Number Systems

MATHEMATICS–IX NUMBER SYSTEMS 1

CHAPTER 1NUMBER SYSTEMS

Points to Remember :1. Number used for counting 1, 2, 3, 4, ... are known as Natural numbers.2. All natural numbers together with zero i.e. 0, 1, 2, 3, 4, ..... are known as whole numbers.3. All natural numbers, zero and negative numbers together i.e. ...., –4, –3, –2, –1, 0, 1, 2, 3, 4, ... are known

as Integers.

4. Rational Numbers : Numbers of the form qp

where p, q both are integers and q 0. For e.g. 14

75

32 ,,

etc.5. Every rational number have either terminating or repeating (recurring) decimal representation.

Terminating Repeating (Recurring)

For eg. etc.5234

134052 .,. For e.g. 303330

31 .......

here, prime factors of denominator are 715

= 2.142857142857...

2 and 5 only. etc.1428572.

6. There are infinitely many rational numbers between any two given rational numbers.

7. Irrational Numbers : Numbers which cannot be written in the form of qp

, where p, q are integers and

q 0.

For e.g. 3 920220222001732 ......,.,,,, etc.

8. Real numbers : Collection of both rational and irrational numbers. For e.g. ,,,,, 520573 etc.

9. Every real number is represented by a unique point on the number line. Also, every point on the numberline represents a unique real number.

10. For every given positive real number x, we can find x geometrically..11. Identities related to square root :

Let p, q be positive real numbers. Then,

(i) q.ppq (ii) 0 q;qp

qp

(iii) qpqpqp )()( (iv) qpqpqp 2)( 2

12. Laws of Radicals : Let x, y > 0 be real numbers and p, q be rationals. then

(i) xp × xq = xp+q (ii) qpqp xxx

(iii) qpqp xx )( (iv) ppp xyy.x )(

2 NUMBER SYSTEMS MATHEMATICS–IX

ILLUSTRATIVE EXAMPLES

Example 1. Find six rational numbers between 3 and 4. —NCERT.Solution. We know that between two rational numbers a and b, such that a < b, there is a rational number

.ba2

A rational number between 3 and 4 is .274)(3

21

Now, a rational number between 3 and 27

is .4

132

7621

273

21

A rational number between 27

and 4 is .4

152

87214

27

21

Also, a rational number between 3 and 4

13 is 8

254

131221

4133

21

A rational number between 4

15 and 4 is 8

314

1615214

415

21

A rational number between is4and831

1663

83231

214

831

21

41663

831

415

27

413

8253

This can be represented on number line as follows :

OR(without using formula)

We have, 1030

101033 and

1040

101044

We need to find six rational numbers between 3 and 4 i.e.

Ans..1036,

1035,

1034,

1033,

1032,

1031arewhich,

1040and

1030

Example 2. Find four rational numbers between 34and

41 .

Solution. 1216

44

34

34and

123

33

41

41

4 rational numbers between .127,

126,

125,

124are

34and

41

1216and

123 i.e.


Example 3. Express 120. in the form of rational number, .qp

Solution. Let 1212120120 ..x ..... ...(1)multiplying both sides by 100, we get

100 x = 12.1212..... ...(2)Subtracting (1) from (2),

100 x – x = 12.1212..... – 0.1212....

99 x = 12 Ans.334

9912

x

Example 4. Represent 3 on the number line.

Solution. Let XOX be number line with O as origin. Let OA = 1 unit. Draw AB OA such that AB = 1 unit.Join OB. Then using Pythagoras theorem, In OAB

units211ABOAOB 2222

Again, draw DC OB such that BC = 1 unit. Join OC.

then, units3(1))2(BCOBOC 2222 .

With O as centre and OC as ra-dius, draw on arc, meeting OX at

P. Then OC = OP = 3 units.Example 5. 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 —NCERT

Solution. (i) True, since collection of real numbers is made up of rational and irrational numbers. (ii) False, no negative number can be the square root of any natural number.(iii) False, for example 5 is a real but not irrational.

Example 6. Write the following in decimal form and say what kind of decimal expansion each as:

(i) 10036

(ii) 111

(iii) 814 (iv) 13

3(v)

112

(vi) 400329

—NCERT

Solution. (i) 36010036 . Terminating decimal expansion.

(ii) Consider ,111

by long division, we have :

0.090909

000000111 . 99

100 99 100 99 1


,........ 0900909090111

which is non-terminating and repeating decimal expansion.

(iii) .833

8184

814

By long division, we have

1254000338..

32 10 8 20 16 40 40 0

,.1254833

terminating decimal expansion.

(iv) Consider, 133

by long division, we have

......2307692300000000313

264039 100 91 90 78 120 117 30 26 40 39 1

........230769230133 ,.2307690 which is non-terminating and repeating decimal expan-

sion.


(v) Consider, 112

, by long division, we have

181818000000211

..

11 90 88 20 11 90 88 20 11

90 88 2

1801818180112 ....... , which is non-terminating and repeating decimal expansion.

(vi) Consider, ,400329

by long division, we have

822500000329400

..

3200 900 800 1000 800 2000 2000 0

,.82250400329

which is terminating decimal expansion.

Example 7. What can be the maximum number of digits be in the repeating block of digits in the decimal

expansion of 171

? Perform the division to check your answer.. —NCERT


Solution......

.17647588235294100000000000000000117

85 150

136 140 136 40 34 60 51 90 85 50 34 160 153

70 68 20 17 30 17 130 119 110 102 80 68 120 119 1

Thus, 176475882352941071 .

The maximum number of digits in the quotient while computing 171

are 15.

Example 8. Look at several examples of rational numbers in the form ,qqp )0( where p and q are integers

with no common factors other than 1 and having terminating decimal representations (expan-sions). Can you guess what property q must satisfy? —NCERT

Solution. Let us consider various such rotational numbers having terminating decimal representation.


625085750

4350

21 .;.;.

etc.0.215200430.088;

125111.56;

2539

from the examples shown above, it can be easily observe that, ‘‘If the denominator of a rationalnumber in standard form has no prime factors other than 2 or 5 or both, then and the only then itcan be represented as a terminating decimal.’’

Example 9. Visualise 3.765 on the number line, using successive magnification. —NCERTSolution. We know that 3.765 lies between 3 and 4. We divide portion of number line between 3 and 4 in 10

equal parts i.e. 3.1, 3.2, ....., 3.9 and then look at the interval [3.7, 3.8] through a magnifying glassand observe that 3.765 lies between 3.7 and 3.8 (see figure).

Now, we imagine that each new intervals [3.1, 3.2], [3.2, 3.3], ...... , [3.9, 4] have been sub-dividedinto 10 equal parts. As before, we can now visualize through the magnifying glass that 3.765 liesin the interval [3.76, 3.77]. (see figure).

Again, 3.765 lies between 3.76 and 3.77. So, let us focus on this portion of the number line, andimagine to divide it again into 10 equal parts. The first mark represents 3.761, second mark repre-sents 3.762, and so on. So, 3.765 is the 5th mark in these subdivisions.

Example 10. Recall, is defined as the ratio of the circumference (say c) of a circle to its diameter (say d).

That is, .dc

This seem to contradict the fact that is irrational. How will you resolve this

contradiction? —NCERTSolution. There is no contradiction. Remember that when you measure a length with a scale or any other

device, you only get on approximate rational value. So, you may not realise that either c or d isirrational.

Example 11. Simplify the following :

(i) 7473 (ii) )37()37(

(iii) 2)35( (iv) 52308


Solution. (i) 84712)7(43 2

(ii) 437)3()7( 22

(iii) 35235)3)(52()3()5( 22 1528

(iv) 645

3028

Example 12. Find 35. geometrically..Solution. Draw AB = 5.3 units and extend it to C such that BC

= 1 unit. Find mid-point O of AC. With O as centre,and OA as radius, draw a semicircle. Draw

,ACBD interesting semicircle at D. Then BD

= 35. units. With B as centre and BD as radius,draw an arc, intersecting AC produced at E. Then,

units.5.3BDBE

Example 13. Find value of a and b, where 23232

ba .

Solution. We have, 2323

2323

2323

72611

292629

)2((3))2(3

22

2

2276

711 ba

76and

711

ba

Example 14. Simplify the following :

(i) 54

52

33 (ii) 41

31

77 (iii) 14 )(3 (iv) 2/5(32)

Solution. (i) 56

542

54

52

333

(ii) 121

1234

41

31

777

(iii) 811

3133 4

4(–1)4 x

(iv) 41

2122)(2 2

2525

52

5

PRACTICE EXERCISE

1. Represent each of the following rational numbers on the number line :

(i) 53

(ii) 47

(iii) – 3.6 (iv) 4.53

2. (i) Find 4 rational numbers between 51

and 32

.

(ii) Find 5 rational numbers between 34and

43 .

3. (i) Represent 2 and 3 on the some number line

(ii) Represent 5 on the number line.4. Without actual division, find which of the following rationals are terminating decimals.

(i) 2413

(ii) 12519

(iii) 356

(iv) 8017

(v) 20031

5. Find the decimal expansions of the following :

(i) 916– (ii) 7

22(iii) 30

11(iv) 175

37

6. Find the decimal representation of .71

Deduce from the decimal representation of ,71

without actual

calculations, the decimal representation of 74

73

72 ,, and 7

5.

7. Express each of the following recurring decimals in the form of a rational number, :qp

(i) 70. (ii) 1230. (iii) 6450. (iv) 5643.

8. (i) Find three irrational numbers between 32

and 54

.

(ii) Insert three irrational numbers between 2 and 3 .

9. Give an example of two irrational numbers whose :(i) difference is a rational number.

(ii) difference is an irrational number.(iii) sum is a rational number.(iv) sum is an irrational number.(v) quotient is a rational number.

(vi) quotient is an irrational number.(vii) Product is a rational number.

(viii) Product is an irrational number.10. Simplify each of the following :

(i) )3(7)3(7 (ii) 2)35(

(iii) 532012 (iv) 2)2332(

(v) )32(43)2(4 (vii) 1842253203

11. Represent the following on the number line :

(i) 42. (ii) 75. (iii) 86. (iv) 29.12. Rationalize the denominator :

(i) 321

(ii) 532

(iii) 3232

(iv) 35

1

13. Rationalize the denominator:

(i) 3251

(ii) 2363

(iii) 5335

1

(iv) 23321

14. Simplify the following :

(i) 3434

3434

(ii)

5353

5353

15. If a and b are rational numbers and if 5523523 ba

, find a and b.

16. If x and y are the rational numbers and ,yx 51515

1515

find x and y.

17. Evaluate the following :

(i) 23 55 (ii) 58 55 (iii) 22 )3( (iv) 2/3(64)

(v) 1/23 )8( (vi) 2–2 43 –

18. Find the value of x if :

(i) 12 13 x (ii) 23

12

913

xx

(iii) 661 37

x

(iv) 1323

23

32

32

xxx

19. Find whether the product of irrational numbers )2(5and)7(3),7(3),2(5 is a rational orirrational number.

20. (i) Given ,.41412 find value of 12

1

(ii) ,.73213 find value of 3232

.

21. (i) Prove that 2 is not a rational number..

(ii) Prove that 32 is an irrational number..


(i) 403020 ... (ii) 430240 .. (iii) 611132 .. (iv) 251614 .. 23. Simplify the following :

(i) 1

1

2222

nn

nn

(ii) xzzyyx 111

24. Assuming that x is positive real number and a, b, c are rational numbers, show that :

(i) 1

c

b

ab

a

ca

c

b

xx

xx

xx

(ii) 1

ac

a

ccb

c

bba

b

a

xx

xx

xx

25. If ,zyx 632 show that 0111

zyx

26. (i) If 154x , find the value of .x

x 22 1

(ii) If 625x , find the value of x

x 1 .


(i) 52

135

232

1

(ii) 26

3226

3436

23

28. Rationalize the denominator of following :

(i) 321

1

(ii) 531

2

29. Prove that :

(i) 298

143

132

121

1

....

(ii) 22

5321

5321

30. Represent the following on the number line :

(i) 13 (ii) 17 (iii) 32 (iv) 21

PRACTICE TEST

M.M : 30 Time : 1 hour

General Instructions :

Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.

1. Find three rational numbers between 34and

43 .

2. Represent 5 on the number line.

3. Rationalise the denominator :

251

4. Find decimal representation of .4516


(i) 2)25( (ii) )322)(3322(3

6. If a and b are rational numbers and ba 33434

, find a and b.

7. Evaluate the following :

(i) 43

47

211

211

(ii)

52

321 /

8. Express 6240. as a rational number in the simplest form.

9. Represent 24. on the number line. Also, give step of constructions.

10. Simplify : 26

3436

2332

6


ANSWERS OF PRACTICE EXERCISE

2. (i) 157

156

155

154 ,,, (ii)

1214

1213

1212

1211

1210 ,,,,

4. (ii), (iv) and (v) 5. (i) 71. (ii) 1428573. (iii) 630. (iv) 142856210.

6. ,.,.,. 4285710732847140

721428570

71

7142850755714280

74 .,.

7. (i) 97

(ii) 49561

(iii) 300137

(iv) 495

1711

8. (i) 0.68010010001 ..., 0.69010110111 .... and 0.7101001000... (ii) 1.501001000 ... and 1.601001000...

9. (i) 35and53 (ii) 5and53 (iii) 53and53

(iv) 52and54 (v) 5and20 (vi) 6and20

(vii) 3and32 (viii) 34and23

10. (i) 46 (ii) 152–8 (iii) 8 (iv) 61230 (v) 23 (vi) 21053

12. (i) 32 (ii) )5(321

(iii) 347 (iv) )35(21

13. (i) 71 )325( (ii) )236(

41

(iii) )533(5301

(iv) )233(261

14. (i) 1338

(ii) 53 15. 1112

1129

b,a 16. x = 3, y = 0

17. (i) 3125 (ii) 125 (iii) 811

(iv) 161

(v) 2 (vi) 144

1

18. (i) 31

(ii) 85

(iii) 25

(iv) 54

19. Rational 20. (i) 0.414 (ii) 13.928 22. (i) 1 (ii) 907

(iii) 231422

(iv) 2971540

23. (i) 2 (ii) 1 26. (i) 62 (ii) 12 27. (i) 0 (ii) 0

28. (i) )622(41

(ii) )152533(7112

ANSWERS OF PRACTICE TEST

1.1212

1211

1210 ,, 3. 25 4. 530. 5. (i) 1027 (ii) 6

6.13

81319

b,a 7. (i) 2

11 (ii) –4 8. 450

11110. 0

14 POLYNOMIALS MATHEMATICS–IX

CHAPTER 2POLYNOMIALS

Points to Remember :

1. A symbol having a fixed numerical value is called a constant. For e.g. 2379 ,,, etc.

2. A symbol which may take different numerical values is known as a variable. We usually denotes variableby x, y, z etc.

3. A combination of constants and variables which are connected by basic mathematical operations, isknown as Algebraic Expression. For e.g. x2 – 7x + 2, xy2 – 3 etc.

4. An algebraic expression in which variable have only whole numbers as a power is called a Polynomial.5. Highest power of the variable is called the degree of the polynomial. For e.g. 7x3 – 9x2 + 7x – 3 is a

polynomial in x of degree 3.6. A polynomial of degree 1 is called a linear polynomial. For e.g. 7x + 3 is a linear polynomial in x.7. A polynomial of degree 2 is called a Quadratic Polynomial. For e.g. 3y2 – 7y + 11 is a Quadratic polyno-

mial in y.8. A polynomial of degree 3 is called a Cubic Polynomial. For e.g. 3t3 – 7t2 + t – 3 is a cubic polynomial in

t.9. According to number of terms, a polynomial having one non-zero term is a monomial, a polynomial

having two non-zero terms is a bionomial and a polynomial have three non-zero terms is a trinomial.10. Remainder Theorem : Let f(x) be a polynomial of degree n 1 and let a be any real number. If f(x) is

divided by linear polynomial (x – a), then the remainder is f(a).11. Factor Theorem : Let f(x) be a polynomial of degree n > 1 and a be any real number.

(i) If f(a) = 0, then x – a is a factor of f(x).(ii) If (x – a) is a factor of f(x) then f(a) = 0.

12. Algebraic Identities : (i) (x + y)2 = x2 + 2xy + y2 (ii) (x – y)2 = x2 – 2xy + y2

(iii) x2 – y2 = (x – y) (x + y) (iv) (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz (v) (x + y)3 = x3 + y3 + 3xy (x + y) (vi) (x – y)3 = x3 – y3 – 3xy (x – y)(vii) x3 – y3 = (x – y) (x2 + xy + y2) (viii) x3 + y3 = (x + y) (x2 – xy + y2) (ix) x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – xz)

13. If x + y + z = 0 then, x3 + y3 + z3 = 3xyz.


Example 1. Which of the following expressions are Polynomials? In case of a polynomial, give degree ofpolynomial.

(i) x4 – 3x3 + 7x2 + 3 (ii) 673 3 yy (iii) 3x

(iv) x

x 1 (v) 7 (vi)

xx / 232

MATHEMATICS–IX POLYNOMIALS 15

Solution. (i) is a polynomial of degree 4.(ii) is a polynomial of degree 3.(v) is a polynomial of degree 0.

Example 2. Verify whether the following are zeros of the polynomial, indicated against them.

(i) 3113)(

x,xxp (ii) 545)( x,xxp

(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, lmx

(vii) p(x) = 3x2 – 1, 3

231 ,x

(viii) p(x) = 2x + 1, 21

x –NCERT

Solution. (i) We have, p(x) = 3x + 1

At 0111313

31

31

p,x

31

is a zero of p(x).

(ii) We have, p(x) = 5x –

At 04545

54

54

p,x

54

is not a zero of p(x).

(iii) We have, 1)( 2 xxp

At x = 1, p(1) = (1)2 – 1 = 1 – 1 = 0also, at x = – 1, p(–1) = (–1)2 – 1 = 1 – 1 = 0 1, – 1 both are zeros of p(x).(iv) We have, p(x) = (x + 1) (x – 2)At x = – 1, p(–1) = (–1 + 1) (–1 –2) = 0 × (–3) = 0also, at x = 2, p(2) = (2 + 1) (2 – 2) = 3 × 0 = 0 –1, 2 both are zeros of p(x).(v) We have, p(x) = x2

At x = 0, p(0) = 02 = 0 0 is a zero of p(x).(vi) We have, p(x) = lx + m

At, 0

mmm

lml

lmp,

lmx

lm

is a zero of p(x).

(vii) We have, p(x) = 3x2 – 1


At, 01113131

313

31

31

2

p,x

and at, 031413431

323

32

32

2

p,x

31 is a zero of p(x), but

32 is not a zero of p(x).

(viii) We have p(x) = 2x + 1

At 02111212

21

21

p,x

21

is not a zero of p(x).

Example 3. 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. —NCERT

Solution. We know, finding a zero of p(x), is the same as solving the equation p(x) = 0.(i) p(x) = 0 x + 5 = 0 x = – 5 –5 is a zero of p(x).(ii) p(x) = 0 x – 5 = 0 x = 5 5 is a zero of p(x).

(iii) p(x) = 0 2x + 5 = 0 2x = – 5 25

x

25

is a zero of p(x).

(iv) p(x) = 0 3x – 2 = 0 3x = 2 32

x

32

is a zero of p(x).

(v) p(x) = 0 3x = 0 x = 0 0 is a zero of p(x).(vi) p(x) = 0 ax = 0 x = 0 ( Given a 0) 0 is a zero of p(x).

(vii) p(x) = 0 cx + d = 0 cx = – d cdx

cd

is a zero of p(x).


Example 4. Using remainder theorem, find the remainder when p(x) = 2x3 – 5x2 + 9x – 8 is divided by (x – 3).Solution. Remainder = p(3)

= 2(3)3 – 5(3)2 + 9 (3) – 8= 54 – 45 + 27 – 8 = 81 – 53 = 28 Ans.

Example 5. Find the remainder when 133 23 xxx is divided by :

(i) x + 1 (ii) 21

x (iii) x (iv) x + (v) 5 + 2x —NCERT

Solution. (i) x + 1 = 0 x = – 1 by remainder theorem, the required remainder is p (–1).Now, p(x) = x3 + 3x2 + 3x + 1 p(–1) = (–1)3 + 3(–1)2 + 3(–1) + 1

= –1 + 3 – 3 + 1 = 0 Remainder = 0

(ii) .xx210

21

by remainder theorem, the required remainder is .p

21

Now, 1213

213

21

21 23

p

8812611

23

43

81

827

Remainder 827

(iii) x = 0 by remainder theorem, the required remainder is p(0).Now, p(0) = (0)3 + 3(0)2 + 3(0) + 1 = 1 Remainder = 1(iv) x + = 0 x = – by remainder theorem, the required remainder is p(–).Now, p(–) = (–)3 + 3(–)2 + 3 (–) + 1

= –3 + 32 – 3 + 1 Remainder = –3 + 32 – 3 + 1.

(v) 5 + 2x = 0 2x = – 5 x = 25

By remainder theorem, the required remainder is .p

25

Now, 1253

253

25

25 23

p

12

15475

8125


827

8860150125

Remainder 827

.

Example 6. Using Factor theorem, show that (x + 2) is a factor of x4 – x2 – 12.Solution. Let p(x) = x4 – x2 – 12.

Now, x + 2 = 0 x = – 2.(x + 2) is a factor of p(x) p(–2) = 0Now, p(–2) = (–2)4 – (–2)2 – 12

= 16 – 4 – 12 = 0,which shows that (x + 2) is a factor of p(x).

Example 7. Use the factor theorem to determine whether g(x) is a factor of p(x) is each of the following cases:(i) p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2(iii) p(x) = x3 – 4x2 + x + 6, g(x) = x – 3 —NCERT

Solution. (i) In order to prove that g(x) = x + 1 is a factor of p(x) = 2x3 + x2 – 2x – 1, it is sufficient to show thatp(–1) = 0.Now, p(–1) = 2 (–1)3 + (–1)2 – 2(–1) – 1

= – 2 + 1 + 2 – 1 = 0 g(x) is a factor of p(x).(ii) In order to prove that g(x) = x + 2 is a factor of p(x) = x3 + 3x2 + 3x + 1, it is sufficient to showthat p(–2) = 0.Now, p (–2) = (–2)3 + 3(–2)2 + 3(–2) + 1

= – 8 + 12 – 6 + 1 = – 1 0. g(x) is not a factor of p(x).(iii) In order to prove that g(x) = x – 3 is a factor of p(x) = x3 – 4x2 + x – 6, it is sufficient to showthat p(3) = 0Now, p(3) = (3)3 – 4(3)2 + 3 – 6 = 27 – 36 + 3 – 6 = – 12 0. g(x) is not a factor of p(x).

Example 8. If x3 + ax2 + bx + 6 has x – 2 as a factor and leaves a remainder 3, when divided by x – 3, find thevalue of a and b.

Solution. Let p(x) = x3 + ax2 + bx + 6Since, x–2 is a factor of p(x) p(2) = 0 ( factor theorem) (2)3 + a(2)2 + b(2) + 6 = 0 2a + b = –7 ...(1)Also, p(x) leaves remainder 3, when divided by x – 3. p(3) = 3 ( Remainder theorem) (3)3 + a(3)2 + b(3) + 6 = 3 3a + b = – 10 ...(2)Solving (1) and (2), we get a = – 3, b = – 1 Ans.


