dx +x +a ++ +∫ + = x cos a – sin a log |sin (x + a)| + C, +where, C = – C1 sin a + a cos a, is another arbitrary constant. +(iii) +cos +1 +tan +cos +sin +dx +x dx +x +x +x += ++ ++ +∫ +∫ += +1 +(cos ++ sin ++ cos +– sin ) +2 +cos +sin +x +x +x +x dx +x +x ++ +∫ += +1 +1 +cos +– sin +2 +2 +cos +sin +x +x +dx +dx +x +x ++ ++ +∫ +∫ += +1 +C +1 +cos +sin +2 +2 +2 +cos +sin +x +x – +x dx +x +x ++ ++ ++ +∫ +... (1) +Reprint 2025-26 + +240 +MATHEMATICS +Now, consider +cos +sin +I +cos +sin +x – +x dx +x +x += ++ +∫ +Put cos x + sin x = t so that (cos x – sin x) dx = dt +Therefore +2 +I +log +C +dt +t +t += += ++ +∫ += +2 +log cos +sin +C +x +x ++ ++ +Putting it in (1), we get +1 +2 +C +C +1 ++ ++ +log cos +sin +1 +tan +2 +2 +2 +2 +dx +x +x +x +x += ++ ++ ++ +∫ += +1 +2 +C +C +1 ++ +log cos +sin +2 +2 +2 +2 +x +x +x ++ ++ ++ += +1 +2 +C +C +1 ++ +log cos +sin +C +C +2 +2 +2 +2 +x +x +x +, + ++ ++ += ++ + + + + +EXERCISE 7.2 +Integrate the functions in Exercises 1 to 37: +1. +2 +2 +1 +x +x ++ +2. ( +) +2 +log x +x +3. +1 +log +x +x +x ++ +4. sin +sin (cos ) +x +x +5. sin ( +) cos ( +) +ax +b +ax +b ++ ++ +6. +ax +b ++ +7. +2 +x +x + +8. +2 +1 +2 +x +x ++ +9. +2 +(4 +2) +1 +x +x +x ++ ++ ++ +10. +1 +x – +x +11. +4 +x +x + +, x > 0 +12. +1 +3 +5 +3 +( +1) +x – +x +13. +2 +3 3 +(2 +3 +) +x +x ++ +14. +1 +(log )m +x +x +, x > 0, +1 +≠ +m +15. +2 +9 +4 +x +– +x +16. +2 +3 +x +e ++ +17. +2 +x +x +e +18. +1 +2 +1 +– +tan +x +e +x ++ +19. +2 +2 +1 +1 +x +x +e +– +e ++ +20. +2 +2 +2 +2 +x +– +x +x +– +x +e +– e +e +e ++ +Reprint 2025-26 + +INTEGRALS 241 +21. tan2 (2x – 3) +22. sec2 (7 – 4x) +23. +1 +2 +sin +1 +– x +– x +24. +2cos +3sin +6cos +4sin +x – +x +x +x ++ +25. +2 +2 +1 +cos +(1 +tan ) +x +– +x +26. +cos +x +x +27. +sin 2 cos 2 +x +x +28. +cos +1 +sin +x +x ++ +29. cot x log sin x +30. +sin +1 +cos +x +x ++ +31. ( +) +2 +sin +1 +cos +x +x ++ +32. +1 +1 +cot x ++ +33. +1 +1 +tan +– +x +34. +tan +sin +cos +x +x +x +35. ( +) +2 +1 +log x +x ++ +36. +( +) +2 +( +1) +log +x +x +x +x ++ ++ +37. +( +) +3 +1 +4 +sin tan +1 +– +x +x +x8 ++ +Choose the correct answer in Exercises 38 and 39. +38. +9 +10 +10 +10 log 10 +10 +x +e +x +x +dx +x ++ ++ +∫ + equals +(A) 10x – x10 + C +(B) 10x + x10 + C +(C) (10x – x10)–1 + C +(D) log (10x + x10) + C +39. +2 +2 +equals +sin +cos +dx +x +x +∫ +(A) +tan x + cot x + C +(B) tan x – cot x + C +(C) +tan x cot x + C +(D) tan x – cot 2x + C +7.3.2 Integration using trigonometric identities +When the integrand involves some trigonometric functions, we use some known identities +to find the integral as illustrated through the following example. +Example 7 Find (i) +2 +cos x dx +∫ + (ii) sin 2 cos 3 +x +x dx +∫ + (iii) +3 +sin x dx +∫ +Reprint 2025-26 + +242 +MATHEMATICS +Solution +(i) +Recall the identity cos 2x = 2 cos2 x – 1, which gives +cos2x = 1 +cos 2 +2 +x ++ +Therefore, + = 1 (1+ cos 2 ) +2 +x dx +∫ += 1 +1 +cos 2 +2 +2 +dx +x dx ++ +∫ +∫ += +1 sin 2 +C +2 +4 +x +x ++ ++ +(ii) +Recall the identity sin x cos y = 1 +2 [sin (x + y) + sin (x – y)] +(Why?) +Then + = += +1 +1 cos 5 +cos +C +2 +5 +– +x +x + + ++ ++ + + + + += +1 +1 +cos 5 +cos +C +10 +2 +– +x +x ++ ++ +(iii) +From the identity sin 3x = 3 sin x – 4 sin3 x, we find that +sin3x = 3sin +sin 3 +4 +x – +x +Therefore, +3 +sin x dx +∫ + = 3 +1 +sin +sin 3 +4 +4 +x dx – +x dx +∫ +∫ + = +3 +1 +– +cos +cos 3 +C +4 +12 +x +x ++ ++ +Alternatively, +3 +2 +sin +sin +sin +x dx +x +x dx += +∫ +∫ + = +2 +(1 – cos +) sin +x +x dx +∫ +Put cos x = t so that – sin x dx = dt +Therefore, +3 +sin x dx +∫ + = +( +) +2 +1 – t +dt +−∫ + = +3 +2 +C +3 +t +– +dt +t dt +– t ++ += ++ ++ +∫ +∫ += +3 +1 +cos +cos +C +3 +– +x +x ++ ++ +Remark It can be shown using trigonometric identities that both answers are equivalent. +Reprint 2025-26 + +INTEGRALS 243 +EXERCISE 7.3 +Find the integrals of the functions in Exercises 1 to 22: +1. +sin2 (2x + 5) +2. sin 3x cos 4x +3. cos 2x cos 4x cos 6x +4. +sin3 (2x + 1) +5. sin3 x cos3 x +6. sin x sin 2x sin 3x +7. +sin 4x sin 8x +8. 1 +cos +1 +cos +– +x +x ++ +9. +cos +1 +cos +x +x ++ +10. +sin4 x +11. cos4 2x +12. +2 +sin +1 +cos +x +x ++ +13. +cos 2 +cos 2 +cos +cos +x – +x – +α +α +14. +cos +sin +1 +sin 2 +x – +x +x ++ +15. tan3 2x sec 2x +16. +tan4x +17. +3 +3 +2 +2 +sin +cos +sin +cos +x +x +x +x ++ +18. +2 +2 +cos 2 +2sin +cos +x +x +x ++ +19. +3 +1 +sin +cos +x +x +20. +( +) +2 +cos 2 +cos +sin +x +x +x ++ +21. sin – 1 (cos x) +22. +1 +cos ( +) cos ( +) +x – a +x – b +Choose the correct answer in Exercises 23 and 24. +23. +2 +2 +2 +2 +sin +cos +is equal to +sin +cos +x +x dx +x +x +− +∫ +(A) tan x + cot x + C +(B) tan x + cosec x + C +(C) – tan x + cot x + C +(D) tan x + sec x + C +24. +2 +(1 +) +equals +cos ( +) +x +x +e +x +dx +e x ++ +∫ +(A) – cot (exx) + C +(B) tan (xex) + C +(C) tan (ex) + C +(D) cot (ex) + C +7.4 Integrals of Some Particular Functions +In this section, we mention below some important formulae of integrals and apply them +for integrating many other related standard integrals: +(1) ∫ +2 +2 +1 +– += +log ++ C +2 ++ +– +dx +x +a +a +x +a +x +a +Reprint 2025-26 + +244 +MATHEMATICS +(2) ∫ +2 +2 +1 ++ += +log ++ C +2 +– +– +dx +a +x +a +a +x +a +x +(3) ∫ +– 1 +2 +2 +1 tan +C +dx +x += ++ +a +a +x + a +(4) ∫ +2 +2 +2 +2 = log ++ +– ++ C +– +dx +x +x +a +x +a +(5) ∫ +– 1 +2 +2 = sin ++ C +– +dx +x +a +a +x +(6) ∫ +2 +2 +2 +2 = log ++ ++ ++ C ++ +dx +x +x +a +x +a +We now prove the above results: +(1) +We have +2 +2 +1 +1 +( +) ( +) +x – a +x +a +x – a += ++ += +1 +( +) – ( +) +1 +1 +1 +2 +( +) ( +) +2 +x +a +x – a +– +a +x – a +x +a +a +x – a +x +a + + ++ + + += + + + + ++ ++ + + + + +Therefore, +2 +2 +1 +2 +dx +dx +dx +– +a +x – a +x +a +x – a + + += + + ++ + + +∫ +∫ +∫ += +[ +] +1 +log ( +)| +log ( +)| +C +2 +| x – a ��� +| x +a +a ++ ++ += 1 log +C +2 +x – a +a +x +a + ++ +(2) +In view of (1) above, we have +2 +2 +1 +1 +( +) +( +) +2 +( +) ( +) +– +a +x +a +x +a +a +x +a +x +a +x + + ++ ++ +− += + + ++ +− + + = 1 +1 +1 +2a +a +x +a +x + + ++ + + +− ++ + + +Reprint 2025-26 + +INTEGRALS 245 + Therefore, +2 +2 +– +dx +a +x +∫ + = 1 +2 +dx +dx +a +a +x +a +x + + ++ + + +− ++ + + +∫ +∫ += 1 [ log | +| +log | +|] +C +2 +a +x +a +x +a − +− ++ ++ ++ += +1 log +C +2 +a +x +a +a +x ++ ++ +− +ANote The technique used in (1) will be explained in Section 7.5. +(3) +Put x = a tan θ. Then dx = a sec2 θ dθ. +Therefore, +2 +2 +dx +x +a ++ +∫ + = += +1 +1 +1 +1 +θ +θ +C +tan +C +– x +d +a +a +a +a += ++ += ++ +∫ +(4) +Let x = a secθ. Then dx = a secθ tanθ dθ. +Therefore, +2 +2 +dx +x +a +− +∫ + = +2 +2 +2 +secθ tanθ θ +sec θ +a +d +a +a +− +∫ += +1 +secθ θ +log secθ + tanθ + C +d = +∫ += +2 +1 +2 +log +1 +C +x +x – +a +a ++ ++ += +2 +2 +1 +log +log +C +x +x – a +a ++ +− ++ += +2 +2 +log ++ C +x +x – a ++ +, where C = C1 – log |a| +(5) +Let x = a sinθ. Then dx = a cosθ dθ. +Therefore, + +2 +2 +dx +a +x +− +∫ + = +2 +2 +2 +θ +θ +θ +cos +sin +a +d +a – a +∫ += +1 +θ = θ + C = sin +C +– +x +d +a + +∫ +(6) +Let x = a tanθ. Then dx = a sec2θ dθ. +Therefore, +2 +2 +dx +x +a ++ +∫ + = +2 +2 +2 +2 +θ +θ +θ +sec +tan +a +d +a +a ++ +∫ + = +1 +θ +θ +secθ θ = log (sec +tan ) +C +d ++ ++ +∫ +Reprint 2025-26 + +246 +MATHEMATICS += +2 +1 +2 +log +1 +C +x +x +a +a ++ ++ ++ += +2 +1 +log +log +C +x +x +a +|a| +2 ++ ++ +− ++ += +2 +log +C +x +x +a2 ++ ++ ++ +, where C = C1 – log |a| +Applying these standard formulae, we now obtain some more formulae which +are useful from applications point of view and can be applied directly to evaluate +other integrals. +(7) +To find the integral +2 +dx +ax +bx +c ++ ++ +∫ +, we write +ax2 + bx + c = +2 +2 +2 +2 +2 +4 +b +c +b +c +b +a x +x +a +x +– +a +a +a +a +a + + + + + + + + ++ ++ += ++ ++ + + + + + + + + + + + + + + + + + + +Now, put +2 +b +x +t +a ++ += so that dx = dt and writing +2 +2 +2 +4 +c +b +– +k +a +a += ± +. We find the +integral reduced to the form +2 +2 +1 +dt +a +t +k +± +∫ + depending upon the sign of +2 +2 +4 +c +b +– +a +a + + + + + + +and hence can be evaluated. +(8) +To find the integral of the type +, proceeding as in (7), we +obtain the integral using the standard formulae. +(9) +To find the integral of the type +2 +px +q +dx +ax +bx +c ++ ++ ++ +∫ +, where p, q, a, b, c are +constants, we are to find real numbers A, B such that +2 ++ += A +( +) + B = A (2 +) + B +d +px +q +ax +bx +c +ax +b +dx ++ ++ ++ +To determine A and B, we equate from both sides the coefficients of x and the +constant terms. A and B are thus obtained and hence the integral is reduced to +one of the known forms. +Reprint 2025-26 + +INTEGRALS 247 +(10) For the evaluation of the integral of the type +2 +( +) +px +q dx +ax +bx +c ++ ++ ++ +∫ +, we proceed +as in (9) and transform the integral into known standard forms. +Let us illustrate the above methods by some examples. +Example 8 Find the following integrals: +(i) +2 +16 +dx +x − +∫ +(ii) +2 +2 +dx +x +x +− +∫ +Solution +(i) +We have +2 +2 +2 +16 +4 +dx +dx +x +x – += +− +∫ +∫ + = +4 +log +C +8 +4 +x – +x +1 ++ ++ +[by 7.4 (1)] +(ii) +Put x – 1 = t. Then dx = dt. +Therefore, +2 +2 +dx +x +x +− +∫ + = +2 +1 +dt +– t +∫ + = +1 +sin +( ) +C +– +t + +[by 7.4 (5)] += +1 +sin +( – 1) +C +– +x ++ +Example 9 Find the following integrals : +(i) +2 +6 +13 +dx +x +x +− ++ +∫ +(ii) +2 +3 +13 +10 +dx +x +x ++ +− +∫ +(iii) +2 +5 +2 +dx +x +x +− +∫ +Solution +(i) +We have x2 – 6x + 13 = x2 – 6x + 32 – 32 + 13 = (x – 3)2 + 4 +So, +6 +13 +dx +x +x +2 − ++ +∫ + = +( +) +2 +2 +1 +3 +2 +dx +x – ++ +∫ +Let +x – 3 = t. Then dx = dt +Therefore, +6 +13 +dx +x +x +2 − ++ +∫ + = +1 +2 +2 +1 tan +C +2 +2 +2 +– +dt +t +t += ++ ++ +∫ +[by 7.4 (3)] += +1 +1 +3 +tan +C +2 +2 +– +x – ++ +Reprint 2025-26 + +248 +MATHEMATICS +(ii) +The given integral is of the form 7.4 (7). We write the denominator of the integrand, +2 +3 +13 +10 +x +x – ++ + = +2 +13 +10 +3 +3 +3 +x +x +– + + ++ + + + + += +2 +2 +13 +17 +3 +6 +6 +x +– + + + + + + ++ + + + + + + + + + + + + + + +(completing the square) +Thus +3 +13 +10 +dx +x +x +2 + +− +∫ + = +2 +2 +1 +3 +13 +17 +6 +6 +dx +x + + + + ++ +− + + + + + + + + +∫ +Put +13 +6 +x +t ++ += . Then dx = dt. +Therefore, +3 +13 +10 +dx +x +x +2 + +− +∫ + = +2 +2 +1 +3 +17 +6 +dt +t + + +− + + + +∫ += +1 +17 +1 +6 +log +C +17 +17 +3 2 +6 +6 +t – +t ++ +× × ++ +[by 7.4 (i)] += +1 +13 +17 +1 +6 +6 +log +C +13 +17 +17 +6 +6 +x +– +x ++ ++ ++ ++ += +1 +1 +6 +4 +log +C +17 +6 +30 +x +x +− ++ ++ += +1 +1 +3 +2 +1 +1 +log +C +log +17 +5 +17 +3 +x +x +− ++ ++ ++ += +1 +3 +2 +log +C +17 +5 +x +x +− ++ ++ +, where C = +1 +1 +1 +C +log +17 +3 ++ +Reprint 2025-26 + +INTEGRALS 249 +(iii) +We have +2 +2 +5 +2 +5 +5 +dx +dx +x +x +x +x – +2 += + + +− + + + + +∫ +∫ += +2 +2 +1 +5 +1 +1 +5 +5 +dx +x – +– + + + + + + + + + + + + +∫ + (completing the square) +Put +1 +5 +x – +t += . Then dx = dt. +Therefore, +5 +2 +dx +x +x +2 − +∫ + = +2 +2 +1 +5 +1 +5 +dt +t –  + + + + + +∫ += +2 +2 +1 +1 +log +C +5 +5 +t +t –  + ++ ++ + + + + +[by 7.4 (4)] += +2 +1 +1 +2 +log +C +5 +5 +5 +x +x – +x – ++ ++ +Example 10 Find the following integrals: +(i) +2 +2 +6 +5 +x +dx +x +x +2 ++ ++ ++ +∫ +(ii) +2 +3 +5 +4 +x +dx +x – x ++ +− +∫ +Solution +(i) +Using the formula 7.4 (9), we express +x + 2 = +( +) +2 +A +2 +6 +5 +B +d +x +x +dx ++ ++ ++ + = A (4 +6) +B +x + ++ +Equating the coefficients of x and the constant terms from both sides, we get +4A = 1 and 6A + B = 2 or A = 1 +4 and B = 1 +2 . +Therefore, +2 +2 +6 +5 +x +x +x +2 ++ ++ ++ +∫ + = +1 +4 +6 +1 +4 +2 +2 +6 +5 +2 +6 +5 +x +dx +dx +x +x +x +x +2 +2 ++ ++ ++ ++ ++ ++ +∫ +∫ += +1 +2 +1 +1 +I +I +4 +2 ++ + (say) +... (1) +Reprint 2025-26 + +250 +MATHEMATICS +In I1, put 2x2 + 6x + 5 = t, so that (4x + 6) dx = dt +Therefore, +I1 = +1 +log +C +dt +t +t = ++ +∫ += +2 +1 +log | 2 +6 +5| +C +x +x ++ ++ ++ + ... (2) +and +I2 = +2 +2 +1 +5 +2 +2 +6 +5 +3 +2 +dx +dx +x +x +x +x += ++ ++ ++ ++ +∫ +∫ += +2 +2 +1 +2 +3 +1 +2 +2 +dx +x + + + + ++ ++ + + + + + + + + +∫ +Put +3 +2 +x +t ++ += , so that dx = dt, we get +I2 = +2 +2 +1 +2 +1 +2 +dt +t + + ++  + + + +∫ + = +1 +2 +1 +tan +2 +C +1 +2 +2 +– +t + +× +[by 7.4 (3)] += +1 +2 +3 +tan +2 ++ C +2 +– +x + + ++ + + + + + = +( +) +1 +2 +tan +2 +3 + C +– +x + +... (3) +Using (2) and (3) in (1), we get +( +) +2 +1 +2 +1 +1 +log 2 +6 +5 +tan +2 +3 +C +4 +2 +2 +6 +5 +– +x +dx +x +x +x +x +x +2 ++ += ++ ++ ++ ++ ++ ++ ++ +∫ +where, +C = +1 +2 +C +C +4 +2 ++ +(ii) +This integral is of the form given in 7.4 (10). Let us express +x + 3 = +2 +A +(5 +4 +) + B +d +– x – x +dx += A (– 4 – 2x) + B +Equating the coefficients of x and the constant terms from both sides, we get +– 2A = 1 and – 4 A + B = 3, i.e., A = +1 +2 +– + and B = 1 +Reprint 2025-26 + +INTEGRALS 251 +Therefore, +2 +3 +5 +4 +x +dx +x +x ++ +− +− +∫ + = +( +) +2 +2 +4 +2 +1 +2 +5 +4 +5 +4 +– +– x dx +dx +– +x +x +x +x ++ +− +− +− +− +∫ +∫ += +1 +2 +– + I1 + I2 +... (1) +In I1, put 5 – 4x – x2 = t, so that (– 4 – 2x) dx = dt. +Therefore, +I1= ( +) +2 +4 +2 +5 +4 +– +x dx +dt +t +x +x +− += +− +− +∫ +∫ + = +1 +2 +C +t + += +2 +1 +2 5 +4 +C +– x – x + +... (2) +Now consider +I2 = +2 +2 +5 +4 +9 +( +2) +dx +dx +x +x +– x += +− +− ++ +∫ +∫ +Put x + 2 = t, so that dx = dt. +Therefore, +I2 = +1 +2 +2 +2 +sin ++ C +3 +3 +– +dt +t +t += +− +∫ +[by 7.4 (5)] += +1 +2 +2 +sin +C +3 +– x + ++ +... (3) +Substituting (2) and (3) in (1), we obtain +2 +1 +2 +3 +2 +5 – 4 – ++ sin +C +3 +5 +4 +– +x +x +– +x +x +– x – x ++ ++ += ++ +∫ +, where +1 +2 +C +C +C +2 +– += +EXERCISE 7.4 +Integrate the functions in Exercises 1 to 23. +1. +2 +6 +3 +1 +x +x + +2. +2 +1 +1 +4x ++ +3. +( +) +2 +1 +2 +1 +– x ++ +4. +2 +1 +9 +25 +– +x +5. +4 +3 +1 +2 +x +x ++ +6. +2 +6 +1 +x +x +− +7. +2 +1 +1 +x – +x – +8. +2 +6 +6 +x +x +a ++ +9. +2 +2 +sec +tan +4 +x +x + +Reprint 2025-26 + +252 +MATHEMATICS +10. +2 +1 +2 +2 +x +x ++ ++ +11. +2 +1 +9 +6 +5 +x +x ++ ++ +12. +2 +1 +7 +6 +– x – x +13. +( +)( +) +1 +1 +2 +x – +x – +14. +2 +1 +8 +3x – x ++ +15. +( +)( +) +1 +x – a +x – b +16. +2 +4 +1 +2 +3 +x +x +x – ++ ++ +17. +2 +2 +1 +x +x – ++ +18. +2 +5 +2 +1 +2 +3 +x +x +x +− ++ ++ +19. +( +)( +) +6 +7 +5 +4 +x +x – +x – ++ +20. +2 +2 +4 +x +x – x ++ +21. +2 +2 +2 +3 +x +x +x ++ ++ ++ +22. +2 +3 +2 +5 +x +x – x ++ +− +23. +2 +5 +3 +4 +10 +x +x +x ++ ++ ++ +. +Choose the correct answer in Exercises 24 and 25. +24. +2 +equals +2 +2 +dx +x +x ++ ++ +∫ +(A) +x tan–1 (x + 1) + C +(B) +tan–1 (x + 1) + C +(C) +(x + 1) tan–1x + C +(D) +tan–1x + C +25. +2 equals +9 +4 +dx +x +x +− +∫ +(A) +–1 +1 +9 +8 +sin +C +9 +8 +x − + ++ + + + + +(B) +–1 +1 +8 +9 +sin +C +2 +9 +x − + ++ + + + + +(C) +–1 +1 +9 +8 +sin +C +3 +8 +x − + ++ + + + + +(D) +–1 +1 +9 +8 +sin +C +2 +9 +x − + ++ + + + + +7.5 Integration by Partial Fractions +Recall that a rational function is defined as the ratio of two polynomials in the form +P( ) +Q( ) +x +x +, where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0. If the degree of P(x) +is less than the degree of Q(x), then the rational function is called proper, otherwise, it +is called improper. The improper rational functions can be reduced to the proper rational +Reprint 2025-26 + +INTEGRALS 253 +functions by long division process. Thus, if P( ) +Q( ) +x +x is improper, then +1P ( ) +P( ) +T( ) +Q( ) +Q( ) +x +x +x +x +x += ++ +, +where T(x) is a polynomial in x and +1P ( ) +Q( ) +x +x is a proper rational function. As we know +how to integrate polynomials, the integration of any rational function is reduced to the +integration of a proper rational function. The rational functions which we shall consider +here for integration purposes will be those whose denominators can be factorised into +linear and quadratic factors. Assume that we want to evaluate +P( ) +Q( ) +x dx +x +∫ +, where P( ) +Q( ) +x +x +is proper rational function. It is always possible to write the integrand as a sum of +simpler rational functions by a method called partial fraction decomposition. After this, +the integration can be carried out easily using the already known methods. The following +Table 7.2 indicates the types of simpler partial fractions that are to be associated with +various kind of rational functions. +Table 7.2 + S.No. +Form of the rational function +Form of the partial fraction +1. +( – ) ( – ) +px +q +x a +x b ++ +, a ≠ b +A +B +x – a +x – b ++ +2. +2 +( – ) +px +q +x +a ++ +( +) +2 +A +B +x – a +x – a ++ +3. +2 +( – ) ( +) ( +) +px +qx +r +x +a +x – b +x – c ++ ++ +A +B +C +x – a +x – b +x – c ++ ++ +4. +2 +2 +( – ) ( +) +px +qx +r +x +a +x – b ++ ++ +2 +A +B +C +( +) +x – a +x – b +x – a ++ ++ +5. +2 +2 +( – ) ( +) +px +qx +r +x +a +x +bx +c ++ ++ ++ ++ +2 +A +B + C +x +x – a +x +bx +c ++ ++ ++ +, +where x2 + bx + c cannot be factorised further +In the above table, A, B and C are real numbers to be determined suitably. +Reprint 2025-26 + +254 +MATHEMATICS +Example 11 Find +( +1) ( +2) +dx +x +x ++ ++ +∫ +Solution The integrand is a proper rational function. Therefore, by using the form of +partial fraction [Table 7.2 (i)], we write +1 +( +1) ( +2) +x +x ++ ++ + = +A +B +1 +2 +x +x ++ ++ ++ +... (1) +where, real numbers A and B are to be determined suitably. This gives +1 = A (x + 2) + B (x + 1). +Equating the coefficients of x and the constant term, we get +A + B = 0 +and +2A + B = 1 +Solving these equations, we get A =1 and B = – 1. +Thus, the integrand is given by +1 +( +1) ( +2) +x +x ++ ++ + = +1 +–1 +1 +2 +x +x ++ ++ ++ +Therefore, +( +1) ( +2) +dx +x +x ++ ++ +∫ + = +1 +2 +dx +dx +– +x +x ++ ++ +∫ +∫ += log +1 +log +2 +C +x +x ++ +− ++ ++ += +1 +log +C +2 +x +x ++ ++ ++ +Remark The equation (1) above is an identity, i.e. a statement true for all (permissible) +values of x. Some authors use the symbol ‘≡’ to indicate that the statement is an +identity and use the symbol ‘=’ to indicate that the statement is an equation, i.e., to +indicate that the statement is true only for certain values of x. +Example 12 Find +2 +2 +1 +5 +6 +x +dx +x +x ++ +− ++ +∫ +Solution Here the integrand +2 +2 +1 +5 +6 +x +x – x ++ ++ + is not proper rational function, so we divide +x2 + 1 by x2 – 5x + 6 and find that +Reprint 2025-26 + +INTEGRALS 255 +2 +2 +1 +5 +6 +x +x – x ++ ++ + = +2 +5 +5 +5 +5 +1 +1 +( +2) ( +3) +5 +6 +x – +x – +x – +x – +x – x ++ += + ++ +Let +5 +5 +( +2) ( +3) +x – +x – +x – + = +A +B +2 +3 +x – +x – ++ +So that +5x – 5 = A (x – 3) + B (x – 2) +Equating the coefficients of x and constant terms on both sides, we get A + B = 5 +and 3A + 2B = 5. Solving these equations, we get A = – 5 and B = 10 +Thus, +2 +2 +1 +5 +6 +x +x – x ++ ++ + = +5 +10 +1 +2 +3 +x – +x – +− ++ +Therefore, +2 +2 +1 +5 +6 +x +dx +x – x ++ ++ +∫ + = +1 +5 +10 +2 +3 +dx +dx +dx +x – +x – +− ++ +∫ +∫ +∫ += x – 5 log |x – 2| + 10 log |x – 3| + C. +Example 13 Find +2 +3 +2 +( +1) ( +3) +x +dx +x +x +− ++ ++ +∫ +Solution The integrand is of the type as given in Table 7.2 (4). We write +2 +3 +2 +( +1) ( +3) +x – +x +x ++ ++ + = +2 +A +B +C +1 +3 +( +1) +x +x +x ++ ++ ++ ++ ++ +So that +3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2 += A (x2 + 4x + 3) + B (x + 3) + C (x2 + 2x + 1 ) +Comparing coefficient of x2, x and constant term on both sides, we get +A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = – 2. Solving these equations, we get +11 +5 +11 +A +B +and C +4 +2 +4 +– +– +, += += += +. Thus the integrand is given by +2 +3 +2 +( +1) ( +3) +x +x +x +− ++ ++ + = +2 +11 +5 +11 +4 ( +1) +4 ( +3) +2 ( +1) +– +– +x +x +x ++ ++ ++ +Therefore, +2 +3 +2 +( +1) ( +3) +x +x +x +− ++ ++ +∫ + = +2 +11 +5 +11 +4 +1 +2 +4 +3 +( +1) +dx +dx +dx +– +x +x +x +− ++ ++ ++ +∫ +∫ +∫ += 11 +5 +11 +log ++1 +log +3 +C +4 +2 ( +1) +4 +x +x +x ++ +− ++ ++ += +11 ++1 +5 +log ++ C +4 ++ 3 +2 ( +1) +x +x +x ++ +Reprint 2025-26 + +256 +MATHEMATICS +Example 14 Find +2 +2 +2 +( +1) ( +4) +x +dx +x +x ++ ++ +∫ +Solution Consider +2 +2 +2 +( +1) ( +4) +x +x +x ++ ++ + and put x2 = y. +Then +2 +2 +2 +( +1) ( +4) +x +x +x ++ ++ + = ( +1) ( +4) +y +y +y ++ ++ +Write +( +1) ( +4) +y +y +y ++ ++ + = +A +B +1 +4 +y +y ++ ++ ++ +So that +y = A (y + 4) + B (y + 1) +Comparing coefficients of y and constant terms on both sides, we get A + B = 1 +and 4A + B = 0, which give +A = +1 +4 +and B +3 +3 +− += +Thus, +2 +2 +2 +( +1) ( +4) +x +x +x ++ ++ + = +2 +2 +1 +4 +3 ( +1) +3 ( +4) +– +x +x ++ ++ ++ +Therefore, +2 +2 +2 +( +1) ( +4) +x dx +x +x ++ ++ +∫ + = +2 +2 +1 +4 +3 +3 +1 +4 +dx +dx +– +x +x ++ ++ ++ +∫ +∫ += +1 +1 +1 +4 +1 +tan +tan +C +3 +3 +2 +2 +– +– x +– +x + +× ++ += +1 +1 +1 +2 +tan +tan +C +3 +3 +2 +– +– x +– +x + ++ +In the above example, the substitution was made only for the partial fraction part +and not for the integration part. Now, we consider an example, where the integration +involves a combination of the substitution method and the partial fraction method. +Example 15 Find ( +) +2 +3sin +2 cos +5 +cos +4 sin +– +d +– +– +φ +φ +φ +φ +φ +∫ +Solution Let y = sinφ +Then +dy = cosφ dφ +Reprint 2025-26 + +INTEGRALS 257 +Therefore, +( +) +2 +3sin +2 cos +5 +cos +4 sin +– +d +– +– +φ +φ +φ +φ +φ +∫ + = +2 +(3 – 2) +5 +(1 +) +4 +y +dy +– +– y +– +y +∫ += +2 +3 +2 +4 +4 +y – +dy +y – +y + +∫ += +( +) +2 +3 +2 +I (say) +2 +y – +y – += +∫ +Now, we write +( +) +2 +3 +2 +2 +y – +y – + = +2 +A +B +2 +( +2) +y +y ++ +− +− +[by Table 7.2 (2)] +Therefore, +3y – 2 = A (y – 2) + B +Comparing the coefficients of y and constant term, we get A = 3 and B – 2A = – 2, +which gives A = 3 and B = 4. +Therefore, the required integral is given by +I = +2 +3 +4 +[ ++ +] +2 +( +2) +dy +y – +y – +∫ + = +2 +3 ++ 4 +2 +( +2) +dy +dy +y – +y – +∫ +∫ += +1 +3 log +2 +4 +C +2 +y +– y + + +− ++ ++ + + +− + + += +4 +3 log sin +2 +C +2 +sin +– +φ − ++ ++ +φ += +4 +3 log (2 +sin ) ++ C +2 +sin +− +φ + +− +φ + (since, 2 – sinφ is always positive) +Example 16 Find +2 +2 +1 +( +2) ( +1) +x +x +dx +x +x ++ ++ ++ ++ +∫ +Solution The integrand is a proper rational function. Decompose the rational function +into partial fraction [Table 2.2(5)]. Write +2 +2 +1 +( +1) ( +2) +x +x +x +x ++ ++ ++ ++ + = +2 +A +B + C +2 +( +1) +x +x +x ++ ++ ++ +Therefore, +x2 + x + 1 = A (x2 + 1) + (Bx + C) (x + 2) +Reprint 2025-26 + +258 +MATHEMATICS +Equating the coefficients of x2, x and of constant term of both sides, we get +A + B =1, 2B + C = 1 and A + 2C = 1. Solving these equations, we get +3 +2 +1 +A +, B +and C +5 +5 +5 += += += +Thus, the integrand is given by +2 +2 +1 +( +1) ( +2) +x +x +x +x ++ ++ ++ ++ + = +2 +2 +1 +3 +5 +5 +5 ( +2) +1 +x +x +x ++ ++ ++ ++ + = +2 +3 +1 +2 +1 +5 ( +2) +5 +1 +x +x +x ++ + + ++ + + ++ ++ + + +Therefore, +2 +2 +1 +( ++1) ( +2) +x +x +dx +x +x ++ ++ ++ +∫ + = +2 +2 +3 +1 +2 +1 +1 +5 +2 +5 +5 +1 +1 +dx +x +dx +dx +x +x +x ++ ++ ++ ++ ++ +∫ +∫ +∫ += +2 +1 +3 +1 +1 +log +2 +log +1 +tan +C +5 +5 +5 +– +x +x +x ++ ++ ++ ++ ++ +EXERCISE 7.5 +Integrate the rational functions in Exercises 1 to 21. +1. +( +1) ( +2) +x +x +x ++ ++ +2. +2 +1 +9 +x – +3. +3 +1 +( +1) ( +2) ( +3) +x – +x – +x – +x – +4. +( +1) ( +2) ( +3) +x +x – +x – +x – +5. +2 +2 +3 +2 +x +x +x ++ ++ +6. +2 +1 +(1 +2 ) +– x +x +– +x +7. +2 +( +1) ( –1) +x +x +x ++ +8. +2 +( +1) ( +2) +x +x – +x + +9. +3 +2 +3 +5 +1 +x +x – x +x ++ +− ++ +10. +2 +2 +3 +( +1) (2 +3) +x +x – +x +− ++ +11. +2 +5 +( +1) ( +4) +x +x +x ++ +− +12. +3 +2 +1 +1 +x +x +x ++ ++ +− +13. +2 +2 +(1 +) (1 +) +x +x +− ++ +14. +2 +3 +1 +( +2) +x – +x + +15. +4 +1 +1 +x − +16. +1 +( +1) +n +x x + + [Hint: multiply numerator and denominator by x n – 1 and put xn = t ] +17. +cos +(1 – sin ) (2 – sin ) +x +x +x +[Hint : Put sin x = t] +Reprint 2025-26 + +INTEGRALS 259 +18. +2 +2 +2 +2 +( +1) ( +2) +( +3) ( +4) +x +x +x +x ++ ++ ++ ++ +19. +2 +2 +2 +( +1) ( +3) +x +x +x ++ ++ +20. +4 +1 +( +1) +x x – +21. +1 +( +1) +xe – +[Hint : Put ex = t] +Choose the correct answer in each of the Exercises 22 and 23. +22. +( +1) ( +2) +x dx +x +x +− +− +∫ + equals +(A) +2 +( +1) +log +C +2 +x +x +− ++ +− +(B) +2 +( +2) +log +C +1 +x +x +− ++ +− +(C) +2 +1 +log +C +2 +x +x +− + + ++ + + +− + + +(D) log ( +1) ( +2) +C +x +x +− +− ++ +23. +2 +( +1) +dx +x x + +∫ +equals +(A) +2 +1 +log +log ( ++1) + C +2 +x +x +− +(B) +2 +1 +log +log ( ++1) + C +2 +x +x ++ +(C) +2 +1 +log +log ( ++1) + C +2 +x +x +− ++ +(D) +2 +1 log +log ( ++1) + C +2 +x +x ++ +7.6 Integration by Parts +In this section, we describe one more method of integration, that is found quite useful in +integrating products of functions. +If u and v are any two differentiable functions of a single variable x (say). Then, by +the product rule of differentiation, we have +( +) +d +uv +dx + = +dv +du +u +v +dx +dx ++ +Integrating both sides, we get +uv = +dv +du +u +dx +v +dx +dx +dx ++ +∫ +∫ +or +dv +u +dx +dx +∫ + = +du +uv – v +dx +dx +∫ +... (1) +Let +u = f (x) and dv +dx = g(x). Then +du +dx = f ′(x) and v = +( ) +g x dx +∫ +Reprint 2025-26 + +260 +MATHEMATICS +Therefore, expression (1) can be rewritten as +( ) ( ) +f x g x dx +∫ + = +( ) +( ) +[ +( ) +] +( ) +f x +g x dx – +g x dx f x dx +′ +∫ +∫∫ +i.e., +( ) ( ) +f x g x dx +∫ + = +( ) +( ) +[ +( ) +( ) +] +f x +g x dx – +f +x +g x dx dx +′ +∫ +∫ +∫ +If we take f as the first function and g as the second function, then this formula +may be stated as follows: +“The integral of the product of two functions = (first function) × (integral +of the second function) – Integral of [(differential coefficient of the first function) +× (integral of the second function)]” +Example 17 Find +cos +x +x dx +∫ +Solution Put f (x) = x (first function) and g (x) = cos x (second function). +Then, integration by parts gives +cos +x +x dx +∫ + = +cos +[ +( ) cos +] +d +x +x dx – +x +x dx dx +dx +∫ +∫ +∫ += +sin +sin +x +x – +x dx +∫ + = x sin x + cos x + C +Suppose, we take +f (x) = cos x and g(x) = x. Then +cos +x +x dx +∫ + = cos +[ +(cos ) +] +d +x +x dx – +x +x dx dx +dx +∫ +∫ +∫ += ( +) +2 +2 +cos +sin +2 +2 +x +x +x +x +dx ++∫ +Thus, it shows that the integral +cos +x +x dx +∫ + is reduced to the comparatively more +complicated integral having more power of x. Therefore, the proper choice of the first +function and the second function is significant. +Remarks +(i) +It is worth mentioning that integration by parts is not applicable to product of +functions in all cases. For instance, the method does not work for +sin +x +x dx +∫ +. +The reason is that there does not exist any function whose derivative is +x sin x. +(ii) +Observe that while finding the integral of the second function, we did not add +any constant of integration. If we write the integral of the second function cos x +Reprint 2025-26 + +INTEGRALS 261 +as sin x + k, where k is any constant, then +cos +x +x dx +∫ + = +(sin +) +(sin +) +x +x +k +x +k dx ++ +− ++ +∫ += +(sin +) +(sin +x +x +k +x dx +k dx ++ +− +− +∫ +∫ += +(sin +) +cos +C +x +x +k +x – kx ++ +− ++ + = +sin +cos +C +x +x +x ++ ++ +This shows that adding a constant to the integral of the second function is +superfluous so far as the final result is concerned while applying the method of +integration by parts. +(iii) +Usually, if any function is a power of x or a polynomial in x, then we take it as the +first function. However, in cases where other function is inverse trigonometric +function or logarithmic function, then we take them as first function. +Example 18 Find log x dx +∫ +Solution To start with, we are unable to guess a function whose derivative is log x. We +take log x as the first function and the constant function 1 as the second function. Then, +the integral of the second function is x. +Hence, +(log .1) +x +dx +∫ + = log +1 +[ +(log ) 1 +] +d +x +dx +x +dx dx +dx +− +∫ +∫ +∫ += +1 +(log ) +– +log +C +x +x +x dx +x +x – x +x +⋅ += ++ +∫ +. +Example 19 Find +x +x e dx +∫ +Solution Take first function as x and second function as ex. The integral of the second +function is ex. +Therefore, +x +x e dx +∫ + = +1 +x +x +x e +e dx +− +⋅ +∫ + = xex – ex + C. +Example 20 Find +1 +2 +sin +1 +– +x +x dx +x +− +∫ +Solution Let first function be sin – 1x and second function be +2 +1 +x +x +− +. +First we find the integral of the second function, i.e., +2 +1 +x dx +x +− +∫ +. +Put t =1 – x2. Then dt = – 2x dx +Reprint 2025-26 + +262 +MATHEMATICS +Therefore, +2 +1 +x dx +x +− +∫ + = +1 +2 +dt +– +t +∫ + = +2 +– +1 +t +x += − +− +Hence, +1 +2 +sin +1 +– +x +x dx +x +− +∫ + = +( +) +1 +2 +2 +2 +1 +(sin +) +1 +( +1 +) +1 +– x +– +x +– +x +dx +x +− +− +− +− +∫ += +2 +1 +1 +sin +C +– +x +x +x +− +− ++ ++ + = +2 +1 +1 +sin +C +x – +x +x +− +− ++ +Alternatively, this integral can also be worked out by making substitution sin–1 x = θ and +then integrating by parts. +Example 21 Find +sin +x +e +x dx +∫ +Solution Take ex as the first function and sin x as second function. Then, integrating +by parts, we have +I +sin +( +cos ) +cos +x +x +x +e +x dx +e +– +x +e +x dx += += ++ +∫ +∫ += – ex cos x + I1 (say) +... (1) +Taking ex + and cos x as the first and second functions, respectively, in I1, we get +I1 = +sin +sin +x +x +e +x – e +x dx +∫ +Substituting the value of I1 in (1), we get +I = – ex cos x + ex sin x – I or 2I = ex (sin x – cos x) +Hence, +I = +sin +(sin +cos ) + C +2 +x +x +e +e +x dx +x – +x += +∫ +Alternatively, above integral can also be determined by taking sin x as the first function +and ex the second function. +7.6.1 Integral of the type +[ +( ) + +( )] +xe +f x +f +x +dx +′ +∫ +We have +I = +[ +( ) + +( )] +xe +f x +f x +dx +′ +∫ + = +( ) ++ +( ) +x +x +e f x dx +e f x dx +′ +∫ +∫ += +1 +1 +I +( ) +, where I = +( ) +x +x +e f +x dx +e f x dx +′ ++ ∫ +∫ +... (1) +Taking f (x) and ex as the first function and second function, respectively, in I1 and +integrating it by parts, we have I1 = f (x) ex – +( ) +C +x +f x e dx +′ ++ +∫ +Substituting I1 in (1), we get +I = +( ) +( ) +( ) +C +x +x +x +e f x +f +x e dx +e f x dx +′ +′ +− ++ ++ +∫ +∫ + = ex f (x) + C +Reprint 2025-26 + +INTEGRALS 263 +Thus, +′ +∫ +[ +( ) +( )] +x +e + f x + f +x +dx = +( ) +C +x +e f x + +Example 22 Find (i) +1 +2 +1 +(tan +) +1 +x +– +e +x +x ++ ++ +∫ +dx (ii) +2 +2 +( ++1) +( +1) +x +x +e +x +∫ + dx +Solution +(i) We have I = +1 +2 +1 +(tan +) +1 +x +– +e +x +dx +x ++ ++ +∫ +Consider f (x) = tan– 1x, then f ′(x) = +2 +1 +1 +x ++ +Thus, the given integrand is of the form ex [ f (x) + f ′(x)]. +Therefore, +1 +2 +1 +I +(tan +) +1 +x +– +e +x +dx +x += ++ ++ +∫ + = ex tan– 1x + C +(ii) We have +2 +2 +( ++1) +I +( +1) +x +x +e +x += ∫ +dx +2 +2 +1+1+1) +[ +] +( +1) +x +x – +e +dx +x += ∫ +2 +2 +2 +1 +2 +[ +] +( +1) +( +1) +x +x – +e +dx +x +x += ++ +∫ + +2 +1 +2 +[ ++ +] ++1 +( +1) +x +x – +e +dx +x +x += ∫ +Consider +1 +( ) +1 +x +f x +x +− += ++ , then +2 +2 +( ) +( +1) +f x +x +′ += ++ +Thus, the given integrand is of the form ex [f (x) + f ′(x)]. +Therefore, +2 +2 +1 +1 +C +1 +( +1) +x +x +x +x +e dx +e +x +x ++ +− += ++ ++ ++ +∫ +EXERCISE 7.6 +Integrate the functions in Exercises 1 to 22. +1. x sin x +2. x sin 3x +3. +x2 ex +4. x log x +5. x log 2x +6. x2 log x +7. +x sin– 1x +8. x tan–1 x +9. x cos–1 x +10. (sin–1x)2 +11. +1 +2 +cos +1 +x +x +x +− +− +12. x sec2 x +13. tan–1x +14. x (log x)2 +15. +(x2 + 1) log x +Reprint 2025-26 + +264 +MATHEMATICS +16. ex (sinx + cosx) 17. +2 +(1 +) +x +x e +x ++ +18. +1 +sin +1 +cos +x +x +e +x + + ++ + + ++ + + +19. +2 +1 +1 +– +xe +x +x + + + + + + +20. +3 +( +3) +( +1) +x +x +e +x +− +− +21. +e2x sin x +22. +1 +2 +2 +sin +1 +– +x +x + + + + ++ + + +Choose the correct answer in Exercises 23 and 24. +23. +3 +2 +x +x e dx +∫ + equals +(A) +3 +1 +C +3 +xe ++ +(B) +2 +1 +C +3 +xe ++ +(C) +3 +1 +C +2 +xe ++ +(D) +2 +1 +C +2 +xe ++ +24. +sec +(1 +tan ) +xe +x +x dx ++ +∫ + equals +(A) ex cos x + C +(B) ex sec x + C +(C) ex sin x + C +(D) ex tan x + C +7.6.2 Integrals of some more types +Here, we discuss some special types of standard integrals based on the technique of +integration by parts : +(i) +2 +2 +x +a dx +− +∫ +(ii) +2 +2 +x +a dx ++ +∫ +(iii) +2 +2 +a +x dx +− +∫ +(i) + Let +2 +2 +I +x +a dx += +− +∫ +Taking constant function 1 as the second function and integrating by parts, we +have +I = +2 +2 +2 +2 +1 +2 +2 +x +x +x +a +x dx +x +a +− +− +− +∫ += +2 +2 +2 +2 +2 +x +x +x +a +dx +x +a +− +− +− +∫ + = +2 +2 +2 +2 +2 +2 +2 +x +a +a +x +x +a +dx +x +a +− ++ +− +− +− +∫ +Reprint 2025-26 + +INTEGRALS 265 += +2 +2 +2 +2 +2 +2 +2 +dx +x +x +a +x +a dx +a +x +a +− +− +− +− +− +∫ +∫ += +2 +2 +2 +2 +2 +I +dx +x +x +a +a +x +a +− +−− +− +∫ +or +2I = +2 +2 +2 +2 +2 +dx +x +x +a +a +x +a +− +− +− +∫ +or +I = ∫ +2 +2 +x – a dx = +2 +2 +2 +2 +2 +– +– +log ++ +– ++ C +2 +2 +x +a +x +a +x +x +a +Similarly, integrating other two integrals by parts, taking constant function 1 as the +second function, we get +(ii) ∫ +2 +2 +2 +2 +2 +2 +2 +1 ++ += ++ ++ +log ++ ++ ++ C +2 +2 +a +x +a dx +x +x +a +x +x +a +(iii) +Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric +substitution x = a secθ in (i), x = a tanθ in (ii) and x = a sinθ in (iii) respectively. +Example 23 Find +2 +2 +5 +x +x +dx ++ ++ +∫ +Solution Note that +2 +2 +5 +x +x +dx ++ ++ +∫ + = +2 +( +1) +4 +x +dx ++ ++ +∫ +Put x + 1 = y, so that dx = dy. Then +2 +2 +5 +x +x +dx ++ ++ +∫ + = +2 +2 +2 +y +dy ++ +∫ += +2 +2 +1 +4 +4 +log +4 +C +2 +2 +y +y +y +y ++ ++ ++ ++ ++ + [using 7.6.2 (ii)] += +2 +2 +1 ( +1) +2 +5 +2 log +1 +2 +5 +C +2 x +x +x +x +x +x ++ ++ ++ ++ ++ + ++ ++ ++ +Example 24 Find +2 +3 +2x +x dx +− +− +∫ +Solution Note that +2 +2 +3 +2 +4 +( +1) +x +x dx +x +dx +− +− += +− ++ +∫ +∫ +Reprint 2025-26 + +266 +MATHEMATICS +Put x + 1 = y so that dx = dy. +Thus +2 +3 +2x +x dx +− +− +∫ + = +2 +4 +y dy +− +∫ += +2 +1 +1 +4 +4 +sin +C +2 +2 +2 +– y +y +y +− ++ ++ +[using 7.6.2 (iii)] += +2 +1 +1 +1 +( +1) +3 +2 +2 sin +C +2 +2 +– +x +x +x +x ++ + + ++ +− +− ++ ++ + + + + +EXERCISE 7.7 +Integrate the functions in Exercises 1 to 9. +1. +2 +4 +x +− +2. +2 +1 +4x +− +3. +2 +4 +6 +x +x ++ ++ +4. +2 +4 +1 +x +x ++ ++ +5. +2 +1 +4x +x +− +− +6. +2 +4 +5 +x +x ++ +− +7. +2 +1 +3x +x ++ +− +8. +2 +3 +x +x ++ +9. +2 +1 +9 +x ++ +Choose the correct answer in Exercises 10 to 11. +10. +2 +1 +x dx ++ +∫ +is equal to +(A) +( +) +2 +2 +1 +1 +log +1 +C +2 +2 +x +x +x +x ++ ++ ++ ++ ++ +(B) +3 +2 2 +2 (1 +) +C +3 +x ++ ++ +(C) +3 +2 2 +2 +(1 +) +C +3 x +x ++ ++ +(D) +2 +2 +2 +2 +1 +1 +log +1 +C +2 +2 +x +x +x +x +x ++ ++ ++ ++ ++ +11. +2 +8 +7 +x +x +dx +− ++ +∫ + is equal to +(A) +2 +2 +1 ( +4) +8 +7 +9log +4 +8 +7 +C +2 x +x +x +x +x +x +− +− ++ ++ +− ++ +− ++ ++ +(B) +2 +2 +1 ( +4) +8 +7 +9log +4 +8 +7 +C +2 x +x +x +x +x +x ++ +− ++ ++ ++ ++ +− ++ ++ +(C) +2 +2 +1 ( +4) +8 +7 +3 +2log +4 +8 +7 +C +2 x +x +x +x +x +x +− +− ++ +− +− ++ +− ++ ++ +(D) +2 +2 +1 +9 +( +4) +8 +7 +log +4 +8 +7 +C +2 +2 +x +x +x +x +x +x +− +− ++ +− +− ++ +− ++ ++ +Reprint 2025-26 + +INTEGRALS 267 +7.7 Definite Integral +In the previous sections, we have studied about the indefinite integrals and discussed +few methods of finding them including integrals of some special functions. In this +section, we shall study what is called definite integral of a function. The definite integral +has a unique value. A definite integral is denoted by +( ) +b +a f x dx +∫ +, where a is called the +lower limit of the integral and b is called the upper limit of the integral. The definite +integral is introduced either as the limit of a sum or if it has an anti derivative F in the +interval [a, b], then its value is the difference between the values of F at the end +points, i.e., F(b) – F(a). +7.8 Fundamental Theorem of Calculus +7.8.1 Area function +We have defined +( ) +b +a f x dx +∫ + as the area of +the region bounded by the curve y = f(x), +the ordinates x = a and x = b and x-axis. Let x +be a given point in [a, b]. Then +( ) +x +a f x dx +∫ +represents the area of the light shaded region +in Fig 7.1 [Here it is assumed that f(x) > 0 for +x ∈ [a, b], the assertion made below is +equally true for other functions as well]. +The area of this shaded region depends upon +the value of x. +In other words, the area of this shaded region is a function of x. We denote this +function of x by A(x). We call the function A(x) as Area function and is given by +A (x) = ∫ +( ) +x +a f x dx +... (1) +Based on this definition, the two basic fundamental theorems have been given. +However, we only state them as their proofs are beyond the scope of this text book. +7.8.2 First fundamental theorem of integral calculus +Theorem 1 Let f be a continuous function on the closed interval [a, b] and let A (x) be +the area function. Then A′(x) = f (x), for all x ∈ [a, b]. +Fig 7.1 +Reprint 2025-26 + +268 +MATHEMATICS +7.8.3 Second fundamental theorem of integral calculus +We state below an important theorem which enables us to evaluate definite integrals +by making use of anti derivative. +Theorem 2 Let f be continuous function defined on the closed interval [a, b] and F be +an anti derivative of f. Then ∫ +( ) +b +a f x dx = [F( )] = +b +a +x + F (b) – F(a). +Remarks +(i) +In words, the Theorem 2 tells us that +( ) +b +a f x dx +∫ += (value of the anti derivative F +of f at the upper limit b – value of the same anti derivative at the lower limit a). +(ii) +This theorem is very useful, because it gives us a method of calculating the +definite integral more easily, without calculating the limit of a sum. +(iii) +The crucial operation in evaluating a definite integral is that of finding a function +whose derivative is equal to the integrand. This strengthens the relationship +between differentiation and integration. +(iv) +In +( ) +b +a f x dx +∫ +, the function f needs to be well defined and continuous in [a, b]. +For instance, the consideration of definite integral +1 +3 +2 +2 +2 ( +–1) +x x +dx +−∫ + is erroneous +since the function f expressed by f(x) = +1 +2 +2 +( +–1) +x x + is not defined in a portion +– 1 < x < 1 of the closed interval [– 2, 3]. +Steps for calculating +( ) +b +a f x dx +∫ +. +(i) +Find the indefinite integral +( ) +f x dx +∫ +. Let this be F(x). There is no need to keep +integration constant C because if we consider F(x) + C instead of F(x), we get +( ) +[F ( ) +C] +[F( ) +C] – [F( ) +C] +F( ) – F( ) +b +b +a +a f x dx +x +b +a +b +a += ++ += ++ ++ += +∫ +. +Thus, the arbitrary constant disappears in evaluating the value of the definite +integral. +(ii) +Evaluate F(b) – F(a) = [F ( )]b +a +x +, which is the value of +( ) +b +a f x dx +∫ +. +We now consider some examples +Reprint 2025-26 + +INTEGRALS 269 +Example 25 Evaluate the following integrals: +(i) +3 +2 +2 x dx +∫ +(ii) +9 +3 +4 +2 +2 +(30 – +) +x +dx +x +∫ +(iii) +2 +1 ( +1) ( +2) +x dx +x +x ++ ++ +∫ +(iv) +3 +4 +0 sin 2 cos2 +t +t dt +π +∫ +Solution +(i) +Let +3 +2 +2 +I +x dx +=∫ +. Since +3 +2 +F ( ) +3 +x +x dx +x += += +∫ +, +Therefore, by the second fundamental theorem, we get +I = +27 +8 +19 +F (3) – F (2) +– +3 +3 +3 += += +(ii) +Let +9 +3 +4 +2 +2 +I +(30 – +) +x +dx +x += ∫ +. We first find the anti derivative of the integrand. +Put +3 +2 +3 +30 – +.Then – 2 +x +t +x dx +dt += += + or +2 +– 3 +x dx +dt += +Thus, +3 +2 +2 +2 +2 +– 3 +(30 – +) +x +dt +dx +t +x += +∫ +∫ + = +2 1 +3 t + + + = +3 +2 +2 +1 +F ( ) +3 +(30 – +) +x +x + + + += + + + + + + +Therefore, by the second fundamental theorem of calculus, we have +I = +9 +3 +2 +4 +2 +1 +F(9) – F(4) +3 +(30 – +) +x + + + + += + + + + + + += +2 +1 +1 +3 (30 – 27) +30 – 8 + + +− + + + + + = 2 1 +1 +19 +3 3 +22 +99 + + +− += + + + + +(iii) +Let +2 +1 +I +( +1) ( +2) +x dx +x +x += ++ ++ +∫ +Reprint 2025-26 + +270 +MATHEMATICS +Using partial fraction, we get +–1 +2 +( +1) ( +2) +1 +2 +x +x +x +x +x += ++ ++ ++ ++ ++ +So +( +1) ( +2) +x dx +x +x ++ ++ +∫ + = – log +1 +2log +2 +F( ) +x +x +x ++ ++ ++ += +Therefore, by the second fundamental theorem of calculus, we have +I = F(2) – F(1) = [– log 3 + 2 log 4] – [– log 2 + 2 log 3] += – 3 log 3 + log 2 + 2 log 4 = +32 +log +27 + + + + + + +(iv) +Let +3 +4 +0 +I +sin 2 cos2 +t +t dt +π += ∫ +. Consider +3 +sin 2 cos2 +∫ +t +t dt +Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = 1 +2 du +So +3 +sin 2 cos2 +∫ +t +t dt = +3 +1 +2 +u du +∫ += +4 +4 +1 +1 +[ +] +sin 2 +F ( ) say +8 +8 +u +t +t += += +Therefore, by the second fundamental theorem of integral calculus +I = +4 +4 +1 +1 +F ( ) – F (0) +[sin +– sin 0] +4 +8 +2 +8 +π +π += += +EXERCISE 7.8 +Evaluate the definite integrals in Exercises 1 to 20. +1. +1 +1( +1) +x +dx +− ++ +∫ +2. +3 +2 +1 dx +x +∫ +3. +2 +3 +2 +1 (4 +– 5 +6 +9) +x +x +x +dx ++ ++ +∫ +4. +sin 2 +0 +4 +x dx +π +∫ +5. +cos 2 +0 +2 +x dx +π +∫ +6. +5 +4 +xe dx +∫ +7. +4 +0 tan x dx +π +∫ +8. +4 +6 +cosec x dx +π +π∫ +9. +1 +0 +2 +1– +dx +x +∫ +10. +1 +2 +01 +dx +x ++ +∫ +11. +3 +2 +2 +1 +dx +x − +∫ +Reprint 2025-26 + +INTEGRALS 271 +12. +2 +2 +0 cos x dx +π +∫ +13. +3 +2 +2 +1 +x dx +x + +∫ +14. +1 +2 +0 +2 +3 +5 +1 +x +dx +x ++ ++ +∫ +15. +2 +1 +0 +x +x e dx +∫ +16. +2 +2 +2 +1 +5 +4 +3 +x +x +x ++ ++ +∫ +17. +2 +3 +4 +0 (2sec +2) +x +x +dx +π ++ ++ +∫ +18. +2 +2 +0 (sin +– cos +) +2 +2 +x +x dx +π∫ +19. +2 +2 +0 +6 +3 +4 +x +dx +x ++ ++ +∫ +20. +1 +0 ( +sin +) +4 +x +x +x e +dx +π ++ +∫ +Choose the correct answer in Exercises 21 and 22. +21. +3 +2 +1 1 +dx +x ++ +∫ + equals +(A) 3 +π +(B) +2 +3 +π +(C) +6 +π +(D) 12 +π +22. +2 +3 +2 +0 4 +9 +dx +x ++ +∫ + equals +(A) 6 +π +(B) 12 +π +(C) +24 +π +(D) 4 +π +7.9 Evaluation of Definite Integrals by Substitution +In the previous sections, we have discussed several methods for finding the indefinite +integral. One of the important methods for finding the indefinite integral is the method +of substitution. +To evaluate +( ) +b +a f x dx +∫ +, by substitution, the steps could be as follows: +1. +Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce +the given integral to a known form. +2. +Integrate the new integrand with respect to the new variable without mentioning +the constant of integration. +3. +Resubstitute for the new variable and write the answer in terms of the original +variable. +4. +Find the values of answers obtained in (3) at the given limits of integral and find +the difference of the values at the upper and lower limits. +Reprint 2025-26 + +272 +MATHEMATICS +ANote In order to quicken this method, we can proceed as follows: After +performing steps 1, and 2, there is no need of step 3. Here, the integral will be kept +in the new variable itself, and the limits of the integral will accordingly be changed, +so that we can perform the last step. +Let us illustrate this by examples. +Example 26 Evaluate +1 +4 +5 +15 +1 +x +x +dx +− ++ +∫ +. +Solution Put t = x5 + 1, then dt = 5x4 dx. +Therefore, +4 +5 +5 +1 +x +x +dx ++ +∫ + = +t dt +∫ + = +3 +2 +2 +3 t = +3 +5 +2 +2 ( +1) +3 +x + +Hence, +1 +4 +5 +15 +1 +x +x +dx +− ++ +∫ + = +1 +3 +5 +2 +– 1 +2 ( +1) +3 +x + + ++ + + + + + + += +( +) +3 +3 +5 +5 +2 +2 +2 (1 +1) – (–1) +1 +3 + + ++ ++ + + + + + + += +3 +3 +2 +2 +2 2 +0 +3 + + +− + + + + + + + = +2 +4 +2 +(2 +2) +3 +3 += +Alternatively, first we transform the integral and then evaluate the transformed integral +with new limits. +Let +t = x5 + 1. Then dt = 5 x4 dx. +Note that, when +x = – 1, t = 0 and when x = 1, t = 2 +Thus, as x varies from – 1 to 1, t varies from 0 to 2 +Therefore +1 +4 +5 +15 +1 +x +x +dx +− ++ +∫ + = +2 +0 +t dt +∫ += +2 +3 +3 +3 +2 +2 +2 +0 +2 +2 2 – 0 +3 +3 +t + + + + += + + + + + + + + + + + + + = 2 +4 +2 +(2 +2) +3 +3 += +Example 27 Evaluate +– 1 +1 +2 +0 +tan +1 +x dx +x ++ +∫ +Reprint 2025-26 + +INTEGRALS 273 +Solution Let t = tan – 1x, then +2 +1 +1 +dt +dx +x += ++ +. The new limits are, when x = 0, t = 0 and +when x = 1, +4 +t +π += +. Thus, as x varies from 0 to 1, t varies from 0 to 4 +π . +Therefore +–1 +1 +2 +0 +tan +1 +x dx +x ++ +∫ += +2 +4 +4 +0 +0 +2 +t +t dt +π +π + + + + + + +∫ + = +2 +2 +1 +– 0 +2 16 +32 + + +π +π += + + + + +EXERCISE 7.9 +Evaluate the integrals in Exercises 1 to 8 using substitution. +1. +1 +2 +0 +1 +x +dx +x + +∫ +2. +5 +2 +0 +sin +cos +d +π +φ +φ φ +∫ +3. +1 +– 1 +2 +0 +2 +sin +1 +x +dx +x + + + + ++ + + +∫ +4. +2 +0 +2 +x +x + +∫ + (Put x + 2 = t2) +5. +2 +2 +0 +sin +1 +cos +x +dx +x +π ++ +∫ +6. +2 +2 +0 +4 – +dx +x +x ++ +∫ +7. +1 +2 +1 +2 +5 +dx +x +x +− ++ ++ +∫ +8. +2 +2 +2 +1 +1 +1 +– +2 +x +e dx +x +x + + + + + + +∫ +Choose the correct answer in Exercises 9 and 10. +9. +The value of the integral +1 +3 3 +1 +1 +4 +3 +( +) +x +x +dx +x +− +∫ + is +(A) 6 +(B) 0 +(C) 3 +(D) 4 +10. +If f (x) = +0 sin +xt +t dt +∫ +, then f ′(x) is +(A) cosx + x sin x +(B) x sinx +(C) x cosx +(D) sinx + x cosx +7.10 Some Properties of Definite Integrals +We list below some important properties of definite integrals. These will be useful in +evaluating the definite integrals more easily. +P0 : +( ) +( ) +b +b +a +a +f x dx +f t dt += +∫ +∫ +P1 : +( ) +– +( ) +b +a +a +b +f x dx +f x dx += +∫ +∫ +. In particular, +( ) +0 +a +a f x dx = +∫ +P2 : +( ) +( ) +( ) +b +c +b +a +a +c +f x dx +f x dx +f x dx += ++ +∫ +∫ +∫ +Reprint 2025-26 + +274 +MATHEMATICS +P3 : +( ) +( +) +b +b +a +a +f x dx +f a +b +x dx += ++ +− +∫ +∫ +P4 : +0 +0 +( ) +( +) +a +a +f x dx +f a +x dx += +− +∫ +∫ +(Note that P4 is a particular case of P3) +P5 : +2 +0 +0 +0 +( ) +( ) +(2 +) +a +a +a +f x dx +f x dx +f +a +x dx += ++ +− +∫ +∫ +∫ +P6 : +2 +0 +0 +( ) +2 +( ) +, if +(2 +) +( ) +a +a +f x dx +f x dx +f +a +x +f x += +− += +∫ +∫ + and + 0 if f (2a – x) = – f (x) +P7 : +(i) +0 +( ) +2 +( ) +a +a +a f x dx +f x dx +− += +∫ +∫ +, if f is an even function, i.e., if f (– x) = f (x). +(ii) +( ) +0 +a +a f x dx +− += +∫ +, if f is an odd function, i.e., if f (– x) = – f (x). +We give the proofs of these properties one by one. +Proof of P0 It follows directly by making the substitution x = t. +Proof of P1 Let F be anti derivative of f. Then, by the second fundamental theorem of +calculus, we have +( ) +F ( ) – F ( ) +– [F ( ) +F ( )] +( ) +b +a +a +b +f x dx +b +a +a +b +f x dx += += +− += − +∫ +∫ +Here, we observe that, if a = b, then +( ) +0 +a +a f x dx = +∫ +. +Proof of P2 Let F be anti derivative of f. Then +( ) +b +a f x dx +∫ + = F(b) – F(a) +... (1) +( ) +c +a f x dx +∫ + = F(c) – F(a) +... (2) +and +( ) +b +c f x dx +∫ + = F(b) – F(c) +... (3) +Adding (2) and (3), we get +( ) +( ) +F( ) – F( ) +( ) +c +b +b +a +c +a +f x dx +f x dx +b +a +f x dx ++ += += +∫ +∫ +∫ +This proves the property P2. +Proof of P3 Let t = a + b – x. Then dt = – dx. When x = a, t = b and when x = b, t = a. +Therefore +( ) +b +a f x dx +∫ + = +( +– ) +a +b f a +b +t dt +− ++ +∫ +Reprint 2025-26 + +INTEGRALS 275 += +( +– ) +b +a f a +b +t dt ++ +∫ + (by P1) += +( +– ) +b +a f a +b +x ++ +∫ +dx by P0 +Proof of P4 Put t = a – x. Then dt = – dx. When x = 0, t = a and when x = a, t = 0. Now +proceed as in P3. +Proof of P5 Using P2, we have +2 +2 +0 +0 +( ) +( ) +( ) +a +a +a +a +f x dx +f x dx +f x dx += ++ +∫ +∫ +∫ +. +Let +t = 2a – x in the second integral on the right hand side. Then +dt = – dx. When x = a, t = a and when x = 2a, t = 0. Also x = 2a – t. +Therefore, the second integral becomes +2 +( ) +a +a +f x dx +∫ + = +0 +– +(2 – ) +a f +a +t dt +∫ + = +0 +(2 – ) +a f +a +t dt +∫ + = +0 +(2 – ) +a f +a +x dx +∫ +Hence +2 +0 +( ) +a f x dx +∫ + = +0 +0 +( ) +(2 +) +a +a +f x dx +f +a +x dx ++ +− +∫ +∫ +Proof of P6 Using P5, we have +2 +0 +0 +0 +( ) +( ) +(2 +) +a +a +a +f x dx +f x dx +f +a +x dx += ++ +− +∫ +∫ +∫ + ... (1) +Now, if +f (2a – x) = f (x), then (1) becomes +2 +0 +( ) +a f x dx +∫ + = +0 +0 +0 +( ) +( ) +2 +( ) +, +a +a +a +f x dx +f x dx +f x dx ++ += +∫ +∫ +∫ +and if +f(2a – x) = – f (x), then (1) becomes +2 +0 +( ) +a f x dx +∫ + = +0 +0 +( ) +( ) +0 +a +a +f x dx +f x dx +− += +∫ +∫ +Proof of P7 Using P2, we have +( ) +a +a f x dx +−∫ + = +0 +0 +( ) +( ) +a +a f x dx +f x dx +− ++ +∫ +∫ +. Then +Let +t = – x in the first integral on the right hand side. +dt = – dx. When x = – a, t = a and when +x = 0, t = 0. Also x = – t. +Therefore +( ) +a +a f x dx +−∫ + = +0 +0 +– +(– ) +( ) +a +a f +t dt +f x dx ++ +∫ +∫ += +0 +0 +(– ) +( ) +a +a +f +x dx +f x dx ++ +∫ +∫ + (by P0) ... (1) +Reprint 2025-26 + +276 +MATHEMATICS +(i) Now, if f is an even function, then f (–x) = f (x) and so (1) becomes +0 +0 +0 +( ) +( ) +( ) +2 +( ) +a +a +a +a +a f x dx +f x dx +f x dx +f x dx +− += ++ += +∫ +∫ +∫ +∫ +(ii) If f is an odd function, then f (–x) = – f (x) and so (1) becomes +0 +0 +( ) +( ) +( ) +0 +a +a +a +a f x dx +f x dx +f x dx +− += − ++ += +∫ +∫ +∫ +Example 28 Evaluate +2 +3 +1 +– +x +x dx +−∫ +Solution We note that x3 – x ≥ 0 on [– 1, 0] and x3 – x ≤ 0 on [0, 1] and that +x3 – x ≥ 0 on [1, 2]. So by P2 we write +2 +3 +1 +– +x +x dx +−∫ + = +0 +1 +2 +3 +3 +3 +1 +0 +1 +( +– ) +– ( +– ) +( +– ) +x +x dx +x +x dx +x +x dx +− ++ ++ +∫ +∫ +∫ += +0 +1 +2 +3 +3 +3 +1 +0 +1 +( +– ) +( – +) +( +– ) +x +x dx +x +x +dx +x +x dx +− ++ ++ +∫ +∫ +∫ += +0 +1 +2 +4 +2 +2 +4 +4 +2 +– 1 +0 +1 +– +– +– +4 +2 +2 +4 +4 +2 +x +x +x +x +x +x + + + + + + ++ ++ + + + + + + + + + + + + += +( +) +1 +1 +1 +1 +1 +1 +– +– +– +4 – 2 – +– +4 +2 +2 +4 +4 +2 + + + + + + ++ ++ + + + + + + + + + + + + += +1 +1 +1 +1 +1 +1 +– +2 +4 +2 +2 +4 +4 +2 ++ ++ +− ++ +− ++ + = 3 +3 +11 +2 +2 +4 +4 +− ++ += +Example 29 Evaluate +2 +4 +– +4 +sin x dx +π +π +∫ +Solution We observe that sin2 x is an even function. Therefore, by P7 (i), we get +2 +4 +– +4 +sin x dx +π +π +∫ + = +2 +4 +0 +2 +sin x dx +π +∫ += +4 +0 +(1 +cos 2 ) +2 +2 +x dx +π +− +∫ + = +4 +0 (1 cos 2 ) +x dx +π +− +∫ +Reprint 2025-26 + +INTEGRALS 277 += +4 +0 +1 +– +sin 2 +2 +x +x +π + + + + + + + = +1 +1 +– +sin +– 0 +– +4 +2 +2 +4 +2 +π +π +π + + += + + + + +Example 30 Evaluate +2 +0 +sin +1 +cos +x +x dx +x +π ++ +∫ +Solution Let I = +2 +0 +sin +1 +cos +x +x +dx +x +π ++ +∫ +. Then, by P4, we have +I = +2 +0 +( +) sin ( +) +1 +cos ( +) +x +x dx +x +π π − +π − ++ +π − +∫ += +2 +0 +( +) sin +1 +cos +x +x dx +x +π π − ++ +∫ + = +2 +0 +sin +I +1 +cos +x dx +x +π +π +− ++ +∫ +or +2 I = π +π sin +cos +x dx +x +1 +2 +0 ++ +∫ +or +I = +2 +0 +sin +2 +1 +cos +x dx +x +π +π ++ +∫ +Put cos x = t so that – sin x dx = dt. When x = 0, t = 1 and when x = π, t = – 1. +Therefore, (by P1) we get +I = +1 +2 +1 +– +2 +1 +dt +t +− +π ++ +∫ += +1 +2 +1 +2 +1 +dt +t +− +π ++ +∫ += +1 +2 +0 1 +dt +t +π ++ +∫ + (by P7, +2 +1 +since +1 +t ++ + is even function) += +2 +1 +– 1 +– 1 +1 +0 +tan +tan +1– tan +0 +– 0 +4 +4 +t +− +π +π + + + + + + +π += π += π += + + + + + + + + +Example 31 Evaluate +1 +5 +4 +1 sin +cos +x +x dx +−∫ +Solution Let I = +1 +5 +4 +1sin +cos +x +x dx +−∫ +. Let f(x) = sin5 x cos4 x. Then +f (– x) = sin5 (– x) cos4 (– x) = – sin5 x cos4 x = – f (x), i.e., f is an odd function. +Therefore, by P7 (ii), I = 0 +Reprint 2025-26 + +278 +MATHEMATICS +Example 32 Evaluate +4 +2 +4 +4 +0 +sin +sin +cos +x +dx +x +x +π ++ +∫ +Solution Let I = +4 +2 +4 +4 +0 +sin +sin +cos +x +dx +x +x +π ++ +∫ +... (1) +Then, by P4 +I = +4 +2 +0 +4 +4 +sin ( +) +2 +sin ( +) +cos ( +) +2 +2 +x +dx +x +x +π +π − +π +π +− ++ +− +∫ + = +4 +2 +4 +4 +0 +cos +cos +sin +x +dx +x +x +π ++ +∫ + ... (2) +Adding (1) and (2), we get +2I = +4 +4 +2 +2 +2 +4 +4 +0 +0 +0 +sin +cos +[ ] +2 +sin +cos +x +x dx +dx +x +x +x +π +π +π ++ +π += += += ++ +∫ +∫ +Hence +I = 4 +π +Example 33 Evaluate +3 +6 1 +tan +dx +x +π +π ++ +∫ +Solution Let I = +3 +3 +6 +6 +cos +1 +tan +cos +sin +x dx +dx +x +x +x +π +π +π +π += ++ ++ +∫ +∫ +... (1) +Then, by P3 +I = +3 +6 +cos +3 +6 +cos +sin +3 +6 +3 +6 +x +dx +x +x +π +π +π +π + + ++ +− + + + + +π +π +π +π + + + + ++ +− ++ ++ +− + + + + + + + + +∫ += +3 +6 +sin +sin +cos +x +dx +x +x +π +π ++ +∫ +... (2) +Adding (1) and (2), we get +2I = +[ ] +3 +3 +6 +6 +3 +6 +6 +dx +x +π +π +π +π +π +π +π += += +− += +∫ +. Hence I +12 +π += +Reprint 2025-26 + +INTEGRALS 279 +Example 34 Evaluate +2 +0 log sin x dx +π +∫ +Solution Let I = +2 +0 log sin x dx +π +∫ +Then, by P4 +I = +2 +2 +0 +0 +log sin +log cos +2 +x dx +x dx +π +π +π + + +− += + + + + +∫ +∫ +Adding the two values of I, we get +2I = +( +) +2 +0 log sin +logcos +x +x dx +π ++ +∫ += +( +) +2 +0 log sin +cos +log 2 +log2 +x +x +dx +π ++ +− +∫ +(by adding and subtracting log2) += +2 +2 +0 +0 +log sin 2 +log2 +x dx +dx +π +π +− +∫ +∫ +(Why?) +Put 2x = t in the first integral. Then 2 dx = dt, when x = 0, t = 0 and when +2 +x +π += +, +t = π. +Therefore +2I = +0 +1 +log sin +log2 +2 +2 +t dt +π +π +− +∫ += +2 +0 +2 +log sin +log2 +2 +2 +t dt +π +π +− +∫ + [by P6 as sin (π – t) = sin t) += +2 +0 log sin +log2 +2 +x dx +π +π +− +∫ + (by changing variable t to x) += I +log2 +2 +π +− +Hence +2 +0 log sin x dx +π +∫ + = – +log2 +2 +π +. +Reprint 2025-26 + +280 +MATHEMATICS +EXERCISE 7.10 +By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. +1. +2 +2 +0 cos x dx +π +∫ +2. +2 +0 +sin +sin +cos +x +dx +x +x +π ++ +∫ +3. +3 +2 +2 +3 +3 +0 +2 +2 +sin +sin +cos +x dx +x +x +π ++ +∫ +4. +5 +2 +5 +5 +0 +cos +sin +cos +x dx +x +x +π ++ +∫ +5. +5 +5 | +2| +x +dx +− ++ +∫ +6. +8 +2 +5 +x +dx +− +∫ +7. +1 +0 +(1 +)n +x +x +dx +− +∫ +8. +4 +0 log (1 +tan ) +x dx +π ++ +∫ +9. +2 +0 +2 +x +x dx +− +∫ +10. +2 +0 (2log sin +log sin 2 ) +x +x dx +π +− +∫ +11. +2 +2 +– +2 +sin x dx +π +π +∫ +12. +0 1 +sin +x dx +x +π ++ +∫ +13. +7 +2 +– +2 +sin x dx +π +π +∫ +14. +2 +5 +0 cos x dx +π +∫ +15. +2 +0 +sin +cos +1 +sin +cos +x +x dx +x +x +π +− ++ +∫ +16. +0 log (1 +cos )x dx +π ++ +∫ +17. +0 +a +x +dx +x +a +x ++ +− +∫ +18. +4 +0 +1 +x +dx +− +∫ +19. +Show that +0 +0 +( ) ( ) +2 +( ) +a +a +f x g x dx +f x dx += +∫ +∫ +, if f and g are defined as f(x) = f(a – x) +and g(x) + g(a – x) = 4 +Choose the correct answer in Exercises 20 and 21. +20. +The value of +3 +5 +2 +2 +( +cos +tan +1) +x +x +x +x +dx +π +−π ++ ++ ++ +∫ + is +(A) 0 +(B) 2 +(C) π +(D) 1 +21. +The value of +2 +0 +4 +3sin +log +4 +3 cos +x +dx +x +π + + ++ + + ++ + + +∫ + is +(A) 2 +(B) 3 +4 +(C) 0 +(D) –2 +Reprint 2025-26 + +INTEGRALS 281 +Miscellaneous Examples +Example 35 Find cos 6 +1 +sin 6 +x +x dx ++ +∫ +Solution Put t = 1 + sin 6x, so that dt = 6 cos 6x dx +Therefore +1 +2 +1 +cos 6 +1 +sin 6 +6 +x +x dx +t dt ++ += +∫ +∫ += +3 +3 +2 +2 +1 +2 +1 +( ) +C = +(1 +sin 6 ) +C +6 +3 +9 +t +x +× ++ ++ ++ +Example 36 Find +1 +4 +4 +5 +( +) +x +x +dx +x +− +∫ +Solution We have +1 +1 +4 +4 +4 +3 +5 +4 +1 +(1 +) +( +) +x +x +x +dx +dx +x +x +− +− += +∫ +∫ +Put +– 3 +3 +4 +1 +3 +1 +1– +, so that +x +t +dx +dt +x +x +− += += += +Therefore +1 +1 +4 +4 +4 +5 +( +) +1 +3 +x +x +dx +t +dt +x +− += +∫ +∫ + = +5 +5 +4 +4 +3 +1 +4 +4 +1 +C = +1 +C +3 +5 +15 +t +x + + +× ++ +− ++ + + + + +Example 37 Find +4 +2 +( +1) ( +1) +x dx +x +x +− ++ +∫ +Solution We have +4 +2 +( +1)( +1) +x +x +x +− ++ + = +3 +2 +1 +( +1) +1 +x +x +x +x ++ ++ +− ++ +− += +2 +1 +( +1) +( +1) ( +1) +x +x +x ++ ++ +− ++ +... (1) +Now express +2 +1 +( +1)( +1) +x +x +− ++ + = +2 +A +B +C +( +1) +( +1) +x +x +x ++ ++ +− ++ +... (2) +Reprint 2025-26 + +282 +MATHEMATICS +So +1 = A (x2 + 1) + (Bx + C) (x – 1) += (A + B) x2 + (C – B) x + A – C +Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1, +which give +1 +1 +A +, B +C +– +2 +2 += += += +. Substituting values of A, B and C in (2), we get +2 +1 +( +1) ( +1) +x +x +− ++ + = +2 +2 +1 +1 +1 +2( +1) +2 ( +1) +2( +1) +x +x +x +x +− +− +− ++ ++ +... (3) +Again, substituting (3) in (1), we have +4 +2 +( +1) ( +1) +x +x +x +x +− ++ ++ + = +2 +2 +1 +1 +1 +( +1) +2( +1) +2 ( +1) +2( +1) +x +x +x +x +x ++ ++ +− +− +− ++ ++ +Therefore +4 +2 +2 +– 1 +2 +1 +1 +1 +log +1 – +log ( +1) – +tan +C +2 +2 +4 +2 +( +1) ( +1) +x +x +dx +x +x +x +x +x +x +x += ++ ++ +− ++ ++ +− ++ ++ +∫ +Example 38 Find +2 +1 +log (log ) +(log ) +x +dx +x + + ++ + + + + +∫ +Solution Let +2 +1 +I +log (log ) +(log ) +x +dx +x + + += ++ + + + + +∫ += +2 +1 +log (log ) +(log ) +x dx +dx +x ++ +∫ +∫ +In the first integral, let us take 1 as the second function. Then integrating it by +parts, we get +I = +2 +1 +log (log ) +log +(log ) +dx +x +x +x dx +x +x +x +− ++ +∫ +∫ += +2 +log (log ) +log +(log ) +dx +dx +x +x +x +x +− ++ +∫ +∫ +... (1) +Again, consider +log +dx +x +∫ +, take 1 as the second function and integrate it by parts, +we have +2 +1 +1 +– +– +log +log +(log ) +dx +x +x +dx +x +x +x +x + + + + + + +=  + + + + + + + + + + + + + +∫ +∫ + ... (2) +Reprint 2025-26 + +INTEGRALS 283 +Putting (2) in (1), we get +2 +2 +I +log (log ) +log +(log ) +(log ) +x +dx +dx +x +x +x +x +x += +− +− ++ +∫ +∫ + = log (log ) +C +log +x +x +x +x +− ++ +Example 39 Find +cot +tan +x +x dx + + ++ + + +∫ +Solution We have +I = +cot +tan +x +x dx + + ++ + + +∫ +tan (1 +cot ) +x +x dx += ++ +∫ +Put tan x = t2, so that sec2 x dx = 2t dt +or +dx = +4 +2 +1 +t dt +t ++ +Then +I = +2 +4 +1 +2 +1 +(1 +) +t +t +dt +t +t + + ++ + + ++ + + +∫ += +2 +2 +2 +4 +2 +2 +2 +1 +1 +1 +1 +( +1) +2 += 2 += 2 +1 +1 +1 +2 +dt +dt +t +t +t +dt +t +t +t +t +t + + + + ++ ++ + + + + ++ + + + + + + ++ + + ++ +− ++ + + + + + + + + +∫ +∫ +∫ +Put +1 +t +t +− = y, so that +2 +1 +1 +t + + ++ + + + + dt = dy. Then +I = +( +) +– 1 +– 1 +2 +2 +1 +2 +2 tan +C = +2 tan +C +2 +2 +2 +t +dy +y +t +y + + +− + + + + += ++ ++ ++ +∫ += +2 +– 1 +– 1 +1 +tan +1 +2 tan +C = +2 tan +C +2 +2tan +t +x +t +x + + +− +− + + ++ ++ + + + + + + + + + + +Example 40 Find +4 +sin 2 cos 2 +9 – cos (2 ) +x +x dx +x +∫ +Solution Let +4 +sin 2 cos 2 +I +9 – cos 2 +x +x dx +x += ∫ +Reprint 2025-26 + +284 +MATHEMATICS +Put cos2 (2x) = t so that 4 sin 2x cos 2x dx = – dt +Therefore +–1 +1 +2 +2 +1 +1 +1 +1 +I +– +– +sin +C +sin +cos 2 +C +4 +4 +3 +4 +3 +9 – +dt +t +x +t +− + + + + += += ++ += − ++ + + + + + + + + +∫ +Example 41 Evaluate +3 +2 +1 +sin( +) +x +x dx +− +π +∫ +Solution Here f (x) = | x sin πx | = +sin +for +1 +1 +3 +sin +for1 +2 +x +x +x +x +x +x +π +−≤ +≤ +− +π +≤ +≤ + +Therefore +3 +2 +1 | +sin +| +x +x dx +− +π +∫ + = +3 +1 +2 +1 +1 +sin +sin +x +x dx +x +x dx +− +π ++ +− +π +∫ +∫ += +3 +1 +2 +1 +1 +sin +sin +x +x dx +x +x dx +− +π +− +π +∫ +∫ +Integrating both integrals on righthand side, we get +3 +2 +1 | +sin +| +x +x dx +− +π +∫ + = += +2 +2 +2 +1 +1 +3 +1 + + +−− +− += ++ + + +π +π +π +π +π + + +Example 42 Evaluate +2 +2 +2 +2 +0 +cos +sin +x dx +a +x +b +x +π ++ +∫ +Solution Let I = +2 +2 +2 +2 +2 +2 +2 +2 +0 +0 +( +) +cos +sin +cos ( +) +sin ( +) +x dx +x dx +a +x +b +x +a +x +b +x +π +π +π − += ++ +π − ++ +π − +∫ +∫ +(using P4) += +2 +2 +2 +2 +2 +2 +2 +2 +0 +0 +cos +sin +cos +sin +dx +x dx +a +x +b +x +a +x +b +x +π +π +π +− ++ ++ +∫ +∫ += +2 +2 +2 +2 +0 +I +cos +sin +dx +a +x +b +x +π +π +− ++ +∫ +Thus +2I = +2 +2 +2 +2 +0 +cos +sin +dx +a +x +b +x +π +π ++ +∫ +Reprint 2025-26 + +INTEGRALS 285 +or +I = +2 +2 +2 +2 +2 +2 +2 +2 +2 +0 +0 +2 +2 +2 +cos +sin +cos +sin +dx +dx +a +x +b +x +a +x +b +x +π +π +π +π += +⋅ ++ ++ +∫ +∫ +(using P6) += +2 +4 +2 +2 +2 +2 +2 +2 +2 +2 +0 +4 +cos +sin +cos +sin +π +π +π + + +π ++ + + ++ ++ + + + + +∫ +∫ +dx +dx +a +x +b +x +a +x +b +x += +2 +2 +2 +4 +2 +2 +2 +2 +2 +2 +0 +4 +sec +cosec +tan +cot +π +π +π + + +π ++ + + ++ ++ + + + + +∫ +∫ +xdx +xdx +a +b +x +a +x +b += +( +) +0 +1 +2 +2 2 +2 +2 +2 +0 +1 +tan +t +cot + + +π +− += += + + ++ ++ + + +∫ +∫ +dt +du +put +x +and +x +u +a +b t +a u +b += +1 +0 +–1 +–1 +0 +1 +tan +– +tan +π +π + + + + + + + + + + + + +bt +au +ab +a +ab +b + = +–1 +–1 +tan +tan +π  + ++ + + + + +b +a +ab +a +b = +2 +2 +π +ab +Miscellaneous Exercise on Chapter 7 +Integrate the functions in Exercises 1 to 23. +1. +3 +1 +x +x +− +2. +1 +x +a +x +b ++ ++ ++ +3. +2 +1 +x +ax +x +− + [Hint: Put x = a +t ] +4. +3 +2 +4 +4 +1 +( +1) +x +x + +5. +1 +1 +3 +2 +1 +x +x ++ + [Hint: +1 +1 +1 +1 +3 +2 +3 +6 +1 +1 +1 +x +x +x +x += + + ++ ++ + + + + + + +, put x = t6] +6. +2 +5 +( +1) ( +9) +x +x +x ++ ++ +7. +sin +sin ( +) +x +x +a +− +8. +5 log +4 log +3 log +2 log +x +x +x +x +e +e +e +e +− +− +9. +2 +cos +4 +sin +x +x +− +10. +8 +8 +2 +2 +sin +cos +1 +2sin +cos +x +x +x +− +− +11. +1 +cos ( +) cos ( +) +x +a +x +b ++ ++ +12. +3 +8 +1 +x +x +− +13. +(1 +) (2 +) +x +x +x +e +e +e ++ ++ +14. +2 +2 +1 +( +1) ( +4) +x +x ++ ++ +15. +cos3x elog sinx +16. e3 logx (x4 + 1)– 1 +17. + f ′ (ax + b) [f (ax + b)]n +Reprint 2025-26 + +286 +MATHEMATICS +18. +3 +1 +sin +sin ( +) +x +x + α +19. +1 +1 +x +x +− ++ +20. +2 +sin 2 +1 +cos2 +x +x e +x ++ ++ +21. +2 +2 +1 +( +1) ( +2) +x +x +x +x ++ ++ ++ ++ +22. +– 1 1 +tan +1 +x +x +− ++ +23. +2 +2 +4 +1 log ( +1) +2 log +x +x +x +x + + ++ ++ +− + + +Evaluate the definite integrals in Exercises 24 to 31. +24. +2 +1 +sin +1 +cos +π +π +− + + + + +− + + +∫ +x +x +e +dx +x +25. +4 +4 +4 +0 +sin +cos +cos +sin +x +x +dx +x +x +π ++ +∫ +26. +2 +2 +2 +2 +0 +cos +cos +4 sin +x dx +x +x +π ++ +∫ +27. +3 +6 +sin +cos +sin 2 +x +x dx +x +π +π ++ +∫ +28. +1 +0 +1 +dx +x +x ++ +− +∫ +29. +4 +0 +sin +cos +9 +16 sin 2 +x +x dx +x +π ++ ++ +∫ +30. +1 +2 +0 sin 2 tan +(sin ) +x +x dx +π +− +∫ +31. +4 +1 [| +1| | +2| +| +3|] +x +x +x +dx +− ++ +− ++ +− +∫ +Prove the following (Exercises 32 to 37) +32. +3 +2 +1 +2 +2 +log +3 +3 +( +1) +dx +x +x += ++ ++ +∫ +33. +1 +0 +1 +x +x e dx = +∫ +34. +1 +17 +4 +1 +cos +0 +x +x dx +− += +∫ +35. +3 +2 +0 +2 +sin +3 +x dx +π += +∫ +36. +3 +4 +0 2 tan +1 +log2 +x dx +π += − +∫ +37. +1 +1 +0sin +1 +2 +x dx +− +π += +− +∫ +Choose the correct answers in Exercises 38 to 40 +38. +x +x +dx +e +e− ++ +∫ + is equal to +(A) tan–1 (ex) + C +(B) tan–1 (e–x) + C +(C) log (ex – e–x) + C +(D) log (ex + e–x) + C +39. +2 +cos2 +(sin +cos ) +x +dx +x +x ++ +∫ + is equal to +Reprint 2025-26 + +INTEGRALS 287 +(A) +–1 +C +sin +cos +x +x + ++ +(B) log |sin +cos | +C +x +x ++ ++ +(C) log |sin +cos | +C +x +x +− ++ +(D) +2 +1 +(sin +cos ) ++ +x +x +40. +If f (a + b – x) = f (x), then +( ) +b +a x f x dx +∫ + is equal to +(A) +( +) +2 +b +a +a +b +f b +x dx ++ +− +∫ +(B) +( +) +2 +b +a +a +b +f b +x dx ++ ++ +∫ +(C) +( ) +2 +b +a +b +a +f x dx +−∫ +(D) +( ) +2 +b +a +a +b +f x dx ++ ∫ +Summary +® Integration is the inverse process of differentiation. In the differential calculus, +we are given a function and we have to find the derivative or differential of +this function, but in the integral calculus, we are to find a function whose +differential is given. Thus, integration is a process which is the inverse of +differentiation. +Let +F( ) +( ) +d +x +f x +dx += +. Then we write +( ) +F ( ) +C +f x dx +x += ++ +∫ +. These integrals +are called indefinite integrals or general integrals, C is called constant of +integration. All these integrals differ by a constant. +® Some properties of indefinite integrals are as follows: +1. +[ ( ) +( )] +( ) +( ) +f x +g x +dx +f x dx +g x dx ++ += ++ +∫ +∫ +∫ +2. For any real number k, +( ) +( ) +k f x dx +k +f x dx += +∫ +∫ +More generally, if f1, f2, f3, ... , fn are functions and k1, k2, ... ,kn are real +numbers. Then +1 +1 +2 +2 +[ +( ) +( ) +... +( )] +n +n +k f x +k f +x +k f +x +dx ++ ++ ++ +∫ += +1 +1 +2 +2 +( ) +( ) +... +( ) +n +n +k +f x dx +k +f +x dx +k +f +x dx ++ ++ ++ +∫ +∫ +∫ +® Some standard integrals +(i) +1 +C +1 +n +n +x +x dx +n ++ += ++ ++ +∫ +, n ≠ – 1. Particularly, +C +dx +x += ++ +∫ +Reprint 2025-26 + +288 +MATHEMATICS +(ii) +cos +sin +C +x dx +x += ++ +∫ +(iii) +sin +– cos +C +x dx +x += ++ +∫ +(iv) +2 +sec +tan +C +x dx +x += ++ +∫ +(v) +2 +cosec +– cot +C +x dx +x += ++ +∫ +(vi) +sec +tan +sec +C +x +x dx +x += ++ +∫ +(vii) +cosec +cot +– cosec +C +x +x dx +x += ++ +∫ +(viii) +1 +2 +sin +C +1 +dx +x +x +− += ++ +− +∫ +(ix) +1 +2 +cos +C +1 +dx +x +x +− += − ++ +− +∫ +(x) +1 +2 +tan +C +1 +dx +x +x +− += ++ ++ +∫ +(xi) +1 +2 +cot +C +1 +dx +x +x +− += − ++ ++ +∫ +(xii) +C +x +x +e dx +e += ++ +∫ +(xiii) +C +log +x +x +a +a dx +a += ++ +∫ +(xiv) +1 +log | +| +C +dx +x +x += ++ +∫ +® Integration by partial fractions +Recall that a rational function is ratio of two polynomials of the form +P( ) +Q( ) +x +x , +where P(x) and Q (x) are polynomials in x and Q (x) ≠ 0. If degree of the +polynomial P (x) is greater than the degree of the polynomial Q (x), then we +may divide P (x) by Q (x) so that +1P ( ) +P( ) +T ( ) +Q( ) +Q( ) +x +x +x +x +x += ++ +, where T(x) is a +polynomial in x and degree of P1(x) is less than the degree of Q(x). T(x) +being polynomial can be easily integrated. +1P ( ) +Q( ) +x +x can be integrated by +expressing +1P ( ) +Q( ) +x +x as the sum of partial fractions of the following type: +1. +( +) ( +) +px +q +x +a +x +b ++ +− +− += +A +B +x +a +x +b ++ +− +− +, a ≠ b +2. +2 +( +) +px +q +x +a ++ +− += +2 +A +B +( +) +x +a +x +a ++ +− +− +Reprint 2025-26 + +INTEGRALS 289 +3. +2 +( +) ( +) ( +) +px +qx +r +x +a +x +b +x +c ++ ++ +− +− +− += +A +B +C +x +a +x +b +x +c ++ ++ +− +− +− +4. +2 +2 +( +) ( +) +px +qx +r +x +a +x +b ++ ++ +− +− += +2 +A +B +C +( +) +x +a +x +b +x +a ++ ++ +− +− +− +5. +2 +2 +( +) ( +) +px +qx +r +x +a +x +bx +c ++ ++ +− ++ ++ += +2 +A +B + C +x +x +a +x +bx +c ++ +− ++ ++ +where x2 + bx + c can not be factorised further. +® Integration by substitution +A change in the variable of integration often reduces an integral to one of the +fundamental integrals. The method in which we change the variable to some +other variable is called the method of substitution. When the integrand involves +some trigonometric functions, we use some well known identities to find the +integrals. Using substitution technique, we obtain the following standard +integrals. +(i) +tan +log sec +C +x dx +x += ++ +∫ +(ii) +cot +log sin +C +x dx +x += ++ +∫ +(iii) +sec +log sec +tan +C +x dx +x +x += ++ ++ +∫ +(iv) +cosec +log cosec +cot +C +x dx +x +x += +− ++ +∫ +® Integrals of some special functions +(i) +2 +2 +1 log +C +2 +dx +x +a +a +x +a +x +a +− += ++ ++ +− +∫ +(ii) +2 +2 +1 log +C +2 +dx +a +x +a +a +x +a +x ++ += ++ +− +− +∫ +(iii) +1 +2 +2 +1 tan +C +dx +x +a +a +x +a +− += ++ ++ +∫ +(iv) +2 +2 +2 +2 +log +C +dx +x +x +a +x +a += ++ +− ++ +− +∫ +(v) +1 +2 +2 +sin +C +dx +x +a +a +x +− += ++ +− +∫ +(vi) +2 +2 +2 +2 +log | +| +C +dx +x +x +a +x +a += ++ ++ ++ ++ +∫ +® Integration by parts +For given functions f1 and f2, we have +Reprint 2025-26 + +290 +MATHEMATICS +, i.e., the +integral of the product of two functions = first function × integral of the +second function – integral of {differential coefficient of the first function × +integral of the second function}. Care must be taken in choosing the first +function and the second function. Obviously, we must take that function as +the second function whose integral is well known to us. +® +[ ( ) +( )] +( ) +C +x +x +e +f x +f +x +dx +e f x dx +′ ++ += ++ +∫ +∫ +® Some special types of integrals +(i) +2 +2 +2 +2 +2 +2 +2 +log +C +2 +2 +x +a +x +a dx +x +a +x +x +a +− += +− +− ++ +− ++ +∫ +(ii) +2 +2 +2 +2 +2 +2 +2 +log +C +2 +2 +x +a +x +a dx +x +a +x +x +a ++ += ++ ++ ++ ++ ++ +∫ +(iii) +2 +2 +2 +2 +2 +1 +sin +C +2 +2 +x +a +x +a +x +dx +a +x +a +− +− += +− ++ ++ +∫ +(iv) Integrals of the types +2 +2 +or +dx +dx +ax +bx +c +ax +bx +c ++ ++ ++ ++ +∫ +∫ +can be +transformed into standard form by expressing +ax2 + bx + c = +2 +2 +2 +2 +2 +4 +b +c +b +c +b +a x +x +a +x +a +a +a +a +a + + + + + + + + ++ ++ += ++ ++ +− + + + + + + + + + + + + + + + + + + +(v) Integrals of the types +2 +2 +or +px +q dx +px +q dx +ax +bx +c +ax +bx +c ++ ++ ++ ++ ++ ++ +∫ +∫ +can be +transformed into standard form by expressing +2 +A +( +) +B +A (2 +) +B +d +px +q +ax +bx +c +ax +b +dx ++ += ++ ++ ++ += ++ ++ +, where A and B are +determined by comparing coefficients on both sides. +® We have defined +( ) +b +a f x dx +∫ + as the area of the region bounded by the curve +y = f (x), a ≤ x ≤ b, the x-axis and the ordinates x = a and x = b. Let x be a +Reprint 2025-26 + +INTEGRALS 291 +given point in [a, b]. Then +( ) +x +a f x dx +∫ + represents the Area function A (x). +This concept of area function leads to the Fundamental Theorems of Integral +Calculus. +® First fundamental theorem of integral calculus +Let the area function be defined by A(x) = +( ) +x +a f x dx +∫ + for all x ≥ a, where +the function f is assumed to be continuous on [a, b]. Then A′ (x) = f (x) for all +x ∈ [a, b]. +® Second fundamental theorem of integral calculus +Let f be a continuous function of x defined on the closed interval [a, b] and +let F be another function such that +F( ) +( ) +d +x +f x +dx += + for all x in the domain of +f, then +[ +] +( ) +F( ) +C +F ( ) +F ( ) +b +b +a +a f x dx +x +b +a += ++ += +− +∫ +. +This is called the definite integral of f over the range [a, b], where a and b +are called the limits of integration, a being the lower limit and b the +upper limit. +—v— +Reprint 2025-26" +class_12,8,Application of Integrals,ncert_books/class_12/lemh2dd/lemh202.pdf,"292 +MATHEMATICS +Fig 8.1 +v One should study Mathematics because it is only through Mathematics that +nature can be conceived in harmonious form. – BIRKHOFF v +8.1 Introduction +In geometry, we have learnt formulae to calculate areas +of various geometrical figures including triangles, +rectangles, trapezias and circles. Such formulae are +fundamental in the applications of mathematics to many +real life problems. The formulae of elementary geometry +allow us to calculate areas of many simple figures. +However, they are inadequate for calculating the areas +enclosed by curves. For that we shall need some concepts +of Integral Calculus. +In the previous chapter, we have studied to find the +area bounded by the curve y = f (x), the ordinates x = a, +x = b and x-axis, while calculating definite integral as the +limit of a sum. Here, in this chapter, we shall study a specific +application of integrals to find the area under simple curves, +area between lines and arcs of circles, parabolas and +ellipses (standard forms only). We shall also deal with finding +the area bounded by the above said curves. +8.2 Area under Simple Curves +In the previous chapter, we have studied +definite integral as the limit of a sum and +how to evaluate definite integral using +Fundamental Theorem of Calculus. Now, +we consider the easy and intuitive way of +finding the area bounded by the curve +y = f (x), x-axis and the ordinates x = a and +x = b. From Fig 8.1, we can think of area +under the curve as composed of large +number of very thin vertical strips. Consider +an arbitrary strip of height y and width dx, +then dA (area of the elementary strip)= ydx, +where, y = f (x). +Chapter 8 +APPLICATION OF INTEGRALS +A.L. Cauchy +(1789-1857) +Reprint 2025-26 + +APPLICATION OF INTEGRALS 293 +Fig 8.2 +This area is called the elementary area which is located at an arbitrary position +within the region which is specified by some value of x between a and b. We can think +of the total area A of the region between x-axis, ordinates x = a, x = b and the curve +y = f (x) as the result of adding up the elementary areas of thin strips across the region +PQRSP. Symbolically, we express +A = +A +( ) +b +b +b +a +a +a +d +ydx +f x