Example 9. Factorize the following:(i) 6x2 – 18xy (ii) x3 + 7x2 – x – 7 (iii) 16a2 – 81b2

(iv) x2 + 5x – 24 (v) 9x2 – 22x + 8Solution. (i) 6x (x – 3y)

(ii) x2 (x + 7) – 1 (x + 7) = (x2 – 1) (x + 7) = (x – 1) (x + 1) (x + 7)(iii) (4a)2 – (9b)2 = (4a + 9b) (4a – 9b)

(iv) 8)(3)8(24382 xxxxxx = (x – 3) (x + 8)

(v) 9x2 – 18x – 4x + 8 = 9x (x – 2) – 4(x – 2) = (x – 2) ( 9x – 4)

Example 10. Factorise :(i) 12x2 – 7x + 1 (ii) 2x2 + 7x + 3 (iii) 6x2 + 5x – 6 (iv) 3x2 – x – 4 –NCERT

Solution. (i) 12x2 – 7x + 1here, p + q = coefficient of x = – 7

pq = coefficient of x2 × constant term = 12 × 1 =12 p + q = – 7 = – 4 – 3, pq = 12 = (–4) × (–3) 12x2 – 7x + 1 = 12x2 – 4x – 3x + 1 = 4x (3x – 1) – 1(3x – 1) = (4x – 1) (3x – 1) Ans.(ii) 2x2 + 7x + 3 = 2x2 + x + 6x + 3 = x (2x + 1) + 3 (2x + 1) = (x + 3) (2x + 1)(iii) 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 = 3x (2x + 3) – 2(2x + 3) = (3x – 2) (2x + 3)(iv) 3x2 – x – 4 = 3x2 + 3x – 4x – 4 = 3x(x + 1) – 4(x + 1) = (3x – 4) (x + 1)

Example 11. Factorize x3 – 3x2 – 9x – 5 using factor theorem.Solution. Let p(x) = x3 – 3x2 – 9x – 5.

factors of constant term are ±1 and ±5.Now, p(1) = 13 – 3(1)2 – 9(1) – 5 0

p(–1) = (–1)3 – 3(–1)2 – 9(–1) – 5 0. (x + 1) is a factor of p(x). p(x) = (x + 1) (x2 – 4x – 5)

= (x + 1) [x2 – 5x + x – 5] = (x + 1) [x(x – 5) + 1 (x – 5)] = (x + 1) (x + 1) (x – 5) Ans.

Example 12. Factorise : x3 + 13x2 + 32x + 20 —NCERTSolution. Let p(x) = x3 + 13x2 + 32x + 20

Now, factors of constant term 20 are ± 1, ± 2, ± 5, ± 10 and ± 20.Now, p(1) = (1)3 + 13(1)2 + 32(1) + 20 = 1 + 13 + 32 + 20 = 66 0

p(–1) = (–1)3 + 13(–1)2 + 32(–1) + 20 = –1 + 13 – 32 + 20 = – 33 + 33 = 0 (x + 1) is a factor of x3 + 13x2 + 3x + 20.Now, divide p(x) by x + 1.


p(x) = (x +1) (x2 + 12x + 20) = (x + 1) [x2 + 2x + 10x + 20] = (x + 1) [x(x + 2) + 10 (x + 2)] = (x + 1) (x + 2) (x + 10) Ans.

Example 13. Expand the following :(i) (a – 2b + 3c)2 (ii) (2x + y)3 (iii) (3x – 2y)3

Solution. (i) (a)2 + (–2b)2 + (3c)2 + 2(a) (–2b) + 2(–2b) (3c) + 2 (a) (3c)= a2 + 4b2 + 9c2 – 4ab – 12bc + 6ac

(ii) (2x)3 + (y)3 + 3(2x)(y) [2x + y]= 8x3 + y3 + 6xy (2x + y)= 8x3 + y3 + 12x2y + 6xy2

(iii) (3x)3 – (2y)3 – 3(3x) (2y) [3x – 2y]= 27x3 – 8y3 – 18xy (3x – 2y)= 27x3 – 8y3 – 54x2y + 36xy2

Example 14. Factorize the following :(i) x3 + 27y3 (ii) 27a3 – 64b3 (iii) a3 – 8b3 + 64c3 + 24abc

Solution. (i) (x)3 + (3y)3 = (x + 3y) (x2 – 3xy + 9y2) [ a3 + b3 = (a + b) (a2 – ab + b2)](ii) (3a)3 – (4b)3 = (3a – 4b) (9a2 + 12ab + 16b2) [ a3 – b3 = (a – b) (a2 + ab + b2)](iii) (a)3 + (–2b)3 + (4c)3 – 3 × a × (–2b) × (4c)

= (a – 2b + 4c) (a2 + 4b2 + 16c2 + 2ab + 8bc – 4ac) [ a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)]

Example 15. Verify that : x3 + y3 + z3 – 3xyz ])()()).[((21 222 xzzyyxzyx –NCERT

Solution. Consider, LHS = x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

21

(x + y + z) . 2 (x2 + y2 + z2 – xy – yz – zx)

21

(x + y + z) (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx)

21

(x + y + z) (x2 + x2 + y2 + y2 + z2 + z2 – 2xy – 2yz – 2zx)

21

(x + y + z) [(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 +x2 – 2zx)]

21

(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]

= RHS. Hence verify.Example 16. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz. —NCERTSolution. We have x + y + z = 0

x + y = – zCubing both sides, we get (x + y)3 = (–z)3

x3 + y3 + 3xy (x + y) = –z3

x3 + y3 + 3xy (–z) = – z3 ( x + y = – z)


x3 + y3 + z3 – 3xyz = 0 x3 + y3 + z3 = 3xyz. Hence shown.

Example 17. Give possible expressions for the length and breadth of each of the following rectangles, in whichtheir areas are given :(i) Area : 25a2 – 35a + 12 (ii) 35y2 + 13y – 12 —NCERT

Solution. (i) Given, Area of rectangle = 25a2 – 35a + 12 = 25a2 – 15a – 20a + 12 = 5a (5a – 3) – 4 (5a – 3) = (5a – 3) (5a – 4)

Possible length and breadth are (5a – 3) and (5a – 4) units.(ii) Area of given rectangle = 35y2 + 13y – 12

= 35y2 + 28y – 15y – 12= 7y (5y + 4) – 3 (5y + 4)= (5y + 4) (7y – 3).

Possible length and breadth are (5y + 4) and (7y – 3) units.Example 18. What are the possible expressions for the dimensions of the cuboid whose volumes are given

below:(i) Volume : 3x2 – 12x (ii) Volume : 12ky2 + 8 ky – 20 k —NCERT

Solution. (i) Volume : 3x2 – 12x = 3x (x – 4) = 3 × x × (x – 4) Possible dimensions of cuboid are 3, x and (x – 4) units.(ii) Volume = 12ky2 + 8ky – 20k

= 4 k (3y2 + 2y – 5)= 4k (3y2 – 3y + 5y – 5)= 4k [3y (y – 1) + 5 (y – 1)]= 4k (y – 1) (3y + 5)

Possible dimensions of a cuboid are 4k, (y – 1) and (3y + 5) units.

PRACTICE EXERCISE1. Which of the following expressions are polynomials in one variable?

(i) 3x2 – x3 + 7x + 1 (ii) xx 4 (iii) 272 2 yy

(iv) 1223 yy / (v) 2x4 – 7x3 + 32. Write degree of each the following polynomial:

(i) 7x4 – 9x3 + 2x + 4 (ii) 7 – y2 + y3 (iii) 175 23 tt(iv) 10 (v) 3x2 – 7x + 4

3. Classify each of the following as linear, quadratic and cubic polynomial:(i) 3x3 – 7x (ii) 4 y2 + 3y – 1 (iii) 7r

(iv) 23xx (v) 3yy 4. Find the value of p(x) = 5x3 – x2 + 3x + 4 at

(i) x = 0 (ii) x = 2 (iii) x = – 15. Find the value of p(0), p(2) and p(–3) where p(x) = x3 – x2 + x – 1.6. Verify whether the following are zeros of the polynomial, indicated against them:

(i) 2332)( x;xxp (ii) p(x) = (x + 3) (x – 4) ; x = – 3, 4

(iii) p(x) = x2 + x – 6; x = 3, –2 (iv) p(x) = x – x3 ; x = 0, 1, –1


7. Find the zero of the polynomial in each of the following :(i) p(x) = x – 4 (ii) p(x) = 3x + 4(iii) p(x) = 7x (iv) p(x) = rx + s; r 0, r, s are real numbers.

8. Using Remainder Theorem, find the remainder when :(i) 4x3 – 7x2 + 3x – 2 is divided by x – 1

(ii) x3 – 7x2 + 6x + 4 is divided by x – 3(iii) x3 + 2x2 – x + 3 is divided by x + 3(iv) 4x3 – 4x2 + x – 2 is divided by 2x + 1(v) x3 – ax2 + 5x + a is divided by x – a

(vi) x3 + ax2 – 6x + 2a is divided by x + a9. If 35 is the remainder when 2x2 + ax + 7 is divided by x – 4, find the value of a.

10. Without actual division, prove that 2x3 + 13x2 + x – 70 is exactly divisible by x – 2.11. Show that x – 1 is a factor of x3 – 2x2 – 5x + 6.12. Find the value of p for which the polynomial x4 – 2x3 + px2 + 2x + 8 is exactly divisible by x + 2.13. Find values of a and b so that the polynomial x4 + 7x3 + 4x2 + ax + b is exactly divisible by x – 1

and x + 3.14. The polynomials ax3 – 3x2 + 7 and 2x3 + 7x – 2a are divided by x + 3. If the remainder in each case is same,

find the value of a.15. The polynomials ax3 + 4x2 – 3 and 4x3 + 4x – a when divided by x – 3, leaves the remainder R1 and R2

respectively. Find the value of a if R1 = 3R2.16. Find the integral zeros of x3 – 3x2 – x + 3.17. Find the integral zeros of x3 + 4x2 – x – 4.18. Show that x – 2 is a factor of p(x) = x3 – 6x2 + 15x – 14.19. Show that x + 4 is a factor of x4 + 3x3 – 4x2 + x + 4.20. Find value of a for which x + a is a factor of f(x) = x3 + ax2 + 3x + a + 4.21. Without actual division, prove that 2x4 + x3 – 2x2 + 5x – 6 is exactly divisible by x2 + x – 2.22. For what value of a is the polynomial 2x3 – ax2 + 8x + a + 4 is exactly divisible by 2x + 1.23. If x3 + ax2 + bx – 12 has x – 3 as a factor and leaves a remainder 10 when divided by x + 2, find a and b.24. Factorise the following, using factor theorem :

(i) x3 – 6x2 + 11x – 6 (ii) x3 – 6x2 + 3x + 10(iii) 2x3 – 5x2 – x + 6 (iv) x3 – 3x2 + 3x – 1(v) x3 + 4x2 – 11x – 30 (vi) 4x3 + 9x2 – 19x – 30

25. Factorize 6x3 + 35x2 – 7x – 6, given x + 6 is one of its factor.26. Factorize 2x3 + 5x2 – 124x – 63, given x + 9 is one of its factor.27. Use suitable identity to find the following products :

(i) (x + 4) (x + 6) (ii) (x – 3) (x + 8)

(iii) (2x + 3) (3x – 2) (iv)

25

25 22 tt

28. Evaluate the following products without multiplying directly:(i) 87 × 93 (ii) 106 × 94


29. Factorize the following using appropriate identities :

(i) 144

92

2 yx (ii) 169 2 xx

(iii) 44 yx (iv) 22 25309 yxyx 30. Factorize by splitting the middle term :

(i) x2 – 21x + 108 (ii) 12y2 – y – 6 (iii) t2 – 11t – 42 (iv) 6z2 + 5z – 6(v) 8x2 – 2x – 15 (vi) 20x2 + 13x – 84

31. Expand each of the following using suitable identity :

(i) 2)32( zyx (ii) 2

33

cba

(iii) (3x – 2y)3 (iv) 3

22

yx

32. Evaluate the following, using suitable identity:(i) (101)3 (ii) (99)3

33. Factorize the following:(i) 4x2 + y2 + 9z2 + 4xy – 6yz – 12xz (ii) 64a3 – 144a2b + 108ab2 – 27b3

(iii) x3 + 6x2y + 12xy2 + 8y3 (iv) 8x3 + 125(v) 27a3 – 64b3 (vi) 27a3 + 8b3 + c3 – 18abc

34. Without actually calculating the cubes, find the value of each of the following :(i) (–14)3 + (8)3 + (6)3 (ii) (19)3 + (–11)3 + (–8)3

35. If x + y + z = 0, prove that x3 + y3 + z3 = 3xyz.

36. Prove that: x3 + y3 + z3 – 3xyz ])–()–()–[()(21 222 xzzyyxzyx

37. Using factor theorem, show that a – b, b – c and c – a are the factors of a(b2 – c2) + b (c2 – a2) + c (a2 – b2).38. Factorize the following :

(i) 1 – 2ab – a2 – b2 (ii) x4 + 5x2 + 9

(iii) 583055 2 xx (iv) 2222 )(16)(24)(9 yxyxyx (v) x6 – y6 (vi) x6 + y6

39. Factorize the following :(i) x3 (y – z)3 + y3 (z – x)3 + z2(x – y)3 (ii) (a – 4b)3 + (4b – 3c)3 + (3c – a)3

40. Prove that : (a + b)3 + (b + c)3 + (c + a)3 – 3 (a + b) (b + c) (c + a) = 2 (a3 + b3 + c3 – 3abc)

PRACTICE TESTMM : 30 Time : 1 hourGeneral Instructions :Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.

1. Find the remainder when 3x3 – 8x2 + 9x – 10 is divided by x – 3.2. Find the value of a for which x3 + ax2 – 3x + 14 is exactly divisible by x + 2.3. Factorize : 64a3 – 27b3 – 144a2b + 108ab2

4. Evaluate : (103)3 , using suitable identity.


5. Factorize : 3x2 + 13x – 10

6. Expand the following : (i) (x – 2y – 3z)2 (ii) 3

3

yx

7. Without actually calculating the cubes, evaluate the following :(20)3 + (–15)3 + (–5)3

8. Factorize : x3 – 64y3 – 8z3 – 24xyz.9. Factorize 6x3 + 25x2 + 21x – 10 using factor theorem.

10. Prove that : a3 + b3 + c3 – 3abc = 222 )(()()(21 ac)cbbacba


1. (i), (iii) and (v) 2. (i) 4 (ii) 3 (iii) 3 (iv) 0 (v) 23. (i) Binomial (ii) Trinomial (iii) Monomial (iv) Bionomial (v) Bionomial4. 4, 46, –5 5. –1, 5, –406. (i) yes (ii) yes (iii) no (iv) yes

7. (i) 4 (ii) 34

(iii) 0 (iv) r5

8. (i) – 2 (ii) 175 (iii) –3 (iv) – 4 (v) 6a (vi) 8a

9. a = – 1 12. p = – 9 13. a = – 21, b = 9 14. 511

a 15. 10

109a

16. 1, –1, 3 17. 1, –1, –4 20. a = 2 22. 31

a 23. a = 1, b = – 8

24. (i) (x – 1) (x – 2) (x – 3) (ii) (x + 1) (x – 2) (x – 5) (iii) (x + 1) (x – 2) (2x – 3)(iv) (x – 1) (x – 1) (x – 1) (v) (x + 2) (x – 3) (x + 5) (vi) (x – 2) (x + 3) (4x + 5)

25. (x + 6) (3x + 1) (2x – 1) 26. (x + 9) (x – 7) (2x – 1)

27. (i) x2 + 10 x + 24 (ii) x2 + 5x – 24 (iii) 6x2 + 5x – 6 (iv) 4254 t

28. (i) 8091 (ii) 9964 29. (i)

123

123 yxyx (ii) (3x – 1) (3x – 1)

(iii) ))()(( 22 yxyxyx (iv) (3x + 5y) (3x + 5y)30. (i) (x – 9) (x – 12) (ii) (3y – 2) (4y + 3) (iii) (t + 3) (t – 14)

(iv) (2z + 3) (3z – 2) (iv) (2x – 3) (4x + 5) (vi) (4x – 7) (5x + 12)

31. (i) x2 + 4y2 + 9z2 – 4xy – 12yz + 6xz (ii) acbcabcba 26329

922

2

(iii) 27x3 – 8y3 – 54x2y + 36xy2 (iv) 223

3

236

88 xyyxyx

32. (i) 1030301 (ii) 970299


33. (i) (2x + y – 3z) (2x + y – 3z) (ii) (4a – 3b) (4a – 3b) (4a – 3b) (iii) (x + 2y) (x + 2y) (x + 2y)(iv) (2x + 5) (4x2 – 10x + 25) (v) (3a – 4b) (9a2 + 12ab + 16b2)(vi) (3a + 2b + c) (9a2 + 4b2 + c2 – 6ab – 2bc – 3ac)

34. (i) –2016 (ii) 5016

38. (i) (1 + a + b) (1 – a – b) (ii) (x2 + x + 3) (x2 – x + 3) (iii) )545()25( xx

(iv) (x + 7y) (x + 7y) (v) (x – y) (x + y) (x2 – xy + y2) (x2 + xy + y2)(vi) (x2 + y2) (x4 – x2y2 + y4)

39. (i) 3xyz (y – z) (z – x) (x – y) (ii) 3 (a – 4b) (4b – 3c) (3c – a)


1. 26 2. a = – 3 3. (4a – 3b) (4a – 3b) (4a – 3b)4. 1092727 5. (3x – 2) (x + 5)

6. (i) x2 + 4y2 + 9z2 – 4xy + 12yz – 6xz (ii) 22

33

327xyyxyx

7. 4500 8. (x – 4y – 2z) (x2 + 16y2 + 4z2 + 4xy – 8yz + 2xz)9. (x + 2) (2x + 5) (3x – 1)

26 CO-ORDINATE GEOMETRY MATHEMATICS–IX

CHAPTER 3CO-ORDINATE GEOMETRY


1. Coordinate axes : Two mutually perpendicular lines X´OX and YOY´ known as x-axis and y-axisrespectively, constitutes to form a co-ordinate axes system. These axes interests at point O, known asorigin.

2. Co-ordinate axes divides the plane into four regions, known as Quadrants.3. The position of any point in a plane is determined with reference to x-axis and y-axis.4. The x-coordinate of a point is its perpendicular distance from the y-axis measured along the x-axis. The

x-coordinate is known as abscissa.5. The y-coordinate of a point is its perpendicular distnace from the x-axis measured along the y-axis. The

y-coordinate is known as ordinate.6. Abscissa and ordinate of a point written in the form of ordered pair, (ascissa, ordinate) is known as the

co-ordinate of a point.7. If the point in the plane is given, we can find the ordered pair of its co-ordinate and if the ordered pair of

real numbers is given, we can find the point in the plane corresponding to this ordered pair.

8. Sing Convention :

––coordinatey––coordinatex

IVIIIIIIofSignQuadrant

MATHEMATICS–IX CO-ORDINATE GEOMETRY 27


Example 1. Write the answer of each of the following questions:(i) What is the name of the 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? —NCERTSolution. (i) Rectangular axes/co-ordinate axes

(ii) Quadrant(iii) Origin

Example 2. Write the co-ordinate of points A, B, C, D, E and F.Solution.

Here, the co-ordinate of A are (3, 3) ; of B are (6, 0) ; of C are (–3, 2) ; of D are (5, –4) ; of E are(0, –5) and of F are (–6, –2).

Example 3.

In which quadrant, do the following points lies?

A (3, 7), B (–9, –6), C (10, 15) and D (–5, 9)Solution.In A (3, –7), x co-ordinate is positive and y-coordinate is negative. A lies in IVth quadrant.Similarly, B lies in IIIrd quadrant ; C lies in Ist quadrant and D lies in IInd quadrant.Example 4.Plot the following points with given co-ordinates in a plane. A (4, 3), B (–5, 2), C (0, –5), D (5, 0), E(–5, –3) and F (3, –4).Solution.For plotting A (4, 3), we first move 4 units along OX and then 3 units along OY. Similary. other pointcan be drawn.AMIT B

AJAJ


Example 5. Look at the figure given, and write the following : (i) The co-ordinate of A. (ii) The abscissa of point B.(iii) The ordinate of point C. (iv) The co-ordinate of D. (v) The point whose co-ordinate are (–2, 5).

Solution. (i) Co-ordinate of A are (6, –4) (ii) Abscissa of B is 2.(iii) Ordinate of C is –3 (iv) Co-ordinates of D are (4, 3) (v) Point E have co-ordinate as (–2, 5)

A (4, 3)


Example 6. Plot the points (x, y) given in the following table on the plane choosing suitable units of distanceon the axes.

132517831012

.y

x

—NCERTSolution.

Example 7. Plot the points A (–3, –3), B (5,–3), C (5, 2) and D (–3, 2) on the graph paper. Join them in order andname the figure so formed. Also, find its area.

Solution.


ABCD is a rectangleArea of ABCD = AB × BC = 8 × 5 sq. units = 40 sq. units

Example 8. Graph the following equations(i) x = –2 (ii) y = 3 (iii) y = x + 2

Solution. (i) x = –2. The given equation can be written as 1.x + 0.y = –2 x is fixed as –2 and y may chooseany value. Let us represent following information in a tabular form.

2101222222

yx

(ii) y = 3. Given equation may be written as 0.x + 1.y = 3 y is fixed as 3 and x may choose anyvalue. Let us represent this information in a tabular form.

3333321012

yx

(iii) y = x + 2here, when x = 0, y = 2 ; x = 1, y = 3 ; x = –1, y = 1 etc. Represent this in the tabular form.

43212101

yx


PRACTICE EXERCISE1. State the quadrant in which the following points lie.

(i) A (3, –4) (ii) B (–5, 11) (iii) C (–10, –15) (iv) D (8, 12)(v) E (–11, 5) (vi) F (–100, –200) (vii) G (10, 50) (viii) H (20, –5)

2. Look at the figure given, and write the following :(i) The co-ordinate of P (ii) The ordinate of Q(iii) The abscissa of R (iv) The point given by (4, –3)(v) The point which is at a distance of 3 units from y-axis(vi) Co-ordinate of point T


3. Plot the points P (1, 3), Q (3, 7) and R (5, 11). Are these points collinear?4. Plot the points A (–2, 3), B (8, 3) and C (6, 7). Join them in order. Name the figure obtained. Also find its

area.5. Plot the points A (3, 2), B (11, 8), C (8, 12) and D (0, 6). Join them in order. Name the figure thus obtained.6. Plot the points P (0, –1), Q (2, 1), R (0, 3), S (–2, 1). Join them in order. Name the figure obtained.7. Plot the points A (–2, 1), B (1, 1), C (–4, –3) and D (3, –3). Join them in order. Name the figure thus

obtained.8. Plot the points P (7, 3), Q (3, 0), R (0, –4) and S (4, –1). Join them in order. Name the figure thus obtained.9. Plot the points A (0, 3), B (–4, 1), C (0, –6) and D (4, 1). Join them in order. Name the figure thus obtained.

10. Graph the folloiwng equations :(i) x = 4 (ii) y = –3 (iii) y = x (iv) x + y = 3

PRACTICE TESTM.M : 15 Time : 1/2 hour


All questions carry 3 marks each.