dx += += +∫ +∫ +∫ +The area A of the region bounded by +the curve x = g (y), y-axis and the lines y = c, +y = d is given by +A = +( ) +d +d +c +c +xdy +g y dy += +∫ +∫ +Here, we consider horizontal strips as shown in +the Fig 8.2 +Remark If the position of the curve under consideration is below the x-axis, then since +f (x) < 0 from x = a to x = b, as shown in Fig 8.3, the area bounded by the curve, x-axis +and the ordinates x = a, x = b come out to be negative. But, it is only the numerical +value of the area which is taken into consideration. Thus, if the area is negative, we +take its absolute value, i.e., +( ) +b +a f x dx +∫ +. +Fig 8.3 +Generally, it may happen that some portion of the curve is above x-axis and some is +below the x-axis as shown in the Fig 8.4. Here, A1 < 0 and A2 > 0. Therefore, the area +A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given +by A = |A1| + A2. +Reprint 2025-26 + +294 +MATHEMATICS +Example 1 Find the area enclosed by the circle x2 + y2 = a2. +Solution From Fig 8.5, the whole area enclosed +by the given circle += 4 (area of the region AOBA bounded by +the curve, x-axis and the ordinates x = 0 and +x = a) [as the circle is symmetrical about both +x-axis and y-axis] += +0 +4 +a ydx +∫ + (taking vertical strips) += +2 +2 +0 +4 +a +a +x dx +− +∫ +Since x2 + y2 = a2 gives y = +2 +2 +a +x +± +− +As the region AOBA lies in the first quadrant, y is taken as positive. Integrating, we get +the whole area enclosed by the given circle += +2 +2 +2 +–1 +0 +4 +sin +2 +2 +a +x +a +x +a +x +a + + +− ++ + + + + += + = +2 +2 +4 +2 +2 +a +a + + +π + +=π + + + + + + +Fig 8.5 +Fig 8.4 +Reprint 2025-26 + +APPLICATION OF INTEGRALS 295 +Alternatively, considering horizontal strips as shown in Fig 8.6, the whole area of the +region enclosed by circle += +0 +4 +a xdy +∫ + = +2 +2 +0 +4 +a +a +y dy +− +∫ +(Why?) += +2 +2 +2 +1 +0 +4 +sin +2 +2 +a +a +y +y +a +y +a +− + + +− ++ + + + + += += +2 +2 +4 2 2 +a +a +π = π +Example 2 Find the area enclosed by the ellipse +2 +2 +2 +2 +1 +x +y +a +b ++ += +Solution From Fig 8.7, the area of the region ABA′B′A bounded by the ellipse += +in +4 +, +0, +area of theregion AOBA +the first quadrantbounded +bythecurve x +axisand theordinates x +x +a + + + + +− += += + + +(as the ellipse is symmetrical about both x-axis and y-axis) += +0 +4 +(taking verticalstrips) +a ydx +∫ +Now +2 +2 +2 +2 +x +y +a +b ++ + = 1 gives +2 +2 +b +y +a +x +a +=± +− +, but as the region AOBA lies in the first +quadrant, y is taken as positive. So, the required area is += +2 +2 +0 +4 +a b +a +x dx +a +− +∫ += +2 +2 +2 +–1 +0 +4 +sin +2 +2 +a +b x +a +x +a +x +a +a + + +− ++ + + + + + (Why?) += +2 +1 +4 +0 +sin +1 +0 +2 +2 +b +a +a +a +− + + + + +× + +− + + + + + + + + + + += +2 +4 +2 2 +b a +ab +a +π =π +Fig 8.6 +Fig 8.7 +Reprint 2025-26 + +296 +MATHEMATICS +Alternatively, considering horizontal strips as +shown in the Fig 8.8, the area of the ellipse is += 4 +0 xdy +b∫ + = 4 +2 +2 +0 +a +b +b +y dy +b +− +∫ + (Why?) += += +2 +–1 +4 +0 +sin 1 +0 +2 +2 +a +b +b +b + + + + +× + +− + + + + + + + + + + += +2 +4 +2 2 +a b +ab +b +π =π +EXERCISE 8.1 +1. +Find the area of the region bounded by the ellipse +2 +2 +1 +16 +9 +x +y ++ += . +2. +Find the area of the region bounded by the ellipse +2 +2 +1 +4 +9 +x +y ++ += . +Choose the correct answer in the following Exercises 3 and 4. +3. +Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines +x = 0 and x = 2 is +(A) π +(B) 2 +π +(C) 3 +π +(D) 4 +π +4. +Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is +(A) 2 +(B) 9 +4 +(C) 9 +3 +(D) 9 +2 +Miscellaneous Examples +Example 3 Find the area of the region bounded by the line y = 3x + 2, the x-axis and +the ordinates x = –1 and x = 1. +Fig 8.8 +Reprint 2025-26 + +APPLICATION OF INTEGRALS 297 +Solution As shown in the Fig 8.9, the line +y = 3x + 2 meets x-axis at x = +2 +3 +− and its graph +lies below x-axis for +and above +x-axis for +. +The required area = Area of the region ACBA + +Area of the region ADEA += +2 +1 +3 +2 +1 +3 +(3 +2) +(3 +2) +x +dx +x +dx +− +− +− ++ ++ ++ +∫ +∫ += +2 +1 +2 +2 +3 +2 +1 +3 +3 +3 +2 +2 +2 +2 +x +x +x +x +− +− +− + + + + ++ ++ ++ + + + + + + + + + = 1 +25 +13 +6 +6 +3 ++ += +Example 4 Find the area bounded by +the curve y = cos x between x = 0 and +x = 2π. +Solution From the Fig 8.10, the required +area = area of the region OABO + area +of the region BCDB + area of the region +DEFD. +Thus, we have the required area += +3π +π +2π +2 +2 +3π +π +0 +2 +2 +cos +cos +cos +xdx +xdx +xdx ++ ++ +∫ +∫ +∫ += [ +] +[ +] +[ +] +3 +2 +2 +2 +3 +0 +2 +2 +sin +sin +sin +x +x +x +π +π +π +π +π ++ ++ += 1 + 2 + 1 = 4 +Fig 8.9 +Fig 8.10 +Reprint 2025-26 + +298 +MATHEMATICS +Miscellaneous Exercise on Chapter 8 +1. +Find the area under the given curves and given lines: +(i) y = x2, x = 1, x = 2 and x-axis +(ii) y = x4, x = 1, x = 5 and x-axis +2. +Sketch the graph of y = +3 +x + + and evaluate +. +3. +Find the area bounded by the curve y = sin x between x = 0 and x = 2π. +Choose the correct answer in the following Exercises from 4 to 5. +4. +Area bounded by the curve y = x3, the x-axis and the ordinates x = – 2 and x = 1 is +(A) – 9 +(B) +15 +4 +− +(C) 15 +4 +(D) 17 +4 +5. +The area bounded by the curve y = x |x|, x-axis and the ordinates x = – 1 and +x = 1 is given by +(A) 0 +(B) 1 +3 +(C) 2 +3 +(D) 4 +3 +[Hint : y = x2 if x > 0 and y = – x2 if x < 0]. +Summary +® The area of the region bounded by the curve y = f (x), x-axis and the lines +x = a and x = b (b > a) is given by the formula: Area +( ) +b +b +a +a +ydx +f x dx += += +∫ +∫ +. +® The area of the region bounded by the curve x = φ (y), y-axis and the lines +y = c, y = d is given by the formula: Area +( ) +d +d +c +c +xdy +y dy += += +φ +∫ +∫ +. +Historical Note +The origin of the Integral Calculus goes back to the early period of development +of Mathematics and it is related to the method of exhaustion developed by the +mathematicians of ancient Greece. This method arose in the solution of problems +on calculating areas of plane figures, surface areas and volumes of solid bodies +etc. In this sense, the method of exhaustion can be regarded as an early method +Reprint 2025-26 + +APPLICATION OF INTEGRALS 299 +of integration. The greatest development of method of exhaustion in the early +period was obtained in the works of Eudoxus (440 B.C.) and Archimedes +(300 B.C.) +Systematic approach to the theory of Calculus began in the 17th century. +In 1665, Newton began his work on the Calculus described by him as the theory +of fluxions and used his theory in finding the tangent and radius of curvature at +any point on a curve. Newton introduced the basic notion of inverse function +called the anti derivative (indefinite integral) or the inverse method of tangents. +During 1684-86, Leibnitz published an article in the Acta Eruditorum which +he called Calculas summatorius, since it was connected with the summation of +a number of infinitely small areas, whose sum, he indicated by the symbol ‘∫’. In +1696, he followed a suggestion made by J. Bernoulli and changed this article to +Calculus integrali. This corresponded to Newton’s inverse method of tangents. +Both Newton and Leibnitz adopted quite independent lines of approach which +was radically different. However, respective theories accomplished results that +were practically identical. Leibnitz used the notion of definite integral and what is +quite certain is that he first clearly appreciated tie up between the antiderivative +and the definite integral. +Conclusively, the fundamental concepts and theory of Integral Calculus and +primarily its relationships with Differential Calculus were developed in the work +of P.de Fermat, I. Newton and G. Leibnitz at the end of 17th century. However, +this justification by the concept of limit was only developed in the works of A.L. +Cauchy in the early 19th century. Lastly, it is worth mentioning the following +quotation by Lie Sophie’s: +“It may be said that the conceptions of differential quotient and integral which +in their origin certainly go back to Archimedes were introduced in Science by the +investigations of Kepler, Descartes, Cavalieri, Fermat and Wallis .... The discovery +that differentiation and integration are inverse operations belongs to Newton +and Leibnitz”. +—v— +Reprint 2025-26" +class_12,9,Differential Equations,ncert_books/class_12/lemh2dd/lemh203.pdf,"MATHEMATICS +300 +vHe who seeks for methods without having a definite problem in mind +seeks for the most part in vain. – D. HILBERT v +9.1 Introduction +In Class XI and in Chapter 5 of the present book, we +discussed how to differentiate a given function f with respect +to an independent variable, i.e., how to find f ′(x) for a given +function f at each x in its domain of definition. Further, in +the chapter on Integral Calculus, we discussed how to find +a function f whose derivative is the function g, which may +also be formulated as follows: +For a given function g, find a function f such that +dy +dx = g(x), where y = f(x) ... (1) +An equation of the form (1) is known as a differential +equation. A formal definition will be given later. +These equations arise in a variety of applications, may it be in Physics, Chemistry, +Biology, Anthropology, Geology, Economics etc. Hence, an indepth study of differential +equations has assumed prime importance in all modern scientific investigations. +In this chapter, we will study some basic concepts related to differential equation, +general and particular solutions of a differential equation, formation of differential +equations, some methods to solve a first order - first degree differential equation and +some applications of differential equations in different areas. +9.2 Basic Concepts +We are already familiar with the equations of the type: +x2 – 3x + 3 = 0 +... (1) +sin x + cos x = 0 +... (2) +x + y = 7 +... (3) +Chapter 9 +DIFFERENTIAL EQUATIONS +Henri Poincare +(1854-1912 ) +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +301 +Let us consider the equation: +dy +x +y +dx + + = 0 +... (4) +We see that equations (1), (2) and (3) involve independent and/or dependent variable +(variables) only but equation (4) involves variables as well as derivative of the dependent +variable y with respect to the independent variable x. Such an equation is called a +differential equation. +In general, an equation involving derivative (derivatives) of the dependent variable +with respect to independent variable (variables) is called a differential equation. +A differential equation involving derivatives of the dependent variable with respect +to only one independent variable is called an ordinary differential equation, e.g., +3 +2 +2 +2 d y +dy +dx +dx + + ++  + + + + = 0 is an ordinary differential equation +.... (5) +Of course, there are differential equations involving derivatives with respect to +more than one independent variables, called partial differential equations but at this +stage we shall confine ourselves to the study of ordinary differential equations only. +Now onward, we will use the term ‘differential equation’ for ‘ordinary differential +equation’. +ANote +1. We shall prefer to use the following notations for derivatives: +2 +3 +2 +3 +, +, +dy +d y +d y +y +y +y +dx +dx +dx +′ +′′ +′′′ += += += +2. For derivatives of higher order, it will be inconvenient to use so many dashes +as supersuffix therefore, we use the notation yn for nth order derivative +n +n +d y +dx +. +9.2.1. Order of a differential equation +Order of a differential equation is defined as the order of the highest order derivative of +the dependent variable with respect to the independent variable involved in the given +differential equation. +Consider the following differential equations: +dy +dx = ex +... (6) +Reprint 2025-26 + +MATHEMATICS +302 +2 +2 +d y +y +dx ++ + = 0 +... (7) +3 +3 +2 +2 +3 +2 +d y +d y +x +dx +dx + + + + ++ + + + + + + + + += 0 +... (8) +The equations (6), (7) and (8) involve the highest derivative of first, second and +third order respectively. Therefore, the order of these equations are 1, 2 and 3 respectively. +9.2.2 Degree of a differential equation +To study the degree of a differential equation, the key point is that the differential +equation must be a polynomial equation in derivatives, i.e., y′, y″, y″′ etc. Consider the +following differential equations: +2 +3 +2 +3 +2 +2 +d y +d y +dy +y +dx +dx +dx + + ++ +− ++ + + + + + = 0 +... (9) +2 +2 +sin +dy +dy +y +dx +dx + + + + ++ +− + + + + + + + + + = 0 +... (10) +sin +dy +dy +dx +dx + + ++ + + + + = 0 +... (11) +We observe that equation (9) is a polynomial equation in y″′, y″ and y′, equation (10) +is a polynomial equation in y′ (not a polynomial in y though). Degree of such differential +equations can be defined. But equation (11) is not a polynomial equation in y′ and +degree of such a differential equation can not be defined. +By the degree of a differential equation, when it is a polynomial equation in +derivatives, we mean the highest power (positive integral index) of the highest order +derivative involved in the given differential equation. +In view of the above definition, one may observe that differential equations (6), (7), +(8) and (9) each are of degree one, equation (10) is of degree two while the degree of +differential equation (11) is not defined. +ANote Order and degree (if defined) of a differential equation are always +positive integers. +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +303 +Example 1 Find the order and degree, if defined, of each of the following differential +equations: +(i) +cos +0 +dy +x +dx − += +(ii) +2 +2 +2 +0 +d y +dy +dy +xy +x +y +dx +dx +dx + + ++ +− += + + + + +(iii) +2 +0 +y +y +y +e ′ +′′′ + ++ += +Solution +(i) +The highest order derivative present in the differential equation is dy +dx , so its +order is one. It is a polynomial equation in y′ and the highest power raised to dy +dx +is one, so its degree is one. +(ii) +The highest order derivative present in the given differential equation is +2 +2 +d y +dx +, so +its order is two. It is a polynomial equation in +2 +2 +d y +dx + and dy +dx and the highest +power raised to +2 +2 +d y +dx + is one, so its degree is one. +(iii) +The highest order derivative present in the differential equation is y′′′, so its +order is three. The given differential equation is not a polynomial equation in its +derivatives and so its degree is not defined. +EXERCISE 9.1 +Determine order and degree (if defined) of differential equations given in Exercises +1 to 10. +1. +4 +4 +sin( +) +0 +d y +y +dx +′′′ ++ += +2. y′ + 5y = 0 +3. +4 +2 +2 +3 +0 +ds +d s +s +dt +dt + ++ += + + + + +4. +2 +2 +2 +cos +0 +d y +dy +dx +dx + + + + ++ += + + + + + + + + +5. +2 +2 +cos3 +sin3 +d y +x +x +dx += ++ +6. +2 +( +) +y′′′ + + (y″)3 + (y′)4 + y5 = 0 +7. +y′′′ + 2y″ + y′ = 0 +Reprint 2025-26 + +MATHEMATICS +304 +8. +y′ + y = ex +9. y″ + (y′)2 + 2y = 0 10. y″ + 2y′ + sin y = 0 +11. +The degree of the differential equation +3 +2 +2 +2 +sin +1 +0 +d y +dy +dy +dx +dx +dx + + + + + + ++ ++ ++ = + + + + + + + + + + + + + is +(A) 3 +(B) 2 +(C) 1 +(D) not defined +12. +The order of the differential equation +2 +2 +2 +2 +3 +0 +d y +dy +x +y +dx +dx +− ++ += + is +(A) 2 +(B) 1 +(C) 0 +(D) not defined +9.3. General and Particular Solutions of a Differential Equation +In earlier Classes, we have solved the equations of the type: +x2 + 1 = 0 +... (1) +sin2 x – cos x = 0 +... (2) +Solution of equations (1) and (2) are numbers, real or complex, that will satisfy the +given equation i.e., when that number is substituted for the unknown x in the given +equation, L.H.S. becomes equal to the R.H.S.. +Now consider the differential equation +2 +2 +0 +d y +y +dx ++ += +... (3) +In contrast to the first two equations, the solution of this differential equation is a +function φ that will satisfy it i.e., when the function φ is substituted for the unknown y +(dependent variable) in the given differential equation, L.H.S. becomes equal to R.H.S.. +The curve y = φ (x) is called the solution curve (integral curve) of the given +differential equation. Consider the function given by +y = φ (x) = a sin (x + b), +... (4) +where a, b ∈ R. When this function and its derivative are substituted in equation (3), +L.H.S. = R.H.S.. So it is a solution of the differential equation (3). +Let a and b be given some particular values say a = 2 and +4 +b +π += +, then we get a +function +y = φ1(x) = 2sin +4 +x +π + + ++ + + + + +... (5) +When this function and its derivative are substituted in equation (3) again +L.H.S. = R.H.S.. Therefore φ1 is also a solution of equation (3). +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +305 +Function φ consists of two arbitrary constants (parameters) a, b and it is called +general solution of the given differential equation. Whereas function φ1 contains no +arbitrary constants but only the particular values of the parameters a and b and hence +is called a particular solution of the given differential equation. +The solution which contains arbitrary constants is called the general solution +(primitive) of the differential equation. +The solution free from arbitrary constants i.e., the solution obtained from the general +solution by giving particular values to the arbitrary constants is called a particular +solution of the differential equation. +Example 2 Verify that the function y = e– 3x is a solution of the differential equation +2 +2 +6 +0 +d y +dy +y +dx +dx ++ +− += +Solution Given function is y = e– 3x. Differentiating both sides of equation with respect +to x , we get +3 +3 +x +dy +e +dx +− += − +... (1) +Now, differentiating (1) with respect to x, we have +2 +2 +d y +dx + = 9 e – 3x +Substituting the values of +2 +2 , +d y dy +dx +dx +and y in the given differential equation, we get +L.H.S. = 9 e– 3x + (–3e– 3x) – 6.e– 3x = 9 e– 3x – 9 e– 3x = 0 = R.H.S.. +Therefore, the given function is a solution of the given differential equation. +Example 3 Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution +of the differential equation +2 +2 +0 +d y +y +dx ++ += +Solution The given function is +y = a cos x + b sin x +... (1) +Differentiating both sides of equation (1) with respect to x, successively, we get +dy +dx = – a sinx + b cosx +2 +2 +d y +dx + = – a cos x – b sinx +Reprint 2025-26 + +MATHEMATICS +306 +Substituting the values of +2 +2 +d y +dx + and y in the given differential equation, we get +L.H.S. = (– a cos x – b sin x) + (a cos x + b sin x) = 0 = R.H.S. +Therefore, the given function is a solution of the given differential equation. +EXERCISE 9.2 +In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a +solution of the corresponding differential equation: +1. +y = ex + 1 +: +y″ – y′ = 0 +2. +y = x2 + 2x + C +: +y′ – 2x – 2 = 0 +3. +y = cos x + C +: +y′ + sin x = 0 +4. +y = +2 +1 +x ++ +: +y′ = +2 +1 +xy +x ++ +5. +y = Ax +: +xy′ = y (x ≠ 0) +6. +y = x sin x +: +xy′ = y + x +2 +2 +x +y +− + (x ≠ 0 and x > y or x < – y) +7. +xy = log y + C +: +y′ = +2 +1 +y +xy +− + (xy ≠ 1) +8. +y – cos y = x +: +(y sin y + cos y + x) y′ = y +9. +x + y = tan–1y +: +y2 y′ + y2 + 1 = 0 +10. +y = +2 +2 +a +x +− +x ∈ (–a, a) : +x + y dy +dx = 0 (y ≠ 0) +11. +The number of arbitrary constants in the general solution of a differential equation +of fourth order are: +(A) 0 +(B) 2 +(C) 3 +(D) 4 +12. +The number of arbitrary constants in the particular solution of a differential equation +of third order are: +(A) 3 +(B) 2 +(C) 1 +(D) 0 +9.4. Methods of Solving First Order, First Degree Differential Equations +In this section we shall discuss three methods of solving first order first degree differential +equations. +9.4.1 Differential equations with variables separable +A first order-first degree differential equation is of the form +dy +dx = F(x, y) +... (1) +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +307 +If F(x, y) can be expressed as a product g (x) h(y), where, g(x) is a function of x +and h(y) is a function of y, then the differential equation (1) is said to be of variable +separable type. The differential equation (1) then has the form +dy +dx = h (y) . g(x) +... (2) +If h(y) ≠ 0, separating the variables, (2) can be rewritten as +1 +( ) +h y dy = g(x) dx +... (3) +Integrating both sides of (3), we get +1 +( ) dy +h y +∫ += +( ) +g x dx +∫ +... (4) +Thus, (4) provides the solutions of given differential equation in the form +H(y) = G(x) + C +Here, H (y) and G (x) are the anti derivatives of +1 +( ) +h y and g(x) respectively and +C is the arbitrary constant. +Example 4 Find the general solution of the differential equation +1 +2 +dy +x +dx +y ++ += +− +, (y ≠ 2) +Solution We have +dy +dx = +1 +2 +x +y ++ +− +... (1) +Separating the variables in equation (1), we get +(2 – y) dy = (x + 1) dx +... (2) +Integrating both sides of equation (2), we get +(2 +) +y dy +− +∫ += +( +1) +x +dx ++ +∫ +or +2 +2 +2 +y +y − + = +2 +1 +C +2 +x +x ++ ++ +or +x2 + y2 + 2x – 4y + 2 C1 = 0 +or +x2 + y2 + 2x – 4y + C = 0, where C = 2C1 +which is the general solution of equation (1). +Reprint 2025-26 + +MATHEMATICS +308 +Example 5 Find the general solution of the differential equation +2 +2 +1 +1 +dy +y +dx +x ++ += ++ +. +Solution Since 1 + y2 ≠ 0, therefore separating the variables, the given differential +equation can be written as +2 +1 +dy +y ++ + = +2 +1 +dx +x ++ +... (1) +Integrating both sides of equation (1), we get +2 +1 +dy +y ++ +∫ + = +2 +1 +dx +x ++ +∫ +or +tan–1 y = tan–1x + C +which is the general solution of equation (1). +Example 6 Find the particular solution of the differential equation +2 +4 +dy +xy +dx = − + given +that y = 1, when x = 0. +Solution If y ≠ 0, the given differential equation can be written as +2 +dy +y = – 4x dx +... (1) +Integrating both sides of equation (1), we get +2 +dy +y∫ + = +4 x dx +−∫ +or +1 +y +− + = – 2x2 + C +or +y = +2 +1 +2 +C +x − +... (2) +Substituting y = 1 and x = 0 in equation (2), we get, C = – 1. +Now substituting the value of C in equation (2), we get the particular solution of the +given differential equation as +2 +1 +2 +1 +y +x += ++ +. +Example 7 Find the equation of the curve passing through the point (1, 1) whose +differential equation is x dy = (2x2 + 1) dx (x ≠ 0). +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +309 +Solution The given differential equation can be expressed as +dy* = +or +dy = +1 +2x +dx +x + + ++ + + + + +... (1) +Integrating both sides of equation (1), we get +dy +∫ + = +1 +2x +dx +x + + ++ + + + + +∫ +or +y = x2 + log |x| + C +... (2) +Equation (2) represents the family of solution curves of the given differential equation +but we are interested in finding the equation of a particular member of the family which +passes through the point (1, 1). Therefore substituting x = 1, y = 1 in equation (2), we +get C = 0. +Now substituting the value of C in equation (2) we get the equation of the required +curve as y = x2 + log |x|. +Example 8 Find the equation of a curve passing through the point (–2, 3), given that +the slope of the tangent to the curve at any point (x, y) is +2 +2x +y +. +Solution We know that the slope of the tangent to a curve is given by dy +dx +. +so, +dy +dx = +2 +2x +y +... (1) +Separating the variables, equation (1) can be written as +y2 dy = 2x dx +... (2) +Integrating both sides of equation (2), we get +2 +y dy +∫ + = +2x dx +∫ +or +3 +3 +y = x2 + C +... (3) +* +The notation +dy +dx due to Leibnitz is extremely flexible and useful in many calculation and formal +transformations, where, we can deal with symbols dy and dx exactly as if they were ordinary numbers. By +treating dx and dy like separate entities, we can give neater expressions to many calculations. +Refer: Introduction to Calculus and Analysis, volume-I page 172, By Richard Courant, +Fritz John Spinger – Verlog New York. +Reprint 2025-26 + +MATHEMATICS +310 +Substituting x = –2, y = 3 in equation (3), we get C = 5. +Substituting the value of C in equation (3), we get the equation of the required curve as +3 +2 +5 +3 +y +x += ++ + or +1 +2 +3 +(3 +15) +y +x += ++ +Example 9 In a bank, principal increases continuously at the rate of 5% per year. In +how many years Rs 1000 double itself? +Solution Let P be the principal at any time t. According to the given problem, +dp +dt = +5 +P +100 + +× + + + + +or +dp +dt = P +20 +... (1) +separating the variables in equation (1), we get +P +dp = 20 +dt +... (2) +Integrating both sides of equation (2), we get +log P = +1 +C +20 +t + +or +P = +1 +C +20 +t +e +e⋅ +or +P = +20 +C +t +e + (where +1 +C +C +e += +) +... (3) +Now +P = 1000, when t = 0 +Substituting the values of P and t in (3), we get C = 1000. Therefore, equation (3), +gives +P = 1000 +20 +t +e +Let t years be the time required to double the principal. Then +2000 = 1000 +20 +t +e + ⇒ t = 20 loge2 +EXERCISE 9.3 +For each of the differential equations in Exercises 1 to 10, find the general solution: +1. +1 +cos +1 +cos +dy +x +dx +x +− += + +2. +2 +4 +( 2 +2) +dy +y +y +dx = +− +−< +< +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +311 +3. +1( +1) +dy +y +y +dx + += +≠ +4. +sec2 x tan y dx + sec2 y tan x dy = 0 +5. +(ex + e–x) dy – (ex – e–x) dx = 0 +6. +2 +2 +(1 +) (1 +) +dy +x +y +dx = ++ ++ +7. +y log y dx – x dy = 0 +8. +5 +5 +dy +x +y +dx = − +9. +1 +sin +dy +x +dx +− += +10. +ex tan y dx + (1 – ex) sec2 y dy = 0 +For each of the differential equations in Exercises 11 to 14, find a particular solution +satisfying the given condition: +11. +3 +2 +( +1) dy +x +x +x +dx ++ ++ ++ + = 2x2 + x; y = 1 when x = 0 +12. +2 +( +1) +1 +dy +x x +dx +− += ; y = 0 when x = 2 +13. +cos +dy +a +dx + += + + + + + (a ∈ R); y = 1 when x = 0 +14. +tan +dy +y +x +dx = +; y = 1 when x = 0 +15. +Find the equation of a curve passing through the point (0, 0) and whose differential +equation is y′ = ex sin x. +16. +For the differential equation +( +2) ( +2) +dy +xy +x +y +dx = ++ ++ +, find the solution curve +passing through the point (1, –1). +17. +Find the equation of a curve passing through the point (0, –2) given that at any +point (x, y) on the curve, the product of the slope of its tangent and y coordinate +of the point is equal to the x coordinate of the point. +18. +At any point (x, y) of a curve, the slope of the tangent is twice the slope of the +line segment joining the point of contact to the point (– 4, –3). Find the equation +of the curve given that it passes through (–2, 1). +19. +The volume of spherical balloon being inflated changes at a constant rate. If +initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of +balloon after t seconds. +Reprint 2025-26 + +MATHEMATICS +312 +20. +In a bank, principal increases continuously at the rate of r% per year. Find the +value of r if Rs 100 double itself in 10 years (loge2 = 0.6931). +21. +In a bank, principal increases continuously at the rate of 5% per year. An amount +of Rs 1000 is deposited with this bank, how much will it worth after 10 years +(e0.5 = 1.648). +22. +In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 +hours. In how many hours will the count reach 2,00,000, if the rate of growth of +bacteria is proportional to the number present? +23. +The general solution of the differential equation +x +y +dy +e +dx ++ += + is +(A) ex + e–y = C +(B) ex + ey = C +(C) e–x + ey = C +(D) e–x + e–y = C +9.4.2 Homogeneous differential equations +Consider the following functions in x and y +F1 (x, y) = y2 + 2xy, +F2 (x, y) = 2x – 3y, +F3 (x, y) = cos +y +x + + + + + +, +F4 (x, y) = sin x + cos y +If we replace x and y by λx and λy respectively in the above functions, for any nonzero +constant λ, we get +F1 (λx, λy) = λ2 (y2 + 2xy) = λ2 F1 (x, y) +F2 (λx, λy) = λ (2x – 3y) = λ F2 (x, y) +F3 (λx, λy) = cos +cos +y +y +x +x +λ + + + + += + + + + +λ + + + + = λ0 F3 (x, y) +F4 (λx, λy) = sin λx + cos λy ≠ λn F4 (x, y), for any n ∈ N +Here, we observe that the functions F1, F2, F3 can be written in the form +F(λx, λy) = λn F (x, y) but F4 can not be written in this form. This leads to the following +definition: +A function F(x, y) is said to be homogeneous function of degree n if +F(λx, λy) = λn F(x, y) for any nonzero constant λ. +We note that in the above examples, F1, F2, F3 are homogeneous functions of +degree 2, 1, 0 respectively but F4 is not a homogeneous function. +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +313 +We also observe that +F1(x, y) = +2 +2 +2 +1 +2 +2 +y +y +y +x +x h +x +x +x + + + + ++ += + + + + + + + + +or +F1(x, y) = +2 +2 +2 +2 +1 +x +x +y +y h +y +y + + + + ++ += + + + + + + + + +F2(x, y) = +1 +1 +3 +3 +2 +y +y +x +x h +x +x + + + + +− += + + + + + + + + +or + F2(x, y) = +1 +1 +4 +2 +3 +x +x +y +y h +y +y + + + + +− += + + + + + + + + +F3(x, y) = +0 +0 +5 +cos +y +y +x +x h +x +x + + + + += + + + + + + + + +F4(x, y) ≠ +6 +n +y +x h +x + + + + + +, for any n ∈ N +or +F4 (x, y) ≠ +7 +n +x +y h +y + + + + + +, for any n ∈ N +Therefore, a function F (x, y) is a homogeneous function of degree n if +F(x, y) = +or +n +n +y +x +x g +y h +x +y + + + + + + + + + + + + +A differential equation of the form dy +dx = F (x, y) is said to be homogenous if +F(x, y) is a homogenous function of degree zero. +To solve a homogeneous differential equation of the type +( +) +F +, +dy +x y +dx = + = +y +g +x + + + + + + +... (1) +We make the substitution + y = v.x + ... (2) +Differentiating equation (2) with respect to x, we get +dy +dx = +dv +v +x dx ++ +... (3) +Substituting the value of dy +dx from equation (3) in equation (1), we get +Reprint 2025-26 + +MATHEMATICS +314 +dv +v +x dx ++ + = g (v) +or +dv +x dx = g (v) – v +... (4) +Separating the variables in equation (4), we get +( ) +dv +g v +v +− + = dx +x +... (5) +Integrating both sides of equation (5), we get +( ) +dv +g v +v +− +∫ + = +1 +C +dx +x ++ +∫ +... (6) +Equation (6) gives general solution (primitive) of the differential equation (1) when +we replace v by y +x . +ANote If the homogeneous differential equation is in the form +F( , ) +dx +x y +dy = +where, F (x, y) is homogenous function of degree zero, then we make substitution +x +v +y = + i.e., x = vy and we proceed further to find the general solution as discussed +above by writing +F( , ) +. +dx +x +x y +h +dy +y + + += += + + + + +Example 10 Show that the differential equation (x – y) dy +dx = x + 2y is homogeneous +and solve it. +Solution The given differential equation can be expressed as +dy +dx = +2 +x +y +x +y ++ +− +... (1) +Let +F(x, y) = +2 +x +y +x +y ++ +− +Now +F(λx, λy) = +0 +( +2 ) +( , ) +( +) +x +y +f x y +x +y +λ ++ += λ ⋅ +λ +− +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +315 +Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential +equation is a homogenous differential equation. +Alternatively, +2 +1 +1 +y +dy +x +y +dx +x + + ++ + + +=  + + + +− + + + = +y +g +x + + + + + + +... (2) +R.H.S. of differential equation (2) is of the form g +y +x + + + + and so it is a homogeneous +function of degree zero. Therefore, equation (1) is a homogeneous differential equation. +To solve it we make the substitution +y = vx +... (3) +Differentiating equation (3) with respect to, x we get +dy +dx = +dv +v +x dx ++ +... (4) +Substituting the value of y and dy +dx in equation (1) we get +dv +v +x dx ++ + = 1 +2 +1 +v +v ++ +− +or +dv +x dx = 1 +2 +1 +v +v +v ++ +− +− +or +dv +x dx = +2 +1 +1 +v +v +v ++ ++ +− +or +2 +1 +1 +v +dv +v +v +− ++ ++ + = +dx +x +− +Integrating both sides of equation (5), we get + = +or + = – log |x| + C1 +Reprint 2025-26 + +MATHEMATICS +316 +or +or +or +or +(Why?) +Replacing v by y +x , we get +or +or +2 +2 +1 +1 +2 +1 +2 +log +1 +3 tan +C +2 +3 +y +y +y +x +x +x +x +x +− + + ++ + + ++ ++ += ++ + + + + + + + + +or +2 +2 +1 +1 +2 +log ( +) +2 3 tan +2C +3 +y +x +y +xy +x +x +− ++ + + ++ ++ += ++ + + + + +or +2 +2 +1 +2 +log ( +) +2 3 tan +C +3 +− ++ + + ++ ++ += ++ + + + + +x +y +x +xy +y +x +which is the general solution of the differential equation (1) +Example 11 Show that the differential equation +cos +cos +y +dy +y +x +y +x +x +dx +x + + + + += ++ + + + + + + + + + is +homogeneous and solve it. +Solution The given differential equation can be written as +dy +dx = +cos +cos +y +y +x +x +y +x +x + ++ + + + + + + + + + + +... (1) +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +317 +It is a differential equation of the form +F( , +) +dy +x y +dx = +. +Here +F(x, y) = +cos +cos +y +y +x +x +y +x +x + ++ + + + + + + + + + + +Replacing x by λx and y by λy, we get +F(λx, λy) = +0 +[ cos +] +[F( , )] +cos +y +y +x +x +x y +y +x +x + + +λ ++ + + + + += λ + + +λ + + + +Thus, F(x, y) is a homogeneous function of degree zero. +Therefore, the given differential equation is a homogeneous differential equation. +To solve it we make the substitution +y = vx +... (2) +Differentiating equation (2) with respect to x, we get +dy +dx = +dv +v +x dx ++ +... (3) +Substituting the value of y and dy +dx in equation (1), we get +dv +v +x dx ++ + = +cos +1 +cos +v +v +v ++ +or +dv +x dx = +cos +1 +cos +v +v +v +v ++ − +or +dv +x dx = +1 +cosv +or +cosv dv = dx +x +Therefore +cosv dv +∫ + = +1 dx +x∫ +Reprint 2025-26 + +MATHEMATICS +318 +or +sin v = log |x| + log |C| +or +sin v = log |Cx| +Replacing v by y +x , we get +sin +y +x + + + + + + = log |Cx| +which is the general solution of the differential equation (1). +Example 12 Show that the differential equation 2 +2 +0 +x +x +y +y +y e dx +y +x e +dy + + + + ++ +− += + + +is +homogeneous and find its particular solution, given that, x = 0 when y = 1. +Solution The given differential equation can be written as +dx +dy = 2 +2 +x +y +x +y +x e +y +y e +− +... (1) +Let +F(x, y) = 2 +2 +x +y +x +y +xe +y +ye +− +Then +F(λx, λy) = +0 +2 +[F( , )] +2 +x +y +x +y +xe +y +x y +ye + + + + +λ +− + + + +=λ + + + + +λ + + + +Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given +differential equation is a homogeneous differential equation. +To solve it, we make the substitution +x = vy +... (2) +Differentiating equation (2) with respect to y, we get +dx +dy = ++ +dv +v +y +dy +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +319 +Substituting the value of +and dx +x +dy in equation (1), we get +dv +v +y +dy ++ + = 2 +1 +2 +v +v +v e +e +− +or +dv +y +dy = 2 +1 +2 +v +v +v e +v +e +−− +or +dv +y +dy = +1 +2 ve +− +or +2ev dv = +dy +y +− +or +2 ve +dv +⋅ +∫ + = +dy +y +−∫ +or +2 ev = – log |y| + C +and replacing v by +x +y , we get +2 +x +y +e + log |y| = C +... (3) +Substituting x = 0 and y = 1 in equation (3), we get +2 e0 + log |1| = C ⇒ C = 2 +Substituting the value of C in equation (3), we get +2 +x +y +e + log |y| = 2 +which is the particular solution of the given differential equation. +Example 13 Show that the family of curves for which the slope of the tangent at any +point (x, y) on it is +2 +2 +2 +x +y +xy ++ +, is given by x2 – y2 = cx. +Solution We know that the slope of the tangent at any point on a curve is dy +dx . +Therefore, +dy +dx = +2 +2 +2 +x +y +xy ++ +Reprint 2025-26 + +MATHEMATICS +320 +or +dy +dx = +2 +2 +1 +2 +y +x +y +x ++ +... (1) +Clearly, (1) is a homogenous differential equation. To solve it we make substitution +y = vx +Differentiating y = vx with respect to x, we get +dy +dx = +dv +v +x dx ++ +or +dv +v +x dx ++ + = +2 +1 +2 +v +v ++ +or +dv +x dx = +2 +1 +2 +v +v +− +2 +2 +1 +v dv +v +− + = dx +x +or +2 +2 +1 +v dv +v − + = +dx +x +− +Therefore +2 +2 +1 +v dv +v − +∫ + = +1 dx +x +−∫ +or +log |v2 – 1 | = – log |x | + log |C1| +or +log |(v2 – 1) (x)| = log |C1| +or +(v2 – 1) x = ± C1 +Replacing v by y +x , we get +2 +2 +1 +y +x +x + + +− + + + + + = ± C1 +or +(y2 – x2) = ± C1 x or x2 – y2 = Cx +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +321 +EXERCISE 9.4 +In each of the Exercises 1 to 10, show that the given differential equation is homogeneous +and solve each of them. +1. +(x2 + xy) dy = (x2 + y2) dx +2. +x +y +y +x ++ +′ = +3. +(x – y) dy – (x + y) dx = 0 +4. +(x2 – y2) dx + 2xy dy = 0 +5. +2 +2 +2 +2 +dy +x +x +y +xy +dx = +− ++ +6. +x dy – y dx = +2 +2 +x +y +dx ++ +7. +cos +sin +sin +cos +y +y +y +y +x +y +y dx +y +x +x dy +x +x +x +x + + + + + + + + + + + + ++ += +− + + + + + + + + + + + + + + + + + + + + + + + + +8. +sin +0 +dy +y +x +y +x +dx +x + + +− ++ += + + + + +9. +log +2 +0 +y +y dx +x +dy +x dy +x + + ++ +− += + + + + +10. +For each of the differential equations in Exercises from 11 to 15, find the particular +solution satisfying the given condition: +11. +(x + y) dy + (x – y) dx = 0; y = 1 when x = 1 +12. +x2 dy + (xy + y2) dx = 0; y = 1 when x = 1 +13. + when x = 1 +14. +cosec +0 +dy +y +y +dx +x +x + + +− ++ += + + + + +; y = 0 when x = 1 +15. +2 +2 +2 +2 +0 +dy +xy +y +x dx ++ +− += +; y = 2 when x = 1 +16. +A homogeneous differential equation of the from +dx +x +h +dy +y + + += + + + + can be solved by +making the substitution. +(A) y = vx +(B) v = yx +(C) x = vy +(D) x = v +Reprint 2025-26 + +MATHEMATICS +322 +17. +Which of the following is a homogeneous differential equation? +(A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 +(B) (xy) dx – (x3 + y3) dy = 0 +(C) (x3 + 2y2) dx + 2xy dy = 0 +(D) y2 dx + (x2 – xy – y2) dy = 0 +9.4.3 Linear differential equations +A differential equation of the from +P +dy +y +dx + + = Q +where, P and Q are constants or functions of x only, is known as a first order linear +differential equation. Some examples of the first order linear differential equation are +dy +y +dx + + = sin x +1 +dy +y +dx +x + + ++  + + + + = ex +log +dy +y +dx +x +x + + ++  + + + + = 1 +x +Another form of first order linear differential equation is +1P +dx +x +dy ++ + = Q1 +where, P1 and Q1 are constants or functions of y only. Some examples of this type of +differential equation are +dx +x +dy + += cos y +2 +dx +x +dy +y +− ++ + = y2e – y +To solve the first order linear differential equation of the type +P +dy +y +dx + + = Q +... (1) +Multiply both sides of the equation by a function of x say g (x) to get +g(x) dy +dx + P.(g(x)) y = Q . g (x) +... (2) +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +323 +Choose g (x) in such a way that R.H.S. becomes a derivative of y . g (x). +i.e. +g(x) dy +dx + P. g(x) y = d +dx [y .g (x)] +or +g(x) dy +dx + P. g(x) y = g(x) dy +dx + y g′ (x) +⇒ +P. g(x) = g′ (x) +or +P = +( ) +( ) +g +x +g x +′ +Integrating both sides with respect to x, we get +Pdx +∫ + = +( ) +( ) +g +x dx +g x +′ +∫ +or +P dx +⋅ +∫ + = log(g (x)) +or +g(x) = +P dx +e∫ +On multiplying the equation (1) by g(x) = +P dx +e∫ +, the L.H.S. becomes the derivative +of some function of x and y. This function g(x) = +P dx +e∫ + is called Integrating Factor +(I.F.) of the given differential equation. +Substituting the value of g (x) in equation (2), we get + = +or +d +dx +ye +dx +P∫ + + + + = +Integrating both sides with respect to x, we get + = +Q +P +.e +dx +dx +∫ + + + + +∫ +or +y = e +e +dx +dx +dx +−∫ +∫ + + + + ++ +∫ +P +P +Q +C +. +. +which is the general solution of the differential equation. +Reprint 2025-26 + +MATHEMATICS +324 +Steps involved to solve first order linear differential equation: +(i) +Write the given differential equation in the form +P +Q +dy +y +dx + += + where P, Q are +constants or functions of x only. +(ii) +Find the Integrating Factor (I.F) = +. +(iii) +Write the solution of the given differential equation as +y (I.F) = +In case, the first order linear differential equation is in the form +1 +1 +P +Q +dx +x +dy ++ += +, +where, P1 and Q1 are constants or functions of y only. Then I.F = +1P dy +e + and the +solution of the differential equation is given by +x . (I.F) = +( +) +1 +Q × I.F +C +dy + +∫ +Example 14 Find the general solution of the differential equation +cos +dy +y +x +dx − += +. +Solution Given differential equation is of the form +P +Q +dy +y +dx + += +, where P = –1 and Q = cos x +Therefore +I.F = +Multiplying both sides of equation by I.F, we get + = e–x cos x +or +( +) +x +dy ye +dx +− += e–x cos x +On integrating both sides with respect to x, we get +ye– x = +cos +C +x +e +x dx +− ++ +∫ +... (1) +Let +I = +cos +x +e +x dx +− +∫ += cos +( sin ) ( +) +1 +x +x +e +x +x +e +dx +− +− + +− +− +− + + +− + +∫ +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +325 += +cos +sin +x +x +x e +x e +dx +− +− +− +−∫ += +cos +sin (– +) +cos +( +) +x +x +x +x e +x +e +x +e +dx +− +− +− + + +− +− +− +− + + +∫ += +cos +sin +cos +x +x +x +x e +x e +x e +dx +− +− +− +− ++ +−∫ +or +I = – e–x cos x + sin x e–x – I +or +2I = (sin x – cos x) e–x +or +I = (sin +cos ) +2 +x +x +x e− +− +Substituting the value of I in equation (1), we get +ye– x = +sin +cos +C +2 +x +x +x e− +− + + ++ + + + + +or +y = +sin +cos +C +2 +x +x +x +e +− + ++ + + + + +which is the general solution of the given differential equation. +Example 15 Find the general solution of the differential equation +2 +2 +( +0) +dy +x +y +x +x +dx + += +≠ +. +Solution The given differential equation is +2 +dy +x +y +dx + + = x 2 +... (1) +Dividing both sides of equation (1) by x, we get +2 +dy +y +dx +x ++ + = x +which is a linear differential equation of the type +P +Q +dy +y +dx + += +, where +2 +P +x += + and Q = x. +So +I.F = +2 dx +x +e∫ += e2 log x = +2 +log +2 +x +e +x += +log +( ) +[ +( )] +f x +as e +f x += +Therefore, solution of the given equation is given by +y . x2 = +2 +( ) ( +) +C +x +x +dx + +∫ + = +3 +C +x dx + +∫ +or +y = +2 +2 +C +4 +x +x− ++ +which is the general solution of the given differential equation. +Reprint 2025-26 + +MATHEMATICS +326 +Example 16 Find the general solution of the differential equation y dx – (x + 2y2) dy = 0. +Solution The given differential equation can be written as +dx +x +dy +y +− + = 2y +This is a linear differential equation of the type +1 +1 +P +Q +dx +x +dy + += +, where +1 +1 +P +y += − + and +Q1 = 2y. Therefore +1 +1 +log +log( ) +1 +I.F +dy +y +y +y +e +e +e +y +− +− +− +∫ += += += += +Hence, the solution of the given differential equation is +1 +x +y = +1 +(2 ) +C +y +dy +y + + ++ + + + + +∫ +or +x +y = +(2 +) +C +dy + +∫ +or +x +y = 2y + C +or +x = 2y2 + Cy +which is a general solution of the given differential equation. +Example 17 Find the particular solution of the differential equation +cot ++ +dy +y +x +dx + = 2x + x2 cot x (x ≠ 0) +given that y = 0 when +2 +x +π += +. +Solution The given equation is a linear differential equation of the type +P +Q +dy +y +dx + += +, +where P = cot x and Q = 2x + x2 cot x. Therefore +I.F = e +e +x +x dx +x +cot +log sin +sin +∫ += += +Hence, the solution of the differential equation is given by +y . sin x = ∫(2x + x2 cot x) sin x dx + C +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +327 +or +y sin x = ∫2x sin x dx + ∫x2 cos x dx + C +or +y sin x = +2 +2 +2 +2 +2 +sin +cos +cos +C +2 +2 +x +x +x +x +dx +x +x dx + + + + +− ++ ++ + + + + + + + + +∫ +∫ +or +y sin x = +2 +2 +2 +sin +cos +cos +C +x +x +x +x dx +x +x dx +− ++ ++ +∫ +∫ +or +y sin x = x2 sin x + C +... (1) +Substituting y = 0 and +2 +x +π += + in equation (1), we get +0 = +2 +sin +C +2 +2 +π +π + + + ++ + + + + + + + + +or +C = +2 +4 +−π +Substituting the value of C in equation (1), we get +y sin x = +2 +2 sin +4 +x +x +π +− +or +y = +2 +2 +(sin +0) +4 sin +x +x +x +π +− +≠ +which is the particular solution of the given differential equation. +Example 18 Find the equation of a curve passing through the point (0, 1). If the slope +of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate +(abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point. +Solution We know that the slope of the tangent to the curve is dy +dx . +Therefore, +dy +dx = x + xy +or +dy +xy +dx − + = x +... (1) +This is a linear differential equation of the type +P +Q +dy +y +dx + += +, where P = – x and Q = x. +Therefore, +I.F = +2 +2 +x +x dx +e +e +− +−∫ += +Reprint 2025-26 + +MATHEMATICS +328 +Hence, the solution of equation is given by +2 +2 +x +y e +− +⋅ + = +( +) +2 +2 +( ) +C +x +x +dx +e +− ++ +∫ +... (2) +Let +I = +2 +2 +( ) +x +x +dx +e +− +∫ +Let +2 +2 +x +t +− += , then – x dx = dt or x dx = – dt. + Therefore, I = +2 +2 +– +x +t +t +e dt +e +e +− +− += − += +∫ +Substituting the value of I in equation (2), we get +2 +2 +x +y e +− + = +2 +2 + C +− +− +x +e +or +y = +2 +2 +1 +C +x +e +−+ +... (3) +Now (3) represents the equation of family of curves. But we are interested in +finding a particular member of the family passing through (0, 1). Substituting x = 0 and +y = 1 in equation (3) we get +1 = – 1 + C . e0 or C = 2 +Substituting the value of C in equation (3), we get +y = +2 +2 +1 +2 +x +e +−+ +which is the equation of the required curve. +EXERCISE 9.5 +For each of the differential equations given in Exercises 1 to 12, find the general solution: +1. +2 +sin +dy +y +x +dx + += +2. +2 +3 +x +dy +y +e +dx +��� ++ += +3. +2 +dy +y +x +dx +x ++ += +4. +(sec ) +tan +0 +2 +dy +x y +x +x +dx +π + + ++ += +≤ +< + + + + +5. +2 +cos +tan +dy +x +y +x +dx + += + 0 +2 +x +π + + +≤ +< + + + + +6. +2 +2 +log +dy +x +y +x +x +dx + += +7. +2 +log +log +dy +x +x +y +x +dx +x ++ += +8. +(1 + x2) dy + 2xy dx = cot x dx (x ≠ 0) +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +329 +9. +cot +0 ( +0) +dy +x +y +x +xy +x +x +dx + +− ++ += +≠ +10. +( +) +1 +dy +x +y dx ++ += +11. +y dx + (x – y2) dy = 0 +12. +2 +( +3 +) +( +0) +dy +x +y +y +y +dx ++ += +> +. +For each of the differential equations given in Exercises 13 to 15, find a particular +solution satisfying the given condition: +13. +2 tan +sin ; +0 when +3 +dy +y +x +x y +x +dx +π ++ += += += +14. +2 +2 +1 +(1 +) +2 +; +0 when +1 +1 +dy +x +xy +y +x +dx +x ++ ++ += += += ++ +15. +3 cot +sin 2 ; +2 when +2 +dy +y +x +x y +x +dx +π +− += += += +16. +Find the equation of a curve passing through the origin given that the slope of the +tangent to the curve at any point (x, y) is equal to the sum of the coordinates of +the point. +17. +Find the equation of a curve passing through the point (0, 2) given that the sum of +the coordinates of any point on the curve exceeds the magnitude of the slope of +the tangent to the curve at that point by 5. +18. +The Integrating Factor of the differential equation +2 +2 +dy +x +y +x +dx − += + is +(A) e–x +(B) e–y +(C) 1 +x +(D) x +19. +The Integrating Factor of the differential equation +2 +(1 +) dx +y +yx +dy +− ++ + = +( 1 +1) +−< +< +ay +y + is +(A) +2 +1 +1 +y − +(B) +2 +1 +1 +y − +(C) +2 +1 +1 +y +− +(D) +2 +1 +1 +y +− +Miscellaneous Examples +Example 19 Verify that the function y = c1 eax cos bx + c2 eax sin bx, where c1, c2 are +arbitrary constants is a solution of the differential equation +( +) +2 +2 +2 +2 +2 +0 +d y +dy +a +a +b +y +dx +dx +− ++ ++ += +Reprint 2025-26 + +MATHEMATICS +330 +Solution The given function is +y = eax [c1 cosbx + c2 sinbx] +... (1) +Differentiating both sides of equation (1) with respect to x, we get +dy +dx = +[ +] [ +] +1 +2 +1 +2 +– +sin +cos +cos +sin +ax +ax +e +bc +bx +bc +bx +c +bx +c +bx e +a ++ ++ ++ +⋅ +or +dy +dx = +2 +1 +2 +1 +[( +)cos +( +)sin +] +ax +e +bc +ac +bx +ac +bc +bx ++ ++ +− +... (2) +Differentiating both sides of equation (2) with respect to x, we get +2 +2 +d y +dx + = +2 +1 +2 +1 +[( +) ( +sin +) +( +) ( cos +)] +ax +e +bc +ac +b +bx +ac +bc +b +bx ++ +− ++ +− ++ +2 +1 +2 +1 +[( +) cos +( +) sin +] +. +ax +bc +ac +bx +ac +bc +bx e +a ++ ++ +− += +2 +2 +2 +2 +2 +1 +2 +1 +2 +1 +[( +2 +) sin +( +2 +) cos +] +ax +e +a c +abc +b c +bx +a c +abc +b c +bx +− +− ++ ++ +− +Substituting the values of +2 +2 , +d y dy +dx +dx + and y in the given differential equation, we get +L.H.S. = +2 +2 +2 +2 +2 +1 +2 +1 +2 +1 +[ +2 +)sin +( +2 +)cos +] +ax +e +a c +abc +b c +bx +a c +abc +b c +bx +− +− ++ ++ +− +2 +1 +2 +1 +2 +[( +)cos +( +)sin +] +ax +ae +bc +ac +bx +ac +bc +bx +− ++ ++ +− +2 +2 +1 +2 +( +) +[ +cos +sin +] +ax +a +b +e +c +bx +c +bx ++ ++ ++ += +( +) +2 +2 +2 +2 +2 +2 +1 +2 +2 +1 +2 +2 +2 +2 +2 +2 +2 +1 +2 +1 +2 +1 +1 +1 +2 +2 +2 +sin +( +2 +2 +2 +)cos +ax +a c +abc +b c +a c +abc +a c +b c +bx +e +a c +abc +b c +abc +a c +a c +b c +bx + + +− +− +− ++ ++ ++ + + + + ++ ++ +− +− +− ++ ++ + + += +[0 sin +0cos +] +ax +e +bx +bx +× ++ += eax × 0 = 0 = R.H.S. +Hence, the given function is a solution of the given differential equation. +Example 20 Find the particular solution of the differential equation log +3 +4 +dy +x +y +dx + += ++ + + + + +given that y = 0 when x = 0. +Solution The given differential equation can be written as +dy +dx = e(3x + 4y) +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +331 +or +dy +dx = e3x . e4y +... (1) +Separating the variables, we get +4y +dy +e += e3x dx +Therefore +4y +e +dy +− +∫ += +3x +e dx +∫ +or +4 +4 +y +e− +− + = +3 +C +3 +x +e ++ +or +4 e3x + 3 e– 4y + 12 C = 0 +... (2) +Substituting x = 0 and y = 0 in (2), we get +4 + 3 + 12 C = 0 or C = +7 +12 +− +Substituting the value of C in equation (2), we get +4e3x + 3e– 4y – 7 = 0, +which is a particular solution of the given differential equation. +Example 21 Solve the differential equation +(x dy – y dx) y sin +y +x + + + + + + = (y dx + x dy) x cos +y +x + + + + + +. +Solution The given differential equation can be written as +2 +2 +sin +cos +cos +sin +y +y +y +y +x y +x +dy +xy +y +dx +x +x +x +x + + + + + + + + + + + + +− += ++ + + + + + + + + + + + + + + + + + + + + + + + + +or +dy +dx = +2 +2 +cos +sin +sin +cos +y +y +xy +y +x +x +y +y +xy +x +x +x + + + + ++ + + + + + + + + + + + + +− + + + + + + + + +Dividing numerator and denominator on RHS by x2, we get +dy +dx = +2 +2 +cos +sin +sin +cos +y +y +y +y +x +x +x +x +y +y +y +x +x +x + + + + + + ++ + + + + + + + + + + + + + + + + +− + + + + + + + + +... (1) +Reprint 2025-26 + +MATHEMATICS +332 +Clearly, equation (1) is a homogeneous differential equation of the form dy +y +g +dx +x + + += + + + +. +To solve it, we make the substitution +y = vx +... (2) +or +dy +dx = +dv +v +x dx ++ +or +dv +v +x dx ++ + = +2 +cos +sin +sin +cos +v +v +v +v +v +v +v ++ +− +(using (1) and (2)) +or +dv +x dx = +2 cos +sin +cos +v +v +v +v +v +− +or +sin +cos +cos +v +v +v +dv +v +v +− + + + + + + + = 2 dx +x +Therefore +sin +cos +cos +v +v +v dv +v +v +− + + + + + + +∫ + = +1 +2 +dx +x +∫ +or +1 +tanv dv +dv +v +− +∫ +∫ + = +1 +2 +dx +x +∫ +or +log sec +log | | +v +v +− + = +1 +2log | +| +log | C | +x + +or +2 +sec +log +v +v x + = log |C1| +or +2 +secv +v x = ± C1 +... (3) +Replacing v by y +x in equation (3), we get +2 +sec +( +) +y +x +y +x +x + + + + + + + + + + + + + = C where, C = ± C1 +or +sec y +x + + + + + + = C xy +which is the general solution of the given differential equation. +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +333 +Example 22 Solve the differential equation +(tan–1y – x) dy = (1 + y2) dx. +Solution The given differential equation can be written as +2 +1 +dx +x +dy +y ++ ++ + = +1 +2 +tan +1 +y +y +− ++ +... (1) +Now (1) is a linear differential equation of the form +1P +dx +dy ++ + x = Q1, +where, +P1 = +2 +1 +1 +y ++ + and +1 +1 +2 +tan +Q +1 +y +y +− += ++ +. +Therefore, +I .F = +1 +2 +1 +tan +1 +dy +y +y +e +e +− ++ +∫ += +Thus, the solution of the given differential equation is +1 +tan +y +xe +− + = +1 +1 +tan +2 +tan +C +1 +y +y +e +dy +y +− +− + + ++ + + ++ + + +∫ +... (2) +Let +I = +1 +1 +tan +2 +tan +1 +y +y +e +dy +y +− +− + + + + ++ + + +∫ +Substituting tan–1 y = t so that +2 +1 +1 +dy +dt +y + + += + + ++ + + +, we get +I = +t +t e dt +∫ += t et – ∫1 . et dt = t et – et = et (t – 1) +or +I = +1 +tan +y +e +− +(tan–1y –1) +Substituting the value of I in equation (2), we get +1 +1 +tan +tan +1 +. +(tan +1) +C +y +y +x e +e +y +− +− +− += +− ++ +or +x = +1 +1 +tan +(tan +1) +C +y +y +e +− +− +− +− ++ +which is the general solution of the given differential equation. +Miscellaneous Exercise on Chapter 9 +1. +For each of the differential equations given below, indicate its order and degree +(if defined). +Reprint 2025-26 + +MATHEMATICS +334 +(i) +2 +2 +2 +5 +6 +log +d y +dy +x +y +x +dx +dx + + ++ +− += + + + + + (ii) +3 +2 +4 +7 +sin +dy +dy +y +x +dx +dx + + + + +− ++ += + + + + + + + + +(iii) +4 +3 +4 +3 +sin +0 +d y +d y +dx +dx + + +− += + + + + +2. +For each of the exercises given below, verify that the given function (implicit or +explicit) is a solution of the corresponding differential equation. +(i) xy = a ex + b e–x + x2 +: +2 +2 +2 +2 +2 +0 +d y +dy +x +xy +x +dx +dx ++ +− ++ +− += +(ii) y = ex (a cos x + b sin x) +: +2 +2 +2 +2 +0 +d y +dy +y +dx +dx +− ++ += +(iii) y = x sin 3x +: +2 +2 +9 +6cos3 +0 +d y +y +x +dx ++ +− += +(iv) x2 = 2y2 log y +: +2 +2 +( +) +0 +dy +x +y +xy +dx ++ +− += +3. +Prove that x2 – y2 = c (x2 + y2)2 is the general solution of differential equation +(x3 – 3x y2) dx = (y3 – 3x2y) dy, where c is a parameter. +4. +Find the general solution of the differential equation +2 +2 +1 +0 +1 +dy +y +dx +x +− ++ += +− +. +5. +Show that the general solution of the differential equation +2 +2 +1 +0 +1 +dy +y +y +dx +x +x ++ ++ ++ += ++ ++ + is +given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter. +6. +Find the equation of the curve passing through the point 0, 4 +π + + + + + + whose differential +equation is sin x cos y dx + cos x sin y dy = 0. +7. +Find the particular solution of the differential equation +(1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0. +8. +Solve the differential equation +2 +( +0) +x +x +y +y +y e dx +x e +y +dy y + + + + += ++ +≠ + + +. +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +335 +9. +Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, +given that y = –1, when x = 0. (Hint: put x – y = t) +10. +Solve the differential equation +2 +1( +0) +x +e +y +dx +x +dy +x +x +− + + +− += +≠ + + + + +. +11. Find a particular solution of the differential equation +cot +dy +y +x +dx + + = 4x cosec x +(x ≠ 0), given that y = 0 when +2 +x +π += +. +12. +Find a particular solution of the differential equation (x + 1) dy +dx = 2 e–y – 1, given +that y = 0 when x = 0. +13. +The general solution of the differential equation +0 +y dx +x dy +y +− += + is +(A) xy = C +(B) x = Cy2 +(C) y = Cx +(D) y = Cx2 +14. +The general solution of a differential equation of the type +1 +1 +P +Q +dx +x +dy + += + is +(A) +( +) +1 +1 +P +P +1 +Q +C +dy +dy +y e +e +dy +∫ +∫ += ++ +∫ +(B) +( +) +1 +1 +P +P +1 +. +Q +C +dx +dx +y e +e +dx +∫ +∫ += ++ +∫ +(C) +( +) +1 +1 +P +P +1 +Q +C +dy +dy +x e +e +dy +∫ +∫ += ++ +∫ +(D) +( +) +1 +1 +P +P +1 +Q +C +dx +dx +x e +e +dx +∫ +∫ += ++ +∫ +15. +The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is +(A) x ey + x2 = C +(B) x ey + y2 = C +(C) y ex + x2 = C +(D) y ey + x2 = C +Reprint 2025-26 + +MATHEMATICS +336 +Summary +® An equation involving derivatives of the dependent variable with respect to +independent variable (variables) is known as a differential equation. +® Order of a differential equation is the order of the highest order derivative +occurring in the differential equation. +® Degree of a differential equation is defined if it is a polynomial equation in its +derivatives. +® Degree (when defined) of a differential equation is the highest power (positive +integer only) of the highest order derivative in it. +® A function which satisfies the given differential equation is called its solution. +The solution which contains as many arbitrary constants as the order of the +differential equation is called a general solution and the solution free from +arbitrary constants is called particular solution. +® Variable separable method is used to solve such an equation in which variables +can be separated completely i.e. terms containing y should remain with dy +and terms containing x should remain with dx. +® A differential equation which can be expressed in the form +( , ) or +( , ) +dy +dx +f x y +g x y +dx +dy += += +where, f (x, y) and g(x, y) are homogenous +functions of degree zero is called a homogeneous differential equation. +® A differential equation of the form ++P +Q +dy +y +dx += +, where P and Q are constants +or functions of x only is called a first order linear differential equation. +Historical Note +One of the principal languages of Science is that of differential equations. +Interestingly, the date of birth of differential equations is taken to be November, +11,1675, when Gottfried Wilthelm Freiherr Leibnitz (1646 - 1716) first put in black +and white the identity +2 +1 +2 +y dy +y += +∫ +, thereby introducing both the symbols ∫ and dy. +Leibnitz was actually interested in the problem of finding a curve whose tangents +were prescribed. This led him to discover the ‘method of separation of variables’ +1691. A year later he formulated the ‘method of solving the homogeneous +Reprint 2025-26 + +DIFFERENTIAL EQUATIONS +337 +—v— +differential equations of the first order’. He went further in a very short time +to the discovery of the ‘method of solving a linear differential equation of the +first-order’. How surprising is it that all these methods came from a single man +and that too within 25 years of the birth of differential equations! +In the old days, what we now call the ‘solution’ of a differential equation, was +used to be referred to as ‘integral’ of the differential equation, the word being +coined by James Bernoulli (1654 - 1705) in 1690. The word ‘solution was first +used by Joseph Louis Lagrange (1736 - 1813) in 1774, which was almost hundred +years since the birth of differential equations. It was Jules Henri Poincare +(1854 - 1912) who strongly advocated the use of the word ‘solution’ and thus the +word ‘solution’ has found its deserved place in modern terminology. The name of +the ‘method of separation of variables’ is due to John Bernoulli (1667 - 1748), +a younger brother of James Bernoulli. +Application to geometric problems were also considered. It was again John +Bernoulli who first brought into light the intricate nature of differential equations. +In a letter to Leibnitz, dated May 20, 1715, he revealed the solutions of the +differential equation +x2 y″ = 2y, +which led to three types of curves, viz., parabolas, hyperbolas and a class of +cubic curves. This shows how varied the solutions of such innocent looking +differential equation can be. From the second half of the twentieth century attention +has been drawn to the investigation of this complicated nature of the solutions of +differential equations, under the heading ‘qualitative analysis of differential +equations’. Now-a-days, this has acquired prime importance being absolutely +necessary in almost all investigations. +Reprint 2025-26" +class_12,10,Vector Algebra,ncert_books/class_12/lemh2dd/lemh204.pdf,"MATHEMATICS +338 +vIn most sciences one generation tears down what another has built and what +one has established another undoes. In Mathematics alone each generation +builds a new story to the old structure. – HERMAN HANKEL v +10.1 Introduction +In our day to day life, we come across many queries such +as – What is your height? How should a football player hit +the ball to give a pass to another player of his team? Observe +that a possible answer to the first query may be 1.6 meters, +a quantity that involves only one value (magnitude) which +is a real number. Such quantities are called scalars. +However, an answer to the second query is a quantity (called +force) which involves muscular strength (magnitude) and +direction (in which another player is positioned). Such +quantities are called vectors. In mathematics, physics and +engineering, we frequently come across with both types of +quantities, namely, scalar quantities such as length, mass, +time, distance, speed, area, volume, temperature, work, +money, voltage, density, resistance etc. and vector quantities like displacement, velocity, +acceleration, force, weight, momentum, electric field intensity etc. +In this chapter, we will study some of the basic concepts about vectors, various +operations on vectors, and their algebraic and geometric properties. These two type of +properties, when considered together give a full realisation to the concept of vectors, +and lead to their vital applicability in various areas as mentioned above. +10.2 Some Basic Concepts +Let ‘l’ be any straight line in plane or three dimensional space. This line can be given +two directions by means of arrowheads. A line with one of these directions prescribed +is called a directed line (Fig 10.1 (i), (ii)). +Chapter 10 +VECTOR ALGEBRA +W.R. Hamilton +(1805-1865) +Reprint 2025-26 + +VECTOR ALGEBRA +339 +Now observe that if we restrict the line l to the line segment AB, then a magnitude +is prescribed on the line l with one of the two directions, so that we obtain a directed +line segment (Fig 10.1(iii)). Thus, a directed line segment has magnitude as well as +direction. +Definition 1 A quantity that has magnitude as well as direction is called a vector. +Notice that a directed line segment is a vector (Fig 10.1(iii)), denoted as + or +simply as , and read as ‘vector +’ or ‘vector ’. +The point A from where the vector + starts is called its initial point, and the +point B where it ends is called its terminal point. The distance between initial and +terminal points of a vector is called the magnitude (or length) of the vector, denoted as +| +|, or | |, or a. The arrow indicates the direction of the vector. +ANote Since the length is never negative, the notation | | < 0 has no meaning. +Position Vector +From Class XI, recall the three dimensional right handed rectangular coordinate system +(Fig 10.2(i)). Consider a point P in space, having coordinates (x, y, z) with respect to +the origin O(0, 0, 0). Then, the vector + having O and P as its initial and terminal +points, respectively, is called the position vector of the point P with respect +to O. Using distance formula (from Class XI), the magnitude of + (or ) is given by +| +|= +2 +2 +2 +x +y +z ++ ++ +In practice, the position vectors of points A, B, C, etc., with respect to the origin O +are denoted by , , , etc., respectively (Fig 10.2 (ii)). +Fig 10.1 +Reprint 2025-26 + + MATHEMATICS +340 +A +O +P +a +90° +X +Y +Z +X +A +O +B +P( +) +x,y,z +C +a +b +g +P( +) +x,y,z +r +x +y +z +Direction Cosines +Consider the position vector + of a point P(x, y, z) as in Fig 10.3. The angles α, +β, γ made by the vector with the positive directions of x, y and z-axes respectively, +are called its direction angles. The cosine values of these angles, i.e., cosα, cosβ and +cos γ are called direction cosines of the vector , and usually denoted by l, m and n, +respectively. +Fig 10.3 +From Fig 10.3, one may note that the triangle OAP is right angled, and in it, we +have +. Similarly, from the right angled triangles OBP and +OCP, we may write cos + and cos +y +z +r +r +β = +γ = +. Thus, the coordinates of the point P may +also be expressed as (lr, mr,nr). The numbers lr, mr and nr, proportional to the direction +cosines are called as direction ratios of vector , and denoted as a, b and c, respectively. +Fig 10.2 +Reprint 2025-26 + +VECTOR ALGEBRA +341 +ANote One may note that l2 + m2 + n2 = 1 but a2 + b2 + c2 ≠ 1, in general. +10.3 Types of Vectors +Zero Vector A vector whose initial and terminal points coincide, is called a zero vector +(or null vector), and denoted as . Zero vector can not be assigned a definite direction +as it has zero magnitude. Or, alternatively otherwise, it may be regarded as having any +direction. The vectors + represent the zero vector, +Unit Vector A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. The +unit vector in the direction of a given vector is denoted by ˆa . +Coinitial Vectors Two or more vectors having the same initial point are called coinitial +vectors. +Collinear Vectors Two or more vectors are said to be collinear if they are parallel to +the same line, irrespective of their magnitudes and directions. +Equal Vectors Two vectors + are said to be equal, if they have the same +magnitude and direction regardless of the positions of their initial points, and written +as +. +Negative of a Vector A vector whose magnitude is the same as that of a given vector +(say, +), but direction is opposite to that of it, is called negative of the given vector. +For example, vector + is negative of the vector +, and written as + = – +. +Remark The vectors defined above are such that any of them may be subject to its +parallel displacement without changing its magnitude and direction. Such vectors are +called free vectors. Throughout this chapter, we will be dealing with free vectors only. +Example 1 Represent graphically a displacement +of 40 km, 30° west of south. +Solution The vector + represents the required +displacement (Fig 10.4). +Example 2 Classify the following measures as +scalars and vectors. +(i) 5 seconds +(ii) 1000 cm3 +Fig 10.4 +Reprint 2025-26 + + MATHEMATICS +342 +Fig 10.5 +(iii) 10 Newton +(iv) 30 km/hr +(v) 10 g/cm3 +(vi) 20 m/s towards north +Solution +(i) Time-scalar +(ii) Volume-scalar +(iii) Force-vector +(iv) Speed-scalar +(v) Density-scalar +(vi) Velocity-vector +Example 3 In Fig 10.5, which of the vectors are: +(i) Collinear +(ii) Equal +(iii) Coinitial +Solution +(i) Collinear vectors : +. +(ii) Equal vectors : +(iii) Coinitial vectors : +EXERCISE 10.1 +1. +Represent graphically a displacement of 40 km, 30° east of north. +2. +Classify the following measures as scalars and vectors. +(i) 10 kg +(ii) 2 meters north-west +(iii) 40° +(iv) 40 watt +(v) 10–19 coulomb +(vi) 20 m/s2 +3. +Classify the following as scalar and vector quantities. +(i) time period +(ii) distance +(iii) force +(iv) velocity +(v) work done +4. +In Fig 10.6 (a square), identify the following vectors. +(i) Coinitial +(ii) Equal +(iii) Collinear but not equal +5. +Answer the following as true or false. +(i) + and – are collinear. +(ii) Two collinear vectors are always equal in +magnitude. +(iii) Two vectors having same magnitude are collinear. +(iv) Two collinear vectors having the same magnitude are equal. +Fig 10.6 +Reprint 2025-26 + +VECTOR ALGEBRA +343 +10.4 Addition of Vectors +A vector + simply means the displacement from a point +A to the point B. Now consider a situation that a girl +moves from A to B and then from B to C +(Fig 10.7). The net displacement made by the girl from +point A to the point C, is given by the vector + and +expressed as + = +This is known as the triangle law of vector addition. +In general, if we have two vectors and (Fig 10.8 (i)), then to add them, they are +positioned so that the initial point of one coincides with the terminal point of the other +(Fig 10.8(ii)). +Fig 10.7 +a +b +a +b +(i) +(iii) +A +C +a +b +(ii) +a +b ++ +A +C +B +B +a +b +– +–b +C’ +Fig 10.8 +For example, in Fig 10.8 (ii), we have shifted vector without changing its magnitude +and direction, so that it’s initial point coincides with the terminal point of . Then, the +vector + , represented by the third side AC of the triangle ABC, gives us the sum (or +resultant) of the vectors and i.e., in triangle ABC (Fig 10.8 (ii)), we have + = +Now again, since +, from the above equation, we have +This means that when the sides of a triangle are taken in order, it leads to zero +resultant as the initial and terminal points get coincided (Fig 10.8(iii)). +Reprint 2025-26 + + MATHEMATICS +344 +Now, construct a vector + so that its magnitude is same as the vector +, but the +direction opposite to that of it (Fig 10.8 (iii)), i.e., + = +Then, on applying triangle law from the Fig 10.8 (iii), we have + = + +The vector + is said to represent the difference of +. +Now, consider a boat in a river going from one bank of the river to the other in a +direction perpendicular to the flow of the river. Then, it is acted upon by two velocity +vectors–one is the velocity imparted to the boat by its engine and other one is the +velocity of the flow of river water. Under the simultaneous influence of these two +velocities, the boat in actual starts travelling with a different velocity. To have a precise +idea about the effective speed and direction +(i.e., the resultant velocity) of the boat, we have +the following law of vector addition. +If we have two vectors + represented +by the two adjacent sides of a parallelogram +in magnitude and direction (Fig 10.9), then their +sum + is represented in magnitude and +direction by the diagonal of the parallelogram +through their common point. This is known as +the parallelogram law of vector addition. +ANote From Fig 10.9, using the triangle law, one may note that + = +or + = + (since + ) +which is parallelogram law. Thus, we may say that the two laws of vector +addition are equivalent to each other. +Properties of vector addition +Property 1 For any two vectors +, + = +(Commutative property) +Fig 10.9 +Reprint 2025-26 + +VECTOR ALGEBRA +345 +Proof Consider the parallelogram ABCD +(Fig 10.10). Let + then using +the triangle law, from triangle ABC, we have +Now, since the opposite sides of a +parallelogram are equal and parallel, from +Fig 10.10, we have, + and +. Again using triangle law, from +triangle ADC, we have +Hence + = +Property 2 For any three vectors +, +and +a b +c + + + + = +(Associative property) +Proof Let the vectors + be represented by +, respectively, as +shown in Fig 10.11(i) and (ii). +Fig 10.11 +Then + = +and + = +So + = +Fig 10.10 +Reprint 2025-26 + + MATHEMATICS +346 +a +a +1 +2 +1 +2 +a +–2 +a +a +2 +and + = +Hence + = +Remark The associative property of vector addition enables us to write the sum of +three vectors + without using brackets. +Note that for any vector a , we have + = +Here, the zero vector is called the additive identity for the vector addition. +10.5 Multiplication of a Vector by a Scalar +Let be a given vector and λ a scalar. Then the product of the vector by the scalar +λ, denoted as λ , is called the multiplication of vector by the scalar λ. Note that, λ +is also a vector, collinear to the vector . The vector λ has the direction same (or +opposite) to that of vector according as the value of λ is positive (or negative). Also, +the magnitude of vector λ is |λ| times the magnitude of the vector , i.e., +|λ | = |λ|| | +A geometric visualisation of multiplication of a vector by a scalar is given +in Fig 10.12. +Fig 10.12 +When λ = –1, then λ = – , which is a vector having magnitude equal to the +magnitude of and direction opposite to that of the direction of . The vector – is +called