1. Name the quadrant in which the following points lie :(i) A (–7, 9) (ii) B (–10, – 25) (iii) C (7, –9) (iv) D (11, 7)

2. Plot A (–6, 3), B (6, 0) and C (4, 5). Join these points in order. Name the figure thus obtained.3. Look at the figure and write the following :

(i) The co-ordinate of A.(ii) The co-ordinate of B.(iii) The abscissa of C.(iv) The point whose co-ordinates are (–4, 3)

0 1 4–1–4–1

–2–5

–2

–3–6–7

–3

–4

–5

2 53 6 7 8

1

2

3

4

5

6

B

C

D

A

X

Y


4. Mark the points P (–4, 2), Q (–4, –4), R (3, –4) and S (3, 2) on the graph paper. Join these points in order.Name the figure obtained. Also, find area of the figure obtained.

5. Draw the graph of y = x + 1. Does the point (–7, 6) lie on this line?

ANSWERS OF PRACTICE EXERCISE1. (i) IVth quadrant (ii) IInd qudrant (iii) IIIrd quadrant (iv) Ist quadrant

(v) IInd qudrant (vi) IIIrd quadrant (vii) Ist quadrant (viii) IVth qudrant2. (i) P (–5, 4) (ii) Q (4, 2) (iii) –6 (iv) T

(v) S (vi) T (4, –3)3. yes

4. Triangle, Area = 20 sq. units


5. Rectangle 6. Square


9. Kite

10.

(i) (ii)


(iii) (iv)


1. (i) IInd quadrant (ii) IIIrd qudrant (iii) IVth quadrant (iv) Ist quadrant2. Triangle

3. (i) A (7, 6) (ii) B (2, –5) (iii) –4 (iv) D4. Rectangle, Area = 42 square units


5. No

38 LINEAR EQUATIONS IN TWO VARIABLES MATHEMATICS–IX

CHAPTER 4LINEAR EQUATIONS IN TWO VARIABLES

Points to Remember :1. An equation of the form ax + by + c = 0, where a, b, c are real numbers, such that a and be are not both

zero, is known as linear equation in two variables.2. A linear equation in two variables has infinitely many solutions.3. The graph of linear equation in two variables is always a straight line.4. y = 0 is the equation of x-axis and x = 0 is the equation of y-axis.5. The graph of x = a is a straight line parallel to the y-axis.6. The graph of y = b is a straight line parallel to the x-axis.7. The graph of y = kx passes through the origin.8. Every point on the graph of a linear equation in two variables is a solution of the linear equation. Also,

every solution of the linear equation is a point on the graph of the linear equation.

ILLUSTRATIVE EXAMPLESExample 1. 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) 4x – 7y = 10 (ii) x + 2y

= – 5 (iii) x = –4y

(iv) 4y – 9 = 0 (v) x = 5Solution. (i) 4x – 7y = 10 4x – 7y – 10 = 0

comparing with ax + by + c = 0, a = 4, b = –7, c = –10

(ii) 52

y

x 052

y

x = ax + by + c

on comparing, a = 1, 21

b , c = 5

(iii) x = –4y x + 4y + 0 = 0 = ax + by + c on comparing, a = 1, b = 4, c = 0(iv) 4y – 9 = 0 0.x + 4.y – 9 = 0 = ax + by + c on comparing, a = 0, b = 4, c = –9(v) x = 5 x – 5 = 0, x + 0.y – 5 = 0 on comparing, a = 1, b = 0, c = –5

Example 2. Which one of the following options is true, and why? y = 3x + 5 has (i) a unique solution, (ii) onlytwo solutions, (iii) infinitely many solutions. —NCERT

Solution. Given equation is y = 3x + 5when x = 0, y = 3(0) + 5 = 5 (0, 5) is one of its solution.

when y = 0, then 35

x

MATHEMATICS–IX LINEAR EQUATIONS IN TWO VARIABLES 39

0

35 , is another solution.

when x = 1, then y = 3(1) + 5 = 8 (1, 8) is also its solution.So, its clear that for every infinitely values we can give to x, we have corresponding value of y. This equation has infinitely many solutions.

Example 3. Bhavya and Anisha have a total of Rs. 100. Express this information in the form of an equation.Solution. Let total amount with Bhavya = Rs x

and total amount with Anisha = Rs. yaccording to the question, x + y = 100, which is required equation.

Example 4. Give five integer solutions of the equation 3x + y = 8.Solution. 3x + y = 8 y = 8 – 3x

Now, when x = 0, y = 8 – 3 (0) = 8 when x = 1, y = 8 – 3 (1) = 5 when x = 2, y = 8 – 3 (2) = 2 when x = –1, y = 8 – 3 (–1) = 11 when x = –2, y = 8 – 3 (–2) = 14

Solutions can be represented in the tabular form as follows :

141125821210

yx

Example 5. Write four solutions for each of the following equations :(i) 2x + y = 7 (ii) x + y = 9 (iii) x = 4y —NCERT

Solution. (i) 2x + y = 7 y = 7 – 2xFor x = 0, y = 7 – 2(0) = 7For x = 1, y = 7 – 2(1) = 7 – 2 = 5For x = 2, y = 7 – 2(2) = 7 – 4 = 3 Four solutions of the given equation are (0, 7), (1, 5), (2, 3), (–1, 9).(ii) Given equation x + y = 9 y = 9 – xFor x = 0, y = 9 – (0) = 9For x = 1, y = 9 – (1) = 9 – For x = 2, y = 9 – (2) = 9 – 2For x = –1, y = 9 – (–1) = 9 + The four solutions of the given equation are (0, 9), (1, 9 – ), (2, 9 – 2), (–1, 9 + ).

(iii) Given equation x = 4y 4xy

For, x = 0, 040y

For x = 4, 144y

For x = – 4, 144

y

For x = 8, 248y

The four solutions of a given equation are (0, 0), (4, 1) (–4, –1) and (8, 2).


Example 6. Find a if x = 3, y = 1 is a solution of the equation 3x – y = a.Solution. Given 3x – y = a

As x = 3, y = 1 is a solution of this given equation, it must satisfy it.

3 (3) –1 = a a = 9 – 1 8a Ans.

Example 7. Give the equations of two lines passing through (2, 14). How many more such lines are there, andwhy? —NCERT

Solution. Equations of two lines passing through (2, 14) can be taken as 2x + y = 18 or 3x – y = – 8.Aslo, equations such as 7x – y = 0, 5x + 2y = 38 etc. are also satisfied by the co-ordinates of thepoint (2, 14). So, any line passing through (2, 14) is an example of a linear equation for which (2, 14)is a solution. Thus, there are infinite number of lines through (2, 14).

Example 8. Draw the graph of the following equations : (i) x = 3 (ii) y = –4 (iii) y = x (iv) y = –x (v) 2x + 5y = 10 (vi) 2x – y = 7

Solution. (i) x = 3 (ii) y = –4As x is constant, y may take any value. As y is constant, x may take any value.

2101233333

yx

4444421012

yx

(iii) y = x (iv) y = –x

2101221012

yx

2101221012

yx


(v) 2x + 5y = 10 (vi) 2x – y = 7

5210 x

y

y = 2x – 7

48.0025350

yx

31575410

yx


Example 9. From the choices given, choose the equation whose graph is given(i) y = 2x (ii) y = 2x + 1 (iii) x + y = 0 (iv) y = 2x – 4

Solution. Given points on the graph are (2, 0), (1, –2) (0, –4) and (–1, –6). By trial and error we observe thatall these 4 given points satisfy the equation y = 2x – 4, so, it is a graph of y = 2x – 4.

Example 10. The taxi fare in a city is as follows. For the first km., the fare is Rs. 8 and for the subsequentdistance it is Rs. 5 per km. Write a linear equation for this information and draw its graph.

Solution. Let total distance covered is x km.

Let total fare is Rs. y

Since, fare for Ist km is Rs. 8 and for remaining (x–1) km. is Rs. 5 per km.

By given information, we have 8 × 1 + 5 (x – 1) = y

5x – y = –3 or 35 xy

Now, equation is y = 5x + 3. Giving different values to x, we get corresponding values of y.

Let us represent this in a tabular form.

23181384321

yx


Example 11. If the work done by a body on application of a constant force is directly proportional to thedistance travelled by the body, express this in the form of an equation in two variables and drawthe graph of the same by taking the constant force as 5 units. Also, read from the graph the workdone when the distance travelled by the body is : (i) 2 units (ii) 0 units. —NCERT

Solution. Let x be the distance and y be the work done. Then, according to the given problem, we havey = 5x. ( 5 is the constant force).

Let us not draw the graph of this linear equation in two varibles.

Required table is : 1050210

yx


(i) From the graph, we see that, x = 2 units distance y = 10 units work done.(ii) From the graph, we see that, x = 0 unit distance y = 0 unit work done.

Example 12. Yamini and Fatima, two students of class IX of a school, together contributed Rs. 100 towards thePrime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satis-fies this date. (You may take their contributions as Rs. x and Rs. y). Draw the graph of the same.

—NCERTSolution. Let, Yamini contributed Rs. x and Fatima contributed Rs. y. then, according to the given question,

we have, x + y = 100

Required table is : 406080604020

yx

Every point of the shaded portion including the line x + y = 100, x-axis and y-axis in the firstquadrant represent the solution set.


Example 13. In countries like USA and Canada, temperature is measured in Fahrenheit, wheres in countries likeIndia, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:

32C59F .

(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?(iii) If the temperature is 95°F, what is the temperature in Celsius?(iv) If the temperature is 0°C, what is the temperature in Fahrenheit, and if the temperature is 0°F,

what is the temperature in Celsius?(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes,

find it. —NCERT

Solution. 32C59F .

Required table is : 866850F302010C


(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

(i) The graph of the line 32C59F is shown in the figure.

(ii) From the graph : C = 30° F = 86°(iii) From the graph : F = 95° C = 35°(iv) If C = 0° F = 32° and, If F = 0° C = – 17.8°(v) Yes, clearly from the graph, the temperature which is numerically the same is both Fahrenheit

and Celcius is – 40° F = – 40°C.

PRACTICE EXERCISE1. Which of the following are linear :

(i) 3x (x – 2) = 3x2 + 2x – 7 (ii) (x + 1) (x – 2) = 10(iii) x (x + 1) = –2x2 + 7x + 2 (iv) 2x2 + 7x – 3 = 2x (x + 1)

2. Write each of the following equation in the form ax + by + c = 0

(i) y – 3 = 2 (x – 1) (ii) 53

yx

(iii) 62

3– yx

(iv) 2 (x – 1) –3 (y + 1) = 1

(iii) 632

yx(iv) 3

413

2

–yx 1

3. Write four solutions for each of the following :(i) 2x – y = 3 (ii) x = 3y

4. Find out which of the following equations have x = 2, y = –1 as a solution :

(i) 3x + 2y = 4 (ii) 4x – y = 8 (iii) 32

yx

(iv) 4x – y = 9 (v) 22

yx

(vi) 7x – 4y = 18

5. Find the value of k if the given point lies on the graph defined by each equation :(i) 3x + ky = 2 ; (1, –1) (ii) y + kx = 8 ; (–3, 2)(iii) 2x – 3y = k ; (–1, 5) (iv) 2x – k = 4y ; (2, 0)

6. Show that x = 1, y = –6 ; x = 2, y = –3 and x = 3, y = 0 are all solution of equation 3x – y = 9.7. The equation of a graph is given by 2x + y = 10. Indicate which of the following points lies on the graph.

(i) (6, –2) (ii) (2, 8) (iii) (2, 6)(iv) (–3, 16) (v) (–6, 18) (vi) (–6, 22)

8. Draw the graph of line 2x + 5y = 13. Is the point (9, –1) lies on the line?9. Draw the graph of y = –2x + 4. Find co-ordinates of point at which it intersects the axes.

10. Draw the graph of y = 3x + 2 and y = 3x – 1 using the same pair of axes. Are these two lines parallel?11. Draw the graph of 3x + 2y = 7 and 2x – 3y = 10 using the same pair of axes. Are these two lines

perpendicular?


12. Draw the graph of 3x – y = 5 and 2x + 3y = 7 on the same axes. What is their point of intersection?13. Graph the following equations :

(i) 2y = –x + 3 (ii) 321 y

x(iii) 3x – 4y = 12 (iv) y = |x|

14. Amit invests Rs. 100 at the rate of 5% p.a. simple interest. Assuming rate to be same, find graphically theinterest he will earn after 5 years?

15. Unit of temperature measurement, Fahrenheit and celsius are related by relation 32C59

F .

(i) Draw graph of linear equation above, using celsius for x-axis and Fahrenheit for y-axis.(ii) If the temperature is 35°C, what is the temperature in Fahrenheit?

(iii) If the temperature is 86°F, what is the temperature in celsius?(iv) Is there a temperature which is numerically the same in both units? If yes, find it.

PRACTICE TEST

M.M : 15 Time : ½ hourGeneral Instructions :All questions carry 3 marks each.

1. Give three integral solutions for equation 5x – y = 9.2. Find the value of a so that the equation 4x – ay = 7 have (2, –3) as a solution.3. Give the geometrical representation of x = –3 as an equation.

(i) In one variable (ii) In two variables4. Yamini and Fatima, together contributed Rs. 100 towards PM relief fund. Write a linear equation which

this data satisfies. Draw the graph of the same.5. Draw the graph of 3x + y = 8. Is x = –1, y = 11, a solution of this equation. Also, shade the portion bounded

by this line and both the axes.


1. (i) and (iv)2. (i) 2x – y + 1 = 0 (ii) x – 3y – 15 = 0 (iii) x – 2y – 15 = 0 (iv) 2x–3y–6=0

(v) 3x – 2y – 36 = 0 (vi) 2x – 3y – 9 = 03. (i) (0, –3), (1, –1), (2, 1), (–1, –5) (ii) (3, 1), (6, 2), (9, 3), (12, 4)4. (i), (iv), (v), (vi)5. (i) k = 1 (ii) k = –2 (iii) k = –17 (iv) k = 47. (i), (iii), (iv), (vi)


8. Yes

9. (0, 4) and (2, 0)


10. Yes 11. Yes

12. (2, 1) 13. (iv) y = | x |


14. 15.

From graph, we observe that after From graph, we have : (ii) 95°F 5 years, he will receive Rs. 25. (iii) 30°C (iv) Yes, –40°


1. (1, –4), (2, 1) and (3, 6) 2. 31

a

3. (i) (ii)

--3


4. x + y = 100

where, x Rs. = contribution by Yamini and y Rs. = contribution by Fatima.5. Yes

52 INTRODUCTION TO EUCLID’S GEOMETRY MATHEMATICS–IX

CHAPTER 5INTRODUCTION TO EUCLID’S GEOMETRY

Points to Remember :1. A point, a line and a plane are concepts only and these terms are taken as undefined.2. Axioms (or Postulates) are assumptions which are self evident truths.3. Theorems are the statements which are proved, using axioms, previously proved statements and

deductive reasoning.4. Some of Euclid’s axioms were:

(a) Things which are equal to the same thing are equal to one another.(b) If equals are added (or subtracted) to / from equals, the wholes / remainders are equal.(c) The whole is greater than the part.(d) Things which are double of the same things are equal to one another.

5. Euclid’s Five Postulates:Postulate 1: A straight line may be drawn from any one point to any other line.Postulate 2: A terminated line can be produced indefinitely.Postulate 3: A circle can be drawn with any center 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 ittaken together less than two right angles, then the two straight lines, if produced indefinitely, meet onthat side on which the sum of angles is less than two right angles.

6. Two equivalent versions of Euclid’s fifth Postulate:(a) Play fair axiom:

“Through a given point, not on the line, one and only one line can be drawn parallel to a given line.”

(b) Two distinct intersecting lines cannot be parallel to the same line.


Example 1. Prove that an equilateral triangle can be constructed on any given line segment.Solution. Draw a line segment, says PQ. Using

Euclid’s postulate 3, draw a circle withcenter P and radius PQ. Again, we drawanother circle with center Q and radius QP.The two circles meet at a point, says R.Join P to R and Q to R. now, in PQR,PQ = PR and PQ = QR. So, by axiom 4, weget PQ = QR = PR.Hence, PQR is an equilateral triangle.

AMIT BAJA

J

MATHEMATICS–IX INTRODUCTION TO EUCLID’S GEOMETRY 53

Example 2. If a point C lies between two points A and B such that AC = BC, then prove that AC = 21

AB.

Solution. Since, C lies between AB. AB = AC + BC ( by addition axiom) AB = AC + AC ( AC = BC) AB = 2AC

AC = 21

AB.

which prove the result.Example 3. Let point C be a mid-point of line segment AB. Prove that every line segment has one and only

one mid-point.Solution. Let if possible, D be another mid-point of AB.

AD = DB ....(1)But, it is given that C is the mid-point of AB. AC = CB ...(2)Subtracting (1) from (2), we get

AC – AD = CB – DB DC = – DC 2DC = 0 DC = 0 C and D must coincides.Thus, every line segment has one and only one mid-point.

Example 4. In figure, if AC = BD, then prove that AB = CD.Solution. AC = BD ...(1) ( given )

Also, AC = AB + BC ...(2)( point B lies between A and C)

and, BD = BC + CD ...(3)( point C lies between B and D)

Substituting for AC and BD from (2) and (3) in (1),We get, AB + BC= BC + CD AB + BC – BC = BC + CD – BC ( subtracting BC both sides) AB = CDHence proved.

Example 5. 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 infinite number of lines which pass through two distinct points.(iii) A terminate line can be produced indefinitely on both sides.(iv) If two circles are equal, then their radii are equal.(v) In figure, if AB = PQ and PQ = XY then AB = XY.

—NCERTT

Solution. (i) False, since, through a point infinite number of lines may be drawn.(ii) False, since one and only one line can pass through two distinct points.(iii) True, since a line can be produced infinitely on both the ends.(iv) True, since two circles will be equal only when their radii are the same.(v) True, since AB = PQ, PQ = XY AB = XY. (By transitive property)

54 INTRODUCTION TO EUCLID’S GEOMETRY MATHEMATICS–IX

PRACTICE EXERCISE1. What is the difference between axiom and a theorem?2. Give any three axioms given by Euclid.3. What are Euclid’s five postulates?4. State Play fair’s axiom.5. Give two equivalent versions of Euclid’s fifth postulate.6. If P,Q and R are three points on a line, and Q lies between P and R, then prove that PQ + QR = PR.7. Prove that an equilateral triangle can be constructed on any given segment.8. Prove that, “Two distinct lines cannot have more than one point in common.”

MATHEMATICS–IX LINES AND ANGLES 55

CHAPTER 6LINES AND ANGLES


1. If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and vice-versa. Thisproperty is known as the Linear Pair Axiom.

2. If two lines intersect each other, then the vertically opposite angles are equal.3. If a transversal intersects two parallel lines, then

(a) each pair of corresponding angles is equal.(b) each pair of alternate interior angles is equal.(c) each pair of interior angles on the same side of the transversal is supplementary.

4. If a transversal intersects two line such that, either(a) any one pair of corresponding angles is equal, or(b) any one pair of alternate interior angles is equal, or(c) any one pair of interior angles on the same side of of the transversal is supplementary, then the lines

are parallel.5. Two intersecting lines cannot both be parallel to the same line.6. Lines which are parallel to a given line are parallel to each other.7. The sum of three angles of a triangle is 180°.8. If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior

opposite angles.

ILLUSTRATIVE EXAMPLESExample 1. If two lines intersects, prove that vertically opposite angles are equal.Solution. Given : Two lines AB and CD intersects at O.

To prove : 1 = 3 and 2 = 4Proof : OB stands on line COD.1 + 2 = 180° ( linear pair) ...(1)Also, OD stands on line AOB2 + 3 = 180° ( linear pair) ...(2)from (1) and (2)1 + 2 = 2 + 31 = 3Similarly, 2 = 4 Hence proved

Example 2. The complement of an angle is half of itself. Find the angle and its complement.Solution. Let the given angle be x. Then, its complement is 90° – x.

According to given question,

90° – x = 21

x 902x

x

3902

902

3 x

xAMIT B

AJAJ

56 LINES AND ANGLES MATHEMATICS–IX

x = 60°Hence, the given angle is 60° and its complement is 90° – 60° i.e. 30°.

Example 3. Two supplementary angles are in the ratio 3 : 2. Find the angles.Solution. Let the angles be 3x and 2x.

according to given question,3x + 2x = 180° ( angles are supplementary)

5x = 180° º365

º180 x

723622and1083633areAngles

ºxºx

Ans.

Example 4. What value of x would make AOB a line in given figure if AOC = (3x –10)° andBOC = (7x + 30)°?

Solution. Since, OB and OA are opposite rays BOC + AOC = 180º 7x + 30° + 3x – 10° = 180° 10x + 20° = 180° 10x = 180° – 20°

10x = 160º 10160

x 16x Ans.

Example 5. Rays OA, OB, OC, OD and OE have the common initial point O. Show that :AOB + BOC + COD + DOE + EOA = 360°

Solution. Let us draw a ray OF oposite to ray OA.

B

D

23

4 05

E

F A

C

61

so, 1 + 2 + 3 = 180° and ...(1)4 + 5 + 6 = 180° ( linear pair axiom) ...(2)

adding (1) and (2),(1 + 2 + 3) + (4 + 5 + 6) = 180° + 180°

1 + 2 + (3+ 4) + 5 + 6 = 360° AOB + BOC + COD + DOE + EOA = 360°Hence shown.


Example 6. Prove that if a transversal intersects two parallel lines, then each pair of interior angle on the sameside of the transversal is supplementary.

Solution. Given : A transversal l intersects two parallel lines AB and CD at P and Q.

To prove : 4 + 5 = 180° and 3 + 6 = 180º

Proof : Ray QD stands on line l 1 + 4 = 180° ...(1) ( linear pair axiom)also, 1 = 5 ...(2) ( corresponding angles)from (1) and (2)

4 + 5 = 180°Again, ray QC stands on line l 2 + 3 = 180° ...(3) ( linear pair axiom)and, 2 = 6 ...(4) ( corresponding angles)from (3) and (4)

3 + 6 = 180° Hence proved.Example 7. If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of

interior angles encloses a rectangle.Solution. Given : Two parallel lines AB and CD are intersected by a transversal at P and R respectively.

PQ, RQ, RS and PS are bisectors of APR, PRC, PRD and BPR respectively.

To prove : PQRS is a rectangleProof : Since AB || CD and l is a transversal APR = PRD (alt. interior angles)

21APR = 2

1PRD QPR = PRS

But, these are alternate interior angles. PQ || RS. Similarly QR || PS. PQRS is a parallelogram.


Now, ray PR stands on AB APR + BPR = 180° (linear pair)

21APR +

21BPR = 90°

QPR + SPR = 90° QPS = 90°Thus, PQRS is a parallelogram, one of whose angle is 90° PQRS is a rectangle.Hence proved.

Example 8. In the given figure lines XY and MN intersect at O. If POY = 90° and a : b = 2 : 3, find c.

—NCERTSolution. Since XY is a line a + b + 90° = 180° ( linear pair)

a + b = 180° – 90° a + b = 90°But a : b = 2 : 3. Let a = 2x and b = 3x. 2x + 3x = 90° 5x = 90° x = 18° a = 2x = 2 × 18° = 36° and b = 3x = 3 × 18° = 54°Now, OM and ON are opposite rays. MON is a line.Since ray OX stands on MN MOX + XON = 180° ( linear pair) c + b = 180° c + 54° = 180° C = 180° – 154° = 126° c = 126° Ans.

Example 9. 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. —NCERT

Solution. Since XY is produced to point P, XP is a straight line.Since, YZ stands on XP, XYZ + ZYP = 180° ( linear pair) 64° + ZYP =1 80° ZYP = 180° – 64° = 116°Since ray YQ bisects ZYP

582

116ZYQQYP

Now, XYQ = XYZ + ZYQ XYQ = 64° + 58° = 122°and reflex QYP = 360° – QYP = 360° – 58° = 302° Ans.