the negative (or additive inverse) of vector and we always have + + (– ) = (– ) + = +Also, if +1 += | +| +a +λ + , provided ≠ 0 i.e. is not a null vector, then +|λ | =|λ|| | = +Reprint 2025-26 + +VECTOR ALGEBRA +347 +So, λ represents the unit vector in the direction of . We write it as +ˆa = +1 +| +| +r +r a +a +ANote For any scalar k, 0 = 0. +r +r +k +10.5.1 Components of a vector +Let us take the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x-axis, y-axis and +z-axis, respectively. Then, clearly +| +|= 1, | + |= 1 and | +|= 1 +The vectors +, each having magnitude 1, +are called unit vectors along the axes OX, OY and OZ, +respectively, and denoted by +ˆ +ˆ ˆ +, and +i j +k , respectively +(Fig 10.13). +Now, consider the position vector + of a point P(x, y, z) +as in Fig 10.14. Let P1 be the foot of the perpendicular from P on the plane XOY. +We, thus, see that P1 P is parallel to z-axis. As +ˆ +ˆ ˆ +, and +i j +k are the unit vectors along the +x, y and z-axes, respectively, and by the definition of the coordinates of P, we have +. Similarly, + and +. +Fig 10.13 +Fig 10.14 +Reprint 2025-26 + + MATHEMATICS +348 +Therefore, it follows that + = +and + = +Hence, the position vector of P with reference to O is given by + = +This form of any vector is called its component form. Here, x, y and z are called +as the scalar components of , and +ˆ +ˆ +ˆ +, + and +xi +yj +zk are called the vector components +of along the respective axes. Sometimes x, y and z are also termed as rectangular +components. +The length of any vector = +ˆ +ˆ +ˆ +xi +yj +zk ++ ++ +, is readily determined by applying the +Pythagoras theorem twice. We note that in the right angle triangle OQP1 (Fig 10.14) + = +, +and in the right angle triangle OP1P, we have + = +Hence, the length of any vector = +ˆ +ˆ +ˆ+ +xi +yj +zk += ++ + is given by +| | = +If + are any two vectors given in the component form +1 +2 +3 ˆ +ˆ +ˆ+ +a i +a j +a k ++ + and +1 +2 +3 ˆ +ˆ +ˆ +b i +b j +b k ++ ++ +, respectively, then +(i) +the sum (or resultant) of the vectors + is given by + = +1 +1 +2 +2 +3 +3 ˆ +ˆ +ˆ +( +) +( +) +( +) +a +b i +a +b +j +a +b k ++ ++ ++ ++ ++ +(ii) +the difference of the vector + is given by + = +1 +1 +2 +2 +3 +3 ˆ +ˆ +ˆ +( +) +( +) +( +) +a +b i +a +b +j +a +b k +− ++ +− ++ +− +(iii) +the vectors + are equal if and only if +a1 = b1, a2 = b2 and a3 = b3 +(iv) +the multiplication of vector by any scalar λ is given by +λ = +1 +2 +3 ˆ +ˆ +ˆ +( +) +( +) +( +) +a i +a +j +a k +λ ++ λ ++ λ +Reprint 2025-26 + +VECTOR ALGEBRA +349 +The addition of vectors and the multiplication of a vector by a scalar together give +the following distributive laws: +Let + be any two vectors, and k and m be any scalars. Then +(i) +(ii) +(iii) +Remarks +(i) +One may observe that whatever be the value of λ, the vector λ is always +collinear to the vector . In fact, two vectors + are collinear if and only if +there exists a nonzero scalar λ such that +. If the vectors + are given +in the component form, i.e. = +1 +2 +3 ˆ +ˆ +ˆ +a i +a j +a k ++ ++ + and +, then the +two vectors are collinear if and only if +1 +2 +3 ˆ +ˆ +ˆ +b i +b j +b k ++ ++ + = +1 +2 +3 ˆ +ˆ +ˆ +( +) +a i +a j +a k +λ ++ ++ +⇔ +1 +2 +3 ˆ +ˆ +ˆ +b i +b j +b k ++ ++ + = +1 +2 +3 ˆ +ˆ +ˆ +( +) +( +) +( +) +a i +a +j +a k +λ ++ λ ++ λ +⇔ +1 +1 +b +a += λ +, +2 +2 +3 +3 +, +b +a +b +a += λ += λ +⇔ +1 +1 +b +a = +3 +2 +2 +3 +b +b +a +a += += λ +(ii) +If = +1 +2 +3 ˆ +ˆ +ˆ +a i +a j +a k ++ ++ +, then a1, a2, a3 are also called direction ratios of . +(iii) +In case if it is given that l, m, n are direction cosines of a vector, then +ˆ +ˆ +ˆ +li +mj +nk ++ ++ += +ˆ +ˆ +ˆ +(cos ) +(cos ) +(cos ) +i +j +k +α ++ +β ++ +γ + is the unit vector in the direction of that vector, +where α, β and γ are the angles which the vector makes with x, y and z axes +respectively. +Example 4 Find the values of x, y and z so that the vectors + and + are equal. +Solution Note that two vectors are equal if and only if their corresponding components +are equal. Thus, the given vectors + will be equal if and only if +x = 2, y = 2, z = 1 +Reprint 2025-26 + + MATHEMATICS +350 +Example 5 Let + and +. Is +? Are the vectors + equal? +Solution We have + and +So, +. But, the two vectors are not equal since their corresponding components +are distinct. +Example 6 Find unit vector in the direction of vector +Solution The unit vector in the direction of a vector is given by +. +Now + = +2 +2 +2 +2 +3 +1 +14 ++ ++ += +Therefore +1 +ˆ +ˆ +ˆ +ˆ +(2 +3 +) +14 +a +i +j +k += ++ ++ + = +2 +3 +1 +ˆ +ˆ +ˆ +14 +14 +14 +i +j +k ++ ++ +Example 7 Find a vector in the direction of vector + that has magnitude +7 units. +Solution The unit vector in the direction of the given vector is + = +1 +1 +2 +ˆ +ˆ +ˆ +ˆ +( +2 ) +5 +5 +5 +i +j +i +j +− += +− +Therefore, the vector having magnitude equal to 7 and in the direction of is +7a +∧ = +1 +2 +7 +5 +5 +i +j +∧ +∧ + + +− + + + + + = +7 +14 +ˆ +ˆ +5 +5 +i +j +− +Example 8 Find the unit vector in the direction of the sum of the vectors, +and +. +Solution The sum of the given vectors is +and + = +2 +2 +2 +4 +3 +( 2) +29 ++ ++ − += +Reprint 2025-26 + +VECTOR ALGEBRA +351 +Thus, the required unit vector is + +1 +4 +3 +2 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +(4 +3 +2 ) +29 +29 +29 +29 +i +j +k +i +j +k += ++ +− += ++ +− +Example 9 Write the direction ratio’s of the vector + and hence calculate +its direction cosines. +Solution Note that the direction ratio’s a, b, c of a vector + are just +the respective components x, y and z of the vector. So, for the given vector, we have +a = 1, b = 1 and c = –2. Further, if l, m and n are the direction cosines of the given +vector, then +Thus, the direction cosines are +1 +1 +2 +, +,– +6 +6 +6 + + + + + + +. +10.5.2 Vector joining two points +If P1(x1, y1, z1) and P2(x2, y2, z2) are any two +points, then the vector joining P1 and P2 is the +vector + (Fig 10.15). +Joining the points P1 and P2 with the origin +O, and applying triangle law, from the triangle +OP1P2, we have + = +Using the properties of vector addition, the +above equation becomes + = +i.e. + + = +2 +2 +2 +1 +1 +1 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +( +) +( +) +x i +y j +z k +x i +y j +z k ++ ++ +− ++ ++ += +2 +1 +2 +1 +2 +1 ˆ +ˆ +ˆ +( +) +( +) +( +) +x +x i +y +y +j +z +z k +− ++ +− ++ +− +The magnitude of vector + is given by +| +| = +2 +2 +2 +2 +1 +2 +1 +2 +1 +( +) +( +) +( +) +x +x +y +y +z +z +− ++ +− ++ +− +Fig 10.15 +Reprint 2025-26 + + MATHEMATICS +352 +Example 10 Find the vector joining the points P(2, 3, 0) and Q(– 1, – 2, – 4) directed +from P to Q. +Solution Since the vector is to be directed from P to Q, clearly P is the initial point +and Q is the terminal point. So, the required vector joining P and Q is the vector +, +given by + = +ˆ +ˆ +ˆ +( 1 +2) +( 2 +3) +( 4 +0) +i +j +k +−− ++ −− ++ −− +i.e. + = +ˆ +ˆ +ˆ +3 +5 +4 . +i +j +k +− +− +− +10.5.3 Section formula +Let P and Q be two points represented by the position vectors +, respectively, +with respect to the origin O. Then the line segment +joining the points P and Q may be divided by a third +point, say R, in two ways – internally (Fig 10.16) +and externally (Fig 10.17). Here, we intend to find +the position vector + for the point R with respect +to the origin O. We take the two cases one by one. +Case I When R divides PQ internally (Fig 10.16). +If R divides + such that + = +, +where m and n are positive scalars, we say that the point R divides + internally in the +ratio of m : n. Now from triangles ORQ and OPR, we have + = +and + = +, +Therefore, we have + = + (Why?) +or + = +(on simplification) +Hence, the position vector of the point R which divides P and Q internally in the +ratio of m : n is given by + = +Fig 10.16 +Reprint 2025-26 + +VECTOR ALGEBRA +353 +Case II When R divides PQ externally (Fig 10.17). +We leave it to the reader as an exercise to verify +that the position vector of the point R which divides +the line segment PQ externally in the ratio +m : n +PR +i.e. QR + + += + + + + +m +n + is given by + = +Remark If R is the midpoint of PQ , then m = n. And therefore, from Case I, the +midpoint R of +, will have its position vector as + = +Example 11 Consider two points P and Q with position vectors + and +. Find the position vector of a point R which divides the line joining P and Q +in the ratio 2:1, (i) internally, and (ii) externally. +Solution +(i) +The position vector of the point R dividing the join of P and Q internally in the +ratio 2:1 is + = +(ii) +The position vector of the point R dividing the join of P and Q externally in the +ratio 2:1 is + = +Example 12 Show that the points +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +A(2 +), B( +3 +5 ), C(3 +4 +4 ) +i +j +k +i +j +k +i +j +k +− ++ +− +− +− +− + are +the vertices of a right angled triangle. +Solution We have + + = +ˆ +ˆ +ˆ +(1 +2) +( 3 1) +( 5 1) +i +j +k +− ++ −+ ++ −− +ˆ +ˆ +ˆ +2 +6 +i +j +k += −− +− + + = +ˆ +ˆ +ˆ +(3 1) +( 4 +3) +( 4 +5) +i +j +k +− ++ −+ ++ −+ +ˆ +ˆ +ˆ +2i +j +k += +− ++ +and + = +ˆ +ˆ +ˆ +(2 +3) +( 1 +4) +(1 +4) +i +j +k +− ++ −+ ++ ++ + +ˆ +ˆ +ˆ +3 +5 +i +j +k += −+ ++ +Fig 10.17 +Reprint 2025-26 + + MATHEMATICS +354 +Further, note that + = +Hence, the triangle is a right angled triangle. +EXERCISE 10.2 +1. +Compute the magnitude of the following vectors: + = ˆ +ˆ +; +i +j +k ++ ++ + = +ˆ +ˆ +ˆ +2 +7 +3 ; +i +j +k +− +− + = +1 +1 +1 ˆ +ˆ +ˆ +3 +3 +3 +i +j +k ++ +− +2. +Write two different vectors having same magnitude. +3. +Write two different vectors having same direction. +4. +Find the values of x and y so that the vectors ˆ +ˆ +ˆ +ˆ +2 +3 and +i +j +xi +yj ++ ++ + are equal. +5. +Find the scalar and vector components of the vector with initial point (2, 1) and +terminal point (– 5, 7). +6. +Find the sum of the vectors = +ˆ +ˆ +ˆ +2 +, +i +j +k +− ++ + = +ˆ +ˆ +ˆ +2 +4 +5 +i +j +k +− ++ ++ +and = +ˆ +ˆ +ˆ +6 – 7 +i +j +k +− +. +7. +Find the unit vector in the direction of the vector = +ˆ +ˆ +ˆ +2 +i +j +k ++ ++ +. +8. +Find the unit vector in the direction of vector +, where P and Q are the points +(1, 2, 3) and (4, 5, 6), respectively. +9. +For given vectors, = +ˆ +ˆ +ˆ +2 +2 +i +j +k +− ++ +and = +ˆ +ˆ +ˆ +i +j +k +−+ +− +, find the unit vector in the +direction of the vector +. +10. +Find a vector in the direction of vector +ˆ +ˆ +ˆ +5 +2 +i +j +k +− ++ + which has magnitude 8 units. +11. +Show that the vectors +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +2 +3 +4 and +4 +6 +8 +i +j +k +i +j +k +− ++ +− ++ +− + are collinear. +12. +Find the direction cosines of the vector +ˆ +ˆ +ˆ +2 +3 +i +j +k ++ ++ +. +13. +Find the direction cosines of the vector joining the points A (1, 2, –3) and +B (–1, –2, 1), directed from A to B. +14. +Show that the vector +ˆ +ˆ +ˆ +i +j +k ++ ++ + is equally inclined to the axes OX, OY and OZ. +15. +Find the position vector of a point R which divides the line joining two points P +and Q whose position vectors are +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +2 + and – +i +j +k +i +j +k ++ +− ++ ++ + respectively, in the +ratio 2 : 1 +(i) internally +(ii) externally +Reprint 2025-26 + +VECTOR ALGEBRA +355 +16. +Find the position vector of the mid point of the vector joining the points P(2, 3, 4) +and Q(4, 1, –2). +17. +Show that the points A, B and C with position vectors, = +ˆ +ˆ +ˆ +3 +4 +4 , +i +j +k +− +− + = +ˆ +ˆ +ˆ +2i +j +k +− ++ + and = +ˆ +ˆ +ˆ +3 +5 +i +j +k +− +− +, respectively form the vertices of a right angled +triangle. +18. +In triangle ABC (Fig 10.18), which of the following is not true: +(A) +(B) +(C) +(D) +19. +If + are two collinear vectors, then which of the following are incorrect: +(A) +(B) +(C) the respective components of + are not proportional +(D) both the vectors + have same direction, but different magnitudes. +10.6 Product of Two Vectors +So far we have studied about addition and subtraction of vectors. An other algebraic +operation which we intend to discuss regarding vectors is their product. We may +recall that product of two numbers is a number, product of two matrices is again a +matrix. But in case of functions, we may multiply them in two ways, namely, +multiplication of two functions pointwise and composition of two functions. Similarly, +multiplication of two vectors is also defined in two ways, namely, scalar (or dot) +product where the result is a scalar, and vector (or cross) product where the +result is a vector. Based upon these two types of products for vectors, they have +found various applications in geometry, mechanics and engineering. In this section, +we will discuss these two types of products. +10.6.1 Scalar (or dot) product of two vectors +Definition 2 The scalar product of two nonzero vectors +, denoted by +, is +Fig 10.18 +Reprint 2025-26 + + MATHEMATICS +356 +defined as + = +where, θ is the angle between + (Fig 10.19). +If either + then θ is not defined, and in this case, we +define +Observations +1. + is a real number. +2. +Let +be two nonzero vectors, then + if and only if +are +perpendicular to each other. i.e. +3. +If θ = 0, then +In particular, + as θ in this case is 0. +4. +If θ = π, then +| +| | +| +⋅ += − +r +r +r +r +a b +a +b +In particular, +, as θ in this case is π. +5. +In view of the Observations 2 and 3, for mutually perpendicular unit vectors +ˆ +ˆ ˆ +, +and +, +i +j +k we have +ˆ ˆ +ˆ ˆ +i i +j j +⋅= +⋅ + = ˆ ˆ +1, +k k +⋅ += +ˆ +ˆ ˆ +ˆ +i +j +j k +⋅ += +⋅ + = ˆ ˆ +0 +k i⋅= +6. +The angle between two nonzero vectors +is given by + or +7. +The scalar product is commutative. i.e. + (Why?) +Two important properties of scalar product +Property 1 (Distributivity of scalar product over addition) Let + be +any three vectors, then +Fig 10.19 +Reprint 2025-26 + +VECTOR ALGEBRA +357 +(i) +B +C +A +l +B +l +A +C +(ii) +A +B +C +l +(iv) +l +C +B +A +(iii) +θ +θ +θ +θ +p +p +p +p +a +a +a +a +(90 < < 180 ) +0 +0 +θ +(0 < < 90 ) +0 +0 +θ +(270 < < 360 ) +0 +0 +θ +(180 < < 270 ) +0 +0 +θ +Property 2 Let + be any two vectors, and l be any scalar. Then +If two vectors +are given in component form as +1 +2 +3 ˆ +ˆ +ˆ +a i +a j +a k ++ ++ + and +1 +2 +3 ˆ +ˆ +ˆ +b i +b j +b k ++ ++ +, then their scalar product is given as + = +1 +2 +3 +1 +2 +3 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +( +) ( +) +a i +a j +a k +b i +b j +b k ++ ++ +⋅ ++ ++ += +1 +1 +2 +3 +2 +1 +2 +3 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +( +) +( +) +a i +b i +b j +b k +a j +b i +b j +b k +⋅ ++ ++ ++ +⋅ ++ ++ + + +3 +1 +2 +3 +ˆ +ˆ +ˆ +ˆ +( +) +a k +b i +b j +b k +⋅ ++ ++ += +1 1 +1 2 +1 3 +2 1 +2 2 +2 3 +ˆ +ˆ +ˆ ˆ +ˆ ˆ +ˆ +ˆ ˆ +ˆ ˆ +ˆ +( +) +( +) +( +) +( +) +( +) +( +) +a b i i +a b i +j +a b i k +a b +j i +a b +j j +a b +j k +⋅ ++ +⋅ ++ +⋅ ++ +⋅ ++ +⋅ ++ +⋅ + + +3 1 +3 2 +3 3 +ˆ +ˆ +ˆ ˆ +ˆ +ˆ +( +) +( +) +( +) +a b k i +a b k +j +a b k k +⋅ ++ +⋅ ++ +⋅ +(Using the above Properties 1 and 2) += a1b1 + a2b2 + a3b3 +(Using Observation 5) +Thus + = +1 1 +2 2 +3 3 +a b +a b +a b ++ ++ +10.6.2 Projection of a vector on a line +Suppose a vector + makes an angle θ with a given directed line l (say), in the +anticlockwise direction (Fig 10.20). Then the projection of + on l is a vector +(say) with magnitude +, and the direction of being the same (or opposite) +to that of the line l, depending upon whether cosθ is positive or negative. The vector +Fig 10.20 +Reprint 2025-26 + + MATHEMATICS +358 +is called the projection vector, and its magnitude | | is simply called as the projection +of the vector + on the directed line l. +For example, in each of the following figures (Fig 10.20(i) to (iv)), projection vector +of + along the line l is vector +. +Observations +1. +If ˆp is the unit vector along a line l, then the projection of a vector on the line +l is given by +ˆp +⋅ +. +2. +Projection of a vector on other vector r +b, is given by +ˆ,b +⋅ + or +3. +If θ = 0, then the projection vector of + will be + itself and if θ = π, then the +projection vector of + will be +. +4. +If += 2 +π +θ + or +3 += 2 +π +θ +, then the projection vector of + will be zero vector. +Remark If α, β and γ are the direction angles of vector +1 +2 +3 ˆ +ˆ +ˆ +a i +a j +a k += ++ ++ +, then its +direction cosines may be given as +Also, note that + are respectively the projections of + along OX, OY and OZ. i.e., the scalar components a1, a2 and a3 of the vector , are +precisely the projections of along x-axis, y-axis and z-axis, respectively. Further, if +is a unit vector, then it may be expressed in terms of its direction cosines as +ˆ +ˆ +ˆ +cos +cos +cos +i +j +k += +α + +β + +γ +Example 13 Find the angle between two vectors +with magnitudes 1 and 2 +respectively and when + =1. +Solution Given +. We have +Reprint 2025-26 + +VECTOR ALGEBRA +359 +Example 14 Find angle ‘θ’ between the vectors +. +Solution The angle θ between two vectors +is given by +cosθ = +Now + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +( +) ( +) +1 1 1 +1 +i +j +k +i +j +k ++ +− +⋅ +− ++ += −−= −. +Therefore, we have + cosθ = +1 +3 +− +hence the required angle is +θ = +Example 15 If +, then show that the vectors + are perpendicular. +Solution We know that two nonzero vectors are perpendicular if their scalar product +is zero. +Here + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +(5 +3 ) +( +3 +5 ) +6 +2 +8 +i +j +k +i +j +k +i +j +k +− +− ++ ++ +− += ++ +− +and + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +(5 +3 ) +( +3 +5 ) +4 +4 +2 +i +j +k +i +j +k +i +j +k +− +− +− ++ +− += +− ++ +So + ( +) . ( +) +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +(6 +2 +8 ) (4 +4 +2 ) +24 +8 +16 +0. +i +j +k +i +j +k += ++ +− +⋅ +− ++ += +− +− += +Hence + are perpendicular vectors. +Example 16 Find the projection of the vector +ˆ +ˆ +ˆ +2 +3 +2 +i +j +k += ++ ++ +on the vector +ˆ +ˆ +ˆ +2 +i +j +k += ++ ++ +. +Solution The projection of vector on the vector +r +b is given by + = +2 +2 +2 +(2 1 +3 2 +2 1) +10 +5 +6 +3 +6 +(1) +(2) +(1) +× + × ++ × += += ++ ++ +Example 17 Find +, if two vectors +are such that +and +. +Solution We have += +Reprint 2025-26 + + MATHEMATICS +360 +B +C +A +a +b ++ +a +b += += +2 +2 +(2) +2(4) +(3) +− ++ +Therefore + = +5 +Example 18 If is a unit vector and +, then find +. +Solution Since is a unit vector, +. Also, + +or + = 8 +or +Therefore + = 3 (as magnitude of a vector is non negative). +Example 19 For any two vectors +, we always have + (Cauchy- +Schwartz inequality). +Solution The inequality holds trivially when either +0 or +0 += += +r +r +r +ra +b +. Actually, in such a +situation we have +. So, let us assume that +. +Then, we have + = | cos | +1 +θ ≤ +Therefore +Example 20 For any two vectors +, we always +have +(triangle inequality). +Solution The inequality holds trivially in case either + (How?). So, let +. Then, + += += + (scalar product is commutative) +≤ +(since +| +| +x +x +x +≤ +∀∈R ) +≤ +(from Example 19) += +Fig 10.21 +Reprint 2025-26 + +VECTOR ALGEBRA +361 +Hence +Remark If the equality holds in triangle inequality (in the above Example 20), i.e. + = +, +then + = +showing that the points A, B and C are collinear. +Example 21 Show that the points +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +A( 2 +3 +5 ), B( +2 +3 ) +i +j +k +i +j +k +− ++ ++ ++ ++ + and +ˆ +ˆ +C(7 +) +i +k +− +are collinear. +Solution We have + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +(1 +2) +(2 +3) +(3 +5) +3 +2 +i +j +k +i +j +k ++ ++ +− ++ +− += +− +− +, + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +(7 +1) +(0 +2) +( 1 +3) +6 +2 +4 +i +j +k +i +j +k +− ++ +− ++ −− += +− +− +, + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +(7 +2) +(0 +3) +( 1 +5) +9 +3 +6 +i +j +k +i +j +k ++ ++ +− ++ −− += +− +− + = +Therefore +Hence the points A, B and C are collinear. +ANote In Example 21, one may note that although + but the +points A, B and C do not form the vertices of a triangle. +EXERCISE 10.3 +1. +Find the angle between two vectors +with magnitudes +3 and 2 , +respectively having +. +2. +Find the angle between the vectors +ˆ +ˆ +ˆ +2 +3 +i +j +k +− ++ + and +ˆ +ˆ +ˆ +3 +2 +i +j +k +− ++ +3. +Find the projection of the vector ˆ +ˆ +i +j +− + on the vector ˆ +ˆ +i +j ++ +. +4. +Find the projection of the vector +ˆ +ˆ +ˆ +3 +7 +i +j +k ++ ++ + on the vector +ˆ +ˆ +ˆ +7 +8 +i +j +k +− ++ +. +5. +Show that each of the given three vectors is a unit vector: +1 +1 +1 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +(2 +3 +6 ), +(3 +6 +2 ), +(6 +2 +3 ) +7 +7 +7 +i +j +k +i +j +k +i +j +k ++ ++ +− ++ ++ +− +Also, show that they are mutually perpendicular to each other. +Reprint 2025-26 + + MATHEMATICS +362 +6. +Find +, if +. +7. +Evaluate the product +. +8. +Find the magnitude of two vectors +, having the same magnitude and such +that the angle between them is 60o and their scalar product is 1 +2 +. +9. +Find +, if for a unit vector , +. +10. +If +are such that +is +perpendicular to , then find the value of λ. +11. +Show that + is perpendicular to +, for any two nonzero +vectors +. +12. +If +, then what can be concluded about the vector ? +13. +If +are unit vectors such that +, find the value of +. +14. +If either vector +. But the converse need not be +true. Justify your answer with an example. +15. +If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), +respectively, then find ∠ABC. [∠ABC is the angle between the vectors + and +]. +16. +Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear. +17. +Show that the vectors +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +2 +, +3 +5 +and 3 +4 +4 +i +j +k i +j +k +i +j +k +− ++ +− +− +− +− + form the vertices +of a right angled triangle. +18. +If is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ is unit +vector if +(A) λ = 1 +(B) λ = – 1 +(C) a = |λ| +(D) a = 1/|λ| +10.6.3 Vector (or cross) product of two vectors +In Section 10.2, we have discussed on the three dimensional right handed rectangular +coordinate system. In this system, when the positive x-axis is rotated counterclockwise +Reprint 2025-26 + +VECTOR ALGEBRA +363 +into the positive y-axis, a right handed (standard) screw would advance in the direction +of the positive z-axis (Fig 10.22(i)). +In a right handed coordinate system, the thumb of the right hand points in the +direction of the positive z-axis when the fingers are curled in the direction away from +the positive x-axis toward the positive y-axis (Fig 10.22(ii)). +Fig 10.22 (i), (ii) +Definition 3 The vector product of two nonzero vectors +, is denoted by +and defined as + = +, +where, θ is the angle between +, 0 ≤θ ≤π and ˆn is a +unit vector perpendicular to both +, such that + form a right handed system (Fig 10.23). i.e., the +right handed system rotated from +moves in the direction +of ˆn . +If either +, then θ is not defined and in this case, we define +. +Observations +1. + is a vector. +2. +Let + be two nonzero vectors. Then + if and only if +are +parallel (or collinear) to each other, i.e., + = +Fig 10.23 +Reprint 2025-26 + + MATHEMATICS +364 +In particular, +and +, since in the first situation, θ = 0 and +in the second one, θ = π, making the value of sin θ to be 0. +3. +If +2 +π +θ = + then +. +4. +In view of the Observations 2 and 3, for mutually perpendicular +unit vectors +ˆ +ˆ +ˆ +, +and +i +j +k (Fig 10.24), we have +ˆ +ˆ +i +i +× = +ˆ +ˆ +i +j +× += ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +, +, +k +j +k +i +k +i +j +× += +× = +5. +In terms of vector product, the angle between two vectors +may be +given as +sin θ = +6. +It is always true that the vector product is not commutative, as + = +. +Indeed, +, where + form a right handed system, +i.e., θ is traversed from +, Fig 10.25 (i). While, +, where + form a right handed system i.e. θ is traversed from +, +Fig 10.25(ii). +Fig 10.25 (i), (ii) +Thus, if we assume +to lie in the plane of the paper, then +1 +ˆ +ˆ + and +n +n both +will be perpendicular to the plane of the paper. But, ˆn being directed above the +paper while +1ˆn directed below the paper. i.e. +1ˆ +ˆ +n +n += − +. +Fig 10.24 +Reprint 2025-26 + +VECTOR ALGEBRA +365 +Hence + = += +7. +In view of the Observations 4 and 6, we have +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +, + and +. +j +i +k +k +j +i +i +k +j +× = − +× += − +× += − +8. +If +represent the adjacent sides of a triangle then its area is given as +. +By definition of the area of a triangle, we have from +Fig 10.26, +Area of triangle ABC = 1 AB CD. +2 +⋅ +But + (as given), and CD = + sinθ. +Thus, Area of triangle ABC = +9. +If + represent the adjacent sides of a parallelogram, then its area is given +by +. +From Fig 10.27, we have +Area of parallelogram ABCD = AB. DE. +But + (as given), and +. +Thus, +Area of parallelogram ABCD = +We now state two important properties of vector product. +Property 3 (Distributivity of vector product over addition): If +are any three vectors and λ be a scalar, then +(i) +(ii) +Fig 10.26 + Fig 10.27 +Reprint 2025-26 + + MATHEMATICS +366 +Let + be two vectors given in component form as +1 +2 +3 ˆ +ˆ +ˆ +a i +a j +a k ++ ++ +and +1 +2 +3 ˆ +ˆ +ˆ +b i +b j +b k ++ ++ +, respectively. Then their cross product may be given by + = +1 +2 +3 +1 +2 +3 +ˆ +ˆ +ˆ +i +j +k +a +a +a +b +b +b +Explanation We have + = +1 +2 +3 +1 +2 +3 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +( +) +( +) +a i +a j +a k +b i +b j +b k ++ ++ +× ++ ++ += +1 1 +1 2 +1 3 +2 1 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +( +) +( +) +( +) +( +) +a b i +i +a b i +j +a b i +k +a b +j +i +× ++ +× ++ +× ++ +× ++ +2 2 +2 3 +ˆ +ˆ +ˆ +ˆ +( +) +( +) +a b +j +j +a b +j +k +× ++ +× ++ +3 1 +3 2 +3 3 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +( +) +( +) +( +) +a b k +i +a b k +j +a b k +k +× ++ +× ++ +× +(by Property 1) += +1 2 +1 3 +2 1 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +( +) +( +) +( +) +a b i +j +a b k +i +a b i +j +× +− +× +− +× ++ +2 3 +3 1 +3 2 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +( +) +( +) +( +) +a b +j +k +a b k +i +a b +j +k +× ++ +× +− +× +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +(as +0 and +, + and +) +i +i +j +j +k +k +i +k +k +i +j +i +i +j +k +j +j +k +× = +× += +× += +× += −× +× = −× +× += −× += +1 2 +1 3 +2 1 +2 3 +3 1 +3 2 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +a b k +a b j +a b k +a b i +a b j +a b i +− +− ++ ++ +− +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +(as +, + and +) +i +j +k +j +k +i +k +i +j +× += +× += +× = += +2 3 +3 2 +1 3 +3 1 +1 2 +2 1 ˆ +ˆ +ˆ +( +) +( +) +( +) +a b +a b i +a b +a b +j +a b +a b k +− +− +− ++ +− += +1 +2 +3 +1 +2 +3 +ˆ +ˆ +ˆ +i +j +k +a +a +a +b +b +b +Example 22 Find +Solution We have + = +ˆ +ˆ +ˆ +2 +1 +3 +3 +5 +2 +i +j +k +− += +ˆ +ˆ +ˆ +( 2 +15) +( 4 +9) +(10 – 3) +i +j +k +−− +−− +− ++ +ˆ +ˆ +ˆ +17 +13 +7 +i +j +k += − ++ ++ +Hence + = +2 +2 +2 +( 17) +(13) +(7) +507 +− ++ ++ += +Reprint 2025-26 + +VECTOR ALGEBRA +367 +Example 23 Find a unit vector perpendicular to each of the vectors + and +where +. +Solution We have +A vector which is perpendicular to both +and ++ +− +r +r +r +r +a +b +a +b is given by + = +Now + = +4 +16 +4 +24 +2 6 ++ ++ += += +Therefore, the required unit vector is +| +| +r +rc +c = +1 +2 +1 ˆ +ˆ +ˆ +6 +6 +6 +i +j +k +− ++ +− +ANote There are two perpendicular directions to any plane. Thus, another unit +vector perpendicular to + will be 1 +2 +1 ˆ +ˆ +ˆ +. +6 +6 +6 +i +j +k +− ++ + But that will +be a consequence of +. +Example 24 Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3) +and C(2, 3, 1) as its vertices. +Solution We have +. The area of the given triangle +is +. +Now, + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +0 +1 +2 +4 +2 +1 +2 +0 +i +j +k +i +j +k += − ++ +− +Therefore + = +16 +4 +1 +21 ++ ++ = +Thus, the required area is +1 +21 +2 +Reprint 2025-26 + + MATHEMATICS +368 +Example 25 Find the area of a parallelogram whose adjacent sides are given +by the vectors +Solution The area of a parallelogram with +as its adjacent sides is given +by +. +Now + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +3 +1 +4 +5 +4 +1 +1 +1 +i +j +k +i +j +k += ++ +− +− +Therefore + = +25 1 16 +42 ++ + += +and hence, the required area is +42 . +EXERCISE 10.4 +1. +Find +. +2. +Find a unit vector perpendicular to each of the vector +, where +. +3. +If a unit vector makes angles +ˆ +ˆ +with , +with +3 +4 +i +j +π +π + and an acute angle θ with +ˆk , then find θ and hence, the components of . +4. +Show that +5. +Find λ and µ if +. +6. +Given that + and +. What can you conclude about the vectors +? +7. +Let the vectors + be given as +1 +2 +3 +1 +2 +3 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +, +, +a i +a j +a k b i +b j +b k ++ ++ ++ ++ +1 +2 +3 ˆ +ˆ +ˆ +c i +c j +c k ++ ++ +. Then show that +. +8. +If either + then +. Is the converse true? Justify your +answer with an example. +9. +Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5). +Reprint 2025-26 + +VECTOR ALGEBRA +369 +10. +Find the area of the parallelogram whose adjacent sides are determined by the +vectors +and +. +11. +Let the vectors +be such that +, then + is a unit +vector, if the angle between +is +(A) π/6 +(B) π/4 +(C) π/3 +(D) π/2 +12. +Area of a rectangle having vertices A, B, C and D with position vectors +1 +1 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +– +4 , +4 +2 +2 +i +j +k i +j +k ++ ++ ++ ++ +, +1 +ˆ +ˆ +ˆ +4 +2 +i +j +k +− ++ + and +1 +ˆ +ˆ +ˆ +– +4 +2 +i +j +k +− ++ +, respectively is +(A) 1 +2 +(B) 1 +(C) 2 +(D) 4 +Miscellaneous Examples +Example 26 Write all the unit vectors in XY-plane. +Solution Let + be a unit vector in XY-plane (Fig 10.28). Then, from the +figure, we have x = cos θ and y = sin θ (since | | = 1). So, we may write the vector as += +ˆ +ˆ +cos +sin +i +j +θ + +θ +... (1) +Clearly, +| | = +2 +2 +cos +sin +1 +θ + +θ = +Fig 10.28 +Also, as θ varies from 0 to 2π, the point P (Fig 10.28) traces the circle x2 + y2 = 1 +counterclockwise, and this covers all possible directions. So, (1) gives every unit vector +in the XY-plane. +Reprint 2025-26 + + MATHEMATICS +370 +Example 27 If +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +, 2 +5 , 3 +2 +3 and +6 ++ ++ ++ ++ +− +− +− +i +j +k +i +j +i +j +k +i +j +k are the position +vectors of points A, B, C and D respectively, then find the angle between + and +. +Deduce that + and + are collinear. +Solution Note that if θ is the angle between AB and CD, then θ is also the angle +between + and +. +Now + = Position vector of B – Position vector of A += +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +(2 +5 ) +( +) +4 +i +j +i +j +k +i +j +k ++ +− ++ ++ += + +− +Therefore +| +| = +2 +2 +2 +(1) +(4) +( 1) +3 2 ++ ++ − += +Similarly + = +ˆ +ˆ +ˆ +2 +8 +2 and |CD | 6 2 +− +− ++ += +uuur +i +j +k +Thus +cos θ = += 1( 2) +4( 8) +( 1)(2) +36 +1 +36 +(3 2)(6 2) +− ++ +− ++ − +− += += − +Since 0 ≤ θ ≤ π, it follows that θ = π. This shows that + and + are collinear. +Alternatively, + which implies that + and + are collinear vectors. +Example 28 Let + be three vectors such that + and +each one of them being perpendicular to the sum of the other two, find +. +Solution Given +Now += ++ += += 9 + 16 + 25 = 50 +Therefore + = +50 +5 2 += +Reprint 2025-26 + +VECTOR ALGEBRA +371 +Example 29 Three vectors + satisfy the condition +. Evaluate +the quantity +. +Solution Since +, we have + = 0 +or + = 0 +Therefore + ... (1) +Again, + = 0 +or + ... (2) +Similarly + = – 4. +... (3) +Adding (1), (2) and (3), we have + = – 29 +or +2µ = – 29, i.e., µ = +29 +2 +− +Example 30 If with reference to the right handed system of mutually perpendicular +unit vectors +, then express in the form +is parallel to + is perpendicular to +Solution Let + is a scalar, i.e., +. +Now + = +ˆ +ˆ +ˆ +(2 +3 ) +(1 +) +3 +i +j +k +−λ ++ ++ λ +− +. +Now, since +2 +β +r + is to be perpendicular to αr, we should have +. i.e., +3(2 +3 ) +(1 +) +−λ − ++ λ = 0 +or +λ = 1 +2 +Therefore +1 = 3 +1 +ˆ +ˆ +2 +2 +i +j +− + and +Reprint 2025-26 + + MATHEMATICS +372 +Miscellaneous Exercise on Chapter 10 +1. +Write down a unit vector in XY-plane, making an angle of 30° with the positive +direction of x-axis. +2. +Find the scalar components and magnitude of the vector joining the points +P(x1, y1, z1) and Q(x2, y2, z2). +3. +A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of +north and stops. Determine the girl’s displacement from her initial point of +departure. +4. +If +, then is it true that + ? Justify your answer. +5. +Find the value of x for which +ˆ +ˆ +ˆ +( +) +x i +j +k ++ ++ + is a unit vector. +6. +Find a vector of magnitude 5 units, and parallel to the resultant of the vectors +. +7. +If +, find a unit vector parallel +to the vector +. +8. +Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and +find the ratio in which B divides AC. +9. +Find the position vector of a point R which divides the line joining two points +P and Q whose position vectors are + externally in the ratio +1 : 2. Also, show that P is the mid point of the line segment RQ. +10. +The two adjacent sides of a parallelogram are +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +2 +4 +5 and +2 +3 +i +j +k +i +j +k +− ++ +− +− +. +Find the unit vector parallel to its diagonal. Also, find its area. +11. +Show that the direction cosines of a vector equally inclined to the axes OX, OY +and OZ are ± + + + +1 +3 +1 +3 +1 +3 +, +, +. +12. +Let +. Find a vector which +is perpendicular to both + and , and +. +13. +The scalar product of the vector +ˆ +ˆ +ˆ +i +j +k ++ ++ + with a unit vector along the sum of +vectors +ˆ +ˆ +ˆ +2 +4 +5 +i +j +k ++ +− + and +ˆ +ˆ +ˆ +2 +3 +i +j +k +λ + ++ + is equal to one. Find the value of λ. +14. +If + are mutually perpendicular vectors of equal magnitudes, show that the +vector + is equally inclined to +. +Reprint 2025-26 + +VECTOR ALGEBRA +373 +15. +Prove that +, if and only if + are perpendicular, given +. +Choose the correct answer in Exercises 16 to 19. +16. +If θ is the angle between two vectors +, then + only when +(A) 0 +2 +π +< θ < +(B) 0 +2 +π +≤θ ≤ +(C) 0 < θ < π +(D) 0 ≤ θ ≤ π +17. +Let + be two unit vectors and θ is the angle between them. Then + is +a unit vector if +(A) +4 +π +θ = +(B) +3 +π +θ = +(C) +2 +π +θ = +(D) +2 +3 +π +θ = +18. +The value of +ˆ +ˆ +ˆ +ˆ ˆ +ˆ +ˆ +ˆ +ˆ +.( +) +( +) +( +) +i +j +k +j +i +k +k +i +j +× ++ +⋅ +× ++ +⋅ +× + is +(A) 0 +(B) –1 +(C) 1 +(D) 3 +19. +If θ is the angle between any two vectors +, then +when θ +is equal to +(A) 0 +(B) +4 +π +(C) +2 +π +(D) π +Summary +® Position vector of a point P(x, y, z) is given as +, and its +magnitude by +2 +2 +2 +x +y +z ++ ++ +. +® The scalar components of a vector are its direction ratios, and represent its +projections along the respective axes. +® The magnitude (r), direction ratios (a, b, c) and direction cosines (l, m, n) of +any vector are related as: +, +, +a +b +c +l +m +n +r +r +r += += += +Reprint 2025-26 + + MATHEMATICS +374 +® The vector sum of the three sides of a triangle taken in order is . +® The vector sum of two coinitial vectors is given by the diagonal of the +parallelogram whose adjacent sides are the given vectors. +® The multiplication of a given vector by a scalar λ, changes the magnitude of +the vector by the multiple |λ|, and keeps the direction same (or makes it +opposite) according as the value of λ is positive (or negative). +® For a given vector , the vector + gives the unit vector in the direction +of . +® The position vector of a point R dividing a line segment joining the points +P and Q whose position vectors are + respectively, in the ratio m : n +(i) +internally, is given by +. +(ii) +externally, is given by +. +® The scalar product of two given vectors +having angle θ between +them is defined as +. +Also, when + is given, the angle ‘θ’ between the vectors +may be +determined by +cosθ = +® If θ is the angle between two vectors +, then their cross product is +given as +where ˆn is a unit vector perpendicular to the plane containing +. Such +that +form right handed system of coordinate axes. +® If we have two vectors +, given in component form as +and λ any scalar, +Reprint 2025-26 + +VECTOR ALGEBRA +375 +then + = +1 +1 +2 +2 +3 +3 ˆ +ˆ +ˆ +( +) +( +) +( +) +a +b i +a +b +j +a +b k ++ ++ ++ ++ ++ +; +λ = +1 +2 +3 ˆ +ˆ +ˆ +( +) +( +) +( +) +a i +a +j +a k +λ ++ λ ++ λ +; + = +1 1 +2 2 +3 3 +a b +a b +a b ++ ++ +; +and + = +1 +1 +1 +2 +2 +2 +ˆ +ˆ +ˆ +. +i +j +k +a +b +c +a +b +c +Historical Note +The word vector has been derived from a Latin word vectus, which means +“to carry”. The germinal ideas of modern vector theory date from around 1800 +when Caspar Wessel (1745-1818) and Jean Robert Argand (1768-1822) described +that how a complex number a + ib could be given a geometric interpretation with +the help of a directed line segment in a coordinate plane. William Rowen Hamilton +(1805-1865) an Irish mathematician was the first to use the term vector for a +directed line segment in his book Lectures on Quaternions (1853). Hamilton’s +method of quaternions (an ordered set of four real numbers given as: +ˆ +ˆ +ˆ +ˆ +ˆ ˆ +, , , +a +bi +cj +dk i +j k ++ ++ ++ + following certain algebraic rules) was a solution to the +problem of multiplying vectors in three dimensional space. Though, we must +mention here that in practice, the idea of vector concept and their addition was +known much earlier ever since the time of Aristotle (384-322 B.C.), a Greek +philosopher, and pupil of Plato (427-348 B.C.). That time it was supposed to be +known that the combined action of two or more forces could be seen by adding +them according to parallelogram law. The correct law for the composition of +forces, that forces add vectorially, had been discovered in the case of perpendicular +forces by Stevin-Simon (1548-1620). In 1586 A.D., he analysed the principle of +geometric addition of forces in his treatise DeBeghinselen der Weeghconst +(“Principles of the Art of Weighing”), which caused a major breakthrough in the +development of mechanics. But it took another 200 years for the general concept +of vectors to form. +In the 1880, Josaih Willard Gibbs (1839-1903), an American physicist and +mathematician, and Oliver Heaviside (1850-1925), an English engineer, created +what we now know as vector analysis, essentially by separating the real (scalar) +Reprint 2025-26 + + MATHEMATICS +376 +part of quaternion from its imaginary (vector) part. In 1881 and 1884, Gibbs +printed a treatise entitled Element of Vector Analysis. This book gave a systematic +and concise account of vectors. However, much of the credit for demonstrating +the applications of vectors is due to the D. Heaviside and P.G. Tait (1831-1901) +who contributed significantly to this subject. +—v— +Reprint 2025-26" +class_12,11,Three Dimensional Geometry,ncert_books/class_12/lemh2dd/lemh205.pdf,"THREE DIMENSIONAL GEOMETRY +377 +v The moving power of mathematical invention is not +reasoning but imagination. – A.DEMORGAN v +11.1 Introduction +In Class XI, while studying Analytical Geometry in two +dimensions, and the introduction to three dimensional +geometry, we confined to the Cartesian methods only. In +the previous chapter of this book, we have studied some +basic concepts of vectors. We will now use vector algebra +to three dimensional geometry. The purpose of this +approach to 3-dimensional geometry is that it makes the +study simple and elegant*. +In this chapter, we shall study the direction cosines +and direction ratios of a line joining two points and also +discuss about the equations of lines and planes in space +under different conditions, angle between two lines, two +planes, a line and a plane, shortest distance between two +skew lines and distance of a point from a plane. Most of +the above results are obtained in vector form. Nevertheless, we shall also translate +these results in the Cartesian form which, at times, presents a more clear geometric +and analytic picture of the situation. +11.2 Direction Cosines and Direction Ratios of a Line +From Chapter 10, recall that if a directed line L passing through the origin makes +angles α, β and γ with x, y and z-axes, respectively, called direction angles, then cosine +of these angles, namely, cos α, cos β and cos γ are called direction cosines of the +directed line L. +If we reverse the direction of L, then the direction angles are replaced by their supplements, +i.e., +, + and +. Thus, the signs of the direction cosines are reversed. +Chapter 11 +THREE DIMENSIONAL GEOMETRY +* For various activities in three dimensional geometry, one may refer to the Book +“A Hand Book for designing Mathematics Laboratory in Schools”, NCERT, 2005 +Leonhard Euler +(1707-1783) +Reprint 2025-26 + + MATHEMATICS +378 +Note that a given line in space can be extended in two opposite directions and so it +has two sets of direction cosines. In order to have a unique set of direction cosines for +a given line in space, we must take the given line as a directed line. These unique +direction cosines are denoted by l, m and n. +Remark If the given line in space does not pass through the origin, then, in order to find +its direction cosines, we draw a line through the origin and parallel to the given line. +Now take one of the directed lines from the origin and find its direction cosines as two +parallel line have same set of direction cosines. +Any three numbers which are proportional to the direction cosines of a line are +called the direction ratios of the line. If l, m, n are direction cosines and a, b, c are +direction ratios of a line, then a = λl, b=λm and c = λn, for any nonzero λ ∈ R. +ANote Some authors also call direction ratios as direction numbers. +Let a, b, c be direction ratios of a line and let l, m and n be the direction cosines +(d.c’s) of the line. Then +l +a = m +b = n +k +c = + (say), k being a constant. +Therefore +l = ak, m = bk, n = ck +... (1) +But +l2 + m2 + n2 = 1 +Therefore +k2 (a2 + b2 + c2) = 1 +or +k = +2 +2 +2 +1 +a +b +c +± ++ ++ +Fig 11.1 +Reprint 2025-26 + +THREE DIMENSIONAL GEOMETRY +379 +Hence, from (1), the d.c.’s of the line are +2 +2 +2 +2 +2 +2 +2 +2 +2 +, +, +a +b +c +l +m +n +a +b +c +a +b +c +a +b +c +=± += ± += ± ++ ++ ++ ++ ++ ++ +where, depending on the desired sign of k, either a positive or a negative sign is to be +taken for l, m and n. +For any line, if a, b, c are direction ratios of a line, then ka, kb, kc; k ≠ 0 is also a +set of direction ratios. So, any two sets of direction ratios of a line are also proportional. +Also, for any line there are infinitely many sets of direction ratios. +11.2.1 Direction cosines of a line passing through two points +Since one and only one line passes through two given points, we can determine the +direction cosines of a line passing through the given points P(x1, y1, z1) and Q(x2, y2, z2) +as follows (Fig 11.2 (a)). +Fig 11.2 +Let l, m, n be the direction cosines of the line PQ and let it makes angles α, β and γ +with the x, y and z-axis, respectively. +Draw perpendiculars from P and Q to XY-plane to meet at R and S. Draw a +perpendicular from P to QS to meet at N. Now, in right angle triangle PNQ, ∠PQN= γ +(Fig 11.2 (b). +Therefore, +cosγ = +2 +1 +NQ +PQ +PQ +z +z +− += +Similarly +cosα = +2 +1 +2 +1 +and cos +PQ +PQ +x +x +y +y +− +− +β= +Hence, the direction cosines of the line segment joining the points P(x1, y1, z1) and +Q(x2, y2, z2) are +Reprint 2025-26 + + MATHEMATICS +380 +2 +1 +PQ +x +x +− +, +2 +1 +PQ +y +y +− +, +2 +1 +PQ +z +z +− +where +PQ = +( +) +2 +2 +2 +2 +1 +2 +1 +2 +1 +( +) +( +) +x +x +y +y +z +z +− ++ +− ++ +− +ANote The direction ratios of the line segment joining P(x1, y1, z1) and Q(x2, y2, z2) +may be taken as +x2 – x1, y2 – y1, z2 – z1 or x1 – x2, y1 – y2, z1 – z2 +Example 1 If a line makes angle 90°, 60° and 30° with the positive direction of x, y and +z-axis respectively, find its direction cosines. +Solution Let the d.c.'s of the lines be l , m, n. Then l = cos 900 = 0, m = cos 600 = 1 +2 , +n = cos 300 = 2 +3 . +Example 2 If a line has direction ratios 2, – 1, – 2, determine its direction cosines. +Solution Direction cosines are +2 +2 +2 +) +2 +( +)1 +( +2 +2 +− ++ +− ++ +, +2 +2 +2 +) +2 +( +)1 +( +2 +1 +− ++ +− ++ +− +, +( +) +2 +2 +2 +) +2 +( +1 +2 +2 +− ++ +− ++ +− +or + 2 +1 +2 +, +, +3 +3 +3 +− +− +Example 3 Find the direction cosines of the line passing through the two points +(– 2, 4, – 5) and (1, 2, 3). +Solution We know the direction cosines of the line passing through two points +P(x1, y1, z1) and Q(x2, y2, z2) are given by +2 +1 +2 +1 +2 +1 +, +, +PQ +PQ +PQ +x +x +y +y +z +z +− +− +− +where +PQ = +( +) +2 +1 +2 +2 +1 +2 +2 +1 +2 +) +( +) +( +z +z +y +y +x +x +− ++ +− ++ +− +Here P is (– 2, 4, – 5) and Q is (1, 2, 3). +So +PQ = +2 +2 +2 +(1 +( 2)) +(2 +4) +(3 +( 5)) +−− ++ +− ++ +−− + = +77 +Thus, the direction cosines of the line joining two points is +3 +2 +8 +, +, +77 +77 +77 +− +Reprint 2025-26 + +THREE DIMENSIONAL GEOMETRY +381 +Example 4 Find the direction cosines of x, y and z-axis. +Solution The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis. +Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90° i.e., 1,0,0. +Similarly, direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively. +Example 5 Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are +collinear. +Solution Direction ratios of line joining A and B are +1 – 2, – 2 – 3, 3 + 4 i.e., – 1, – 5, 7. +The direction ratios of line joining B and C are +3 –1, 8 + 2, – 11 – 3, i.e., 2, 10, – 14. +It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel +to BC. But point B is common to both AB and BC. Therefore, A, B, C are +collinear points. +EXERCISE 11.1 +1. +If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its +direction cosines. +2. +Find the direction cosines of a line which makes equal angles with the coordinate +axes. +3. If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ? +4. +Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear. +5. Find the direction cosines of the sides of the triangle whose vertices are +(3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2). +11.3 Equation of a Line in Space +We have studied equation of lines in two dimensions in Class XI, we shall now study +the vector and cartesian equations of a line in space. +A line is uniquely determined if +(i) +it passes through a given point and has given direction, or +(ii) +it passes through two given points. +11.3.1 +Equation of a line through a given point and parallel to given vector +Let be the position vector of the given point A with respect to the origin O of the +rectangular coordinate system. Let l be the line which passes through the point A and +is parallel to a given vector . Let be the position vector of an arbitrary point P on the +line (Fig 11.3). +Reprint 2025-26 + + MATHEMATICS +382 +Then AP +uuur + is parallel to the vector , i.e., +AP +uuur + = λ , where λ is some real number. +But +AP +uuur + = OP – OA +uuur +uuur +i.e. +λ = +− +r +r +r +a +Conversely, for each value of the +parameter λ, this equation gives the position +vector of a point P on the line. Hence, the +vector equation of the line is given by + = +» +r +ra + +b +... (1) +Remark If +ˆ +ˆ +ˆ += ++ ++ +r +b +ai +bj +ck, then a, b, c are direction ratios of the line and conversely, +if a, b, c are direction ratios of a line, then +ˆ +ˆ +ˆ += ++ ++ +r +b +ai +bj +ck will be the parallel to +the line. Here, b should not be confused with | |. +Derivation of cartesian form from vector form +Let the coordinates of the given point A be (x1, y1, z1) and the direction ratios of +the line be a, b, c. Consider the coordinates of any point P be (x, y, z). Then +ˆ +ˆ +ˆ +r +xi +yj +zk += ++ ++ +r +; +1 +1 +1 +ˆ +ˆ +ˆ +a +x i +y j +z +k += ++ ++ +r +and +ˆ +ˆ +ˆ += ++ ++ +r +b +a i +b j +c k +Substituting these values in (1) and equating the coefficients of ˆ +ˆ +,i +j and kˆ , we get +x = x1 + λa; y = y1 + λ b; z = z1+ λ c +... (2) +These are parametric equations of the line. Eliminating the parameter λ from (2), +we get +1 +x – x +a + = +1 +1 +y – y +z – z += +b +c +... (3) +This is the Cartesian equation of the line. +ANote If l, m, n are the direction cosines of the line, the equation of the line is +1 +x – x +l + = +1 +1 +y – y +z – z += +m +n +Example 6 Find the vector and the Cartesian equations of the line through the point +(5, 2, – 4) and which is parallel to the vector +ˆ +ˆ +ˆ +3 +2 +8 +i +j +k ++ +− +. +Solution We have + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +5 +2 +4 +and +3 +2 +8 ++ +− += ++ +− +r +i +j +k +b +i +j +k +Fig 11.3 +Reprint 2025-26 + +THREE DIMENSIONAL GEOMETRY +383 +Therefore, the vector equation of the line is + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +5 +2 +4 +( 3 +2 +8 +) +i +j +k +i +j +k ++ +− ++ λ ++ +− +Now, is the position vector of any point P(x, y, z) on the line. +Therefore, +ˆ +ˆ +ˆ +xi +y j +z k ++ ++ + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +5 +2 +4 +( 3 +2 +8 ) ++ +− ++ λ ++ +− +i +j +k +i +j +k += +ˆ +ˆ +ˆ +(5 +3 ) +(2 +2 ) +( 4 +8 ) ++ λ ++ ++ λ ++ − +−λ +i +j +k +Eliminating λ , we get +5 +3 +x− + = +2 +4 +2 +8 +y +z +− ++ += − +which is the equation of the line in Cartesian form. +11.4 Angle between Two Lines +Let L1 and L2 be two lines passing through the origin +and with direction ratios a1, b1, c1 and a2, b2, c2, +respectively. Let P be a point on L1 and Q be a point +on L2. Consider the directed lines OP and OQ as +given in Fig 11.6. Let θ be the acute angle between +OP and OQ. Now recall that the directed line +segments OP and OQ are vectors with components +a1, b1, c1 and a2, b2, c2, respectively. Therefore, the +angle θ between them is given by +cosθ = +1 2 +1 2 +1 2 +2 +2 +2 +2 +2 +2 +1 +1 +1 +2 +2 +2 +a a +b b +c c +a +b +c +a +b +c ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ ++ +... (1) +The angle between the lines in terms of sin θ is given by +sin θ = +2 +1 +cos +− +θ += +( +)( +) +2 +1 2 +1 2 +1 2 +2 +2 +2 +2 +2 +2 +1 +1 +1 +2 +2 +2 +( +) +1 +a a +b b +c c +a +b +c +a +b +c ++ ++ +− ++ ++ ++ ++ += +( +)( +) ( +) +( +) ( +) +2 +2 +2 +2 +2 +2 +2 +1 +1 +1 +2 +2 +2 +1 2 +1 2 +1 2 +2 +2 +2 +2 +2 +2 +1 +1 +1 +2 +2 +2 +a +b +c +a +b +c +a a +b b +c c +a +b +c +a +b +c ++ ++ ++ ++ +− ++ ++ ++ ++ ++ ++ += +2 +2 +2 +1 +2 +2 +1 +1 +2 +2 +1 +1 +2 +2 +1 +2 +2 +2 +2 +2 +2 +1 +1 +1 +2 +2 +2 +( +) +( +) +( +) +− ++ +− ++ +− ++ ++ ++ ++ +a b +a b +b c +b c +c a +c a +a +b +c +a +b +c +... (2) +Fig 11.4 +Reprint 2025-26 + + MATHEMATICS +384 +ANote In case the lines L1 and L2 do not pass through the origin, we may take +lines +1 +2 +L andL +′ +′ which are parallel to L1 and L2 respectively and pass through +the origin. +If instead of direction ratios for the lines L1 and L2, direction cosines, namely, +l1, m1, n1 for L1 and l2, m2, n2 for L2 are given, then (1) and (2) takes the following form: +cos θ = |l1 l2 + m1m2 + n1n2| (as +2 +2 +2 +1 +1 +1 +1 +l +m +n ++ ++ += +2 +2 +2 +2 +2 +2 +l +m +n += ++ ++ +) +... (3) +and +sin θ = +( +) +2 +2 +2 +1 +2 +2 +1 +1 +2 +2 +1 +1 2 +2 1 +( +) +( +) +l m +l m +m n +m n +n l +n l +− +− +− ++ +− + ... (4) +Two lines with direction ratios a1, b1, c1 and a2, b2, c2 are +(i) +perpendicular i.e. if θ = 90° by (1) +a1a2 + b1b2 + c1c2 = 0 +(ii) +parallel i.e. if θ = 0 by (2) +1 +2 +a +a = +1 +1 +2 +2 +b +c +b +c += +Now, we find the angle between two lines when their equations are given. If θ is +acute the angle between the lines + = +1 +1 +a +b ++ λ + + and = +2 +2 ++ µ +r +ra +b +then +cosθ = +1 +2 +1 +2 +⋅ +r +r +r +r +b +b +b +b +In Cartesian form, if θ is the angle between the lines +1 +1 +x +x +a +− + = +1 +1 +1 +1 +y +y +z +z +b +c +− +− += +... (1) +and +2 +2 +x +x +a +− + = +2 +2 +2 +2 +y +y +z +z +b +c +− +− += +... (2) +where, a1, b1, c1 and a2, b2, c2 are the direction ratios of the lines (1) and (2), respectively, +then +cos θ = +1 2 +1 2 +1 2 +2 +2 +2 +2 +2 +2 +1 +1 +1 +2 +2 +2 +a a +b b +c c +a +b +c +a +b +c ++ ++ ++ ++ ++ ++ +Example 7 Find the angle between the pair of lines given by + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +3 +2 +4 +( +2 +2 ) +i +j +k +i +j +k ++ +− ++ λ ++ ++ +Reprint 2025-26 + +THREE DIMENSIONAL GEOMETRY +385 +and + = +ˆ +ˆ +ˆ +ˆ +ˆ +5 +2 +(3 +2 +6 ) +i +j +i +j +k +− ++ µ ++ ++ +Solution Here 1 +r +b = +ˆ +ˆ +ˆ +2 +2 +i +j +k ++ ++ + and 2 +r +b = +ˆ +ˆ +ˆ +3 +2 +6 +i +j +k ++ ++ +The angle θ between the two lines is given by +cos θ = +1 +2 +1 +2 +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +( +2 +2 ) (3 +2 +6 ) +1 +4 +4 +9 +4 +36 +⋅ ++ ++ +⋅ ++ ++ += ++ ++ ++ ++ +r +r +r +r +b b +i +j +k +i +j +k +b b += +3 +4 +12 +19 +3 7 +21 ++ ++ += +× +Hence +θ = cos–1 19 +21 + + + + + + +Example 8 Find the angle between the pair of lines +3 +3 +x + + = +1 +3 +5 +4 +y +z +− ++ += +and +1 +1 +x + = +4 +5 +1 +2 +y +z +− +− += +Solution The direction ratios of the first line are 3, 5, 4 and the direction ratios of the +second line are 1, 1, 2. If θ is the angle between them, then +cos θ = +2 +2 +2 +2 +2 +2 +3.1 +5.1 +4.2 +16 +16 +8 3 +15 +50 +6 +5 2 +6 +3 +5 +4 +1 +1 +2 ++ ++ += += += ++ ++ ++ ++ +Hence, the required angle is cos–1 8 +3 +15 + + + + + + + + +. +11.5 Shortest Distance between Two Lines +If two lines in space intersect at a point, +then the shortest distance between them is +zero. Also, if two lines in space are parallel, +then the shortest distance between them +will be the perpendicular distance, i.e. the +length of the perpendicular drawn from a +point on one line onto the other line. +Further, in a space, there are lines which +are neither intersecting nor parallel. In fact, +such pair of lines are non coplanar and +are called skew lines. For example, let us +consider a room of size 1, 3, 2 units along +x, y and z-axes respectively Fig 11.5. +Fig 11.5 +Reprint 2025-26 + + MATHEMATICS +386 +l2 +S +T +Q +P +l1 +The line GE that goes diagonally across the ceiling and the line DB passes through +one corner of the ceiling directly above A and goes diagonally down the wall. These +lines are skew because they are not parallel and also never meet. +By the shortest distance between two lines we mean the join of a point in one line +with one point on the other line so that the length of the segment so obtained is the +smallest. +For skew lines, the line of the shortest distance will be perpendicular to both +the lines. +11.5.1 Distance between two skew lines +We now determine the shortest distance between two skew lines in the following way: +Let l1 and l2 be two skew lines with equations (Fig. 11.6) + = +1 +1 ++ λ +r +ra +b + ... (1) +and + = +2 +2 ++ µ +r +ra +b + ... (2) +Take any point S on l1 with position vector 1ar + and T on l2, with position vector +2 +ar . +Then the magnitude of the shortest distance vector +will be equal to that of the projection of ST along the +direction of the line of shortest distance (See 10.6.2). +If PQ +uuur + is the shortest distance vector between l1 +and l2 , then it being perpendicular to both 1 +r +b and 2 +r +b , +the unit vector nˆ along PQ +uuur + would therefore be +ˆn = +1 +2 +1 +2 +| +| +× +× +r +r +r +r +b +b +b +b + ... (3) +Then +PQ +uuur + = d nˆ +where, d is the magnitude of the shortest distance vector. Let θ be the angle between +ST +uur + and PQ +uuur +. Then +PQ = ST |cos θ| +But +cos θ = +PQ ST +| PQ | | ST | +⋅ +uuur uur +uuuur +uur + = +2 +1 +ˆ ( +) +ST +⋅ +− +r +r +d n +a +a +d +(since +2 +1 +ST +) += +− +uur +r +r +a +a += +1 +2 +2 +1 +1 +2 +( +) ( +) +ST +× +⋅ +− +× +r +r +r +r +r +r +b +b +a +a +b +b +[From (3)] +Fig 11.6 +Reprint 2025-26 + +THREE DIMENSIONAL GEOMETRY +387 +Hence, the required shortest distance is +d = PQ = ST |cos θ| +or +d = +( +× +) . ( +) +| +× +| +1 +2 +2 +1 +1 +2 +r +r +r +r +r +r +b +b +a +a +b +b +× +Cartesian form +The shortest distance between the lines +l1 : +x +x +a +− +1 +1 + = +y +y +b +z +z +c +− += +− +1 +1 +1 +1 +and +l2 : +x +x +a +− +2 +2 + = +y +y +b +z +z +c +− += +− +2 +2 +2 +2 +is +x +x +y +y +z +z +a +b +c +a +b +c +b c +b c +c a +c a +a b +2 +1 +2 +1 +2 +1 +1 +1 +1 +2 +2 +2 +1 2 +2 1 +2 +1 +2 +2 1 +2 +1 2 +− +− +− +− ++ +− ++ +( +) +( +) +( +−a b +2 1 +2) +11.5.2 Distance between parallel lines +If two lines l1 and l2 are parallel, then they are coplanar. Let the lines be given by +... (1) +and +… (2) +where, 1ar is the position vector of a point S on l1 and +2 +ar +is the position vector of a point T on l2 Fig 11.7. +As l1, l2 are coplanar, if the foot of the perpendicular +from T on the line l1 is P, then the distance between the +lines l1 and l2 = |TP|. +Let θ be the angle between the vectors ST +uur + and . +Then +ST +× += +uur +r +b + +ˆ ... +(3) +where nˆ is the unit vector perpendicular to the plane of the lines l1 and l2. +But +ST +uur + = +2 +1 +− +r +r +a +a +Fig 11.7 +Reprint 2025-26 + + MATHEMATICS +388 +Therefore, from (3), we get +2 +1 +( +) +× +− +r +r +r +b +a +a = +ˆ +| PT +r +b +n (since PT = ST sin θ) +i.e., +2 +1 +| +( +)| +× +− +r +r +r +b +a +a + = | +| PT 1⋅ +r +b + (as | +| +ˆn = 1) +Hence, the distance between the given parallel lines is +d = +Example 9 Find the shortest distance between the lines l1 and l2 whose vector +equations are + = +ˆ +ˆ +ˆ +ˆ +ˆ +(2 +) +i +j +i +j +k ++ ++ λ +− ++ +... (1) +and + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +2 +(3 +5 +2 +) +i +j +k +i +j +k ++ +− ++ µ +− ++ +... (2) +Solution Comparing (1) and (2) with = 1 +1 ++ λ +r +ra +b and +2 +2 +r +a +b +µ += ++ +r +r +r + respectively, +we get +1ar = +1 +ˆ +ˆ +ˆ +ˆ +ˆ +, +2 ++ += +− ++ +r +i +j +b +i +j +k +2ar = 2 ˆi + ˆj – ˆk and 2 +r +b = 3 ˆi – 5 ˆj + 2 ˆk +Therefore +2 +1 +− +r +r +a +a = +ˆ +ˆi +k +− +and +1 +2 +× +r +r +b +b = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +( 2 +) +( 3 +5 +2 +) +i +j +k +i +j +k +− ++ +× +− ++ += +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +2 +1 +1 +3 +7 +3 +5 +2 +i +j +k +i +j +k +− += +− +− +− +So +1 +2 | +| +× +r +r +b +b = +9 +1 +49 +59 ++ + += +Hence, the shortest distance between the given lines is given by +d = +1 +2 +2 +1 +1 +2 +( b +b ) . ( a +a ) +| b +b +| +× +− +× +r +r +r +r +r +r + +59 +10 +59 +| +7 +0 +3 +| += ++ +− += +Example 10 Find the distance between the lines l1 and l2 given by + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +2 +4 +( 2 +3 +6 +) +i +j +k +i +j +k ++ +− ++ λ ++ ++ +and + = +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +3 +3 +5 +( 2 +3 +6 +) +i +j +k +i +j +k ++ +− ++ µ ++ ++ +Reprint 2025-26 + +THREE DIMENSIONAL GEOMETRY +389 +Solution The two lines are parallel (Why? ) We have +1ar = +ˆ +ˆ +ˆ +2 +4 +i +j +k ++ +− +, +2ar = +ˆ +ˆ +ˆ +3 +3 +5 +i +j +k ++ +− + and = +ˆ +ˆ +ˆ +2 +3 +6 +i +j +k ++ ++ +Therefore, the distance between the lines is given by +d = +2 +1 +( +) +| | +× +− +r +r +r +r +b +a +a +b + = +ˆ +ˆ +ˆ +2 +3 +6 +2 +1 +1 +4 +9 +36 +i +j +k +− ++ ++ +or += +ˆ +ˆ +ˆ +| +9 +14 +4 +| +293 +293 +7 +49 +49 +i +j +k +− ++ +− += += +EXERCISE 11.2 +1. +Show that the three lines with direction cosines +12 +3 +4 +4 +12 +3 +3 +4 12 +, +, +; +, +, +; +, +, +13 +13 +13 +13 +13 +13 +13 +13 +13 +− +− +− + are mutually perpendicular. +2. +Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the +line through the points (0, 3, 2) and (3, 5, 6). +3. +Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line +through the points (– 1, – 2, 1), (1, 2, 5). +4. +Find the equation of the line which passes through the point (1, 2, 3) and is +parallel to the vector +ˆ +ˆ +ˆ +3 +2 +2 +i +j +k ++ +− +. +5. +Find the equation of the line in vector and in cartesian form that passes through +the point with position vector +ˆ +ˆ +2 +4 +i +j +k +− ++ +and is in the direction +ˆ +ˆ +ˆ +2 +i +j +k ++ +− +. +6. Find the cartesian equation of the line which passes through the point (– 2, 4, – 5) +and parallel to the line given by +3 +4 +8 +3 +5 +6 +x +y +z ++ +− ++ += += +. +7. +The cartesian equation of a line is +5 +4 +6 +3 +7 +2 +x +y +z +− ++ +− += += +. Write its vector form. +8. +Find the angle between the following pairs of lines: +(i) +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +2 +5 +(3 +2 +6 ) += +− ++ ++ λ ++ ++ +rr +i +j +k +i +j +k and +ˆ +ˆ +ˆ +ˆ +ˆ +7 +6 +( +2 +2 ) += +− ++ µ ++ ++ +rr +i +k +i +j +k +Reprint 2025-26 + + MATHEMATICS +390 +(ii) +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +3 +2 +( +2 ) += ++ +− ++ λ +− +− +rr +i +j +k +i +j +k and +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +2 +56 +(3 +5 +4 ) += +− +− ++ µ +− +− +rr +i +j +k +i +j +k +9. +Find the angle between the following pair of lines: +(i) +2 +1 +3 +2 +4 +5 +and +2 +5 +3 +1 +8 +4 +x +y +z +x +y +z +− +− ++ ++ +− +− += += += += +− +− +(ii) +5 +2 +3 +and +2 +2 +1 +4 +1 +8 +x +y +z +x +y +z +− +− +− += += += += +10. +Find the values of p so that the lines 1 +7 +14 +3 +3 +2 +2 +x +y +z +p +− +− +− += += +and 7 +7 +5 +6 +3 +1 +5 +x +y +z +p +− +− +− += += + are at right angles. +11. +Show that the lines +5 +2 +7 +5 +1 +x +y +z +− ++ += += +− + and 1 +2 +3 +x +y +z += += + are perpendicular to +each other. +12. +Find the shortest distance between the lines +ˆ +ˆ +ˆ +( +2 +) += ++ ++ +rr +i +j +k + +ˆ +ˆ +ˆ +( +) +i +j +k +λ +− ++ + and +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +2 +(2 +2 ) += +− +− ++ µ ++ ++ +rr +i +j +k +i +j +k +13. +Find the shortest distance between the lines +1 +1 +1 +7 +6 +1 +x +y +z ++ ++ ++ += += +− + and +3 +5 +7 +1 +2 +1 +x +y +z +− +− +− += += +− +14. +Find the shortest distance between the lines whose vector equations are +ˆ +ˆ +ˆ +( +2 +3 ) += ++ ++ +rr +i +j +k + +ˆ +ˆ +ˆ +( +3 +2 ) +i +j +k +λ +− ++ +and +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +4 +5 +6 +(2 +3 +) += ++ ++ ++ µ ++ ++ +rr +i +j +k +i +j +k +15. +Find the shortest distance between the lines whose vector equations are +ˆ +ˆ +ˆ +(1 +) +( +2) +(3 +2 ) += +− ++ +− ++ +− +rr +t i +t +j +t k and +ˆ +ˆ +ˆ +( +1) +(2 +1) +(2 +1) += ++ ++ +− +− ++ +rr +s +i +s +j +s +k +Miscellaneous Exercise on Chapter 11 +1. +Find the angle between the lines whose direction ratios are a, b, c and +b – c, c – a, a – b. +2. +Find the equation of a line parallel to x-axis and passing through the origin. +Reprint 2025-26 + +THREE DIMENSIONAL GEOMETRY +391 +3. If the lines +1 +2 +3 +1 +1 +6 +and +3 +2 +2 +3 +1 +5 +x +y +z +x +y +z +k +k +− +− +− +− +− +− += += += += +− +− + are perpendicular, +find the value of k. +4. +Find the shortest distance between lines +ˆ +ˆ +ˆ +ˆ +ˆ +ˆ +6 +2 +2 +( +2 +2 +) += ++ ++ ++ λ +− ++ +rr +i +j +k +i +j +k +and +ˆ +ˆ +ˆ +ˆ +ˆ +4 +(3 +2 +2 +) += − +− ++ µ +− +− +rr +i +k +i +j +k . +5. +Find the vector equation of the line passing through the point (1, 2, – 4) and +perpendicular to the two lines: +7 +10 +16 +19 +3 +8 +− += +− ++ += +− +z +y +x +and +15 +3 +x − + = +29 +5 +8 +5 +y +z +− +− += +− +. +Summary +® Direction cosines of a line are the cosines of the angles made by the line +with the positive directions of the coordinate axes. +® If l, m, n are the direction cosines of a line, then l2 + m2 + n2 = 1. +® Direction cosines of a line joining two points P(x1, y1, z1) and Q(x2, y2, z2) are +2 +1 +2 +1 +2 +1 +, +, +PQ +PQ +PQ +x +x +y +y +z +z +− +− +− +where PQ = +( +) +2 +1 +2 +2 +1 +2 +2 +1 +2 +) +( +) +( +z +z +y +y +x +x +− ++ +− ++ +− +® Direction ratios of a line are the numbers which are proportional to the +direction cosines of a line. +® If l, m, n are the direction cosines and a, b, c are the direction ratios of a line +then +l = +2 +2 +2 +c +b +a +a ++ ++ +; m = +2 +2 +2 +c +b +a +b ++ ++ +; n = +2 +2 +2 +c +b +a +c ++ ++ +® Skew lines are lines in space which are neither parallel nor intersecting. +They lie in different planes. +® Angle between skew lines is the angle between two intersecting lines +drawn from any point (preferably through the origin) parallel to each of the +skew lines. +® If l1, m1, n1 and l2, m2, n2 are the direction cosines of two lines; and θ is the +acute angle between the two lines; then +cosθ = |l1l2 + m1m2 + n1n2| +Reprint 2025-26 + + MATHEMATICS +392 +® If a1, b1, c1 and a2, b2, c2 are the direction ratios of two lines and θ is the +acute angle between the two lines; then +cosθ = +1 +2 +1 +2 +1 +2 +2 +2 +2 +2 +2 +2 +1 +1 +1 +2 +2 +2 +a a +b b +c c +a +b +c +a +b +c ++ ++ ++ ++ ++ ++ +® Vector equation of a line that passes through the given point whose position +vector is and parallel to a given vector is += ++ λ +r +r +r +r +a +b . +® Equation of a line through a point (x1, y1, z1) and having direction cosines l, m, n is +1 +1 +1 +x +x +y +y +z +z +l +m +n +− +− +− += += +® The vector equation of a line which passes through two points whose position +vectors are and is +( +) += ++ λ +− +r +r +r +r +r +a +b +a . +® If θ is the acute angle between +1 +1 += ++ λ +r +r +r +r +a +b and +2 +2 += ++ λ +r +r +r +r +a +b , then +1 +2 +1 +2 +cos +| +| | +| +⋅ +θ = +r +r +r +r +b b +b +b +® If +1 +1 +1 +1 +1 +1 +n +z +z +m +y +y +l +x +x +− += +− += +− + and +2 +2 +2 +2 +2 +2 +n +z +z +m +y +y +l +x +x +− += +− += +− +are the equations of two lines, then the acute angle between the two lines is +given by cos θ = |l1l2 + m1m2 + n1n2|. +® Shortest distance between two skew lines is the line segment perpendicular +to both the lines. +® Shortest distance between +1 +1 += ++ λ +r +r +r +r +a +b and +2 +2 += ++ µ +r +r +r +r +a +b is +1 +2 +2 +1 +1 +2 +( +) ( +– +) +| +| +× +⋅ +× +r +r +r +r +r +r +b +b +a +a +b +b +® Shortest distance between the lines: +1 +1 +1 +1 +1 +1 +x +x +y +y +z +z +a +b +c +− +− +− += += + and +2 +2 +2 +2 +x +x +y +y +a +b +− +− += + = +2 +2 +z +z +c +− + is +Reprint 2025-26 + +THREE DIMENSIONAL GEOMETRY +393 +2 +1 +2 +1 +2 +1 +1 +1 +1 +2 +2 +2 +2 +2 +2 +1 2 +2 1 +1 2 +2 1 +1 2 +2 1 +( +) +( +) +( +) +x +x +y +y +z +z +a +b +c +a +b +c +b c +b c +c a +c a +a b +a b +− +− +− +− ++ +− ++ +− +® Distance between parallel lines +1 += ++ λ +r +r +r +r +a +b and +2 += ++ µ +r +r +r +r +a +b is +2 +1 +( +) +| +| +× +− +r +r +r +r +b +a +a +b +—v— +Reprint 2025-26" +class_12,12,Linear Programming,ncert_books/class_12/lemh2dd/lemh206.pdf,"394 +MATHEMATICS +vThe mathematical experience of the student is incomplete if he never had +the opportunity to solve a problem invented by himself. – G. POLYA v +12.1 Introduction +In earlier classes, we have discussed systems of linear +equations and their applications in day to day problems. In +Class XI, we have studied linear inequalities and systems +of linear inequalities in two variables and their solutions by +graphical method. Many applications in mathematics +involve systems of inequalities/equations. In this chapter, +we shall apply the systems of linear inequalities/equations +to solve some real life problems of the type as given below: +A furniture dealer deals in only two items–tables and +chairs. He has Rs 50,000 to invest and has storage space +of at most 60 pieces. A table costs Rs 2500 and a chair +Rs 500. He estimates that from the sale of one table, he +can make a profit of Rs 250 and that from the sale of one +chair a profit of Rs 75. He wants to know how many tables and chairs he should buy +from the available money so as to maximise his total profit, assuming that he can sell all +the items which he buys. +Such type of problems which seek to maximise (or, minimise) profit (or, cost) form +a general class of problems called optimisation problems. Thus, an optimisation +problem may involve finding maximum profit, minimum cost, or minimum use of +resources etc. +A special but a very important class of optimisation problems is linear programming +problem. The above stated optimisation problem is an example of linear programming +problem. Linear programming problems are of much interest because of their wide +applicability in industry, commerce, management science etc. +In this chapter, we shall study some linear programming problems and their solutions +by graphical method only, though there are many other methods also to solve such +problems. +Chapter 12 +LINEAR PROGRAMMING +L. Kantorovich +Reprint 2025-26 + +LINEAR PROGRAMMING 395 +12.2 Linear Programming Problem and its Mathematical Formulation +We begin our discussion with the above example of furniture dealer which will further +lead to a mathematical formulation of the problem in two variables. In this example, we +observe +(i) +The dealer can invest his money in buying tables or chairs or combination thereof. +Further he would earn different profits by following different investment +strategies. +(ii) +There are certain overriding conditions or constraints viz., his investment is +limited to a maximum of Rs 50,000 and so is his storage space which is for a +maximum of 60 pieces. +Suppose he decides to buy tables only and no chairs, so he can buy 50000 ÷ 2500, +i.e., 20 tables. His profit in this case will be Rs (250 × 20), i.e., Rs 5000. +Suppose he chooses to buy chairs only and no tables. With his capital of Rs 50,000, +he can buy 50000 ÷ 500, i.e. 100 chairs. But he can store only 60 pieces. Therefore, he +is forced to buy only 60 chairs which will give him a total profit of Rs (60 × 75), i.e., +Rs 4500. +There are many other possibilities, for instance, he may choose to buy 10 tables +and 50 chairs, as he can store only 60 pieces. Total profit in this case would be +Rs (10 × 250 + 50 × 75), i.e., Rs 6250 and so on. +We, thus, find that the dealer can invest his money in different ways and he would +earn different profits by following different investment strategies. +Now the problem is : How should he invest his money in order to get maximum +profit? To answer this question, let us try to formulate the problem mathematically. +12.2.1 Mathematical formulation of the problem +Let x be the number of tables and y be the number of chairs that the dealer buys. +Obviously, x and y must be non-negative, i.e., +0 +... (1) +(Non-negative constraints) +... (2) +0 +x +y +≥ + + +≥ +The dealer is constrained by the maximum amount he can invest (Here it is +Rs 50,000) and by the maximum number of items he can store (Here it is 60). +Stated mathematically, +2500x + 500y ≤50000 (investment constraint) +or +5x + y ≤100 +... (3) +and +x + y ≤60 (storage constraint) +... (4) +Reprint 2025-26 + +396 +MATHEMATICS +The dealer wants to invest in such a way so as to maximise his profit, say, Z which +stated as a function of x and y is given by +Z = 250x + 75y (called objective function) +... (5) +Mathematically, the