Example 10. In the given figure, if AB || CD, CD || EF and y : z = 3 : 7, find x. —NCERT

Solution. Since, AB || CD and CD || EF AB || CD || EF.Now, CD || EF and PR is a transversal. PQD = PRF 180° – y = z y + z = 180°also, y : z = 3 : 7. Let y = 3a and z = 7a 3a + 7a = 180° 10a = 180° a = 18° y = 3 × 18° = 54° and z = 7 × 18° = 126°Now, AB || CD and PQ is a transversal. x + y = 180° ( consecutive interior angles are supplementary) x + 54° = 180° x = 180° – 54° = 126° x = 126° Ans.

Example 11. In the given figure, X = 62°, XYZ = 54°. If YO and ZO are the bisectors of XYZ and XZYrespectively of XYZ, find OZY and YOZ. —NCERT

Solution. In XYZ, YXZ + XYZ + XZY = 180° ( angle sum property of a triangle) 62° + 54° + XZY = 180° XZY = 180° – 62° – 54° = 64°Since, YO and ZO are bisectors of XYZ and XZY,

275421XYZ

21OYZ

and, 326421XZY

21OZY

In OYZ, we have YOZ + OYZ + OZY = 180° ( angle sum property of a triangle) YOZ + 27° + 32° = 180° YOZ = 180° – 27° – 32° = 121°Hence, OZY = 32° and YOZ = 121° Ans.


Example 12. In the given figure, if PQ PS, PQ || SR, SQR = 28° and QRT = 65°, then find the values of xand y. —NCERT

Solution. QRT = RQS + QSR ( Exterior angle property in SRQ) 65° = 28° + QSR QSR = 65° – 28° = 37°Since, PQ || SR and the transversal PS intersects then at P and S respectively. PSR + SPQ = 180° ( sum of consecutive interior angles is 180°) (PSQ + QSR) + 90° = 180° y + 37° + 90° = 180° y = 180° – 37° – 90° = 53°Now, in the right SPQ, we have PQS + PSQ = 90° x + 53°= 90° x = 90° – 53° = 37°Hence, x = 37° and y = 53° Ans.

Example 13. In each of the following figures, AB || CD, find x.

O

25°

B

DC

A

50°

x°

(i)

Solution. (i) Through O, Draw EOF || AB || CD.then, 1 + 2 = xNow, EO || AB and BO is the transversal. 1 + 50º = 180º( Interior angles on the same side of the transversal)

O25°

B

DC

A

E F

50°

1

2


1 = 180° – 50° 1 = 130° ...(1)Again, EO || CD and OD is a transversal 2 + 25° = 180° 2 = 180° – 25° 2 = 155° ...(2)adding (1) and (2)1 + 2 = 130° + 155° = 285°

i.e. º285x Ans.

(ii) Produce AB to intersect CE and F.

CA

x°

105°

D

B

E

125° GF

Now, ABE + EBF = 180° ( Linear pair) 125° + EBF = 180° EBF = 180° – 125° = 55°Again, CD || FG and CF is a transversalDCF + CFG = 180° ( Interior angles on same side of transversal)105° + CFG = 180ºCFG = 180° – 105° = 75°Also, BFE = CFG = 75° ( vertically opposite angles)Now, In BEF,BEF + EBF + BFE = 180° ( angle sum property) x° + 55° + 75° = 180° xº + 130° = 180° x = 180º – 130°

50x Ans.

Example 14. The side QR of PQR is produced to a point S. If the bisectors of PQR and PRS meet at point

T, then prove that QTR = 21QPR. —NCERT

Solution. Side QR of PQR is produced to S, Ext. PRS =P + Q ( Exterior angle sum property of a triangle)

21

Ext PRS = 21P + 2

1Q

2 = 21P + 1 ...(1)

R1

1 22

SQ

P T

Again, In QRT., Ext. TRS = T + 1 ( same as above) 2 = T + 1 ...(2)


Equating (1) and (2), we get

21P + 1 = T + 1

T = 21P or QTR = 2

1QPR

Hence Proved.Example 15. Bisectors of exterior angles of ABC (obtained by producing sides AB and AC) meet at O. Prove

that BOC = 90° – 21A.

Solution. Given : ABC in which bisectors of exterior angles meet at O (as shown)

To prove : BOC = 90 – 21A

Proof : In OBC1 + 2 + 0 = 180º ( angle sum property of a triangle)

21CBD + 2

1BCE + O = 180°

CBD + BCE + 2O = 360° (180º – B) + (180º – C) + 2O = 360° B + C = 2O 180 – A = 2O

O = 21

(180 – A)

BOC = 90° – 21A. Hence proved.

PRACTICE EXERCISE1. If an angle is 24° less than its complement. find its measure.2. An angle is 40° less than one-third of its supplement. Find the angle and its supplement.3. Two supplementary angles are in the ratio 11 : 7. Find them.4. Find the angle whose supplement is four times its complement.5. Find the measure of an angle if, three times its supplement is 60° more than six times its complement.6. In the following figure, it is given that 2a – 5b = 10º find a and b.

a b

A B7. In the following figures, what value of x will make AOB a straight line?

AMIT BAJA

J


8. In the following figure, COA = 90º and AOB is a straight line. Find x and y.

9. Prove that the bisectors of the angles of a linear pair are at right angles.10. If the bisectors of two adjacent angles form a right angle then prove that their non-common arms are in

the same straight line.11. In given figure, AOB is a line. Ray OD is perpendicular to AB . OC is another ray lying between OA and

OD. Prove that DOC = 21

(BOC – AOC)

A B

D

90°

12. In the following figure, AB and CD intersects at O and BOE = 70°. Find value of a, b and c.

13. In the given figure 1 : 2 = 5 : 4 and AB || CD. Find all the labelled angles.

87


14. Find missing x in the following diagrams. Given AB || CD.

140°

85°

x°

E

BA

C D(v)

15. In the following figure AB || CD. Find the values of a, b and c.

16. In the following figure, show that AB || CD.

160°

20°30°

50°

E F

C D

BA


17. In the following figure, AB || CD and CD || EF. Also EA AB. If BEF = 65°, find the value of a, b and c.

18. The angles of a triangle are

50

2x ,

60

3x and (2x – 15)°, find the angles.

19. If a transversal cuts two parallel lines and is perpendicular to one of them, show that it will be perpen-dicular to the other also.

20. If two parallel lines are intersected by a transversal, show that the bisectors of any corresponding anglesare parallel.

21. If two lines are intersected by a transversal in such a way that the bisectors of a pair of correspondingangles are parallel, show that the lines are parallel to each other. —NCERT

22. Prove that the bisectors of a pair of alternate angles of two parallel lines are themselves parallel.23. Prove that if the arms of an angle are, respectively, parallel to the arms of another angle, then the angles

either have equal measure or they are supplementary.24. 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. —NCERT

Q

S

P

R

DA

B

C

25. If two straight lines are perpendicular to the same line, prove that they are parallel to each other.26. One of the angles of a triangle is 55°. Find the remaining two angles, if their diffrence is 35°.27. In the given figures find x :


28. Sides QP and RQ of PQR are produced to points S and T respectively. If SPR = 140° and POT = 105°,find PRQ.

29. A square ABCD is surmounted by an equilateral triangle EDC. Find x.

C

E

x°

D

A B

30. In the following figure, AB || DC. If 2yx and zy

94

. Find the values of x, y and z.

yx

zC

D

B

APRACTICE TEST

MM : 30 Time : 1 hourGeneral Instructions :Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.

1. What value of x would make AOB a straight line?

BA(3 + 3)°x

(2 –15)°x

x°O

C

D


2. If two parallel lines are intersected by a transversal, prove that the bisectors of two interior alternateangles are parallel.

3. Find the value of x°.

4. Find the value of x°, it is given that AB || CD.

5. If two straight lines are perpendicular to the same line, prove that they are parallel to each other.6. In the given figure, show that AB || CD.

145°

35°45°

80°

A B

C D

FE

7. In the following figure, AB || CD, find a and b.

B

DG135°

Fa

b

E55°

C

A

8. Angles A, B and C of a triangle satisfy B – A = 30° and C – B = 45°. Find all the angles.9. Prove that the sum of three angles of a triangle is 180°.


10. In the given figure, the sides AB and AC of ABC are produced to points E and D respectively. Ifbisectors BO and CO of CBE and BCD respectively meet at O, then prove that :

BOC = 90° – 21BAC

CB

E D

O

A

ANSWERS OF PRACTICE EXERCISE1. 33° 2. 15°, 165° 3. 110°, 70° 4. 60° 5. 20°6. 130°, 50° 7. (i) x = 20° (ii) x = 33° (iii) x = 15° 8. x = 33°, y = 11°

12. a = 22°, b = 44°, c = 92° 13. 1 = 3 = 5 = 7 = 80°, 2 = 4 = 6 = 8 = 100°14. (i) x = 115° (ii) x = 40° (iii) x = 55° (iv) x = 80° (v) x = 135°15. a = 100°, b = 35°, c = 45° 17. a = 25°, b = 115°, c = 115°18. 65°, 70°, 45° 26. a = 80°, b = 45°27. (i) 65° (ii) 95° 28. 65°29. x = 45° 30. 24°, 48°, 108°


1. x = 32° 3. x = 110° 4. 290°

7. a = 55°, b = 80° 8. A = 25°, B = 55°, C = 100°

MATHEMATICS–IX TRIANGLES 69

CHAPTER 7TRIANGLE

Points to Remember :1. Two figures are congruent, if they are of same shape and same size.2. If two triangles ABC and XYZ are congruent under the correspondence A X, B Y and C Z, then

symbolically, ABC XYZ3. SAS Congruence Rule : 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.4. ASA Congruence Rule : If two angles and the included side of one triangle are equal to two angles and

the side of the other triangle, then the two triangles are congruent.5. AAS Congruence Rule : 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.6. RHS congruence Rule : 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.7. SSS Congruence rule : If three sides of one triangle are equal to the three sides of anoter triangle, then

the two triangles are congruent.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 60°.11. Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular

line segment is the shortest.12. In a triangle, angle opposite to the longer side is greater.13. In a triangle, side opposite the greater angle is longer.14. Sum of any two sides of a triangle is greater than the third side.15. Difference between any two sides of a triangle is less than its third side.

ILLUSTRATIVE EXAMPLESExample 1. In the given figure, AB = AC and ACD = 110°, find A.

110°

DCB

A

Solution. Since AB = AC B = C (angles opposite to equal sides are equal)NowACB = 180° – 110° ( linear pair) ACB = 70° B = ACB = 70°Now, A B + C = 180º (angle sum property of a triangle)

70 TRIANGLES MATHEMATICS–IX

A + 70° + 70° = 180°

A = 180° – 140° 40A Ans.

Example 2. In given figure, AB = CF, EF = BD and AFE = DBC. Prove that AFE CBD.Solution. Given, AB = CF

Add BF both sides, we get AB + BF = CF + BF AF = CBNow, In AFE and CBD, AF = BC (proved above)

D

CFB

E

A

AFE = CBD (given)and EF = BD (given)So, by SAS congruence rule,AFE CBD

Example 3. In quadrilateral ABCD, AC = AD and AB bisects A. (see figure). Show that ABC ABD.What can you say about BC and BD? —NCERT

Solution. In ABC and ABD, we have AC = AD (given)

CAB = BAD ( AB bisects A)and, AB = AB (common side) ABC ABD (SAS congruence condition)also, BC = BD (cpct)

Example 4. ABCD is a quadrilateral in which AD = BC and DAB = CBA (see figure).Prove that :(i) ABD BAC(ii) BD = AC(iii) ABD = BAC —NCERT

Solution. In ABD and BAC, we have AD = BC (given)DAB = CBA (given) AB = AB (common side)

ABD BAC, (SAS congruence condition)which proves (i) part BD = ACand ABD = BAC, which proves (ii) and (iii) (cpct)

Example 5. If the bisector of the vertical angle of a triangle bisects the base, prove that the triangle isisosceles.

Solution. Given ABC in which AD is the bisector of A which meets BC in D such that BD = DC.To prove : AB = ACConstruction : Produce AD to E, such that AD = DE.Join E to C


Proof : In ABD and ECD BD = DC (given) AD = DE (by construction)ADB = EDC (vertically opp. angles)

C

E

B

A

3

D

1 2

ABD ECD (by SAS congruence condition) AB = EC and 1 = 3 (cpct)also, 1 = 2 (AD is angle bisector) 2 = 3 EC = AC (sides opp. to equal angles) AB = AC ( EC = AB)Hence, ABC is isosceles.

Example 6. In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is doublethe smallest side.

Solution. Given PQR in which Q = 90° and QRP = 2QPRTo prove : PR = 2QRConstruction : Produce RQ to S such that RQ = QS. Join P to S.Proof : In PSQ and PQR

PQ = PQ (common side) PQS = PQR (= 90° each)

QS = QR (by construction) PSQ PQR (SAS congruence condition) PS = PR and SPQ = RPQ = xº (say) (cpct) SPR = 2x° = PRQ = PRSNow, In PSR, SPR = PRD = 2x° PS = SR PR = SR ( PS = PR) PR = 2QR ( SR = 2QR)Hence proved.

Example 7. In the given figure, PQRS is a quadrilateral and T and U are respectively points on PS and RS suchthat PQ = QR, PQT = RQU and TQS = UQS. Prove that QT = QU.

Solution. PQT = RQU (given) ....(1)TQS = UQS (given) .....(2)adding, (1) and (2) we getPQT + TQS = RQU + UQSor PQS = RQSNow, In PQS and RQS PQ = RQ (given)PQS = UQS (proved above) QS = QS (common side) PQS RQS (SAS congruence condition)So, 1 = 2 (cpct)

12

S

UR

Q

PTT

Again, In TQS and UQSTQS = UQS (given)1 = 2 (proved above)


QS = QS (common side) TQS UQS (ASA congruence condition)Hence, QT = QU (cpct)

Example 8. Line l is the bisector of an A, and B is any point on . BP and BQ are perpendiculars from B tothe arms of A (see figure). —NCERT

Show that :(i) APB AQB(ii) BP = BQ or B is equidistant from the arms of A.

Solution. In APB and AQB, we haveAPB = AQB ( each = 90°)PAB = QAB ( AB bisects PAQ)

AB = AB ( common side) APB AQB, (AAS congruence condition)which proves (i) part.and, BP = PQ (cpct)which proves (ii) part.

Example 9. In the given figure, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. —NCERT

Solution. In ABC and ADE, we have AB = AD (given)

BAC = DAE( BAD = EAC BAD + DAC = EAC + DAC BAC = DAE)

and, AC = AE (given) ABC ADE (SAS congruence condition) BC = DE (cpct)Hence shown.

Example 10. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M andproduced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that(i) AMC BMD(ii) DBC is a right angle(iii) DBC ACB

(iv) .AB21CM

—NCERTSolution. (i) In AMC and BMD, we have

AM = BM ( M is the mid-point of AB) AMC = BMD (Vertically opp. angles)

CM = MD (given) AMC BMD (SAS congruence condition)


(ii) Now, AMC BMD BD = CA and BDM = ACM ...(1) (cpct)Thus, transversal CD cuts CA and BD at C and D respectively such that the alternate anglesBDM and ACM are equal. BD || CA. CBD + BCA = 180°

( consecutive interior angles on the same side of a transversal are supplementary). CBD + 90° = 180° CBD = 90° or, DBC = 90°(iii) Now, in DBC and ACB, we have

BD = CA (from (i))DBC = ACB ( each = 90°) BC = BC (common side)

DBC ACB (SAS congruence condition)(iv) CD = AB (cpct)

AB21CD

21

AB.21CM Hence proved.

Example 11. In the given figure, AB = AC, DB = DC. Prove that ABD = ACD.Solution. Join A to D

In ABD and ACD, we have AB = AC (given) BD = CD (given) AD = AD (common side)

C

D

A

B ABD ACD (SSS congruence condition)So, ABD = ACD (cpct)

Example 12. In ABC, D is the mid point of base BC. DC and DF are perpendiculars to AB and AC respectively,such that DE = DF. Prove that B = C.

Solution. In BED and CFD, we have DE = DF (given) BD = DC (D is mid point of BC)BED = CFD (= 90° each) BED CFD (RHS congruence condition)So, B = C (cpct)

Example 13. 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 : —NCERT(i) OB = OC(ii) AO bisects A.

Solution. (i) In ABC, we have AB = AC B = C ( angles opposite to equal sides are equal)

C21B

21

OBC = OCB

( OB and OC bisect B and C respectively. )C21OCBandB

21OBC

OB = OC ( sides opposites to equal angles are equal).


(ii) Now, In ABO and ACO, we have, AB = AC (given)

OBC = OCB (proved above) OB = OC (proved above)

ABO ACO (SAS congruence condition) BAO = CAO (cpct) AO bisects BAC. Hence proved.

Example 14. ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respec-tively (see figure). Show that these altitudes are equal. —NCERT

Solution. In ABC, AB = AC ACB = ABC ECB = FBC Now, In BEC and CFB, we haveBEC = CFB (each = 90°)ECB = FBC (proved above)

BC = BC (common side) BEF CFB (AAS congruence condition) BE = CF (cpct)Hence shown.

Example 15. ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ABD =ACD. —NCERT

Solution. In ABC, we have AB = AC ABC = ACB ...(1)

( angles opposite to equal sides are equal)Also, In BCD, we have BD = CD DBC = DCB ...(2)

( same reason as above)Adding (1) and (2), we get,ABC + DBC = ACB + DCB ABD = ACD. Hence shown.

Example 16. ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (seefigure). Show that BCD is a right angle. —NCERT

Solution. In ABC, we have AB = AC ACB = ABC ...(1)

( angles opp. to equal sides are equal).Now, AB = AD (given) AD = AC ( AB = AC).Thus, In ADC, we have AD = AC ACD = ADC ...(2)Adding (1) and (2), we get

ACB + ACD = ABC + ADC BCD = ABC + BDC ( ADC = BDC)BCD + BCD = ABC + BDC + BCD

( adding BCD both sides)2BCD = 180° ( angle sum property)BCD = 90°Hence, BCD is a right angle.


Example 17. ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on thesame side of BC. (see figure). If AD is extended to intersect BC at P, show that : —NCERT (i) ABD ACD (ii) ABP ACP(iii) AP bisects A as well as D. (iv) AP is the perpendicular bisector of BC.

Solution. (i) In ABD and DBC, we have AB = AC (given) BD = DC (given) AD = AD (common side)

ABD DBC (SSS congruence condition)(ii) In ABP and ACP, we have

AB = AC (given) BAP = CAP

( ABD ACD BAD = DAC BAP = PAC) AP = AP (common side) ABP ACP (SAS congruence condition)(iii) Since ABD ACD BAD = DAC AD bisects A AP bisects A ...(1)In BDP and CDP, we have

BD = CD (given) BP = PC ( ABP ACP BP = PC) DP = DP (Common side)

BDP CDP (SSS congruence condition) BDP = PDC DP bisects D AP bisects D ...(2)from (1) and (2), we getAP bisects A as well as D.(iv) Since AP stands on BC. APB + APC = 180° ( linear pair)But, APB = APC

APB = APC

902

180

Also, BP = PC (proved above) AP is perpendicular bisector of BC.

Example 18. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ andQR and median PN of PQR (see figure). Show that : —NCERT (i) ABM PQN(ii) ABC PQR


Solution. Since, BC = QR

QNBMQR21BC

21

...(1)

Now, In ABM and PQN, we have AB = PQ (given) BM = QN (from (1)) AM = PN (given)

ABM PQN, (SSS congruence condition)which proves (i) part. B = Q (cpct) ...(2)Now, In ABC and PQR, we have

AB = PQ (given) B = Q (from (2))

BC = QR (given) ABC PQR (SAS congruence condition)which proves (ii) part.

Example 19. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that thetriangle ABC is isosceles. —NCERT

Solution. In BCF and CBE, we haveBFC = CEB ( each = 90°) hyp. BC = hyp. BC (common side)

FC = EB BCF CBE ( RHS congruence condition) FBC = ECB (cpct)Now, In ABC

ABC = ACB ( FBC = ECB) AB = ACABC is an isosceles triangle.

Example 20. In the given figure, the line segment joining the mid points M and N of opposite sides AB and DCof quadrilateral ABCD is perpendicular to both these sides. Prove that the other two sides ofquadrilateral are equal.

Solution. Join M to D and M to C.In MND and MNC

DN = CN (given)MND = MNC (each = 90°) MN = MN (common side)

MND MNC (RHS congruence condition) DM = CM and DMN = CMN ....(1)Now, AMN = BMN (= 90° each) ....(2)Subtracting (1) from (2), we get

AMN – DMN = BMN – CMN

M BA

CND

AMD = BMCNow, In AMD and BMC, we have

AM = BM (given)


MD = MC (proved above)AMD = BMC (proved above)

AMD BMC (SAS congruence condition)So, AD = BC (c.p.c.t.)i.e. other two sides of the quadrilateral are equal.

Example 21. In figure, B < A and C < D. Show that AD < BC. —NCERT

Solution. Since, B < A and C < D AO < BO and OD < OC ( side opp. to greater angle is larger)Adding these results, we get

AO + OD < BO + OC AD < BC. Hence shown.

Example 22. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure).Show that A > C and B > D. —NCERT

Solution. Join A to C and B to D.Since, AB is the smallest side of quadrilateral ABCD. In ABC, we have BC > AB 8 > 3 ...(1)

( angle opp. to longer side is greater).Since, CD is the longest side of quadrilateral ABCD, In ACD, we have CD > AD 7 > 4 ...(2)

( angle opp. to longer side is greater).Adding (1) and (2), 8 + 7 > 3 + 4 A > CAgain, In ABD, we have

AD > AB ( AB is the shortest side) 1 > 6 ...(3)In BCD, we have

CD > BC ( CD is the longest side) 2 > 5 ...(4)Adding (3) and (4), we get

1 + 2 > 5 + 6 B > DThus, A > C and B > D. Hence shown.

Example 23. In the given figure, PR > PQ and PS bisects QPR. Prove that PSR > PSQ. —NCERT


Solution. In PQR, we have PR > PQ (given) PQR > PRQ ( Angle opposite to larger side is greater) PQR + 1 > PRQ + 1 (adding 1 both sides) PQR + 1 > PRQ + 2 ...(1)

( PS is the bisector of P 1 = 2)Now, In PQS and PSR, we havePQR + 1 + PSQ = 180° and PRQ + 2 + PSR = 180°PQR + 1 = 180° – PSQ and PRQ + 2 = 180° – PSR 180° – PSQ > 180° – PSR (Using (1)) – PSQ > – PSR PSQ < PSR i.e. PSR > PSQ.Hence proved.

Example 24. In the given figure, ABC is a triangle and D is any point in its interior. Show that :BD + DC < AB + AC

Solution. Extend BD to meet AC in E.In ABE, AB + AE > BE

( sum of any two sides is greater than the third side) AB + AE > BD + DE ....(1)In CDE, DE + EC > DC ....(2)

A

D E

CBfrom (1) and (2), we get

AB + AE + DE + EC > BD + DE + DC AB + (AE + EC) > BD + DC AB + AC > BD + DC BD + DC > AB + ACHence shown.

Example 25. In the given figure PQRS is a qudrilateral in which diagonals PR and QS intersect in O. Show that (i) PQ + QR + RS + SP > PR + QS(ii) PQ + QR + RS + SP < 2 (PR + QS)

Solution. In PSR, we havePS + SR > PR ....(1)( sum of any two sides is greater than the third side)

In PQR, PQ + QR > PR ....(2)adding (1) and (2), we get

PQ + QR + PS + SR > 2PR ....(3) O

RS

QPIn PSQ, PS + PQ > SQ ....(4)In RSQ, RS + RQ > SQ ....(5)adding (4) and (5), we get

PQ + RQ + RS + SP > 2SQ ....(6)adding (3) and (6), we get

2 (PQ + QR + RS + SP) > 2 (PR + SQ) PQ + QR + RS + SP > PR + SQ ....(7)which proves (i) part


Again, In OPQ, OP + OQ > PQ ....(8)In OQR, OR + OQ > RQ ....(9)In OSR, OS + OR > SR ....(10)In OPS, OP + OS > PS ....(11)

adding (8), (9), (10), (11), we get2 (OP + OQ + OR + OS) > PQ + QR + RS + SP

or 2 (PQ + QS) > PQ + QR + RS + SPor PQ + QR + RS + SP < 2 (PR + QS)which proves (ii) part.

PRACTICE EXERCISE1. In PQR, P = 80° and PQ = PR. Find Q and R. (Ans. 50°, 50°)2. Prove that the measure of each angle of an equilateral triangle is 60°.3. In ABC, BAC = 50° and AB = AC. If ACD = x°, find the value of x. (Ans. x = 115°)

x°DCB

A

50°

4. If the altitude from one vertex of a triangle bisects the opposite side, prove that the triangle is isosceles.5. Prove that the perpendiculars drawn from the vertices of equal angles of an isosceles triangle to the

opposite sides are equal.6. Prove that medians of an equilateral triangle are equal.

7. If S is the mid-point of the hypotenuse PR of a right-angled PQR. Prove that QS = 21

PR

8. In the given figure, it is given that AE = AD and CE = BD. Prove that ABE ACD.

C

A

B

E D

O

9. In the given figure, D and E are the points on the base BC of ABC such that BD = CE, AD = AE andADE = AED. Prove that ABE ACD.

C

A

B ED


10. Equilateral triangles ABD and ACE are drawn on the sides of a ABC. Prove that CD = BE.11. BD is the bisector of ABC. P is any point on BD. Prove that the perpendiculars drawn from P to AB and

BC are equal.12. In the given figure, AD is the median and BE and CF are drawn perpendicular on AD and AD produced

from B and C respectively. Prove that BE = CF.

13. In an isosceles ABC, in which AB = AC, the bisector of B and C meet at I. Show that(i) BI = CI (ii) AI is the bisector of A

14. If the altitudes AD, BE and CF of a ABC are equal, prove that ABC is equilateral.15. Prove that diagonals of a rhombus bisect each other at right angle.16. A point O is taken inside a rhombus ABCD such that its distances from the angular points A and C are

equal. Show that BO and DO are in one and the same straight line.17. In the given figure, m || n and E is the mid-point of AB. Prove that E is also the mid-point of any line

segment CD having its end points at m and n respectively.

18. In the given figure, C is the mid-point of AB. If ACD = BCE and CBD = CAE, prove that DC = EC.


19. In the given figure, EDC is an equilateral triangle and DABC is a square. Prove that :(i) EA = EB (ii) EBC = 15°

E

CD

A B20. In the given figure, CE AB and DF AB, and CE = DF. Prove that OC = OD.

A

C

E BO

F

D21. Prove that in a right triangle, the hypotenuse is the longest side. —NCERT22. In the given figure, D is a point on the side of ABC such that AD = AC. Prove that AB > AD.

A

D CB23. Prove that the sum of three altitudes of a triangle is less than the sum of the three sides of a triangle.24. AD is a median of ABC. Prove that AB + AC > 2AD.25. In the given figure, D is a point on side AB of ABC and E is a point such that BD = DE. Prove that :

AC + BC > AEA

E D

B C


PRACTICE TESTM.M : 30 Time : 1 hourGeneral Instructions :Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.

1. Prove that the angles opposite to the equal sides of a triangle are equal.2. In ABC, D is the mid-point of BC. If DL AB and DM AC such that DL = DM, prove that AB = AC.3. In the given figure, CPD = BPD and AD is the bisector of BAC, prove that CAP BAP.

DP

C

B

A

4. In the given figure, ABCD is a square. E and F are the mid-points of the sides AD and BC respectively.Prove that BE = DF.

C

F

BA

E

D

5. In the given figure AB and CD are respectively the smallest and the longest sides of a quadrilateralABCD. Prove that A > C and B > D.

6. ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show thatBCD = 90°

C

D

A

B


7. Show that of all the line segments drawn from a given point not on it, the perpendicular line segment isthe shortest.

8. AD and BE are respectively altitudes of an isosceles triangle ABC with AC = BC. Prove that AE = BD.

9. ABC is a right angled triangle at C. M is the mid-point of AB. Prove that CM = 21

AB

10. Prove that “Two triangles are congruent if any two angles and the included side of one triangle is equalto any two angles and the included side of the other triangle.”

84 QUADRILATERALS MATHEMATICS–IX

CHAPTER 8QUADRILATERALS

Points to Remember :1. The sum of the angles of a quadrilateral is 360°.2. A diagonal of a parallelogram divides it into two congruent triangles.3. In a Parallelogram:

(i) opposite sides are equal(ii) opposite angles are equal(iii) diagonals bisect each other

4. A quadrilateral is a parallelogram, if(i) opposite sides are equal, or(ii) opposite angles are equal, or(iii) diagonals bisect each other, or(iv) a pair of opposite sides is equal and parallel.

5. Diagonals of a rhombus bisect each other at a right angle and vice-versa.6. Diagonals of a rectangle bisect each other and are equal, and vice-versa.7. Diagonals of a square bisect each other at right angles and are equal, and vice-versa.8. Mid-point Theorem : 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.9. A line through the mid-point of a side of a triangle parallel to another side bisects the third side.

10. The quadrilateral formed by joining the mid-points of hte sides of a quadrilateral, in order, is a parallelo-gram.


Example 1. Four angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4. Find them.Solution. Let the four angles be x, 2x, 3x and 4x.

Since, sum of four angles of a quadrilateral is 360°,x + 2x + 3x + 4x = 360°

10x = 360°, or x = 36° four angles are 36°, 2 × 36°, 3 × 36°, 4 × 36°

i.e. 36°, 72°, 108°, 144° Ans.Example 2. In the given figure, PQRS is a trapezium in which PQ || SR. If P = 60° and Q = 75°, find S

and R.

MATHEMATICS–IX QUADRILATERALS 85

Solution. P + S = 180° andQ + R = 180° ( interior angles on the same side of transversal are supplementarly) 60° + S = 180°and 75° + R = 180° S = 180 – 60° and R = 180° – 75° S = 120° and R = 105° Ans.

Example 3. Show that the diagnomals of a rhombus are perpendicular to each other.Solution. Let ABCD be a given Rhombus.

here, AB = BC = CD = DANow, In AOD and COD,

OA = OC ( diagonals of parallelogram bisect each other) OD = OD (common side) AD = CD

condition)congruence(SSSΔCODΔAOD AOD = COD (c.p.c.t.)But, AOD + COD = 180° (linear pair) 2AOD = 180° AOD = 90°So, the diagonals of a rhombus are perpendicular to each other.

Example 4. If the diagonals of a parallelogram are equal, then show that it is a rectangle. —NCERTSolution. Given : A parallelogram ABCD in which AC = BD.

To prove : ABCD is a rectangle.Proof : In ABC and DCB, we have

AB = DC (opp. sides of parallelogram) BC = BC (common side) AC = DB (given)

ABC DCB (SSS congruence condition) ABC = DCB ...(1) (cpct)But AB || DC and BC cuts them. ACB + DCB = 180° (sum of consecutive interior angles is 180°) 2ABC = 180° ABC = 90°Thus ABC = DCB = 90° ABCD is a parallelogram one of whose angle is 90°.Hence, ABCD is a rectangle.

Example 5. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.—NCERT

Solution. Given : A quadrilateral ABCD in which the diagonals AC and BD intersect at O such that AO = OC,BO = OD and AC BD.To prove : ABCD is a rhombus.Proof : In AOD and COB, we have

AO = OC (given) OD = OB given)AOD = COB (vertically opp. angles)

AOD COB (SAS congruence condition) OAD = OCB ...(i) (cpct)


Now, line AC intersects AD and BC at A and C respectively such that AOD = OCB(proved in (i))i.e. alternate interior angles are equal. AD || BC.Similarly, AB || CDHence, ABCD is a parallelogram.Again, In AOD and COD, we have

OA = OC (given)AOD = COD (each = 90°)

OD = OD (common side) AOD COD (SAS congruence condition) AD = CD ...(2) (cpct)Now, ABCD is a parallelogram, AB = CD and AD = BC (opp. sides of a parallelogram are equal) AB = CD = AD = BC ( using (2))Hence, quadrilateral ABCD is a rhombus.

Example 6. Show that the diagonals of a square are equal and bisect each other at right angles. —NCERTSolution. Given : A square ABCD.

To prove : AC = BD, AC BD and OA = OC, OB = OD.Proof : Since ABCD is square, AB || DC and AD || BC.Now, AB || DC and transversal AC intersectsthem at A and C respectively.BAC = DCA (alternate interior angles are equal) BAO = DCO ...(1)Again AB || DC and BD intersects them at B and D respectively. ABD = CDB (alternate interior angles are equal) ABO = CDO ...(2)Now, In AOB and COD, we have

BAO = DCO (from (1)) AB = CD (opp. sides of a parallelogram are equal)

ABO = CDO (from (2)) AOB COD (ASA congruence condition) OA = OC and OB = OD (cpct)Hence, the diagonals bisect each other.Again, In ADB and BCA, we have

AD = BC (sides of a square are equal)BAD = ABC (each 90°) AB = AB (common side)

ADB BCA (SAS congruence condition) AC = BD (cpct)Hence, diagonals are equal.Now, In AOB and AOD, we have,

OB = OD (diagonals of a parallelogram bisect each other) AB = AD ( sides of a square are equal) AO = AO (common side)

AOB AOD (SSS congruence condition) AOB AOD (cpct)


but, AOB AOD = 180°

AOB AOD = 2

180 = 90°

AO BD AC BD.Hence, diagonals intersect at right angles. Hence proved.

Example 7. Show that if the diagonals of a quadrilateral one equal and bisect each other at righ angles, thenit is a square. —NCERT

Solution. Given : A quadrilateral ABCD in which the diagonals AC = BD, AO = OC, BO = OD and AC BD.To prove : Quadrilateral ABCD is a square.Proof : In AOD and COB, we have

AO = OC (given) OD = OB (given)AOD = COB (vertically opp. angles)

AOD COB (SAS congruence condition) OAD = OCB ...(1) (cpct)Now, line AC intersects AD and BC at A and C respectively such that OAD = OCB, i.e.alternate interior angles are equal. AD || BCSimilarly, AB || CD.Hence, ABCD is a parallelogram.Now, In AOB and AOD, we have

AO = AO (common side)AOB = AOD (each = 90°) OB = OD ( diagonals of a parallelogram bisect each other)

AOB AOD (SAS congruence condition) AB = AD (cpct)But, AB = CD and AD = BC (opposite sides of a parallelogram are equal) AB = BC = CD = AD ...(2)Now, In ABD and BAC, we have

AB = AB (common side) AD = BC (opp. sides of parallelogram are equal) BD = AC (given)

ABD BAC (SSS congruence condition) DAB = CBA (cpct)But, DAB + CBA = 180°

DAB CBA = 2

180 = 90°

Thus, ABCD is a parallelogram whose all the sides are equal and one of the angle is 90°. ABCD is a square.Hence proved.


Example 8. Prove that bisectors of a parallelogram form a rectangle.Solution. Given : A paralleogram ABCD in which AR, BR, CP and DP are the bisectors of A, B, C and

D respectively forming quadrilateral PQRS.

To prove : PQRS is a rectangle.Proof : DCB + ABC = 180° (co-interior angles of parallelogram are supplementary)

90ABC21DCB

21

9021 ...(1)Also, In CQB, 1 + 2 + CQB = 180° ...(2)from (1) and (2), we get

CQB = 180° – 90° = 90° RQP = 90° ( CQB = RQP, vertically opp. angles)Similarly, it can be shown,

QRP = RSP = SPQ = 90°So, Quadrilateral PQRS is a rectangle.

Example 9. In the given figure, ABCD is a parallelogram in which P and Q are the mid points of AB and CDrespectively. If AQ and DP intersects at S and PC and BQ intersects at R, show that the quadrilat-eral PRQS is a parallelogram.

Solution. AB || DC AP || QC

Also, AB = DC QCAPDC21AB

21

AP || QC and AP = QC APCQ is a parallelogram.So, AQ || PC or SQ || PR.Similarly, it can be easily shown that PS || QR.Thus, In quadrilateral PRQS,

SQ || PR and PS || RQ.So, PRQS is a parallelogram.


Example 10. ABCD is a parallelogram and line segments AP and CQ bisects the A and C respectively. Showthat AP || CQ.

Solution.

here, A = C ( opp. angles of a parallelogram are equal)

C21A

21

1 = 2 ...(1)Now, AB || DC and CQ is a transversal 2 = 3 ...(2)From (1) and (2), 1 = 3Thus, transversal AB intersects AP and CQ at A and Q, such that 1 = 3 i.e. correspondingangles are equal.Hence, AP || CQ.

Example 11. Diagonals AC of a parallelogram ABCD bisects A (see figure). Show that :(i) it bisects C also.(ii) ABCD is a rhombus.

Solution. (i) Given : A parallelogram ABCD, in which diagonal AC bisects A.To prove : (i) AC bisects C i.e. 3 = 4 (ii) ABCD is a rhombusProof : Since ABCD is a parallelogram, AB || DC.Now, AB || DC and AC intersects them 1 = 3 ...(1) ( alternate interior angles)Again, AD || BC and AC intersects them. 2 = 4 ...(2) ( alternate interior angles)but, 1 = 2 ...(3) (given)from (1), (2) and (3), we get

3 = 4Hence, AC bisects C.(ii) To prove : ABCD is a rhombus.from (i) part, We have 1 = 2 = 3 = 4Now, in ABC, 1 = 4 AB = BC (sides opp. to equal angles in a triangle are equal)Similarly, In ADC, we have 2 = 3 AD = DC.Also, ABCD is a parallelogram. AB = CD and AD = BCcombining these, we get AB = BC = CD = DA.Hence, ABCD is a rhombus.


Example 12. ABCD is a rhombus. Show that diagonal AC bisects A as well as C and diagonal BD bisectsB as well as D. —NCERT

Solution. Given : A rhombus ABCDTo prove : (i) Diagonal AC bisects A as well as C.(ii) Diagonal BD bisects B as well as D.Proof : ADC, AD = DC (sides of a rhombus are equal) DAC = DCA ...(1)

(angles opp. to equal sides of a triangle are equal)Now, AB || DC and AC intersects them BCA = DAC ...(2) (alternate angles)from (1) and (2), we get

DCA = BCA ...(3) AC bisects C.In ABC, AB = BC (sides of a rhombus are equal)from (3) and (4), we get

BAC = DAC AC bisects A.Hence, diagonal AC bisects A as well as C.Similarly, diagonal BD bisects B as well as D.

Example 13. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (seefigure). Show that :(i) APD CQB (ii) AP = CQ (iii) AQB CPD(iv) AQ = CP (v) APCQ is a parallelogram. —NCERT

Solution. Given : ABCD is a parallelogram P and Q are points on the diagonal BD such that DP = BQ.To prove : (i) APD CQB(ii) AP = CQ(iii) AQB CPD (iv) AQ = CP(v) APCQ is a parallelogram.Construction : Join A to C to meet BD in O.Proof : We know that the diagonals of parallelogram bisect each other.Now, AC and BD bisect each other at O. OB = ODBut BQ = DP (given) OB – BQ = OD – DP OQ = OPThus, in quadrilateral APCQ, diagonals AC and PQ are such that OQ = OP and OA = OC i.e., thediagonals AC and PQ bisects each other.Hence, APCQ is a parallelogram, which proves (v) part.(i) Now, In APD and CQB, we have


AD = CB (opp. sides of a a parallelogram ABCD) AP = CQ (opp. sides of a a parallelogram APCQ) DP = BQ (given)

APD CQB (SSS congruence condition)(ii) AP = CQ (cpct)(iii) In AQB and CPD, we have

AB = CD (opp. sides of a a parallelogram ABCD) AQ = CP (opp. sides of a a parallelogram APCQ) BQ = DP (given)

AQB CPD (SSS congruence condition)(iv) Since, AQB CPD AQ = CP (cpct)

Example 14. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BDrespectively. (see figure). Show that :(i) APB CQD(ii) AP = CQ —NCERT

Solution. (i) Since, ABCD is a parallelogram, DC || AB.Now, DC || AB and transversal BD intersects them at B and D respectively.ABD = BDC (alt. int. angles)Now, In APB and CQD, we have ABP = QDC (ABD = BDC) APB = CQD (each = 90°) AB = CD (opp. sides of a parallelogram)APB CQD (AAS congruence condition)(ii) Since, APB CQD AP = CQ (cpct)

Example 15. ABCD is a trapezium in which AB || CD and AD = BC (see figure).

Show that :(i) A = B(ii) C = D(iii) ABC BAD(iv) diagonal AC = diagonal BD

—NCERTSolution. Given : ABCD is a trapezium, in which AB || CD and AD = BC.

To prove : (i) A = B (ii) C = D(iii) ABC BAD (iv) diagonal AC = diagonal BDConstruction : Produce AB and draw a line CE || AD.Proof : (i) Since AD || CE and transversal AE cuts them at A and E respectively. A + E = 180° ...(1)


Since AB || CD and AD || CE, AECD is a parallelogram. AD = CE BC = CE (AD = BC (given))Thus, In BCE, we have

BC = CECBE = CEB 180° –B = E B + E = 180° ...(2)from (1) and (2), we get,

A + E = B + E A = B(ii) Since A = B BAD = ABD

180° – BAD = 180° – ABD ADB = BCD D = C i.e. C = D

(iii) In ABC and BAD, we have BC = AD (given) AB = AB (common)A = B (proved above)

ABC BAD (SAS Congruence condition)(iv) Since, ABC BAD AC = BD (cpct)Hence proved.

Example 16. Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a quadrilat-eral is a parallelogram.

Solution. Given : A quadrilateral ABCD, in whcih P, Q, R, S are the mid-points of AB, BC, CD and DArespectively.To prove : Quadrilteral PQRS is a parallelogram.Construction : Join A to C.Proof : In ABC, P and Q are mid-points of AB and BCrespectively.

PQ || AC and AC21PQ ( mid-point theorem)

Again, In DAC, R and S are mid-points of sides CD and ADrespectively.

SR || AC and AC21SR [ mid-point theorem]

Now, PQ || AC and SR || AC PQ || SR

Again, SRPQSRAC21PQ

PQ || SR and PQ = SRHence, PQRS is a parallelogram.


Example 17. ABCD is a rhombus and P, Q, R and S are respectively the mid-points of the sides AB, BC, CD andDA respectively. Show that the quadrilateral PQRS is a rectangle. —NCERT

Solution. Given : ABCD is a rhombus in which P, Q, R and S are the mid-points of AB, BC, CD and DArespectively. PQ, QR, RS and SP are joined to obtain a quadrilateral PQRS.To prove : PQRS is a rectangle.Construction : Join A to C.Proof : In ABC, P and Q are the mid-points of AB and BC respectively.

PQ || AC and PQ = 21

AC ...(1)

( mid-point theorem)Similarly, In ADC, R and S are the mid-points of CD and DA respectively.

SR || AC and SR = 21

AC ...(2)

( mid-point theorem)from (1) and (2), we get

PQ || SR and PQ = SR PQRS is a parallelogram.Now, AB = BC (sides of rhombus are equal)

21

AB = 21

BC PB = BQ 3 = 4 ( angles opp. to equal sides of a triangle are equal)

Now, In APS and CQR, we have AP = CQ (half of equal sides AB and BC) AS = CR (half of equal sides AD and CD) PS = QR (opp. sides of parallelogram PQRS)

APS CQR (SSS congruence condition) 1 = 2 (cpct)Now, 1 + SPQ + 3 = 180° ( linear pair) 1 + SPQ + 3 = 2 + PQR + 4but, 1 = 2 and 3 = 4 (proved above) SPQ = PQR ...(3)Now, SP || RQ and PQ intersects them, SPQ + PQR = 180° ...(4)from (3) and (4), we get

2SPQ = 180° SPQ = 90°Thus, PQRS is a parallelogram whose one angle is 90°.Hence, PQRS is a rectangle.

Example 18. ABCD is a rectangle and P, Q, R, S are mid-points of the sides AB, BC, CD and DA respectively.Show that the quadrilateral PQRS is a rhombus. —NCERT


Solution. Given : ABCD is a rectangle, in which P, Q, R and S are the mid-points of sides AB, BC, CD and DArespectively. PQ, QR, RS and SP are joined to obtain a quadrilateral PQRS.To prove : PQRS is a rhombus.Construction : Join A to C.Proof : In ABC, P and Q are the mid-points of sides AB and BC.

PQ || AC and PQ = 21

AC ...(1) ( mid-point theorem)

Similarly, In ACD, R and S are the mid-points of sides CD and DA.

RS || AC and RS = 21

AC ...(2) ( mid-point theorem)

from (1) and (2), we get PQ || SR and PQ = SR PQRS is a parallelogram.Now, AD = BC (opp. sides of a rectangle ABCD)

BQASBC21AD

21

Now, In APS and BPQ, we have AP = BP ( P is the mid-point of AB)

PAS = PBQ (each = 90°) AS = BQ (proved above)

APS BPQ (SAS congruence condition) PS = PQ (cpct)Thus, PQRS is a parallelogram in which adjacent sides are equal. PQRS is a rhombus.

Example 19. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line isdrawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point ofBC. —NCERT

Solution. Given : A trapezium ABCD, in which AD || BC. E is the mid-point of AD and EF || AB.

To prove : F is the mid-point of BC.Construction : Join B to D. Let it intersect EF in G.Proof : In DAB, E is the mid-point of AD (given)

EG || AB ( EF || AB) By converse of mid-point theorem, G is the mid-point of DB.Now, In BCD, G is the mid-point of BD ( proved above)

GF || DC ( AB || DC, EF || AB DC || EF) By converse of mid-point theorem, F is the mid point of BC.


Example 20. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallelto BC intersects AC at D. Show that :

(i) D is the mid-point of AC (ii) MD AC (iii) AB21MACM

—NCERTSolution. Given : ABC is right angled at C, M is the mid-point of hyp. AB . MD || BC.

To prove : (i) D is the mid-point of AC(ii) MD AC

(iii) AB21MACM

Proof : (i) In ABC, M is the mid-point of AB and MD || BC. D is the mid-point of ACi.e. AD = DC ...(1)(ii) Since MD || BC ADM = ACB (corresponding angles) ADM = 90° ( ACB = 90°, MD || BC)But, ADM + CDM = 180° ( linear pair) 90° + CDM = 180° CDM = 90°Thus, ADM = CDM = 90° ...(2) MD AC.(iii) Now, In AMD and CMD, we have

AD = CD (from (1))ADM = CDM (from (2))

MD = MD (common side) AMD CMD (SAS congruence condition) MA MC (cpct)

Also, AB,21MA Since M is the mid-point of AC.

Hence, CM = MA 21

AB.