given problems now reduces to: +Maximise Z = 250x + 75y +subject to the constraints: +5x + y ≤100 +x + y ≤60 +x ≥ 0, y ≥0 +So, we have to maximise the linear function Z subject to certain conditions determined +by a set of linear inequalities with variables as non-negative. There are also some other +problems where we have to minimise a linear function subject to certain conditions +determined by a set of linear inequalities with variables as non-negative. Such problems +are called Linear Programming Problems. +Thus, a Linear Programming Problem is one that is concerned with finding the +optimal value (maximum or minimum value) of a linear function (called objective +function) of several variables (say x and y), subject to the conditions that the variables +are non-negative and satisfy a set of linear inequalities (called linear constraints). +The term linear implies that all the mathematical relations used in the problem are +linear relations while the term programming refers to the method of determining a +particular programme or plan of action. +Before we proceed further, we now formally define some terms (which have been +used above) which we shall be using in the linear programming problems: +Objective function Linear function Z = ax + by, where a, b are constants, which has +to be maximised or minimized is called a linear objective function. +In the above example, Z = 250x + 75y is a linear objective function. Variables x and +y are called decision variables. +Constraints The linear inequalities or equations or restrictions on the variables of a +linear programming problem are called constraints. The conditions x ≥ 0, y ≥ 0 are +called non-negative restrictions. In the above example, the set of inequalities (1) to (4) +are constraints. +Optimisation problem A problem which seeks to maximise or minimise a linear +function (say of two variables x and y) subject to certain constraints as determined by +a set of linear inequalities is called an optimisation problem. Linear programming +problems are special type of optimisation problems. The above problem of investing a +Reprint 2025-26 + +LINEAR PROGRAMMING 397 +given sum by the dealer in purchasing chairs and tables is an example of an optimisation +problem as well as of a linear programming problem. +We will now discuss how to find solutions to a linear programming problem. In this +chapter, we will be concerned only with the graphical method. +12.2.2 Graphical method of solving linear programming problems +In Class XI, we have learnt how to graph a system of linear inequalities involving two +variables x and y and to find its solutions graphically. Let us refer to the problem of +investment in tables and chairs discussed in Section 12.2. We will now solve this problem +graphically. Let us graph the constraints stated as linear inequalities: +5x + y ≤100 +... (1) +x + y ≤60 +... (2) +x ≥0 +... (3) +y ≥0 +... (4) +The graph of this system (shaded region) consists of the points common to all half +planes determined by the inequalities (1) to (4) (Fig 12.1). Each point in this region +represents a feasible choice open to the dealer for investing in tables and chairs. The +region, therefore, is called the feasible region for the problem. Every point of this +region is called a feasible solution to the problem. Thus, we have, +Feasible region The common region determined by all the constraints including +non-negative constraints x, y ≥ 0 of a linear programming problem is called the feasible +region (or solution region) for the problem. In Fig 12.1, the region OABC (shaded) is +the feasible region for the problem. The region other than feasible region is called an +infeasible region. +Feasible solutions Points within and on the +boundary of the feasible region represent +feasible solutions of the constraints. In +Fig 12.1, every point within and on the +boundary of the feasible region OABC +represents feasible solution to the problem. +For example, the point (10, 50) is a feasible +solution of the problem and so are the points +(0, 60), (20, 0) etc. +Any point outside the feasible region is +called an infeasible solution. For example, +the point (25, 40) is an infeasible solution of +the problem. +Fig 12.1 +Reprint 2025-26 + +398 +MATHEMATICS +Optimal (feasible) solution: Any point in the feasible region that gives the optimal +value (maximum or minimum) of the objective function is called an optimal solution. +Now, we see that every point in the feasible region OABC satisfies all the constraints +as given in (1) to (4), and since there are infinitely many points, it is not evident how +we should go about finding a point that gives a maximum value of the objective function +Z = 250x + 75y. To handle this situation, we use the following theorems which are +fundamental in solving linear programming problems. The proofs of these theorems +are beyond the scope of the book. +Theorem 1 Let R be the feasible region (convex polygon) for a linear programming +problem and let Z = ax + by be the objective function. When Z has an optimal value +(maximum or minimum), where the variables x and y are subject to constraints described +by linear inequalities, this optimal value must occur at a corner point* (vertex) of the +feasible region. +Theorem 2 Let R be the feasible region for a linear programming problem, and let +Z = ax + by be the objective function. If R is bounded**, then the objective function +Z has both a maximum and a minimum value on R and each of these occurs at a +corner point (vertex) of R. +Remark If R is unbounded, then a maximum or a minimum value of the objective +function may not exist. However, if it exists, it must occur at a corner point of R. +(By Theorem 1). +In the above example, the corner points (vertices) of the bounded (feasible) region +are: O, A, B and C and it is easy to find their coordinates as (0, 0), (20, 0), (10, 50) and +(0, 60) respectively. Let us now compute the values of Z at these points. +We have +* +A corner point of a feasible region is a point in the region which is the intersection of two boundary lines. +** +A feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a +circle. Otherwise, it is called unbounded. Unbounded means that the feasible region does extend +indefinitely in any direction. +Vertex of the +Corresponding value +Feasible Region +of Z (in Rs) +O (0,0) +0 +C (0,60) +4500 +B (10,50) +6250 +A (20,0) + 5000 +Maximum +← +Reprint 2025-26 + +LINEAR PROGRAMMING 399 +We observe that the maximum profit to the dealer results from the investment +strategy (10, 50), i.e. buying 10 tables and 50 chairs. +This method of solving linear programming problem is referred as Corner Point +Method. The method comprises of the following steps: +1. +Find the feasible region of the linear programming problem and determine its +corner points (vertices) either by inspection or by solving the two equations of +the lines intersecting at that point. +2. +Evaluate the objective function Z = ax + by at each corner point. Let M and m, +respectively denote the largest and smallest values of these points. +3. +(i) +When the feasible region is bounded, M and m are the maximum and +minimum values of Z. +(ii) In case, the feasible region is unbounded, we have: +4. +(a) M is the maximum value of Z, if the open half plane determined by +ax + by > M has no point in common with the feasible region. Otherwise, Z +has no maximum value. +(b) Similarly, m is the minimum value of Z, if the open half plane determined by +ax + by < m has no point in common with the feasible region. Otherwise, Z +has no minimum value. +We will now illustrate these steps of Corner Point Method by considering some +examples: +Example 1 Solve the following linear programming problem graphically: +Maximise Z = 4x + y +... (1) +subject to the constraints: +x + y ≤50 +... (2) +3x + y ≤90 +... (3) +x ≥ 0, y ≥0 +... (4) +Solution The shaded region in Fig 12.2 is the feasible region determined by the system +of constraints (2) to (4). We observe that the feasible region OABC is bounded. So, +we now use Corner Point Method to determine the maximum value of Z. +The coordinates of the corner points O, A, B and C are (0, 0), (30, 0), (20, 30) and +(0, 50) respectively. Now we evaluate Z at each corner point. +Reprint 2025-26 + +400 +MATHEMATICS +Fig 12.2 +Hence, maximum value of Z is 120 at the point (30, 0). +Example 2 Solve the following linear programming problem graphically: +Minimise Z = 200 x + 500 y +... (1) +subject to the constraints: +x + 2y ≥10 +... (2) +3x + 4y ≤24 +... (3) +x ≥ 0, y ≥0 +... (4) +Solution The shaded region in Fig 12.3 is the feasible region ABC determined by the +system of constraints (2) to (4), which is bounded. The coordinates of corner points +Corner Point Corresponding value +of Z +(0, 0) +0 +(30, 0) +120 ← +Maximum +(20, 30) +110 +(0, 50) +50 +Corner Point Corresponding value +of Z +(0, 5) +2500 +(4, 3) +2300 +(0, 6) +3000 +Minimum +← +Fig 12.3 +Reprint 2025-26 + +LINEAR PROGRAMMING 401 +A, B and C are (0,5), (4,3) and (0,6) respectively. Now we evaluate Z = 200x + 500y +at these points. +Hence, minimum value of Z is 2300 attained at the point (4, 3) +Example 3 Solve the following problem graphically: +Minimise and Maximise Z = 3x + 9y +... (1) +subject to the constraints: +x + 3y ≤60 +... (2) +x + y ≥10 +... (3) +x ≤y +... (4) +x ≥ 0, y ≥0 +... (5) +Solution First of all, let us graph the feasible region of the system of linear inequalities +(2) to (5). The feasible region ABCD is shown in the Fig 12.4. Note that the region is +bounded. The coordinates of the corner points A, B, C and D are (0, 10), (5, 5), (15,15) +and (0, 20) respectively. +Fig 12.4 +Corner +Corresponding value of +Point + Z = 3x + 9y +A (0, 10) +90 +B (5, 5) +60 +C (15, 15) +180 +D (0, 20) +180 +Minimum +Maximum +(Multiple +optimal +solutions) +← +}← +We now find the minimum and maximum value of Z. From the table, we find that +the minimum value of Z is 60 at the point B (5, 5) of the feasible region. +The maximum value of Z on the feasible region occurs at the two corner points +C (15, 15) and D (0, 20) and it is 180 in each case. +Remark Observe that in the above example, the problem has multiple optimal solutions +at the corner points C and D, i.e. the both points produce same maximum value 180. In +such cases, you can see that every point on the line segment CD joining the two corner +points C and D also give the same maximum value. Same is also true in the case if the +two points produce same minimum value. +Reprint 2025-26 + +402 +MATHEMATICS +Example 4 Determine graphically the minimum value of the objective function +Z = – 50x + 20y +... (1) +subject to the constraints: +2x – y ≥ – 5 +... (2) +3x + y ≥ 3 +... (3) +2x – 3y ≤ 12 +... (4) +x ≥0, y ≥ 0 +... (5) +Solution First of all, let us graph the feasible region of the system of inequalities (2) to +(5). The feasible region (shaded) is shown in the Fig 12.5. Observe that the feasible +region is unbounded. +We now evaluate Z at the corner points. +From this table, we find that – 300 is the smallest value of Z at the corner point +(6, 0). Can we say that minimum value of Z is – 300? Note that if the region would +have been bounded, this smallest value of Z is the minimum value of Z (Theorem 2). +But here we see that the feasible region is unbounded. Therefore, – 300 may or may +not be the minimum value of Z. To decide this issue, we graph the inequality +– 50x + 20y < – 300 (see Step 3(ii) of corner Point Method.) +i.e., +– 5x + 2y < – 30 +and check whether the resulting open half plane has points in common with feasible +region or not. If it has common points, then –300 will not be the minimum value of Z. +Otherwise, –300 will be the minimum value of Z. +Fig 12.5 +Corner Point +Z = – 50x + 20y +(0, 5) +100 +(0, 3) +60 +(1, 0) +–50 +(6, 0) +– 300 +smallest +← +Reprint 2025-26 + +LINEAR PROGRAMMING 403 +As shown in the Fig 12.5, it has common points. Therefore, Z = –50 x + 20 y +has no minimum value subject to the given constraints. +In the above example, can you say whether z = – 50 x + 20 y has the maximum +value 100 at (0,5)? For this, check whether the graph of – 50 x + 20 y > 100 has points +in common with the feasible region. (Why?) +Example 5 Minimise Z = 3x + 2y +subject to the constraints: +x + y ≥8 +... (1) +3x + 5y ≤15 +... (2) +x ≥ 0, y ≥0 +... (3) +Solution Let us graph the inequalities (1) to (3) (Fig 12.6). Is there any feasible region? +Why is so? +From Fig 12.6, you can see that +there is no point satisfying all the +constraints simultaneously. Thus, the +problem is having no feasible region and +hence no feasible solution. +Remarks From the examples which we +have discussed so far, we notice some +general features of linear programming +problems: +(i) +The feasible region is always a +convex region. +(ii) +The maximum (or minimum) +solution of the objective function occurs at the vertex (corner) of the feasible +region. If two corner points produce the same maximum (or minimum) value +of the objective function, then every point on the line segment joining these +points will also give the same maximum (or minimum) value. +EXERCISE 12.1 +Solve the following Linear Programming Problems graphically: +1. +Maximise Z = 3x + 4y +subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0. +Fig 12.6 +Reprint 2025-26 + +404 +MATHEMATICS +2. +Minimise Z = – 3x + 4 y +subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0. +3. +Maximise Z = 5x + 3y +subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0. +4. +Minimise Z = 3x + 5y +such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0. +5. +Maximise Z = 3x + 2y +subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0. +6. +Minimise Z = x + 2y +subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0. +Show that the minimum of Z occurs at more than two points. +7. +Minimise and Maximise Z = 5x + 10 y +subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0. +8. +Minimise and Maximise Z = x + 2y +subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0. +9. +Maximise Z = – x + 2y, subject to the constraints: +x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0. +10. +Maximise Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0. +Summary +® A linear programming problem is one that is concerned with finding the optimal +value (maximum or minimum) of a linear function of several variables (called +objective function) subject to the conditions that the variables are +non-negative and satisfy a set of linear inequalities (called linear constraints). +Variables are sometimes called decision variables and are non-negative. +Historical Note +In the World War II, when the war operations had to be planned to economise +expenditure, maximise damage to the enemy, linear programming problems +came to the forefront. +The first problem in linear programming was formulated in 1941 by the Russian +mathematician, L. Kantorovich and the American economist, F. L. Hitchcock, +Reprint 2025-26 + +LINEAR PROGRAMMING 405 +both of whom worked at it independently of each other. This was the well +known transportation problem. In 1945, an English economist, G.Stigler, +described yet another linear programming problem – that of determining an +optimal diet. +In 1947, the American economist, G. B. Dantzig suggested an efficient method +known as the simplex method which is an iterative procedure to solve any +linear programming problem in a finite number of steps. +L. Katorovich and American mathematical economist, T. C. Koopmans were +awarded the nobel prize in the year 1975 in economics for their pioneering +work in linear programming. With the advent of computers and the necessary +softwares, it has become possible to apply linear programming model to +increasingly complex problems in many areas. +—v— +Reprint 2025-26" +class_12,13,Probability,ncert_books/class_12/lemh2dd/lemh207.pdf,"406 +MATHEMATICS +vThe theory of probabilities is simply the Science of logic +quantitatively treated. – C.S. PEIRCE v +13.1 Introduction +In earlier Classes, we have studied the probability as a +measure of uncertainty of events in a random experiment. +We discussed the axiomatic approach formulated by +Russian Mathematician, A.N. Kolmogorov (1903-1987) +and treated probability as a function of outcomes of the +experiment. We have also established equivalence between +the axiomatic theory and the classical theory of probability +in case of equally likely outcomes. On the basis of this +relationship, we obtained probabilities of events associated +with discrete sample spaces. We have also studied the +addition rule of probability. In this chapter, we shall discuss +the important concept of conditional probability of an event +given that another event has occurred, which will be helpful +in understanding the Bayes' theorem, multiplication rule of +probability and independence of events. We shall also learn +an important concept of random variable and its probability +distribution and also the mean and variance of a probability distribution. In the last +section of the chapter, we shall study an important discrete probability distribution +called Binomial distribution. Throughout this chapter, we shall take up the experiments +having equally likely outcomes, unless stated otherwise. +13.2 Conditional Probability +Uptill now in probability, we have discussed the methods of finding the probability of +events. If we have two events from the same sample space, does the information +about the occurrence of one of the events affect the probability of the other event? Let +us try to answer this question by taking up a random experiment in which the outcomes +are equally likely to occur. +Consider the experiment of tossing three fair coins. The sample space of the +experiment is +S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} +Chapter 13 +PROBABILITY +Pierre de Fermat +(1601-1665) +Reprint 2025-26 + +PROBABILITY 407 +Since the coins are fair, we can assign the probability 1 +8 to each sample point. Let +E be the event ‘at least two heads appear’ and F be the event ‘first coin shows tail’. +Then +E = {HHH, HHT, HTH, THH} +and +F = {THH, THT, TTH, TTT} +Therefore +P(E) = P ({HHH}) + P ({HHT}) + P ({HTH}) + P ({THH}) += 1 +1 +1 +1 +1 +8 +8 +8 +8 +2 ++ ++ ++ += + (Why ?) +and +P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT}) += 1 +1 +1 +1 +1 +8 +8 +8 +8 +2 ++ ++ ++ += +Also +E ∩ F = {THH} +with +P(E ∩ F) = P({THH}) = 1 +8 +Now, suppose we are given that the first coin shows tail, i.e. F occurs, then what is +the probability of occurrence of E? With the information of occurrence of F, we are +sure that the cases in which first coin does not result into a tail should not be considered +while finding the probability of E. This information reduces our sample space from the +set S to its subset F for the event E. In other words, the additional information really +amounts to telling us that the situation may be considered as being that of a new +random experiment for which the sample space consists of all those outcomes only +which are favourable to the occurrence of the event F. +Now, the sample point of F which is favourable to event E is THH. +Thus, Probability of E considering F as the sample space = 1 +4 , +or +Probability of E given that the event F has occurred = 1 +4 +This probability of the event E is called the conditional probability of E given +that F has already occurred, and is denoted by P (E|F). +Thus +P(E|F) = 1 +4 +Note that the elements of F which favour the event E are the common elements of +E and F, i.e. the sample points of E ∩ F. +Reprint 2025-26 + + 408 +MATHEMATICS +Thus, we can also write the conditional probability of E given that F has occurred as +P(E|F) = +Numberof elementaryeventsfavourableto E +F +Numberof elementaryeventswhicharefavourableto F +∩ += +(E +F) +(F) +n +n +∩ +Dividing the numerator and the denominator by total number of elementary events +of the sample space, we see that P(E|F) can also be written as +P(E|F) = +(E +F) +P(E +F) +(S) +(F) +P(F) +(S) +n +n +n +n +∩ +∩ += +... (1) +Note that (1) is valid only when P(F) ≠ 0 i.e., F ≠ φ (Why?) +Thus, we can define the conditional probability as follows : +Definition 1 If E and F are two events associated with the same sample space of a +random experiment, the conditional probability of the event E given that F has occurred, +i.e. P (E|F) is given by +P(E|F) = P(E +F) +P(F) +∩ + provided P(F) ≠ 0 +13.2.1 Properties of conditional probability +Let E and F be events of a sample space S of an experiment, then we have +Property 1 P(S|F) = P(F|F) = 1 +We know that +P(S|F) = P(S +F) +P(F) +1 +P(F) +P(F) +∩ += += +Also +P(F|F) = P(F +F) +P(F) 1 +P(F) +P(F) +∩ += += +Thus +P(S|F) = P(F|F) = 1 +Property 2 If A and B are any two events of a sample space S and F is an event +of S such that P(F) ≠ 0, then +P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F) +Reprint 2025-26 + +PROBABILITY 409 +In particular, if A and B are disjoint events, then +P((A∪B)|F) = P(A|F) + P(B|F) +We have +P((A∪B)|F) = P[(A +B) +F] +P(F) +∪ +∩ += P[(A +F) +(B +F)] +P(F) +∩ +∪ +∩ +(by distributive law of union of sets over intersection) += +P(A +F)+P(B +F)– P(A +B +F) +P(F) +∩ +∩ +∩ +∩ += P(A +F) +P(B +F) +P[(A +B) +F] +P(F) +P(F) +P(F) +∩ +∩ +∩ +∩ ++ +− += P(A|F) + P(B|F) – P((A ∩B)|F) +When A and B are disjoint events, then +P((A ∩ B)|F) = 0 +⇒ +P((A ∪ B)|F) = P(A|F) + P(B|F) +Property 3 P(E′|F) = 1 − P(E|F) +From Property 1, we know that P(S|F) = 1 +⇒ +P(E ∪ E′|F) = 1 + since S = E ∪ E′ +⇒ +P(E|F) + P (E′|F) = 1 + since E and E′ are disjoint events +Thus, +P(E′|F) = 1 − P(E|F) +Let us now take up some examples. +Example 1 If P(A) = 7 +13 , P(B) = 9 +13 and P(A ∩ B) = 4 +13 , evaluate P(A|B). +Solution We have +4 +P(A +B) +4 +13 +P(A|B)= +9 +P(B) +9 +13 +∩ += += +Example 2 A family has two children. What is the probability that both the children are +boys given that at least one of them is a boy ? +Reprint 2025-26 + + 410 +MATHEMATICS +Solution Let b stand for boy and g for girl. The sample space of the experiment is +S = {(b, b), (g, b), (b, g), (g, g)} +Let E and F denote the following events : +E : ‘both the children are boys’ +F : ‘at least one of the child is a boy’ +Then +E = {(b,b)} and F = {(b,b), (g,b), (b,g)} +Now +E ∩ F = {(b,b)} +Thus +P(F) = 3 +4 and P (E ∩ F )= 1 +4 +Therefore +P(E|F) = +1 +P(E +F) +1 +4 +3 +P(F) +3 +4 +∩ += += +Example 3 Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and +then one card is drawn randomly. If it is known that the number on the drawn card is +more than 3, what is the probability that it is an even number? +Solution Let A be the event ‘the number on the card drawn is even’ and B be the +event ‘the number on the card drawn is greater than 3’. We have to find P(A|B). +Now, the sample space of the experiment is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} +Then +A = {2, 4, 6, 8, 10}, B = {4, 5, 6, 7, 8, 9, 10} +and +A ∩ B = {4, 6, 8, 10} +Also +P(A) = 5 +7 +4 +, P(B) = +and P(A +B) +10 +10 +10 +∩ += +Then +P(A|B) = +4 +P(A +B) +4 +10 +7 +P(B) +7 +10 +∩ += += +Example 4 In a school, there are 1000 students, out of which 430 are girls. It is known +that out of 430, 10% of the girls study in class XII. What is the probability that a student +chosen randomly studies in Class XII given that the chosen student is a girl? +Solution Let E denote the event that a student chosen randomly studies in Class XII +and F be the event that the randomly chosen student is a girl. We have to find P (E|F). +Reprint 2025-26 + +PROBABILITY 411 +Now + P(F) = 430 +0.43 +1000 = + and +43 +P(E +F)= +0.043 +1000 +∩ += + (Why?) +Then + P(E|F) = P(E +F) +0.043 +0.1 +P(F) +0.43 +∩ += += +Example 5 A die is thrown three times. Events A and B are defined as below: +A : 4 on the third throw +B : 6 on the first and 5 on the second throw +Find the probability of A given that B has already occurred. +Solution The sample space has 216 outcomes. +Now +A = +(1,1,4) (1,2,4) ... (1,6,4) (2,1,4) (2,2,4) ... (2,6,4) +(3,1,4) (3,2,4) ... (3,6,4) (4,1,4) (4,2,4) ...(4,6,4) +(5,1,4) (5,2,4) ... (5,6,4) (6,1,4) (6,2,4) ...(6,6,4) + + + + + + + + + + +B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)} +and +A ∩ B = {(6,5,4)}. +Now +P(B) = +6 +216 and P (A ∩ B) = 1 +216 +Then +P(A|B) = +1 +P(A +B) +1 +216 +6 +P(B) +6 +216 +∩ += += +Example 6 A die is thrown twice and the sum of the numbers appearing is observed +to be 6. What is the conditional probability that the number 4 has appeared at least +once? +Solution Let E be the event that ‘number 4 appears at least once’ and F be the event +that ‘the sum of the numbers appearing is 6’. +Then, +E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)} +and +F = {(1,5), (2,4), (3,3), (4,2), (5,1)} +We have +P(E) = 11 +36 and P(F) = 5 +36 +Also +E∩F = {(2,4), (4,2)} +Reprint 2025-26 + + 412 +MATHEMATICS +Therefore +P(E∩F) = +2 +36 +Hence, the required probability +P(E|F) = +2 +P(E +F) +2 +36 +5 +P(F) +5 +36 +∩ += += +For the conditional probability discussed above, we have considered the elemen- +tary events of the experiment to be equally likely and the corresponding definition of +the probability of an event was used. However, the same definition can also be used in +the general case where the elementary events of the sample space are not equally +likely, the probabilities P(E∩F) and P(F) being calculated accordingly. Let us take up +the following example. +Example 7 Consider the experiment of tossing a coin. If the coin shows head, toss it +again but if it shows tail, then throw a die. Find the +conditional probability of the event that ‘the die shows +a number greater than 4’ given that ‘there is at least +one tail’. +Solution The outcomes of the experiment can be +represented in following diagrammatic manner called +the ‘tree diagram’. +The sample space of the experiment may be +described as +S = {(H,H), (H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)} +where (H, H) denotes that both the tosses result into +head and (T, i) denote the first toss result into a tail and +the number i appeared on the die for i = 1,2,3,4,5,6. +Thus, the probabilities assigned to the 8 elementary +events +(H, H), (H, T), (T, 1), (T, 2), (T, 3) (T, 4), (T, 5), (T, 6) +are 1 1 1 +1 +1 +1 +1 +1 +, +, +, +, +, +, +, +4 4 12 12 12 12 12 12 respectively which is +clear from the Fig 13.2. +Fig 13.1 +Fig 13.2 +Reprint 2025-26 + +PROBABILITY 413 +Let F be the event that ‘there is at least one tail’ and E be the event ‘the die shows +a number greater than 4’. Then +F = {(H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)} +E = {(T,5), (T,6)} and E ∩ F = {(T,5), (T,6)} +Now +P(F) = P({(H,T)}) + P ({(T,1)}) + P ({(T,2)}) + P ({(T,3)}) ++ P ({(T,4)}) + P({(T,5)}) + P({(T,6)}) += 1 +1 +1 +1 +1 +1 +1 +3 +4 +12 +12 +12 +12 +12 +12 +4 ++ ++ ++ ++ ++ ++ += +and +P(E ∩ F) = P ({(T,5)}) + P ({(T,6)}) = 1 +1 +1 +12 +12 +6 ++ += +Hence +P(E|F) = +1 +P(E +F) +2 +6 +3 +P(F) +9 +4 +∩ += += +EXERCISE 13.1 +1. +Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and +P(E ∩ F) = 0.2, find P(E|F) and P(F|E) +2. +Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32 +3. +If P(A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find +(i) P(A ∩ B) +(ii) P(A|B) +(iii) P(A ∪ B) +4. +Evaluate P(A ∪ B), if 2P(A) = P(B) = 5 +13 and P(A|B) = 2 +5 +5. +If P(A) = 6 +11 , P(B) = 5 +11 and P(A ∪ B) +7 +11 += +, find +(i) P(A∩B) +(ii) P(A|B) +(iii) P(B|A) +Determine P(E|F) in Exercises 6 to 9. +6. +A coin is tossed three times, where +(i) E : head on third toss , F : heads on first two tosses +(ii) E : at least two heads , F : at most two heads +(iii) E : at most two tails , F : at least one tail +Reprint 2025-26 + + 414 +MATHEMATICS +7. +Two coins are tossed once, where +(i) +E : tail appears on one coin, +F : one coin shows head +(ii) +E : no tail appears, +F : no head appears +8. +A die is thrown three times, +E : 4 appears on the third toss, +F : 6 and 5 appears respectively +on first two tosses +9. +Mother, father and son line up at random for a family picture +E : son on one end, +F : father in middle +10. +A black and a red dice are rolled. +(a) Find the conditional probability of obtaining a sum greater than 9, given +that the black die resulted in a 5. +(b) Find the conditional probability of obtaining the sum 8, given that the red die +resulted in a number less than 4. +11. +A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5} +Find +(i) P(E|F) and P(F|E) +(ii) P(E|G) and P(G|E) +(iii) P((E ∪ F)|G) and P((E ∩ F)|G) +12. +Assume that each born child is equally likely to be a boy or a girl. If a family has +two children, what is the conditional probability that both are girls given that +(i) the youngest is a girl, (ii) at least one is a girl? +13. +An instructor has a question bank consisting of 300 easy True / False questions, +200 difficult True / False questions, 500 easy multiple choice questions and 400 +difficult multiple choice questions. If a question is selected at random from the +question bank, what is the probability that it will be an easy question given that it +is a multiple choice question? +14. +Given that the two numbers appearing on throwing two dice are different. Find +the probability of the event ‘the sum of numbers on the dice is 4’. +15. +Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the +die again and if any other number comes, toss a coin. Find the conditional probability +of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’. +In each of the Exercises 16 and 17 choose the correct answer: +16. +If P(A) = 1 +2 , P(B) = 0, then P(A|B) is +(A) 0 +(B) 1 +2 +(C) not defined +(D) 1 +Reprint 2025-26 + +PROBABILITY 415 +17. +If A and B are events such that P(A|B) = P(B|A), then +(A) A ⊂ B but A ≠ B +(B) A = B +(C) A ∩ B = φ +(D) P(A) = P(B) +13.3 Multiplication Theorem on Probability +Let E and F be two events associated with a sample space S. Clearly, the set E ∩ F +denotes the event that both E and F have occurred. In other words, E ∩ F denotes the +simultaneous occurrence of the events E and F. The event E ∩ F is also written as EF. +Very often we need to find the probability of the event EF. For example, in the +experiment of drawing two cards one after the other, we may be interested in finding +the probability of the event ‘a king and a queen’. The probability of event EF is obtained +by using the conditional probability as obtained below : +We know that the conditional probability of event E given that F has occurred is +denoted by P(E|F) and is given by +P(E|F) = P(E +F) ,P(F) 0 +P(F) +∩ +≠ +From this result, we can write +P(E ∩ F) = P(F) . P(E|F) +... (1) +Also, we know that +P(F|E) = P(F +E) ,P(E) 0 +P(E) +∩ +≠ +or +P(F|E) = P(E +F) +P(E) +∩ + (since E ∩ F = F ∩ E) +Thus, +P(E ∩ F) = P(E). P(F|E) +.... (2) +Combining (1) and (2), we find that +P(E ∩ F) = P(E) P(F|E) += P(F) P(E|F) provided P(E) ≠ 0 and P(F) ≠ 0. +The above result is known as the multiplication rule of probability. +Let us now take up an example. +Example 8 An urn contains 10 black and 5 white balls. Two balls are drawn from the +urn one after the other without replacement. What is the probability that both drawn +balls are black? +Solution Let E and F denote respectively the events that first and second ball drawn +are black. We have to find P(E ∩ F) or P (EF). +Reprint 2025-26 + + 416 +MATHEMATICS +Now +P(E) = P (black ball in first draw) = 10 +15 +Also given that the first ball drawn is black, i.e., event E has occurred, now there +are 9 black balls and five white balls left in the urn. Therefore, the probability that the +second ball drawn is black, given that the ball in the first draw is black, is nothing but +the conditional probability of F given that E has occurred. +i.e. +P(F|E) = 9 +14 +By multiplication rule of probability, we have +P (E ∩ F) = P(E) P(F|E) += 10 +9 +3 +15 +14 +7 +× += +Multiplication rule of probability for more than two events If E, F and G are +three events of sample space, we have +P(E ∩ F ∩ G) = P(E) P(F|E) P(G|(E ∩ F)) = P(E) P(F|E) P(G|EF) +Similarly, the multiplication rule of probability can be extended for four or +more events. +The following example illustrates the extension of multiplication rule of probability +for three events. +Example 9 Three cards are drawn successively, without replacement from a pack of +52 well shuffled cards. What is the probability that first two cards are kings and the +third card drawn is an ace? +Solution Let K denote the event that the card drawn is king and A be the event that +the card drawn is an ace. Clearly, we have to find P (KKA) +Now +P(K) = 4 +52 +Also, P(K|K) is the probability of second king with the condition that one king has +already been drawn. Now there are three kings in (52 − 1) = 51 cards. +Therefore +P(K|K) = 3 +51 +Lastly, P(A|KK) is the probability of third drawn card to be an ace, with the condition +that two kings have already been drawn. Now there are four aces in left 50 cards. +Reprint 2025-26 + +PROBABILITY 417 +Therefore +P(A|KK) = 4 +50 +By multiplication law of probability, we have +P(KKA) = P(K) P(K|K) P(A|KK) += 4 +3 +4 +2 +52 +51 +50 +5525 +× +× += +13.4 Independent Events +Consider the experiment of drawing a card from a deck of 52 playing cards, in which +the elementary events are assumed to be equally likely. If E and F denote the events +'the card drawn is a spade' and 'the card drawn is an ace' respectively, then +P(E) = 13 +1 +4 +1 +and P(F) +52 +4 +52 +13 += += += +Also E and F is the event ' the card drawn is the ace of spades' so that +P(E ∩F) = 1 +52 +Hence +P(E|F) = +1 +P(E +F) +1 +52 +1 +P(F) +4 +13 +∩ += += +Since P(E) = 1 +4 = P (E|F), we can say that the occurrence of event F has not +affected the probability of occurrence of the event E. +We also have +P(F|E) = +1 +P(E +F) +1 +52 +P(F) +1 +P(E) +13 +4 +∩ += += += +Again, P(F) = 1 +13 = P(F|E) shows that occurrence of event E has not affected +the probability of occurrence of the event F. +Thus, E and F are two events such that the probability of occurrence of one of +them is not affected by occurrence of the other. +Such events are called independent events. +Reprint 2025-26 + + 418 +MATHEMATICS +Definition 2 Two events E and F are said to be independent, if +P(F|E) = P (F) provided P (E) ≠ 0 +and +P (E|F) = P (E) provided P (F) ≠ 0 +Thus, in this definition we need to have P (E) ≠ 0 and P(F) ≠ 0 +Now, by the multiplication rule of probability, we have +P(E ∩ F) = P(E) . P (F|E) +... (1) +If E and F are independent, then (1) becomes +P(E ∩ F) = P(E) . P(F) +... (2) +Thus, using (2), the independence of two events is also defined as follows: +Definition 3 Let E and F be two events associated with the same random experiment, +then E and F are said to be independent if +P(E ∩ F) = P(E) . P (F) +Remarks +(i) +Two events E and F are said to be dependent if they are not independent, i.e. if +P(E ∩ F ) ≠ P(E) . P (F) +(ii) +Sometimes there is a confusion between independent events and mutually +exclusive events. Term ‘independent’ is defined in terms of ‘probability of events’ +whereas mutually exclusive is defined in term of events (subset of sample space). +Moreover, mutually exclusive events never have an outcome common, but +independent events, may have common outcome. Clearly, ‘independent’ and +‘mutually exclusive’ do not have the same meaning. +In other words, two independent events having nonzero probabilities of occurrence +can not be mutually exclusive, and conversely, i.e. two mutually exclusive events +having nonzero probabilities of occurrence can not be independent. +(iii) +Two experiments are said to be independent if for every pair of events E and F, +where E is associated with the first experiment and F with the second experiment, +the probability of the simultaneous occurrence of the events E and F when the +two experiments are performed is the product of P(E) and P(F) calculated +separately on the basis of two experiments, i.e., P (E ∩ F) = P (E) . P(F) +(iv) +Three events A, B and C are said to be mutually independent, if +P(A ∩ B) = P(A) P(B) +P(A ∩ C) = P(A) P(C) +P(B ∩ C) = P(B) P(C) +and +P(A ∩ B ∩ C) = P(A) P(B) P(C) +Reprint 2025-26 + +PROBABILITY 419 +If at least one of the above is not true for three given events, we say that the +events are not independent. +Example 10 A die is thrown. If E is the event ‘the number appearing is a multiple of +3’ and F be the event ‘the number appearing is even’ then find whether E and F are +independent ? +Solution We know that the sample space is S = {1, 2, 3, 4, 5, 6} +Now +E = { 3, 6}, F = { 2, 4, 6} and E ∩ F = {6} +Then +P(E) = 2 +1 +3 +1 +1 +, P(F) +and P(E + F) +6 +3 +6 +2 +6 += += += +∩ += +Clearly +P(E ∩ F) = P(E). P (F) +Hence +E and F are independent events. +Example 11 An unbiased die is thrown twice. Let the event A be ‘odd number on the +first throw’ and B the event ‘odd number on the second throw’. Check the independence +of the events A and B. +Solution If all the 36 elementary events of the experiment are considered to be equally +likely, we have +P(A) = 18 +1 +36 +2 += + and +18 +1 +P(B) +36 +2 += += +Also +P(A ∩ B) = P (odd number on both throws) += 9 +1 +36 +4 += +Now +P(A) P(B) = 1 +1 +1 +2 +2 +4 +× += +Clearly +P(A ∩ B) = P(A) × P(B) +Thus, +A and B are independent events +Example 12 Three coins are tossed simultaneously. Consider the event E ‘three heads +or three tails’, F ‘at least two heads’ and G ‘at most two heads’. Of the pairs (E,F), +(E,G) and (F,G), which are independent? which are dependent? +Solution The sample space of the experiment is given by +S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} +Clearly +E = {HHH, TTT}, F= {HHH, HHT, HTH, THH} +Reprint 2025-26 + + 420 +MATHEMATICS +and +G = {HHT, HTH, THH, HTT, THT, TTH, TTT} +Also +E ∩ F = {HHH}, E ∩ G = {TTT}, F ∩ G = {HHT, HTH, THH} +Therefore +P(E) = 2 +1 +4 +1 +7 +, P(F) +, P(G) +8 +4 +8 +2 +8 += += += += +and +P(E∩F) = 1 +1 +3 +, P(E +G) +, P(F +G) +8 +8 +8 +∩ += +∩ += +Also +P(E) . P(F) = 1 +1 +1 +1 +7 +7 +, P(E) P(G) +4 +2 +8 +4 +8 +32 +× += +⋅ += +× += +and +P(F) . P(G) = 1 +7 +7 +2 +8 +16 +× += +Thus +P(E ∩ F) = P(E) . P(F) +P(E ∩ G) ≠P(E) . P(G) +and +P(F ∩ G) ≠P (F) . P(G) +Hence, the events (E and F) are independent, and the events (E and G) and +(F and G) are dependent. +Example 13 Prove that if E and F are independent events, then so are the events +E and F′. +Solution Since E and F are independent, we have +P(E ∩ F) = P(E) . P(F) +....