PRACTICE EXERCISE

1. In the given figure, PQRS is a rectangle whose diagonals PR and QS intersect at O. If OPQ = 32°, findOQR. (Ans. : 58°)

2. Prove that, if in a parallelogram diagonals are equal and perpendicular to each other, then it is a squqre.3. In the given figure, PQRS is a parallelogram and X and Y are the points on the diagonal QS such that

SX = QY. Prove that PYRS is a parallelogram.


4. Let ABC and DEF be two triangles drawn in such a way that AB || DE ; AB = DE; BC || EF and BC = EF.Show that AC || DF and AC = DF.

5. In a parallelogram ABCD, the bisectors of consecutive angles A and B intersect at O. Prove thatAOB = 90°.

6. In the given figure, ABCD is a parallelogram and P is the mid-point of AD. A line through D, drawnparallel to PB, meets AB produced at Q and BC at R, prove that :(i) AQ = 2DC (ii) DQ = 2DR

7. In a parallelogram ABCD, if A = (4x + 20)° and B = (3x – 15)°, find the value of x and the measure ofecah angle of a parallelogram. (Ans. x = 25 and angles are 60°, 120°, 60°, 120°)

8. A ABC is given. If lines are drawn through A, B, C parallel respectively to the sides BC, CA and AB

forming XYZ, as shown. Show that YZ.21BC

9. In a parallelogram ABCD, the bisector of A also bisects the side BC at E. Show that AD = 2AB.10. If a transversal cuts two parallel lines, prove that the bisectors of the interior angles form a rectangle.


11. ABC is a triangle right angled at B and D is the mid-point of AC. DE is drawn perpendicular to BC. Prove

that AC.21BD

12. If P, Q and R are respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC,prove that PQR is also equilateral triangle.

13. ABCD is a parallelogram in which E and F are mid-points of the sides AB and CD respectively. Prove thatthe line segments CE and AF trisect the diagonal BD.

14. Let ABCD be a trapezium in which AB || DC and let E be the midpoint of AD. Let F be a point on BC suchthat EF || AB. Prove that:(i) F is the mid-point of BC

(ii) DC)(AB21EF

15. Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to theparallel sides and is equal to half their difference.

16. Show that the quadrilateral formed by joining the midpoints of pairs of adjacent sides of a rhombus is arectangle.

17. Show that the quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangleis a rhombus.

18. Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other.19. Show that the four triangles formed by joining the mid-points of the three sides of a triangle are congru-

ent to each other.20. In ABC, AD is the median, and E is the mid-point of AD . BE produced meets AC in F. Prove that

AC31AF .

21. The diagonals of a quadrilateral are perpendicular. Prove that the quadrilateral, formed by joining themid-points of its sides is a rectangle.

22. D, E and F are respectively the mid-points of the sides BC, AC and AB of an isosceles triangle ABC, inwhich AB = AC. Prove that AD is perpendicular to EF and is bisected by it.

23. In the given figure, ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and ABrespectively intersecting at P, Q and R. Prove that perimeter of PQR is double the perimeter of ABC.


24. ABC is any triangle. D is a point of AB such that AB41AD and E is a point on AC such that

AC.41AE Prove that BC.

41DE

25. BX and CY are perpendiculars to a line passing through the vertex A of a triangle ABC. If Z is the mid-point of BC. Prove that XZ = YZ.

PRACTICE TESTMM : 30 Time : 1 hourGeneral Instructions :Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.

1. In parallelogram ABCD, AB = 10 cm and BC = 8 cm. If B = 120°, find :(i) DA (ii) C (iii) D

2. ABCD is a parallelogram and AB is produced to X so that AB = BX. Prove that DX and BC bisect eachother.

3. Two adjacent angles of a parallelogram are (2x + 15)° and (3x – 25)°. Find the value of x and also themeasure of the angles.

4. ABCD is a trapezium, in which AB || DC. X and Y are respectively mid-points of AD and BC. If AB = 13 cmand CD = 9 cm, find XY.

5. In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angle.6. ABCD is a parallelogram whose diagonals intersect each other at O. Through O, XY is drawn as shown.

Prove that OX = OY.

7. PQRS is a parallelogram. PO and QO are respectively the angle bisectors to P and Q. Line AOB isdrawn parallel to PQ. Prove that :(i) PA = QB (ii) AO = OB


8. Prove that in a right angled triangle, the median bisecting the hypotenuse is half of the hypotenuse.9. 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) AB.21MACM

10. Prove that the quadrilateral formed by joining mid-points of the sides of a square is again a square.

ANSWERS OF PRACTICE EXERCISE1. (i) 8 cm (ii) 60° (iii) 120°3. x = 38, angles are 89°, 91°, 89°, 91°4. 11 sq. units

100 AREAS OF PARALLELOGRAMS AND TRIANGLES MATHEMATICS–IX

CHAPTER 9AREAS OF PARALLELOGRAMS AND TRIANGLES

Points to Remember :1. Two congruent figures must have equal areas. However, two figures having equal areas need not to be

congruent.2. Two figures are said to be on the same base and between the same parallels, if they have a common base

and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to thebase.

3. Parallelograms on the same base and between the same parallels are equal in area.4. Area of parallelogram = Base × corresponding height.5. Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels.6. Two triangles on the same base (or equal bases) and between the same parallels are equal in area.7. Two triangles having the same base (or equal bases) and equal areas lie betwen the same parallels.

8. Area of Triangle 21

× Base × corresponding height.

9. Area of a Rhombus 21

× product of diagonals.

10. Area of a Trapezium 21

× (sum of the parallel sides) × (distance between them).

11. A median of a triangle divides it into two triangles of equal area.12. The diagonals of a parallelogram divides it into four triangles of equal area.

ILLUSTRATIVE EXAMPLESExample 1. In the given figure, ABCD is a parallelogram, AB = 12 cm, altitude DF = 7.2 cm and DE = 6 cm, find

the perimeter of parallelogram ABCD.Solution. Area of parallelogram ABCD = AB × DE ...(1)

also, Area of parallelogram ABCD = BC × DF ...(2)From (1) and (2), we get

AB × DE = BC × DF 12 × 6 = BC × 7.2

cm107.2

612BC

Perimeter of parallelogram ABCD = 2 (AB + BC) = 2 (12 + 10) cm

= 2 (22) cm = 44 cm Ans.

Example 2. P, Q, R, S are respectively, the mid-points of sides AB, BC, CD and DA of parallelogram ABCD.Show that the quadrilateral PQRS is a parallelogram and its area is half the area of the parallelo-gram ABCD.

MATHEMATICS–IX AREAS OF PARALLELOGRAMS AND TRIANGLES 101

Solution. Join A to C and Q to S

Now, In ABC, P and Q are respectively mid-points of AB and BC.

PQ || AC and AC21PQ ...(1) ( mid-point theorem)

Similarly, SR || AC and AC21SR ...(2)

From (1) and (2), we getPQ || SR and PQ = SR

So, PQRS is a parallelogram.Now, SPQ and parallelogram ABQS stand on same base SQ and between same parallels AB andQS,

ar (PQS) ABQS)gram(paralleloar21

...(3)

Similarly, ar (SRQ) SQCD)gram(paralleloar21

...(4)

adding (3) and (4), we get,

ar (PQS) + ar (SRQ) SQCD)]gram(paralleloarABQS)gram(parallelo[ar21

ABCD)gram(paralleloar21PQRS)gram(paralleloar

which proves the result.Example 3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.

Show that ar (APB) = ar (BQC). —NCERTSolution. APB and parallelogram ABCD stand on the same base AB and lie between the same parallels AB

and DC.

ar (APB) = ar(ABCD)21

...(1)

Similarly, BQC and parallelogram stand on the same base BC and lie between the same parallels BC and AD.

(ABCD)ar21(BQC)ar ...(2)

from (1) and (2), we getar (APB) = ar (BQC). Hence proved.

Example 4. A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS andjoined it to points P and Q. In how many parts the fields is divided? What are the shapes of theseparts? The farmer wants to sow wheat and pulses in equal portions of the field seperately. Howshould she do it? —(NCERT)


Solution. Clearly, the field parallelogram PQRS is divided into 3 parts. Each part is in shape of a triangle.Since APQ and parallelogram PQRS stand on the same base PQ and lie between the sameparallels PQ and SR.

ar (APQ) = 21

. ar (PQRS) ...(1)

Clearly, ar (APS) + ar (AQR) = ar (PQRS) – ar (APQ)

= ar (PQRS) – 21

ar (PQRS) ( using (1))

= 21

ar (PQRS) ...(2)

From (1) and (2), we getar (APS) + ar (AQR) = ar (APQ)

Thus, the farmer should sow wheat and pulses either as [(APS and AQR) or APQ] or as[APQ or (APS and AQR)].

Example 5. Show that a median of a triangle divides it into two triangles of equal area. —NCERTSolution. Let ABC be a given triangle and AD is a median. Draw AE BC.

Since, D is mid point of BC, so we have,BD = DC ...(1)

Now, ar (ABD) AEBD21

...(2)

and, ar (ADC) AECD21

AEBD21

...(3)

( BD = DC)From (2) and (3), we get

ar (ABD) = ar (ADC)which proves the desired result.

Example 6. E is any point on median AD of a ABC. Show that ar (ABE) = ar (ACE) —NCERTSolution. Given : AD is a median of ABC and E is any point on AD.

To prove : ar (ABE) = ar (ACE)Proof : AD is the median of ABC ar (ABD) = ar (ACD) ...(1) also, ED is the median of EBC, ar (BED) = ar (CED) ...(2)Subtracting (2) from (1), we get

ar (ABD) – ar (BED) = ar (ACD) – ar (CED) ar (ABE) = ar (ACE). Hence shown.

Example 7. In a ABC, E is the mid-point of median AD. Show that ar (BED) = 41

ar (ABC). —NCERTT

Solution. Given : ABC in which E is the mid-point of median AD.

To prove : ar (BED) 41

. ar (ABC)


Proof : Since AD is a median of ABC and median divides a triangle into two triangles of equalarea. ar (ABD) = ar (ADC)

ar (ABD) = 21

ar (ABC) ...(1)

In ABD, BE is a median, ar (BED) = ar (BAE)

ar (BED) = (ABD)ar21

(ABC)ar21

21 ( using (1))

= (ABC)ar41

Hence shown.Example 8. If the medians of a ABC intersect at G, show that

ar (AGB) = ar (AGC) = ar (BGC) 31

ar (ABC)

Solution. We know that a median of a triangle divides it into two triangles of equal areas.In ABC, AD is the median. ar (ABD) = ar (ACD) ...(1)Again, In GBC, ar (GBD) = ar (GCD) ...(2)Subtracting (2) from (1),

ar (ABD) – ar (GBD) = ar (ACD) – ar (GCD) ar (ABG) = ar (AGC) ...(3)Similarly, ar (AGB) = ar (BGC) ...(4)From (3) and (4), we get,

ar (ABG) = ar (AGC) = ar (BGC)Now, ar (ABC) = ar (ABG) + ar (AGC) + ar (BGC)

= 3 ar (ABG)

, ar (ABC) ABC)(ar31

Hence, ar(AGB) = ar (AGC) = ar (BGC) ABC)(ar31

Example 9. PQRS and ABRS are parallelograms and X is any point on the side BR. Show that : (i) ar (parallelogram PQRS) = ar (parallelogram ABRS)

(ii) ar (AXS) = 21

ar (parallelogram PQRS)


Solution. Since, parallelogram PQRS and ABRS stand on same base SR and between same parallels SR andPB, ar (parallelogram PQRS) = ar (parallelogram ABRS)

Again, ar (ASX) = 21

ar (parallelogram ABRS)

( ASX and parallelogram ABRS standon same base AS and between parallellines AS and BR)But, ar (parallelogram ABRS) = ar (paral-lelogram PQRS)

ar (ASX) 21

ar (parallelogram PQRS)

Hence proved.Example 10. In ABC, D, E and F are mid-points of the sides BC, CA and AB respectively. Prove that :

(i) BDEF is a parallelogram

(ii) ar (DEF) 41

ar (ABC)

(iii) ar (BDEF) 21

ar (ABC).

Solution. (i) FE || BC or FE || BDand DE || BF ( mid point theorem) BDEF is a parallelogram.(ii) Since a diagonal of a parallelogram divides it into two triangles equal in area. ar (BDF) = ar (DEF)Similarly, ar (DCE) = ar (DEF)

and ar (DEF) = ar (AFE)Combining all these, we getar (BDF) = ar (DEF) = ar (DCE) = ar (AFE)But, ar (ABC) = ar (BDF) + ar (DEF) + ar (DCE) + ar (AFE) = 4 ar (DEF)

ar (DEF) ABC)(ar.41

(iii) Again, 4ar (DEF) = ar (ABC)

2 . ar (DEF) = 21

ar (ABC)

ar (BDEF) = 21

. ar (ABC)

Hence proved.


Example 11. In the given figure, ABC and ABD are two triangles on the same base AB. If line segment CD isbisected by AB at O, show that ar (ABC) = ar (ABD). —NCERT

Solution. Given : ABC and ABD are two triangles on the same base AB. A line segment CD is bisected by ABat O. i.e. OC = OD.To prove : ar (ABC) = ar (ABD)Proof : In ACD, we have OC = OD (given) AO is a median. ar (AOC) = ar (AOD) ...(1)

( median divides a triangle in two triangles of equal areas).Similarly, In BCD, BO is the median. ar (BOC) = ar (BOD) ...(2)adding (1) and (2), we get

ar (AOC) + ar (BOC) = ar (AOD) + ar (BOD) ar (ABC) = ar (ABD).

Example 12. In the given figure, diagonals AC and BD of a quadrilateral ABCD intersect at O such thatOB = OD. If AB = CD, then show that : (i) ar (DOC) = ar (AOB) (ii) ar (DCB) = ar (ACB)(iii) DA || CB or ABCD is a parallelogram. —NCERT

Solution. (i) Draw DN AC and BM AC.In DON and BOMDNO = BMO (each = 90°)DON = BOM (vert. opp. angles) OD = OB (given) DON BOM (AAS congruence condition) ar (DON) = ar (BOM) ...(1)

( congruent triangles have equal area)Again, In DCN and BAM

DNC = BMA (each = 90°) DC = AB (given) DN = BM (DON BOM DN = BM)

DCN BAM (RHS congruence condition) ar (DCN) = ar (BAM) ...(2) ( congruent triangles have equal areas)adding (1) and (2), we getar (DON) + ar (DCN) = ar (BOM) + ar (BAM) ar (DOC = ar (AOB)(ii) Since, ar (DOC) = ar (AOB) ar (DOC) + ar (BOC) = ar (AOB) + ar (BOC)

AMIT BAJA

J


ar (DCB) = ar (ACB)(iii) DCB and ACB have equal areas and have the same base. So, these triangles lie betweenthe same parallels. DA || CB. i.e. ABCD is a parallelogram.

Example 13. A point O inside a rectangle ABCD is joined to the vertices. Prove that :ar (AOD) + ar (BOC) = ar (AOB) + as (COD)

Solution. Draw POQ || AD and ROS || AB.Since, POQ || AD and DC cuts them.ADC = PQC = 90° ( corr. angles)i.e. OQ CD. Similarly, OR AD, OP AB and OS BC.Consider, ar (AOD) + ar (BOC)

OSBC21ORAD

21

OS)(ORAD21

BC)AD(

ABAD21RSAD

21

ABCD)(rect.ar21

...(1)

Again, ar (AOB) + ar (COD)

OQCD21OPAB

21

OQ)(OPAB21

( AB = CD)

ADAB21PQAB

21

ABCD)(rect.ar21

...(2)

From (1) and (2), we getar (AOD) + ar (BOC) = ar (AOB) + ar (COD)

Example 14. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel toCP meets CB produced at Q and then parallelogram PBQR is completed (see figure). Show thatar (ABCD) = ar (PBQR).

Solution. Join A to C and P and Q. Since AC and PQ are diagonals of parallelogram ABCD and parallelogramBPQR respectively.


ar (ABC) = 21

ar (ABCD) ...(1)

and, ar (PBQ) = 21

ar (BPRQ) ...(2)

Now, ACQ and AQP are in the same base AQ and CP. ar (ACQ) = ar (AQP) ar (ACQ) – ar (ABQ) = ar (AQP) – ar (ABQ)

( Subtracting ar (ABQ) from both sides) ar (ABC) = ar (BPQ)

ar(BPRQ)21(ABCD)ar

21

( using (1) and (2))

ar (ABCD) = ar (BPRQ) Hence proved.Example 15. In the given figure, PQ is a line parallel to side BC of ABC. If BX || CA and CY || BA meet the line

PQ produced in X and Y respectively, show that : ar (ABX) = ar (ACY).

Solution. Parallelogram XBCQ and ABX stand on the same base BX and between the same parallels BXand CA, we have

ar (ABX) 21

ar (parallelogram XBCQ) ...(1)

Also, parallelogram BCYP and ACY stand on same base CY and between the same parallels CYand BA, we have

ar (ACY) 21

ar (parallelogram BCYP) ...(2)

But, parallelogram XBCQ and parallelogram BCYP stand on same base BC and between sameparallels BC and XY, ar (parallelogram XBCQ) = ar (parallelogram BCYP) ...(3)from (1), (2) and (3), we get,

ar (ABX) = ar (ACY)Hence proved.

Example 16. In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced atF. Show that(i) ar (ACB) = ar (ACF)(ii) ar (AEDF) = ar (ABCDE) —NCERT

Solution. (i) Since ACD and ACF are on the same baseAC and betweeen the same parallels AC and BF, ar (ACB) = ar (ACF)(ii) Now, ar (ACB) = ar (ACF)adding ar (ACDE) both sides, we get


ar (ACB) + ar (ACDE) = ar (ACF) + ar (ACDE) ar (AEDF) = ar (ABCDE)Hence proved.

Example 17. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) =ar (BOC). Prove that ABCD is a trapezium. —NCERT

Solution. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that,ar (AOD) = ar (BOC) ...(1)adding ar (ODC) on both sides, we getar (AOD) + ar (ODC) = ar (BOC) + ar (ODC) ar (ADC) = ar (BDC)

BMDC21ALDC

21

AL = BM AB || DC. (distance between two parallel lines is same)Hence ABCD is a trapezium.

Example 18. In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilateralsABCD and DCPR are trapeziums.

Solution. ar (BDP) = ar (ARC) ...(1) (given)and ar (DPC) = ar (DRC) ...(2) (given)subtracting (2) from (1), we getar (BDP) – ar (DPC) = ar (ARC) – ar (DRC) ar (BDC) = ar (ADC) DC || ABHence, ABCD is a trapezium. ar (DRC) = ar (DPC) (given)on subtracting ar (DLC) from both sides, we getar (DRC) – ar (DLC) = ar (DPC) – ar (DLC) ar (DLR) = ar (CLP)on adding ar (RLP) to both sides, we get ar (DLR) + ar (RLP) = ar (CLP) + ar (RLP) ar (DRP) = ar (CRP) RP || DCHence, DCPR is a trapezium.

Example 19. Let P, Q, R and S be respectively the mid-points of the sides AB, BC, CD and DA of quadrilateral

ABCD. Show that PQRS is a parallelogram such that ar (parallelogram PQRS) 21

ar (quad.

ABCD).Solution. Join A to C and A to R.

In ABC, P and Q are mid-points of AB and BC respectively.

PQ || AC and AC.21PQ

In DAC, S and R are mid-points of AD and DC respectively.

SR || AC and AC.21SR

Thus, PQ || SR and PQ = SR.


PQRS is a parallelogram.Now, median AR divides ACD into the triangles of equal area.

ar (ARD) = 21

. ar (ACD) ...(1)

median RS divides ARD into two triangles of equal area.

ar (DSR) = 21

ar (ARD) ...(2)

from (1) and (2), we get, ar (DSR) =21

ar (ARD) = 21

(21

ar (ACD)) = 41

ar (ACD)

Similarly, ar (BQP) = 41

ar (ABC).

ar (DSR) + ar (BQP) = 41

[ar (ACD) + ar (ABC)]

ar (DSR) + ar (BQP) = 41

ar (quad. ABCD) ...(3)

Similarly, ar (CRQ) + ar (ASP) = 41


Adding (3) and (4), we get

ar (DSR) + ar (BQP) + ar (CRQ) + ar (ASP) = 21


But, ar (DSR) + ar (BQP) + ar (CRQ) + ar (ASP) + ar (parallelogram ABCD) = ar (quad ABCD) ...(6)

Subtracting (5) from (6), we get

ar (|| PQRS) = 21

ar (quad. ABCD). Hence proved.

Example 20. Prove that, of all the parallelograms of given sides, the parallelogram which is a rectangle has thegreatest area.

Solution. In parallelogram ABCD of sides a and b, let h be the height corresponding to the base a.

Now, In DAE,h < b ( b, being the hypotenuse, is the longest side of the triangle)

multiplying both sides by a, we getah < ab

i.e. ar (parallelogram ABCD) < ar (rect ABCD),which proves the result.


PRACTICE EXERCISE

1. PQRS is a parallelogram in which SX PQ and QY PS. If PS = 18 cm, SX =1 2 cm and QY = 8 cm, findthe value of PQ. (Ans. PQ = 12 cm)

2. Show that the line segment joining the mid-points of a pair of opposite sides of a parallelogram dividesit into two equal parallelograms.

3. In the given figure, O is a point in the interior of parallelogram ABCD. Show that :

(i) ar (AOB) + ar (OCD) 21

ar (ABCD)

(ii) ar (AOD) + ar (OBC) = ar (AOB) + ar (OCD)

4. The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect ADat X and BC at Y. Show that XY divides the parallelogram into two parts of equal area.

5. Find the area of rhombus, the lengths of whose diagonals are 8 cm and 6 cm. Also, find length of the sideof the rhombus. (Ans. Area = 24 cm2, side = 5 cm)

6. Find the area of a trapezium, whose parallel sides are 11 cm and 7 cm respectively and the distancebetween these sides is 6 cm. (Ans. Area = 54 cm2)

7. BD is one of the diagonals of a quadrilateral ABCD. If AM BD and CN BD. Show that :

ar (quad. ABCD) 21

× BD × (AM + CN)

8. In the given figure, ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersects at O.Prove that ar (AOD) = ar (BOC).

9. In the given figure, D is the midpoint of side AB of ABC and X is any point on BC. If CY || XD meets AB

in Y, prove that : ar (BXY) 21

ar (ABC)


10. Show that the diagonals of a parallelogram divides it into four triangle of equal area.11. If each diagonal of a quadrilateral seperates it into two triangles of equal area then show that the

quadrilateral is a parallelogram.12. E is any point on the median AD of ABC. Show that ar (ABE) = ar (ACE).

13. In a triangle ABC, E is the mid point of median AD. Show that ar (BED) 41

ar (ABC).

14. In the given figure, the side AB of parallelogram ABCD is produced to a point P and a line through A,parallel to CP, meets CB produced in Q and the parallelogram BQRP is completed. Show thatar (parallelogram ABCD) = ar (parallelogram BQRP).

15. Diagonal BD of a quadrilateral ABCD bisects the other diagonal AC in O. Prove that :ar (ABD) = ar (CBD)

16. Let P, Q, R and S be respectively the mid points of the sides AB, BC, CD and DA of quadrilateral ABCD.

Show that PQRS is a parallelogram such that ar (parallelogram PQRS) 21

ar (quad. ABCD).

17. In the given figure, ABCD is a parallelogram and E is any point on BC. Prove that :ar (ABE) + ar (DEC) = ar (EDA)

18. The medians CF and BE intersect at G. Prove that ar (GBC) = ar (quad. AFGE).19. D and E are points on sides AB and AC respectively of ABC such that ar (DBC) = ar (EBC). Prove that

DE || BC.20. ABCD is a parallelogram, E and F are the mid-points of BC and CD respectively. Prove that :

ar (AEF) 83

ar (parallelogram ABCD).


PRACTICE TEST

MM : 15 Time : 1 hour


Each Questions carries 3 marks.

1. Show that a median of a triangle divides it into two triangles of equal areas.2. ABCD is a parallelogram and O is any point in its interior. Prove that :

ar (AOB) + ar (COD) = ar (BOC) + ar (AOD)3. Triangles ABC and DBC are on the same base BC with A, D on opposite sides of line BC, such that

ar (ABC) = ar (DBC). Show that BC bisects AD.4. D, E and F are the mid-points of the sides BC, CA and AB respectively of ABC. Prove that BDEF is a

parallelogram whose area is half of ABC. Also, show that ar (DEF) = 41

ar (ABC).

5. In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

MATHEMATICS–IX CIRCLES 113

CHAPTER 10CIRCLES

Points to Remember :1. A circle is a collection of all the 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

centre) 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. There is one and only one circle passing through three non-collinear points.7. Equal chords of a circle (or of congruent circles) are equidistant from the centre (or corresponding

centres).8. Chords equidistant from the centre (or corresponding centres) of a circle (or of congruent circles) are

equal.9. 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.10. Congruent arcs of a circle subtend equal angles at the centre.11. 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.12. Angles in the same segment of a circle are equal.13. Angle in a semicircle is a right angle.14. 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.15. The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.16. If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.


Example 1. Give a method to find the centre of given circle.

Solution. Let A, B and C be any three distinct pointson the given circle. Join A to B and B to C.Draw perpendicular bisectors PQ and RSof AB and BC respectively to meet at apoint O.

Then, O is the centre of the circle.

114 CIRCLES MATHEMATICS–IX

Example 2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chordsare equal. —NCERT

Solution. Given : Two congruent circle C(O, r) and C(O, r) such that AOB = COD.To prove : CDAB Proof : In AOB and COD

OA = OC (each = r) OB = OD (each = r)AOB = COD (given) AOB COD (SAS congruence condition) CDAB (cpct)

Example 3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector ofthe common chord. —NCERT

Solution. Given : Two circles, with centres O and O intersect at two points A and B so that AB is thecommon chord of the two circles and OO is the line segment joining the centres of the two circles.Let OO intersect AB at M.To prove : OO is the perpendicular bisector of AB.Construction : Draw line segments OA, OB, OA and OB. Proof : In OAO and OBO, we have

OA = OB (Radii of same circle) OA = OB (Radii of same circle) OO = OO (Common side)

OAO OBO (SSS congruence condition) AOO = BOO (cpct) AOM = BOM ...(1)Now, In AOM and BOM, we have

OA = OB (Radii of same circle)AOM = BOM (from (1))

OM = OM (common side) BOM (SAS congruence condition) AM = BM and AMO = BMO (cpct)But, AMO + BMO = 180° 2AMO = 180° AMO = 90°Thus, AM = BM and AMO = BMO = 90°Hence, OO is the perpendicular bisector of AB.

Example 4. Find the length of a chord which is at a distance of 8 cm from the centre of a circle of radius 17 cm.Solution. Let AB be a chord of a circle with centre O and radius 17 cm.

Draw OC AB. Join O to C.Then, OC = 8 cm . OA = 17 cmIn right triangle OAC, using pythagoras theorem

OA2 = OC2 + AC2

172 = 82 + AC2

AC2 = 172 – 82

AC2 = 225 AC = 15 cmSince, perpendicular from the centre of a circle to a chord bisects a chord, we have

AB = 2 AC = 2 × 15 cm = 30 cm Ans.


Example 5. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centresis 4 cm. Find the length of the common chord. —NCERT

Solution. Clearly, the common chord AOB is the diameter of the cirlce with radius 3 cm.

Length of common chord AOB = 2 × 3 cm = 6 cm Ans.Example 6. AB and CD are two parallel chords of a circle which are on opposite sides of the centre such that

AB = 10 cm, CD = 24 cm and the distance between them is 17 cm. Find the radius of the circle.Solution. Draw ON AB and OM CD.

Since, ON AB, OM CD and AB || CD M, O, N are collinear points. MN = 17 cmLet ON = x cm, then OM = (17 – x) cm. Now, we know that perpendicular from the centre of a circle to achord bisects the chord,

andcm,5cm1021AB

21AN

cm12cm2421CD

21CM

In ONA, OA2 = ON2 + AN2

r2 = x2 + (5)2 ...(1)Again, In OCN, OC2 = OM2 + CM2

r2 = (17 – x)2 + (12)2 ...(2)from (1) and (2), we get

2222 (12)x)(17(5) x

1443428925 22 xxx 34x = 408 x = 12Putting x = 12 in (1),

r2 = (12)2 + (5)2 = 144 + 25 = 169 r = 13Hence, radius of circle is 13 cm. Ans.

Example 7. In a circle of radius 5 cm, AB and AC are two chords such that AB = AC = 6 cm. Find the length ofchord BC.

Solution. Given, OA = OC = 5 cmand AB = AC = 6 cmSince, points A and C are equidistant from A, so AO is the perpendicular bisector of BC.ADB = 90°Now, In right ADC,

AC2 = AD2 + CD2


(6)2 = AD2 + CD2

CD2 = 36 – AD2 ...(1)Also, In BDO,

OB2 = BD2 + OD2

(5)2 = BD2 + (AO – AD)2

25 = BD2 + (5 – AD)2

BD2 = 25 – (5 – AD)2

CD2 = 25 – (5 – AD)2 ( BD = CD) ...(2)from (1) and (2), we get,

36 – AD2 = 25 – (5 – AD)2

36 – AD2 = 25 – 25 – AD2 + 10 AD AD = 3.6 cmusing in eq. (1), CD2 = 36 – (3.6)2

= 36 – 12.96 = 23.04

4.823.04CD cm BC = 2CD = 2 × 4.8 cm

= 9.6 cm Ans.Example 8. Prove that the line joining the mid-points of two parallel chords of a circle passes through the

centre of the circle.Solution. Given : M and N are the mid-points of two parallel chords AB and CD respectively of circle with

centre O.To prove : MON is a straight line.Construction : Join OM, ON and draw OE || AB || CD.Proof : Since, the line segment joining the centre of a circle to themid point of a chord is perpendicular to the chord OM ABand ON CDNow, OM AB and AB || OE OM OE

EOM = 90°Also, OM CD and CD || OE ON OE

EON = 90° EOM + EON = 90° + 90° = 180°Hence, MON is a straight line.

Example 9. In the given figure, there are two concentric circles with common centre O. l is a line intersectingthese circles at A, B, C and D. Show that AB = CD.

Solution. Draw OM l. We know that perpendicular from the cen-tre of a circle to a chord bisects a chord.Now, BC is a chord of smaller circle andOM BC. BM = CM ...(1)Again, AD is a chord of bigger circle and OM AD. AM = DM ...(2)Subtracting (1) from (2), we get

AM – BM = DM – CM AB = CD. Hence proved.


Example 10. If two equal chords of a circle intersect within the circle, prove that the segments of one chord areequal to corresponding segments of the other chord. —NCERT

Solution. Given : AB and CD are chords of a circle with centre O. AB and CD intersect at P and AB = CD.To prove : (i) AP = PD (ii) PB = CPConstruction : Draw OM AB and ON CD.Join O to P.

Proof : AB21MBAM

( perpendicular from centre bisects the chord)

also, CN = ND = 21

CD

( perpendicular from centre bisects the chord)

But AB = CD 21

AB = 21

CD

AM = ND and MB = CN ...(1)Now, in OMP and ONP, we have

OM = ON (equal chords of a circle are equidistant from the centre).OMP = ONP (each = 90°)

OP = OP (common side) OMP ONP (RHS congruence condition) MP = PN ...(2) (cpct)Adding (1) and (2), we get

AM + MP = ND + PN AP = PDSubtracting (2) from (1), we get MB – MP = CN – PN PB = CPHence proved.

Example 11. A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sittingat 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. —NCERT

Solution. Let ABC is an equilateral triangle of side 2x metres.

Clearly, metresmetres2

22

BCBM xx

In right ABM, AM2 = AB2 – BM2

22222 34)()2( xxxxx

AM = x3 m

Now, OM = AM – OA = )203( x metresIn right OBM, we have OB2 = BM2 + OM2

(20)2 = x2 + 2)203( x 400 = x2 + 3x2 + 400 – 40 3 x 4x2 – 40 3 x = 0 4x (x – 10 3 ) = 0

x = 0 or x – 10 3 = 0


But x 0 3100310 xx mNow, BC = 2 MB = 2x = 2 × 10 3 m = 20 3 mHence, the length of each string = 20 3 m Ans.

Example 12. If O is the centre of the circle, find the value of x, in each of the following figures:

(i) (ii) (iii)Solution. (i) BAC = BDC = 25° ( angles in same segment are equal)

Now, In BCD, DBC + BDC + x = 180° ( angles sum property of a triangle) 75° + 25° + x = 180° 100° + x = 180° x = 180° – 100° = 80° Ans.(ii) Since, OB = OA (radii of same circle) OBA is an isosceles triangle OBA = BAO = 25° ...(1)Similarly, OAC is an isosceles. OCA = OAC = 30° ...(2)adding (1) and (2), we get

OAB + OAC = 25° + 30° BAC = 55°Now, BOC = 2BAC ( 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) x = 2 × 55° = 110° Ans.(iii) Reflex AOC = 360° – 120° = 240°

ABC = AOC.reflex21

( same as above)

12024021

x = 120° Ans.Example 13. In the given figure, AB is a diameter of a circle with centre O and chord CD = radius OC. If AC and

BD when produced meet at P, prove that APB = 60°.Solution. Join O to D and B to C.

Now, CD = OC = OD (radii of same circle) OCD is equilateral

COD = 60°

and 306021COD

21CBD

( angle made by at centre = 2 × angle at any point on its remaining part).


Now, BCA + BCP = 180° ( linear pair)But, BCA = 90° ( angle in semi-circle) 90° + BCP = 180° BCP = 90°Now, in BCP, BCP + CBP + CPB = 180° 90° + 30° + CPB = 180° CPB = 180° – 120° = 60° APB = 60° ( CPB = APB)Hence proved.

Example 14. In the following figures, if O is the centre of the circle, find x.

Solution. (i) ACB = 90° ( angle is a semi circle)ACB = 180° – 135° ( opposite angles of cyclic quadrilateral are supplementary)

Now, In ABC, CAB + ACB + ABC = 180° ( angle sum property) x + 90° + 45° = 180° x = 45° Ans.(ii) Take any point P on the major arc.

Now, AOC.21APC

( 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)

6012021

.

Also, APC + ABC = 180° ( opp. angles of cyclic quadrilateral are supplementary) 60° + ABC = 180° ABC = 180° – 60° = 120°Now, ABC + DBC = 180° ( linear pair) 120° + DBC = 180° DBC = 60° x = 60° Ans.

Example 15. Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.Solution. Given : A cyclic quadrilateral ABCD in which AP, BP, CR and DR are the bisectors of A, B, C

and D respectively, forming a quadrilateral PQRS.To prove : PQRS is a cyclic quadrilateral.


Proof : In APB, APB + PAB + PBA = 180° ( angle sum property of a triangle)Also, InDRC, CRD + RCD + RDC = 180° ( same as above)

180B21A

21APB ...(1)

and 180D21C

21CRD ...(2)

Adding (1) and (2), we get

360D)CBA(21CRDAPB

360)(36021CRDAPB (A + B + C + D = 360°)

APB + CRD = 180°Thus, two opposite angles of quadrilateral PQRS are supplementary. Quadrilateral PQRS is cyclic.

Example 16. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quad-rilateral, prove that it is a rectangle. —NCERT

Solution. Given : Diagonals AC and BD of a cyclic quadrilateral are diameter of the circle through thevertices A, B, C and D of the quadrilateral ABCD.To prove : ABCD is a rectangle.Proof : Since AC is a diameter. ABC = 90° ( angle in a semi-circles is 90°) also, quadrilateral ABCD is a cyclic. ADC = 180° – ABC ADC = 180° – 90° = 90°Similarly, BAC = BCD = 90°.Now, each angle of a cyclic quadrilateral ABCD is 90°. ABCD is a rectangle.

Example 17. If the non-parallel sides of a trapezium are equal, prove that it is cyclic. —NCERTSolution. Given : A trapezium ABCD in which AB || DC and AD = BC.

To prove : ABCD is a cyclic trapezium. Construction : Draw DE AB and CF AB.Proof : In order to prove that ABCD is a cyclic trapezium, it issufficient to prove that B + D = 180°.Now, In DEA and CFB, we have

AD = BC (given)DEA = CFB (each = 90°) DE = CF (distance between two parallel lines is always equal)

DEA CFB (RHS congruence condition) A = B and ADE = BCF (cpct)Now, ADE = BCF 90° + ADE = 90° + BCF EDC + ADE = FCD + BCF (EDC = 90°, FCD = 90°) ADC = BCD D = CThus, A = B and C = D.


Now, A + B + C + D = 360° ( sum of angles of a quadrilateral is 360°) 2B + 2D = 360°

1802

360DB

Hence, ABCD is a cyclic trapezium.Example 18. 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 figure). Prove that ACP = QCD.—NCERT

Solution. Since angles in the same segment of a circle are equal. ACP = ABP ...(1)and QCD = QBD ...(2)But, ABP = QBD ...(3)

(vertically opposite angles) from (1), (2) and (3), we get

ACP = QCD

Example 19. Two circles are drawn taking two sides of a triangle as diameters, prove that the point of intersec-tion of these circles lie on the third side. —NCERT

Solution. Given : Two circles are drawn with sides AB and AC of ABC as diameters. The circles intersectat D.To prove : D lies on BC.Construction : Join A to D.Proof : Since AB and AC are diameters of the circles, ADB = 90° and ADC = 90°

( angles in a semi-circle is 90°)Adding, we get, ADB + ADC = 90° + 90° = 180° BDC is a straight line.Hence, D lies on BC.

Example 20. ABC and ADC are two right triangles with common hypotenuse AC. Prove that CAD = CBD.Solution. ABC and ADC are right angled triangles with common hypotenuse AC. Draw a circle with AC

as diameter passing through B and D. Join B to D. —NCERT

Clearly, CAD = CBD. ( angles in the same segment are equal)Hence proved.

Example 21. Prove that a cyclic parallelogram is a rectangle. —NCERTSolution. Given : ABCD is parallelogram inscribed in a circle.

To prove : ABCD is a rectangle.Proof : Since ABCD is a cyclic quadrilateral, A + C = 180° ...(1)But, A = C ...(2)

(opposite angles of a parallelogram are equal)


from (1) and (2), we get, A + A = 180° 2A = 180° A = 90°i.e. A = C = 90°Similarly, B = D = 90° ABCD is a parallelogram whose each angle is equal to 90°. ABCD is a rectangle.

Example 22. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point ofintersection of its diagonals. —NCERT

Solution. Given : ABCD is a rhombus. AC and BD are its two diagonalswhich bisect each other at right angles.To prove : A circle drawn on AB as a diameter will passthrough O.Construction : From O, draw PQ || AD and EF || AB,

Proof : Since, AB = DC 21

AB = 21

DC

AQ = DP ( Q and P are mid-points of

AB and DC respectively)Similarly, AE = OQ AQ = OQ = QB A circle drawn with Q as a centre and radius AQ passes through A, O and B, which provesthe desired result.

Example 23. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.Prove that AE = AD. —NCERT

Solution. Since ABCE is a cyclic quadrilateral, AED + ABC = 180° ...(1)Now, CDE is a straight line. ADE + ADC = 180° ...(2)

( ADC and ABC are opposite angles of a parallelogram i.e. ADC = ABC)From (1) and (2), we get

AED + ABC = ADE + ABC AED = ADE In AED, AED = ADE AD = AE (sides opposite to equal angles are equal)Hence proved.

Example 24. AC and BD are chords of a circle which bisect each other. Prove that :(i) AC and BD are diameters (ii) ABCD is a rectangle. —NCERT

Solution. (i) Let AB and CD be two chords of a circle with center O.Let they bisect each other at O.Join AC, BD, AD and BC.


Now, In AOC and BOD, we have OA = OB ( O is mid-point of AB)

AOC = BOD (vertically opp. angles) OC = OD ( O is mid-point of CD)

AOC BOD (SAS congruence condition) AC = BD (cpct)

= ...(1)Similarly, from AOD and BOC, we have

= ...(2)Adding (1) and (2), we get,

CD divides the circle into two equal parts CD is a diameter.Similarly, AB is a diameter.(i) Since, AOC BOD (proved above) OAC i.e. BAC = OBD i.e. ABD AC || BD.Again, AOD COB (proved above) AD || CB ABCD is a cyclic parallelogram. DAC = DBA ...(3) ( opp. angles of a parallelogram)also, ABCD is a cyclic quadrilateral, DAC + DBA = 180° ...(4)from (3) and (4), we get

902

180DBADAC

Hence, ABCD is a rectangle.Example 25. Bisectors of angles A, B and C of a triangle ABC intersect the circumcircle at D, E and F respec-

tively. Prove that the angles of the DEF are A,2190 B

2190 and C.

2190 —NCERTT

Solution. We have, D = EDF = EDA + ADF = EBA + FCA( EDA and EBA are in the same segmentare in the same segment of a circle)

EDA = EBA.Similarly, ADF and FCA are the angles in the same segment,

ADF = FCA

C)B(21C

21B

21

A)(18021

[ A + B + C = 180°]


A2190

Similarly, other two angles of DEF are

C.2190andB

2190

Hence proved.

PRACTICE EXERCISE1. Show how to complete a circle if an arc of the circle is given.2. The radius of a circle is 13 cm and the length of one of its chord is 10 cm. Find the distance of the chord

from the centre.3. AB and CD are two parallel chords of a circle which are on the opposite sides of the centre such that

AB = 8 cm and CD = 6 cm. Also, radius of circle is 5 cm. Find the distance between the two chords.4. Two chords AB and AC of a circle are equal. Prove that the centre of the circle lies on the angle bisector

of BAC.5. If two circles intersect in two points, prove that the line through their centres is the perpendicular

bisector of the common chord.6. If a diameter of a circle bisects each of the two chords of the circle, prove that the chords are parallel.7. In the given figure, two equal chords AB and CD of a circle with centre O, when produced meet at a point

P. Prove that (i) BP = DP (ii) AP = CP.

8. Two circles whose centres are O and O intersects at P. Through P, a line l parallel to OO, intersecting thecircles at C and D, is drawn. Prove that CD = 2.OO.

9. In the given figure, O is the centre of the circle and MO bisects AMC. Prove that AB = CD.

10. Show that if two chords of a circle bisect each other, they must be the diameters of the circle.11. In the given figure, OD is perpendicular to the chord AB of a circle with centre O. If BC is a diameter, show

that AC || OD and AC = 2OD.


12. Prove that two different circles cannot intersect each other at more than two points.13. Two equal circles intersect in P and Q. A straight line through P meets the circle in X and Y. Prove

that QX = QY.

14. 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.

15. AB and AC are two equal chords of a circle whose centre is O. If OD AB and OE AC, prove thatADE is an isosceles triangle.

16. Prove that angle is a semi-circle is a right angle.17. Prove that the angles in the same segment of a circle are equal.18. Prove that the angle formed by a chord in the major segment is acute.19. Prove that the angle formed by a chord in the minor segment is obtuse.20. If O is the centre of a circle, find the value of x in the following figures:

(i) (ii) (iii)

(iv) (v) (vi)


21. In the given figure, two circles intersect at P and Q. PR and PS are respectively the diameters of the circle.Prove that the points R, Q, S are collinear.

22. Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter, bisects thethird side of the triangle.

23. In the given figure, O is the centre of the circle. Prove that a + b = c.

24. In an isosceles triangle ABC with AB = AC, a circle passing through B and C intersects the sides AB andAC at D and E respectively. Prove that DE || BC.

25. In the given figure, PQ is a diameter of a circle with center O. If PQR = 65°, SPR = 35° and PQT = 50°,find :(i) QPR (ii) QPT (iii) PRS


26. In the given figure ABC is isosceles with AB = AC and ABC = 55°. Find BDC and BEC.

27. Find the angles marked with a letter. O is the centre of the circle.

(i) (ii) (iii)

(iv) (v) (vi)28. In the following figure, find x and y.

29. Prove that every cyclic parallelogram is a rectangle.30. If two non-parallel sides of a trapezium are equal, prove that it is cyclic.31. Prove that cyclic trapezium is always isosceles and its diagonals are equal.


32. In an isosceles ABC with AB = AC, a circle passing through B and C intersects the sides AB and ACat D and E respectively. Prove that DE || BC.

33. In the given figure, ABCD is a parallelogram. A circle through A, B, C intersects CD produced at E. Provethat AD = AE.

34. The bisectors of the opposite angles A and C of a cyclic quadrilateral ABCD intersect the circle at thepoints E and F respectively. Prove that EF is a diameter of the circle.

35. Prove that the angle bisectors of the angles formed by producing opposite sides of a cyclic quadrilateral(provided the are not parallel) intersect at a right angle.


PRACTICE TESTGeneral Instructions :MM : 30 Time : 1 hourQ. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.

1. In the given figure, A, B, C and D are four points on the circle. AC and BD intersect at a point E such thatBEC = 120°, and ECD = 20°. Find BAC.

2. Prove that the line joining the mid-points of the two parallel chords of a circle passes through the centreof the circle.

3. Find the value of x and y:

4. If O is the centre of the circle, find the value of x.

5. If two intersecting circles have a common chord of length 16 cm, and if the radii of two circles are 10 cmand 17 cm, find the distance between their centres.

6. If two non-parallel sides of a trapezium are equal, prove that it is cyclic.7. In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB.

Also, CO is joined and produced to meet the circle in D. If ACD = b° and AOD = a°, prove thata = 3b°.


8. In the given figure, ABCD is a cyclic quadrilateral. A circle passing through A and B meets AD and BC inE and F respectively. Prove that EF || DC.

9. Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateal is also cyclic.10. The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point

on the remaining part of the circle. Prove it.

ANSWERS OF PRACTICE EXERCISE2. 12 cm3. 7 cm

14. 6 cm20. (i) 100° (ii) 50° (iii) 55° (iv) 35° (v) 30° (vi) 50°25. (i) 15° (ii) 40° (iii) 40°26. (i) 70° (ii) 110°27. (i) a = 50° (ii) b = 40° (iii) c = 35°

(iv) a = 45°, b = 64°, c = 58° (v) x = 40°, y = 32°, z = 40° (vi) a = 41°, b = 41°, c = 41°28. x = 40°, y = 25°


1. 100° 3. x = 75°, y = 105° 4. 55° 5. 21 cm

MATHEMATICS–IX CONSTRUCTIONS 131

CHAPTER 11CONSTRUCTIONS

Points to Remember :Using a graduated scale and a compass, we can construct the following :

1. Perpendicular bisector of a line segement.2. Bisector of an angle.3. Angle of measures 60°, 90°, 45° etc.4. A triangle given its base, a base angle and the sum of the other two sides.5. A triangle give its base, a base angle and the difference of the other two sides.6. A triangle given its perimeter and its base angles.