(1) +From the venn diagram in Fig 13.3, it is clear +that E ∩ F and E ∩ F′ are mutually exclusive events +and also E =(E ∩ F) ∪ (E ∩ F′). +Therefore +P(E) = P(E ∩ F) + P(E ∩ F′) +or +P(E ∩ F′) = P(E) − P(E ∩ F) += P(E) − P(E) . P(F) +(by (1)) += P(E) (1−P(F)) += P(E). P(F′) +Hence, E and F′ are independent +(E +F )’ +∩ +(E +F) +’∩ +E +F +S +(E +F) +∩ +(E +F ) +’ +’ +∩ +Fig 13.3 +Reprint 2025-26 + +PROBABILITY 421 +ANote In a similar manner, it can be shown that if the events E and F are +independent, then +(a) +E′ and F are independent, +(b) +E′ and F′ are independent +Example 14 If A and B are two independent events, then the probability of occurrence +of at least one of A and B is given by 1– P(A′) P(B′) +Solution We have +P(at least one of A and B) = P(A ∪ B) += P(A) + P(B) − P(A ∩ B) += P(A) + P(B) − P(A) P(B) += P(A) + P(B) [1−P(A)] += P(A) + P(B). P(A′) += 1− P(A′) + P(B) P(A′) += 1− P(A′) [1− P(B)] += 1− P(A′) P (B′) +EXERCISE 13.2 +1. +If P(A) +3 +5 += + and P (B) +1 +5 += +, find P (A ∩ B) if A and B are independent events. +2. Two cards are drawn at random and without replacement from a pack of 52 +playing cards. Find the probability that both the cards are black. +3. +A box of oranges is inspected by examining three randomly selected oranges +drawn without replacement. If all the three oranges are good, the box is approved +for sale, otherwise, it is rejected. Find the probability that a box containing 15 +oranges out of which 12 are good and 3 are bad ones will be approved for sale. +4. +A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on +the coin’ and B be the event ‘3 on the die’. Check whether A and B are +independent events or not. +5. +A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, +‘the number is even,’ and B be the event, ‘the number is red’. Are A and B +independent? +6. +Let E and F be events with P(E) +3 +5 += +, P(F) +3 +10 += + and P (E ∩ F) = 1 +5 . Are +E and F independent? +Reprint 2025-26 + + 422 +MATHEMATICS +7. +Given that the events A and B are such that P(A) = 1 +2 , P(A ∪ B) = 3 +5 and +P(B) = p. Find p if they are (i) mutually exclusive (ii) independent. +8. +Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find +(i) P(A ∩ B) +(ii) P(A ∪ B) +(iii) P(A|B) +(iv) P(B|A) +9. +If A and B are two events such that P(A) = 1 +4 , P (B) = 1 +2 and P(A ∩ B) = 1 +8 , +find P (not A and not B). +10. +Events A and B are such that P (A) = 1 +2 , P(B) = 7 +12 and P(not A or not B) = 1 +4 . +State whether A and B are independent ? +11. +Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. +Find +(i) P(A and B) +(ii) P(A and not B) +(iii) P(A or B) +(iv) P(neither A nor B) +12. +A die is tossed thrice. Find the probability of getting an odd number at least once. +13. +Two balls are drawn at random with replacement from a box containing 10 black +and 8 red balls. Find the probability that +(i) both balls are red. +(ii) first ball is black and second is red. +(iii) one of them is black and other is red. +14. +Probability of solving specific problem independently by A and B are 1 +2 and 1 +3 +respectively. If both try to solve the problem independently, find the probability +that +(i) the problem is solved +(ii) exactly one of them solves the problem. +15. +One card is drawn at random from a well shuffled deck of 52 cards. In which of +the following cases are the events E and F independent ? +(i) E : ‘the card drawn is a spade’ +F : ‘the card drawn is an ace’ +(ii) E : ‘the card drawn is black’ +F : ‘the card drawn is a king’ +(iii) E : ‘the card drawn is a king or queen’ +F : ‘the card drawn is a queen or jack’. +Reprint 2025-26 + +PROBABILITY 423 +16. +In a hostel, 60% of the students read Hindi newspaper, 40% read English +newspaper and 20% read both Hindi and English newspapers. A student is +selected at random. +(a) Find the probability that she reads neither Hindi nor English newspapers. +(b) If she reads Hindi newspaper, find the probability that she reads English +newspaper. +(c) If she reads English newspaper, find the probability that she reads Hindi +newspaper. +Choose the correct answer in Exercises 17 and 18. +17. +The probability of obtaining an even prime number on each die, when a pair of +dice is rolled is +(A) 0 +(B) 1 +3 +(C) +1 +12 +(D) +1 +36 +18. +Two events A and B will be independent, if +(A) A and B are mutually exclusive +(B) P(A′B′) = [1 – P(A)] [1 – P(B)] +(C) P(A) = P(B) +(D) P(A) + P(B) = 1 +13.5 Bayes' Theorem +Consider that there are two bags I and II. Bag I contains 2 white and 3 red balls and +Bag II contains 4 white and 5 red balls. One ball is drawn at random from one of the +bags. We can find the probability of selecting any of the bags (i.e. 1 +2 ) or probability of +drawing a ball of a particular colour (say white) from a particular bag (say Bag I). In +other words, we can find the probability that the ball drawn is of a particular colour, if +we are given the bag from which the ball is drawn. But, can we find the probability that +the ball drawn is from a particular bag (say Bag II), if the colour of the ball drawn is +given? Here, we have to find the reverse probability of Bag II to be selected when an +event occurred after it is known. Famous mathematician, John Bayes' solved the problem +of finding reverse probability by using conditional probability. The formula developed +by him is known as ‘Bayes theorem’ which was published posthumously in 1763. +Before stating and proving the Bayes' theorem, let us first take up a definition and +some preliminary results. +13.5.1 Partition of a sample space +A set of events E1, E2, ..., En is said to represent a partition of the sample space S if +(a) Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, 3, ..., n +Reprint 2025-26 + + 424 +MATHEMATICS +Fig 13.4 +(b) E1 ∪ Ε2 ∪ ... ∪ En= S and +(c) P(Ei) > 0 for all i = 1, 2, ..., n. +In other words, the events E1, E2, ..., En represent a partition of the sample space +S if they are pairwise disjoint, exhaustive and have nonzero probabilities. +As an example, we see that any nonempty event E and its complement E′ form a +partition of the sample space S since they satisfy E ∩ E′ = φ and E ∪ E′ = S. +From the Venn diagram in Fig 13.3, one can easily observe that if E and F are any +two events associated with a sample space S, then the set {E ∩ F′, E ∩ F, E′ ∩ F, E′ ∩ F′} +is a partition of the sample space S. It may be mentioned that the partition of a sample +space is not unique. There can be several partitions of the same sample space. +We shall now prove a theorem known as Theorem of total probability. +13.5.2 Theorem of total probability +Let {E1, E2,...,En} be a partition of the sample space S, and suppose that each of the +events E1, E2,..., En has nonzero probability of occurrence. Let A be any event associated +with S, then +P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + ... + P(En) P(A|En) += +1 +P(E )P(A|E ) +n +j +j +j=∑ +Proof Given that E1, E2,..., En is a partition of the sample space S (Fig 13.4). Therefore, +S = E1 ∪ E2 ∪ ... ∪ En + ... (1) +and +Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, ..., n +Now, we know that for any event A, +A = A ∩ S += A ∩ (E1 ∪ E2 ∪ ... ∪ En) += (A ∩ E1) ∪ (A ∩ E2) ∪ ...∪ (A ∩ En) +Also A ∩ Ei and A ∩ Ej are respectively the subsets of Ei and Ej. We know that +Ei and Ej are disjoint, for i +j +≠ +, therefore, A ∩ Ei and A ∩ Ej are also disjoint for all +i ≠ j, i, j = 1, 2, ..., n. +Thus, +P(A) = P [(A ∩ E1) ∪ (A ∩ E2)∪ .....∪ (A ∩ En)] += P (A ∩ E1) + P (A ∩ E2) + ... + P (A ∩ En) +Now, by multiplication rule of probability, we have +P(A ∩ Ei) = P(Ei) P(A|Ei) as P (Ei) ≠ 0∀i = 1,2,..., n +Reprint 2025-26 + +PROBABILITY 425 +Therefore, +P (A) = P (E1) P (A|E1) + P (E2) P (A|E2) + ... + P (En)P(A|En) +or +P(A) = +1 +P(E )P(A|E ) +n +j +j +j=∑ +Example 15 A person has undertaken a construction job. The probabilities are 0.65 +that there will be strike, 0.80 that the construction job will be completed on time if there +is no strike, and 0.32 that the construction job will be completed on time if there is a +strike. Determine the probability that the construction job will be completed on time. +Solution Let A be the event that the construction job will be completed on time, and B +be the event that there will be a strike. We have to find P(A). +We have +P(B) = 0.65, P(no strike) = P(B′) = 1 − P(B) = 1 − 0.65 = 0.35 +P(A|B) = 0.32, P(A|B′) = 0.80 +Since events B and B′ form a partition of the sample space S, therefore, by theorem +on total probability, we have +P(A) = P(B) P(A|B) + P(B′) P(A|B′) + = 0.65 × 0.32 + 0.35 × 0.8 + = 0.208 + 0.28 = 0.488 +Thus, the probability that the construction job will be completed in time is 0.488. +We shall now state and prove the Bayes' theorem. +Bayes’ Theorem If E1, E2 ,..., En are n non empty events which constitute a partition +of sample space S, i.e. E1, E2 ,..., En are pairwise disjoint and E1∪ E2∪ ... ∪ En = S and +A is any event of nonzero probability, then +P(Ei|A) = +1 +P(E )P(A|E ) +P(E )P(A|E ) +i +i +n +j +j +j=∑ + for any i = 1, 2, 3, ..., n +Proof By formula of conditional probability, we know that +P(Ei|A) = P(A +E ) +P(A) +i +∩ += P(E )P(A|E ) +P(A) +i +i (by multiplication rule of probability) += +1 +P(E )P(A|E ) +P(E )P(A|E ) +i +i +n +j +j +j=∑ + (by the result of theorem of total probability) +Reprint 2025-26 + + 426 +MATHEMATICS +Remark The following terminology is generally used when Bayes' theorem is applied. +The events E1, E2, ..., En are called hypotheses. +The probability P(Ei) is called the priori probability of the hypothesis Ei +The conditional probability P(Ei |A) is called a posteriori probability of the +hypothesis Ei. +Bayes' theorem is also called the formula for the probability of ""causes"". Since the +Ei's are a partition of the sample space S, one and only one of the events Ei occurs (i.e. +one of the events Ei must occur and only one can occur). Hence, the above formula +gives us the probability of a particular Ei (i.e. a ""Cause""), given that the event A has +occurred. +The Bayes' theorem has its applications in variety of situations, few of which are +illustrated in following examples. +Example 16 Bag I contains 3 red and 4 black balls while another Bag II contains 5 red +and 6 black balls. One ball is drawn at random from one of the bags and it is found to +be red. Find the probability that it was drawn from Bag II. +Solution Let E1 be the event of choosing the bag I, E2 the event of choosing the bag II +and A be the event of drawing a red ball. +Then +P(E1) = P(E2) = 1 +2 +Also +P(A|E1) = P(drawing a red ball from Bag I) = 3 +7 +and +P(A|E2) = P(drawing a red ball from Bag II) = 5 +11 +Now, the probability of drawing a ball from Bag II, being given that it is red, +is P(E2|A) +By using Bayes' theorem, we have +P(E2|A) = +2 +2 +1 +1 +2 +2 +P(E )P(A|E ) +P(E )P(A|E )+P(E )P(A|E ) = +1 +5 +35 +2 11 +1 +3 +1 +5 +68 +2 +7 +2 11 +× += +× ++ +× +Example 17 Given three identical boxes I, II and III, each containing two coins. In +box I, both coins are gold coins, in box II, both are silver coins and in the box III, there +is one gold and one silver coin. A person chooses a box at random and takes out a coin. +If the coin is of gold, what is the probability that the other coin in the box is also of gold? +Reprint 2025-26 + +PROBABILITY 427 +Solution Let E1, E2 and E3 be the events that boxes I, II and III are chosen, respectively. +Then +P(E1) = P(E2) = P(E3) = 1 +3 +Also, let A be the event that ‘the coin drawn is of gold’ +Then +P(A|E1) = P(a gold coin from bag I) = 2 +2 = 1 +P(A|E2) = P(a gold coin from bag II) = 0 +P(A|E3) = P(a gold coin from bag III) = 1 +2 +Now, the probability that the other coin in the box is of gold += the probability that gold coin is drawn from the box I. += P(E1|A) +By Bayes' theorem, we know that +P(E1|A) = +1 +1 +1 +1 +2 +2 +3 +3 +P(E )P(A|E ) +P(E )P(A|E )+P(E )P(A|E )+P(E )P(A|E ) += +1 1 +2 +3 +1 +1 +1 +1 +3 +1 +0 +3 +3 +3 +2 +× += +× + × + × +Example 18 Suppose that the reliability of a HIV test is specified as follows: +Of people having HIV, 90% of the test detect the disease but 10% go undetected. Of +people free of HIV, 99% of the test are judged HIV–ive but 1% are diagnosed as +showing HIV+ive. From a large population of which only 0.1% have HIV, one person +is selected at random, given the HIV test, and the pathologist reports him/her as +HIV+ive. What is the probability that the person actually has HIV? +Solution Let E denote the event that the person selected is actually having HIV and A +the event that the person's HIV test is diagnosed as +ive. We need to find P(E|A). +Also E′ denotes the event that the person selected is actually not having HIV. +Clearly, {E, E′} is a partition of the sample space of all people in the population. +We are given that +P(E) = 0.1% +0.1 +0.001 +100 += += +Reprint 2025-26 + + 428 +MATHEMATICS +P(E′) = 1 – P(E) = 0.999 +P(A|E) = P(Person tested as HIV+ive given that he/she +is actually having HIV) += 90% +90 +0.9 +100 += += +and +P(A|E′) = P(Person tested as HIV +ive given that he/she +is actually not having HIV) += 1% = 1 +100 = 0.01 +Now, by Bayes' theorem +P(E|A) = +P(E)P(A|E) +P(E)P(A|E)+P(E )P(A|E ) +′ +′ += +0.001 0.9 +90 +0.001 0.9 0.999 0.01 +1089 +× += +× ++ +× += 0.083 approx. +Thus, the probability that a person selected at random is actually having HIV +given that he/she is tested HIV+ive is 0.083. +Example 19 In a factory which manufactures bolts, machines A, B and C manufacture +respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are +respectively defective bolts. A bolt is drawn at random from the product and is found +to be defective. What is the probability that it is manufactured by the machine B? +Solution Let events B1, B2, B3 be the following : +B1 : the bolt is manufactured by machine A +B2 : the bolt is manufactured by machine B +B3 : the bolt is manufactured by machine C +Clearly, B1, B2, B3 are mutually exclusive and exhaustive events and hence, they +represent a partition of the sample space. +Let the event E be ‘the bolt is defective’. +The event E occurs with B1 or with B2 or with B3. Given that, +P(B1) = 25% = 0.25, P (B2) = 0.35 and P(B3) = 0.40 +Again P(E|B1) = Probability that the bolt drawn is defective given that it is manu- +factured by machine A = 5% = 0.05 +Similarly, +P(E|B2) = 0.04, P(E|B3) = 0.02. +Reprint 2025-26 + +PROBABILITY 429 +Hence, by Bayes' Theorem, we have +P(B2|E) = +2 +2 +1 +1 +2 +2 +3 +3 +P(B )P(E|B ) +P(B )P(E|B )+P(B )P(E|B )+P(B )P(E|B ) += +0.35 +0.04 +0.25 +0.05 +0.35 +0.04 +0.40 +0.02 +× +× ++ +× ++ +× += 0.0140 +28 +0.0345 +69 += +Example 20 A doctor is to visit a patient. From the past experience, it is known that +the probabilities that he will come by train, bus, scooter or by other means of transport +are respectively 3 1 1 +2 +, +, +and +10 5 10 +5 . The probabilities that he will be late are 1 1 +1 +, , and +4 3 +12, +if he comes by train, bus and scooter respectively, but if he comes by other means of +transport, then he will not be late. When he arrives, he is late. What is the probability +that he comes by train? +Solution Let E be the event that the doctor visits the patient late and let T1, T2, T3, T4 +be the events that the doctor comes by train, bus, scooter, and other means of transport +respectively. +Then +P(T1) = +2 +3 +4 +3 +1 +1 +2 +, P(T ) +,P(T ) +and P(T ) +10 +5 +10 +5 += += += +(given) +P(E|T1) = Probability that the doctor arriving late comes by train = 1 +4 +Similarly, P(E|T2) = 1 +3 , P(E|T3) = 1 +12 and P(E|T4) = 0, since he is not late if he +comes by other means of transport. +Therefore, by Bayes' Theorem, we have +P(T1|E) = Probability that the doctor arriving late comes by train += +1 +1 +1 +1 +2 +2 +3 +3 +4 +4 +P(T )P(E|T ) +P(T )P(E|T )+P(T )P(E|T )+P(T )P(E|T )+P(T )P(E|T ) += +3 +1 +10 +4 +3 +1 +1 +1 +1 +1 +2 +0 +10 +4 +5 +3 +10 +12 +5 +× +× ++ +× ++ +× ++ +× + = 3 +120 +1 +40 +18 +2 +× += +Hence, the required probability is 1 +2 . +Reprint 2025-26 + + 430 +MATHEMATICS +Example 21 A man is known to speak truth 3 out of 4 times. He throws a die and +reports that it is a six. Find the probability that it is actually a six. +Solution Let E be the event that the man reports that six occurs in the throwing of the +die and let S1 be the event that six occurs and S2 be the event that six does not occur. +Then +P(S1) = Probability that six occurs = 1 +6 +P(S2) = Probability that six does not occur = 5 +6 +P(E|S1) = Probability that the man reports that six occurs when six has +actually occurred on the die += Probability that the man speaks the truth = 3 +4 +P(E|S2) = Probability that the man reports that six occurs when six has +not actually occurred on the die += Probability that the man does not speak the truth +3 +1 +1 +4 +4 += − += +Thus, by Bayes' theorem, we get +P(S1|E) = Probability that the report of the man that six has occurred is +actually a six += +1 +1 +1 +1 +2 +2 +P(S )P(E |S ) +P(S )P(E|S )+P(S )P(E|S ) += +1 +3 +1 +24 +3 +6 +4 +1 +3 +5 +1 +8 +8 +8 +6 +4 +6 +4 +× += +× += +× ++ +× +Hence, the required probability is 3. +8 +Remark A random variable is a real valued function whose domain is the sample +space of a random experiment. +For example, let us consider the experiment of tossing a coin two times in succession. +The sample space of the experiment is S = {HH, HT, TH, TT}. +Reprint 2025-26 + +PROBABILITY 431 +If X denotes the number of heads obtained, then X is a random variable and for +each outcome, its value is as given below : +X(HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0. +More than one random variables can be defined on the same sample space. For +example, let Y denote the number of heads minus the number of tails for each outcome +of the above sample space S. +Then +Y(HH) = 2, Y (HT) = 0, Y (TH) = 0, Y (TT) = – 2. +Thus, X and Y are two different random variables defined on the same sample +space S. +EXERCISE 13.3 +1. +An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is +noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn +are put in the urn and then a ball is drawn at random. What is the probability that +the second ball is red? +2. + A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black +balls. One of the two bags is selected at random and a ball is drawn from the bag +which is found to be red. Find the probability that the ball is drawn from the +first bag. +3. +Of the students in a college, it is known that 60% reside in hostel and 40% are +day scholars (not residing in hostel). Previous year results report that 30% of all +students who reside in hostel attain A grade and 20% of day scholars attain A +grade in their annual examination. At the end of the year, one student is chosen +at random from the college and he has an A grade, what is the probability that the +student is a hostlier? +4. +In answering a question on a multiple choice test, a student either knows the +answer or guesses. Let 3 +4 be the probability that he knows the answer and 1 +4 +be the probability that he guesses. Assuming that a student who guesses at the +answer will be correct with probability 1 +4 . What is the probability that the stu- +dent knows the answer given that he answered it correctly? +5. +A laboratory blood test is 99% effective in detecting a certain disease when it is +in fact, present. However, the test also yields a false positive result for 0.5% of +the healthy person tested (i.e. if a healthy person is tested, then, with probability +0.005, the test will imply he has the disease). If 0.1 percent of the population +Reprint 2025-26 + + 432 +MATHEMATICS +actually has the disease, what is the probability that a person has the disease +given that his test result is positive ? +6. +There are three coins. One is a two headed coin (having head on both faces), +another is a biased coin that comes up heads 75% of the time and third is an +unbiased coin. One of the three coins is chosen at random and tossed, it shows +heads, what is the probability that it was the two headed coin ? +7. +An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 +truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. +One of the insured persons meets with an accident. What is the probability that +he is a scooter driver? +8. +A factory has two machines A and B. Past record shows that machine A produced +60% of the items of output and machine B produced 40% of the items. Further, +2% of the items produced by machine A and 1% produced by machine B were +defective. All the items are put into one stockpile and then one item is chosen at +random from this and is found to be defective. What is the probability that it was +produced by machine B? +9. +Two groups are competing for the position on the Board of directors of a +corporation. The probabilities that the first and the second groups will win are +0.6 and 0.4 respectively. Further, if the first group wins, the probability of +introducing a new product is 0.7 and the corresponding probability is 0.3 if the +second group wins. Find the probability that the new product introduced was by +the second group. +10. +Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and +notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and +notes whether a head or tail is obtained. If she obtained exactly one head, what +is the probability that she threw 1, 2, 3 or 4 with the die? +11. +A manufacturer has three machine operators A, B and C. The first operator A +produces 1% defective items, where as the other two operators B and C pro- +duce 5% and 7% defective items respectively. A is on the job for 50% of the +time, B is on the job for 30% of the time and C is on the job for 20% of the time. +A defective item is produced, what is the probability that it was produced by A? +12. +A card from a pack of 52 cards is lost. From the remaining cards of the pack, +two cards are drawn and are found to be both diamonds. Find the probability of +the lost card being a diamond. +13. +Probability that A speaks truth is 4 +5 . A coin is tossed. A reports that a head +appears. The probability that actually there was head is +Reprint 2025-26 + +PROBABILITY 433 +(A) 4 +5 +(B) 1 +2 +(C) +1 +5 +(D) 2 +5 +14. +If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the +following is correct? +(A) +P(B) +P(A | B) +P(A) += +(B) P(A|B) < P(A) +(C) P(A|B) ≥ P(A) +(D) None of these +Miscellaneous Examples +Example 22 Coloured balls are distributed in four boxes as shown in the following table: +Box +Colour + Black White Red + Blue +I +3 +4 +5 +6 +II +2 +2 +2 +2 +III +1 +2 +3 +1 +IV +4 +3 +1 +5 +A box is selected at random and then a ball is randomly drawn from the selected +box. The colour of the ball is black, what is the probability that ball drawn is from the +box III? +Solution Let A, E1, E2, E3 and E4 be the events as defined below : +A : a black ball is selected +E1 : box I is selected +E2 : box II is selected +E3 : box III is selected +E4 : box IV is selected +Since the boxes are chosen at random, +Therefore +P(E1) = P(E2) = P(E3) = P(E4) = 1 +4 +Also +P(A|E1) = 3 +18 , P(A|E2) = 2 +8 , P(A|E3) = 1 +7 and P(A|E4) = 4 +13 +P(box III is selected, given that the drawn ball is black) = P(E3|A). By Bayes' +theorem, +Reprint 2025-26 + + 434 +MATHEMATICS +P(E3|A) = +3 +3 +1 +1 +2 +2 +3 +3 +4 +4 +P(E ) P(A|E ) +P(E )P(A|E ) +P(E )P(A|E )+P(E )P(A|E ) +P(E )P(A|E ) +⋅ ++ ++ += +1 +1 +4 +7 +0.165 +1 +3 +1 +1 +1 +1 +1 +4 +4 +18 +4 +4 +4 +7 +4 +13 +× += +× ++ +× ++ +× ++ +× +Example 23 A and B throw a die alternatively till one of them gets a ‘6’ and wins the +game. Find their respective probabilities of winning, if A starts first. +Solution Let S denote the success (getting a ‘6’) and F denote the failure (not getting +a ‘6’). +Thus, +P(S) = 1 +5 +, P(F) +6 +6 += +P(A wins in the first throw) = P(S) = 1 +6 +A gets the third throw, when the first throw by A and second throw by B result into +failures. +Therefore, +P(A wins in the 3rd throw) = P(FFS) = +5 +5 +1 +P(F)P(F)P(S)= 6 +6 +6 +× +× += +2 +5 +1 +6 +6 + +× + + + + +P(A wins in the 5th throw) = P (FFFFS) = + + + + +5 +6 +1 +6 +4 +and so on. +Hence, +P(A wins) = 1 +6 +5 +6 +1 +6 +5 +6 +1 +6 +2 +4 ++ + + + + + + ++  + + + + + + ++ ... += +1 +6 +25 +1 36 +− + = 6 +11 +P(B wins) = 1 – P (A wins) = +6 +5 +1 11 11 +− += +Remark If a + ar + ar2 + ... + arn–1 + ..., where |r| < 1, then sum of this infinite G.P. +is given by +. +1 +a +r +− + (Refer A.1.3 of Class XI Text book). +Reprint 2025-26 + +PROBABILITY 435 +Example 24 If a machine is correctly set up, it produces 90% acceptable items. If it is +incorrectly set up, it produces only 40% acceptable items. Past experience shows that +80% of the set ups are correctly done. If after a certain set up, the machine produces +2 acceptable items, find the probability that the machine is correctly setup. +Solution Let A be the event that the machine produces 2 acceptable items. +Also let B1 represent the event of correct set up and B2 represent the event of +incorrect setup. +Now +P(B1) = 0.8, P(B2) = 0.2 +P(A|B1) = 0.9 × 0.9 and P(A|B2) = 0.4 × 0.4 +Therefore +P(B1|A) = +1 +1 +1 +1 +2 +2 +P(B ) P(A|B ) +P(B ) P(A|B ) + P(B ) P(A|B ) += +0.8× 0.9× 0.9 +648 +0.95 +0.8× 0.9× 0.9 + 0.2× 0.4× 0.4 +680 += += +Miscellaneous Exercise on Chapter 13 +1. +A and B are two events such that P (A) ≠ 0. Find P(B|A), if +(i) A is a subset of B +(ii) A ∩ B = φ +2. A couple has two children, +(i) Find the probability that both children are males, if it is known that at least +one of the children is male. +(ii) Find the probability that both children are females, if it is known that the +elder child is a female. +3. +Suppose that 5% of men and 0.25% of women have grey hair. A grey haired +person is selected at random. What is the probability of this person being male? +Assume that there are equal number of males and females. +4. +Suppose that 90% of people are right-handed. What is the probability that +at most 6 of a random sample of 10 people are right-handed? +5. +If a leap year is selected at random, what is the chance that it will contain 53 +tuesdays? +6. +Suppose we have four boxes A,B,C and D containing coloured marbles as given +below: +Reprint 2025-26 + + 436 +MATHEMATICS +Box +Marble colour +Red +White +Black +A +1 +6 +3 +B +6 +2 +2 +C +8 +1 +1 +D +0 +6 +4 + One of the boxes has been selected at random and a single marble is drawn from +it. If the marble is red, what is the probability that it was drawn from box A?, box B?, +box C? +7. +Assume that the chances of a patient having a heart attack is 40%. It is also +assumed that a meditation and yoga course reduce the risk of heart attack by +30% and prescription of certain drug reduces its chances by 25%. At a time a +patient can choose any one of the two options with equal probabilities. It is given +that after going through one of the two options the patient selected at random +suffers a heart attack. Find the probability that the patient followed a course of +meditation and yoga? +8. +If each element of a second order determinant is either zero or one, what is the +probability that the value of the determinant is positive? (Assume that the indi- +vidual entries of the determinant are chosen independently, each value being +assumed with probability 1 +2 ). +9. +An electronic assembly consists of two subsystems, say, A and B. From previ- +ous testing procedures, the following probabilities are assumed to be known: +P(A fails) = 0.2 +P(B fails alone) = 0.15 +P(A and B fail) = 0.15 +Evaluate the following probabilities +(i) P(A fails|B has failed) +(ii) P(A fails alone) +10. +Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. +One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. +The ball so drawn is found to be red in colour. Find the probability that the +transferred ball is black. +Choose the correct answer in each of the following: +11. +If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then +(A) A ⊂ B +(B) B ⊂ A +(C) B = φ +(D) A = φ +Reprint 2025-26 + +PROBABILITY 437 +12. +If P(A|B) > P(A), then which of the following is correct : +(A) P(B|A) < P(B) +(B) P(A ∩ B) < P(A) . P(B) +(C) P(B|A) > P(B) +(D) P(B|A) = P(B) +13. +If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then +(A) P(B|A) = 1 +(B) P(A|B) = 1 +(C) P(B|A) = 0 +(D) P(A|B) = 0 +Summary +The salient features of the chapter are – +® The conditional probability of an event E, given the occurrence of the event F +is given by +P(E +F) +P(E | F) +P(F) +∩ += +, P(F) ≠ 0 +® 0 ≤ P (E|F) ≤ 1, +P (E′|F) = 1 – P (E|F) +P ((E ∪ F)|G) = P (E|G) + P (F|G) – P ((E ∩ F)|G) +® P (E ∩ F) = P (E) P (F|E), P (E) ≠ 0 +P (E ∩ F) = P (F) P (E|F), P (F) ≠ 0 +® If E and F are independent, then +P (E ∩ F) = P (E) P (F) +P (E|F) = P (E), P (F) ≠ 0 +P (F|E) = P (F), P(E) ≠ 0 +® Theorem of total probability +Let {E1, E2, ...,En) be a partition of a sample space and suppose that each of +E1, E2, ..., En has nonzero probability. Let A be any event associated with S, +then +P(A) = P(E1) P (A|E1) + P (E2) P (A|E2) + ... + P (En) P(A|En) +® Bayes' theorem If E1, E2, ..., En are events which constitute a partition of +sample space S, i.e. E1, E2, ..., En are pairwise disjoint and E1 4 E2 4 ... 4 En = S +and A be any event with nonzero probability, then +P(E +A +P(E )P(A|E ) +P(E )P(A|E ) +i +i +i +j +j +j +n +| +)= +=∑ +1 +Reprint 2025-26 + + 438 +MATHEMATICS +Historical Note +The earliest indication on measurement of chances in game of dice appeared +in 1477 in a commentary on Dante's Divine Comedy. A treatise on gambling +named liber de Ludo Alcae, by Geronimo Carden (1501-1576) was published +posthumously in 1663. In this treatise, he gives the number of favourable cases +for each event when two dice are thrown. +Galileo (1564-1642) gave casual remarks concerning the correct evaluation +of chance in a game of three dice. Galileo analysed that when three dice are +thrown, the sum of the number that appear is more likely to be 10 than the sum 9, +because the number of cases favourable to 10 are more than the number of +cases for the appearance of number 9. +Apart from these early contributions, it is generally acknowledged that the +true origin of the science of probability lies in the correspondence between two +great men of the seventeenth century, Pascal (1623-1662) and Pierre de Fermat +(1601-1665). A French gambler, Chevalier de Metre asked Pascal to explain +some seeming contradiction between his theoretical reasoning and the +observation gathered from gambling. In a series of letters written around 1654, +Pascal and Fermat laid the first foundation of science of probability. Pascal solved +the problem in algebraic manner while Fermat used the method of combinations. +Great Dutch Scientist, Huygens (1629-1695), became acquainted with the +content of the correspondence between Pascal and Fermat and published a first +book on probability, ""De Ratiociniis in Ludo Aleae"" containing solution of many +interesting rather than difficult problems on probability in games of chances. +The next great work on probability theory is by Jacob Bernoulli (1654-1705), +in the form of a great book, ""Ars Conjectendi"" published posthumously in 1713 +by his nephew, Nicholes Bernoulli. To him is due the discovery of one of the most +important probability distribution known as Binomial distribution. The next +remarkable work on probability lies in 1993. A. N. Kolmogorov (1903-1987) is +credited with the axiomatic theory of probability. His book, ‘Foundations of +probability’ published in 1933, introduces probability as a set function and is +considered a ‘classic!’. +—v— +Reprint 2025-26"