ILLUSTRATIVE EXAMPLESExample 1. Draw a line segment AB = 7 cm. Construct its perpendicular bisector. Give step of constructions

also.Solution.

Step of constructions:(i) Draw a line segment AB = 7 cm

(ii) Taking A and B as centres and radius more than AB,21

draw arcs on both sides of the linesegment AB.

(iii) Let these arcs intersect each other at P and Q. Join P to Q.(iv) Let PQ intersect AB at M. The line PMQ is the reqired perpendicular bisector of AB.

Example 2. Construct an angle of 60° at the initial point of a given ray.Solution. Step of constructions:

(i) Draw the ray AB.(ii) Taking A as centre and some suitable radius, draw an

arc of a circle, which intersects AB, say at a point D.(iii) Taking D as centre and with the same radius as before,

draw an arc intersecting the previously drawn arc, sayat a point E.

(iv) Draw the ray AC passing through E.Then, CAB is the required angle of 60°.

132 CONSTRUCTIONS MATHEMATICS–IX

Example 3. Construct the bisector of a given angle ABC = 70°.Solution. Step of constructions:

(i) Draw ABC = 70° using a protractor.(ii) Taking B as centre and any radius, draw an arc to

intersect the rays BA and BC, say at E and Drespectively.

(iii) Taking E and D as centres and with the radius more

than DE,21

draw arcs to intersect each other, say

at F.(iv) Draw the ray BF. This ray BF is the required bisector of the ABC.

Example 4. Construct a ABC in which BC = 5 cm, B = 60° and the sum of the other two sides is 7 cm.Solution.

Step of Constructions:(i) Draw BC = 5 cm.

(ii) Draw XBC = 60°(iii) On the ray BX, mark off point D such that BD = 7 cm.(iv) Join D to C.(v) Draw perpendicular bisector EF of CD. Let it intersects BD at A.

(vi) Join A to C.ABC is the required triangle.

Example 5. Construct a ABC in which BC = 5.5 cm, B = 30° and AB – AC = 2cm.Solution.

MATHEMATICS–IX CONSTRUCTIONS 133

Step of constructions:(i) Draw BC = 5.5 cm.

(ii) Draw XBC = 30°(iii) On the ray BX, mark off point D such that BD = 2 cm.(iv) Join D to C.(v) Draw perpendicular bisector EF of CD. Let it intersects BX at A.

(vi) Join A to C.ABC is the required triangle.

Example 6. Construct a ABC in which B = 45°, C = 60° and AB + BC + AC = 13 cm.Solution. Step of constructions:

(i) Draw XY = 13 cm.(ii) Draw MXY = 45° and NYX = 60°

(iii) Draw angle bisectors of MXY and NYX, meeting at a point, say A.(iv) Draw perpendicular bisector of XA and YA, meeting XY at B and C respectively.(v) Join A to B and A to C.

ABC is the required triangle.

PRACTICE EXERCISE1. Construct the angles of the following measurements :

(i) 90° (ii) 2122 (iii) 75° (iv) 15° (v) 135°

2. Construct an equilateral triangle XYZ, in which XY = 5.4 cm.3. Construct a triangle ABC, in which BC = 5.8 cm, ACB = 45° and AB + AC = 7 cm.4. Construct a triangle ABC, in which AC = 5 cm, A = 60° and sum of the sides AB and BC is 8.5 cm.5. Construct a triangle ABC in which AC = 5.5 cm, A = 30° and BC – AB = 1.6 cm.6. Construct a triangle ABC in which AB = 4.2 cm, B = 45° and BC – AC = 2 cm.7. Construct a triangle ABC in which B = 60°, C = 45° and AB + BC + AC = 11 cm.8. Construct a triangle PQR in which Q = 45°, R = 30° and PQ + QR + PR = 9.5 cm.9. Construct a right-angled triangle whose perimeter is 12 cm and one acute angle is equal to 60°.

10. Construct a ABC such that BC = 6 cm, AB = 6 cm and median AD = 4 cm.

134 CONSTRUCTIONS MATHEMATICS–IX

PRACTICE TEST



Each Questions carry 5 marks.

1. Construct an angle of 90° at the initial point of a ray AB and justify your construction.2. Construct a ABC in which BC = 8 cm, B = 60°, and AB + AC = 13.5 cm.3. Construct a ABC in which BC = 6 cm, B = 60° and AC – AB = 2.2 cm.4. Construct a PQR in which Q = 30°, R = 45° and PQ + QR + PR = 10 cm.

MATHEMATICS–IX HERON’S FORMULA 135

CHAPTER 12HERON’S FORMULA


1. Area of right triangle heightBase21

2. Area of an Equilateral triangle 2(side)43

3. Area of a isosceles triangle ,aba 2244

where, a is base and b represents equal sides.

4. Heron’s Formula : If a, b, c denote the lengths of the sides of a triangle, then its Area,))()(( csbsassA

where,2

cbas

5. Area of a quadrilateral can be calculated by dividing the quadrilateral into triangles and using heron’sformula for calculating area of each triangle.

ILLUSTRATIVE EXAMPLESExample 1. Find the area of a triangle whose sides are 13 cm, 14 cm and 15 cm respectively.Solution. Let a = 13 cm, b = 14 cm, c = 15 cm

cm212

1514132

cbas

Area ))()(( csbsass

cm.sq5)1214)(121)(1321(21

cmsq.32722273cmsq.67821 = 2× 2 × 3 × 7 sq. cm. = 84 sq. cm.

Area of triangle = 84 sq. cm. Ans.Example 2. 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 it perimeter is 180 cm. —NCERTSolution. Given perimeter of equilateral triangle = 180 cm.

Side of an equilateal triangle is a cm, then its perimeter is 3a cm. 3a = 180 a = 60 cm

Now, cm.902

1802

6060602

aaas

Using heron’s formula,Area ))()(( asasass

2cm60)60)(9060)(9090(90

22 cm1000033333cm30303090

Ans.cm3900 2 2cm310033

136 HERON’S FORMULA MATHEMATICS–IX

Example 3. A triangular park ABC has sides 120 m, 80 m and 50 m (see figure). A gardener Dhania has to puta fence all around it and also plant grass inside. How much area does she need to plant? Find thecost of fencing it with barbed wire at the rate of Rs 20 per metre leaving a space 3 m wide for a gateon one side.

Solution.

(i) For Area of the park :Let a = 120 m, b = 80 m, c = 50 m

m1252

50801202

cbas

Area ))()(( csbsass

2)50125)(80125)(120125(125 m

22 m15375m75455125 Also, Perimeter of the park = AB + BC + AC = 250 m length of wire needed for fencing = 250 m – 3 m (to left for gate)

= 247 mand cost of fencing = 247 × 20 Rs = Rs. 4940 Ans.

Example 4. Triangular side walls of a flyover have been used for advertisements. The sides of the walls are122 m, 22 m and 120 m (see figure). The advertisements yield an earning of Rs. 5000 per m2 per year.A company hired one of its walls for 3 months, how much rent did it pay? —NCERT


Solution. We will calculate area of triangular wallhere, a = 122 m, b = 22 m, c = 120 m

mmmcbas 1322

2642

120221222

))()((Area csbsass

2m120)(13222)(132122)(132132

2m1211010132

222 m1320m121110m121110101211 Now, Rent charges = Rs. 5000 per m2 per year Rent charged from company for 3 months

12313205000Rs.

= Rs. 16,50,000 Ans.Example 5. There is a slide in a park. One of the its side walls has been painted, in colour with a message

‘‘KEEP THE PARK GREEN AND CLEAN’’, (see figure). If the sides of the wall are 15 m, 11 m and6 m, find the area painted in colour. —NCERT

Solution. here, a = 15 m, b = 11m, c = 6m

m16m2

32m2

611152

cbas

Area ))()(( csbsass

22 m105116m6)11)(1615)(1616(16

222 m220m254m52544 Area painted in a colour = Area of side wall

= Ans.m220 2

Example 6. An isosceles triangle has perimeter 30 m and each of the equal sides is 12 cm. Find the area of thetriangle. —NCERT

Solution. here, a = b = 12 cm.also, a + b + c = 30 m 12 + 12 + c = 30 cm c = 6 cm


cm15cm2

30cm2

cbas

Area of the triangle

))()(( csbsass

2cm6)12)(1512)(1515(15

22 cm333335cm93315

Ans.cm159 2 2cm3533Example 7. The perimeter of a triangle is 450 m and its sides are in the ratio of 13 : 12 : 5. Find the area of the

triangle.Solution. Let the sides of the triangle by 13 x, 12 x and 5x.

Given, perimeter of triangle = 450 m 13 x + 12 x + 5 x = 450 30x = 450 x = 15 Sides are 13 × 15 m, 12 × 15 m and 5 × 15 mi.e. 195 m, 180 m and 75 m.Let a = 195 m, b = 180 m, c = 75 m

m225m2

4502

1751801952

cbas

Area ))()(( csbsass

msq.75)180)(225195)(225225(225

m.sq.1504530225

32551531521515 sq. m.

= 15 × 15 × 2 × 3 × 5 sq. m. = 6750 sq. m Ans.

Example 8. A park, in the shape of a quadrilateral ABCD, has C = 90°, AB = 9 m, BC = 12 m, CD = 5m andAD = 8 m. How much area does it occupy? —NCERT

Solution. Area of 22 m30m51221CDBC

21ΔBCD

Now, In BCD, BD2 = BC2 + CD2

m13m169m25144m512BD 22

For ABD, let a = 13 m, b = 8 m, c = 9 m.

Now, m15m2

30m2

98132

cbas

Area of ))()((ΔABD csbsass

2m9)8)(1513)(1515(15


22 m753322m67215

(approx.)m5.96m356m3532 222 = 35.4 m2 (approx.)

Required Area = ar (ABD) + ar (BCD) = 35.4 m2 + 30 m2 = 65.4 m2 (approx.) Ans.

Example 9. A parallelogram, the length of whose sides are 60 m and 25 m has one diagonal 65 m long. Find thearea of the parallelogram.

Solution. Let ABCD be the given parallelogram.Area of parallelogram ABCD

= area of ABC + area of ACD = 2 area of ABC

Let, a = 60 m, b = 65 m, c = 25 m

mms 752

256560

area of ABC ))()(( csbsass

225)65)(7560)(7575(75 m

250101575 m = 5 × 3 × 5 × 2 × 5 m2 = 750 m2

Area of parallelogram ABCD= 2 × area of ABC = 2 × 750 m2

= 1500 m2 Ans.Example 10. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are

26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of theparallelogram. —NCERT

Solution. For the triangle ;Let a = 26 cm, b = 28 cm, c = 30 cm

then, cm42cm2

842

cm30cm28cm262

cbas

Now, Area of triangle ))()(( csbsass

2cm30)28)(4226)(4242(42

2cm12141642

2cm3227244732

2cm7744332222 = 2 × 2 × 3 × 4 × 7 cm2 = 336 cm2

For the parallelogram :Area = base × height

baseAreaheight ( Area of parallelogram = Area of triangle)

Ans.cm.12 cm28

336


Example 11. Find the area of a quadrilateral ABCD, in which B = 90°, AB = 15 cm, BC = 36 cm, CA = 24 cm andDA = 21 cm.

Solution. Join A to C.In ABC, B = 90°

Area of ABC 2153621 cm

= 270 cm2

also, In ABC,AC2 = AB2 + BC2 = 152 + 362 = 225 + 1296 = 1521

cm39cm1521AC Now, For area of ACD, we will use hero’s formula.let, a = 39 cm, b = 24 cm, c = 21 cm.

cm422

2124392

cbas

Area of ACD ))()(( csbsass

2cm21)24)(4239)(42(4242

222 cm218.232cm3126cm2118342 Area of ABCD = 270 cm2 + 218.232 cm2 = 488.232 cm2 Ans.

Example 12. Radha made a picture of an aeroplane with coloured paper as shown in the figure. Find the totalarea of the paper used. —NCERT

Solution. Area I = area of isosceles triangle with a = 1 cm and b = 5 cm

2244

aba

22 cm(1)25441


(approx)cm2.5cm499 22

Area II = area of rectangle = length × breadth = 6.5 × 1 cm2 = 6.5 cm2

Area III = Area of Trapezium = 3 × area of equilateral triangle with a = 1 cm

= 2222 cm4

5.196cm4

1.7323cm(1)433

= 1.299 cm2 1.3 cm2 (approx)

Area of IV and V 22 cm9cm1.56212

Total area of paper used = area I + area II + area III + area IV + area V = 2.5 cm2 + 6.5 cm2 + 1.3 cm2 + 9 cm2

= 19.3 cm2 (approx) Ans.Example 13. An Umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure),

each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for theumbrella? —NCERT

Solution. In one triangular piece, a = 20 cm, b = 50 cm, c = 50 cm.

Now, cm60cm2

1202

cm505020

s

Area ))()(( csbsass

2cm50)50)(6020)(60(6060

22 cm6200cm10104060

Area of 5, Ist coloured triangles = 5 × 200 6 cm2 = 1000 6 cm2

and area of 5, IInd coloured triangles = 5 × 200 6 cm2 = 1000 6 cm2

Example 14. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm andsides 6 cm each is to be made of three different shades as shown in figure. How much paper ofeach shade has been used in it. —NCERT

Solution. Each diagonal of square = 32 cm

cm16cm3221CO

Area of each shaded portion I and II

heightbase21

2cm256 2cm163221

And, ara of IIIrd portion with sides 6 cm, 6 cm and 8 cm.

22222 cm(8)4(6)484

4 aba


22 cm802cm641442

Ans.cm17.92 2 2cm58Example 15. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being

9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50 p per cm2.—NCERT

Solution. For each triangular tile,Let a = 9 cm, b = 28 cm, c = 35 cm

cm36cm2

72cm2

352892

cbas

Area of each triangular tile ))()(( csbsass

2cm35)28)(369)(36(3636

222 cm636cm6326cm932436

Total area of floral design 2cm63616 Cost of polishing the 16 triangular tiles

Ans.705.45Rs. 6288Rs.6361621Rs.

Example 16. Find the area of a trapezium ABCD in which AB || DC, AB = 77 cm, BC = 25 cm, CD = 60 cm andDA = 26 cm.

Solution. Draw DE || BC and DF ABthe, DE = BC= 25 cm.AE = AB – EB = AB – DC = 77 cm – 60 cm = 17 cm.In DAE,Let a = 17 cm, b = 25 cm, c = 26 cm

cm34cm2

2625172

cbas

Area of DAE )()()( csbsass

2cm26)25)(3417)(34(3434

22 cm204cm4317 ...(1)

Also, area of DFAE21ΔDAE

2cmDF1721

...(2)

from (1) and (2), we get 204DF1721

cm24cm172042DF

area of trapezium ABCD DFDC)(AB21


2cm2460)(7721

= 1644 cm2 Ans.

PRACTICE EXERCISE1. Find the area of the triangle whose sides are :

(i) 9 cm, 12 cm and 15 cm (ii) 150 cm, 120 cm and 2 m(iii) 6 m, 11 m and 15 m (iv) 26 cm, 28 cm, 30 cm

2. Find area of a equilateral triangle of side 16 cm.3. An isosceles triangle has a perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the

triangle.4. Calculate the area of the shaded portion, in the given figure.

5. The sides of a triangle are in the ratio 13:14:15 and its perimeter is 210 cm. Find its area.

6. If the area of an equilateral triangle is 349 cm2, find its perimeter..

7. Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find(i) The area of the field.(ii) the length of perpendicular from the opposite vertex on the side measuring 154 m.

8. ABC is a right triangle right angled at A with AB = 6 cm, AC = 8cm. A circle with centre O and radius r isinscribed in the triangle. Find r.

9. The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, find the length of theperpendicular on the side of length 50 m from the opposite vertex. Calculate also the cost of watering itat Rs. 3.50 per 100 sq. m.

10. A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest altitude.11. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5

cm.12. The area of a rhombus is 72 cm2. If one of the diagonals is 18 cm long, find the length of the other

diagonal.13. The diagonal of a four sided field is 40 m. The perpendiculars from the opposite vertices on this diagonal

are 20 m and 14 m. Find the area of the field.14. A field is in the shape of a trapezium whose parallel sides are 24 m and 52 m. The non-parallel sides are

26 m and 30 m. Find the area of the field.


15. Find area of the following polygons :

(i) (ii)

PRACTICE TESTMM : 20 Time : 1 hour


Each Question carry 4 marks.

1. The perimeter of a right triangle is 450 cm and its sides are in the ratio 13 : 12 : 5. Find the area of thetriangle.

2. Find the area of a rhombus, one of whose sides is 25 cm and one of whose diagonal is 48 cm.3. In a ABC, AB = 4 cm, BC = 9 cm and AC = 7 cm. Find the length of the perpendicular from A to BC.4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm,

28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.5. A field is in the shape of a trapezium whose parallel sides are 55 m, 40 m and non-parallel sides are 20 m,

25 m. Find the area of the field.


1. (i) 54 cm2 (ii) 8969 cm2 (iii) 2m220 (iv) 336 cm2

2. 2cm364 3. 2cm159 4. 384 cm2 5. 2100 cm2

6. 42 cm 7. (i) 2772 m2 (ii) 36 m 8. 2 cm9. 67.2 cm, Rs. 58.80 10. 939.14 cm2, 15.39 cm 11. 15.2 cm2 12. 8 cm

13. 680 m2 14. 912 m2 15. (i) 249.4 cm2 (ii) 444 m2


1. 9000 m2 2. 336 cm2 3. cm534

4. 12 cm 9. 950 m2

MATHEMATICS–IX SURFACE AREAS AND VOLUMES 1

CHAPTER 13SURFACE AREAS AND VOLUMES


1. Cuboid

(i) Volume = lbh

(ii) Curved surface area = 2h (l + b)

(iii) Total surface area = 2 (lb + bh + lh)

(iv) Diagonal 222 hbl

2. Cube

(i) Volume = a3

(ii) Curved surface area = 4a2

(iii) Total surface area = 6a2

(iv) Diagonal a.3

3. Cylinder

(i) Volume = r2h

(ii) Curved surface area = 2rh

(iii) Total surface area = 2r (r + h)

4. Cone

(i) Volume hr 2

31

(ii) slant height, 22 rhl

(iii) curved surface area lr

(iv) Total surface area )( rlr

2 SURFACE AREAS AND VOLUMES MATHEMATICS–IX

5. Sphere

(i) Volume 3

34 r

(ii) Total surface area = 4 r2

6. Hemi-sphere

(i) Volume 3

32 r

(ii) Curved surface area = 2r2

(iii) Total surface area = 3r2

ILLUSTRATIVE EXAMPLESExample 1. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm

high.(i) Which box has the greater lateral surface area and by how much?(ii) Which box has the smaller surface area and by how much? —NCERT

Solution. (i) Lateral surface area (L1) of cubical box = 4 × (edge)2

= 4 × (10)2 cm2

= 400 cm2

and, lateral surface area (L2) of cuboidal box= 2 (length + breadth)× height= 2 (12.5 + 10) × 8 cm2 = 360 cm2

Clearly, L1 > L2.Now, L1 – L2 = 400 cm2 – 360 cm2 = 40 cm2

The cubical box has larger lateral surface area and is greater by 40 cm2.(ii) Total surface area of the cubical box (S1) = 6 (edge)2

= 6 × (10)2 cm2 = 600 cm2

Total surface area of the cuboidal box (S2)= 2 (lb + lh + bh)= 2 (12.5 × 10 + 10 × 8 + 8 × 12.5) cm2

= 2 (125 + 80 + 100) cm2 = 610 cm2

Clearly, S2 > S1. S2 – S1 = 610 cm2 – 600 cm2 = 10 cm2

Thus, the cuboidal box has greater surface area and is greater by 10 cm2.

Example 2. Two cubes each of 15 cm edge are joined end to end. Find the surface area of the resulting cuboid.Solution. here, l = length of resulting cuboid = 15 cm + 15 cm = 30 cm

b = breadth of resulting cuboid = 15 cmh = height of resulting cuboid = 15 cm


Surface area of resulting cuboid= 2 (lb + bh + lh)= 2 (30 × 15 + 15 × 15 + 30 × 15) cm2

= 2 (450 + 225 + 450) cm2 = 2 (1125) cm2

= 2250 cm2 Ans.Example 3. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top.

Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box.(ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs. 20. –NCERT

Solution. We have, Length; l = 1.5 m, Breadth, b = 1.25 m and Depth = Height, h = 0.65 m(i) Since the plastic box is open at the top, Plastic sheet required for making such box

= 2 (l + b) × h + lb= 2 (1.5 + 1.25) × 0.65 m2 + 1.5 × 1.25 m2

= 2 × 2.75 × 0.65 m2 + 1.875 m2

= 3.575 m2 + 1.875 m2 = 5.45 m2

(ii) Cost of 1 m2 of sheet = Rs. 20 Total cost of 5.45 m2 of sheet = Rs. 5.45 × 20 = Rs. 109 Ans.

Example 4. Parveen wanted to make a temporary shelter for her car, by making a box-like structure withtarpaulin that covers all the four sides and the top of the car (with the front face as a flap which canbe rolled up). Assuming that the stitching margins are very small, and therefore negligible, howmuch tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m ×3 m? –NCERT

Solution. here, l = 4m, b = 3m, h = 2.5 m.Since there is no tarpaulin for the floor. Tarpaulin required = [2 (l + b) × h + lb]

= [2 (4 + 3) × 2.5 + 4 × 3] m2

= (2 × 7 × 2.5 + 12) m2

= (35 + 12) m2 = 47 m2 Ans.Example 5. The sum of length, breadth and height of a cuboid is 21 cm and the length of its diagonal is 12 cm.

Find the surface area of the cuboid.Solution. Let the length, breadth and height of the cuboid be l cm, b cm and h cm respectively.

Then, l + b + h = 21 ...(1)Now, diagonal = 12 cm

12222 hbl l2 + b2 + h2 = 144 ...(2)Now, l + b + h = 21Squaring both sides, we get

(l + b + h)2 = (21)2

l2 + b2 + h2 + 2lb + 2bh + 2lh = 441 144 + 2 (lb + bh + lh) = 441 ( using (2)) 2 (lb + bh + lh) = 441 – 144 = 297 Surface area of the cuboid is 297 cm2 Ans.


Example 6. Aggarwal sweets stall was placing an order for making card board boxes for packing their sweets.Two size of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smallerof dimensions 15 cm × 12 cm × 5 cm. 5% of the total surface area is required extra, for all theoverlaps. If the cost of cardboard is Rs. 5 for 1000 cm2, find the cost of cardboard required forsupplying 300 boxes of each kind.

Solution. Surface area of Ist box = 2 (25 × 20 + 20 × 5 + 25 × 5) cm2

= 2 (500 + 100 + 125) cm2 = 1450 cm2

Surface area of IInd box = 2 (15 × 12 + 12 × 5 + 15 × 5) cm2

= 2 (180 + 60 + 75) cm2 = 630 cm2

Total combined surface area = 1450 cm2 + 630 cm2 = 2080 cm2

Area of overlaps = 5% of 2080 cm2 22 cm104cm2080100

5

Total surface area of 2 boxes (one of each kind)= (2080 + 104) cm2 = 2184 cm2

Surface area of 300 boxes of each kind= 300 × 2184 cm2 = 655200 cm2

Now, cost of cardboard for 1000 cm2 = Rs. 5

Cost of cardboard for 1000

5Rs.cm1 3

Cost of cardboard for 655200 cm2 6552001000

5Rs.

= Rs. 3276 Ans.Example 7. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved

surface of the pillar at the rate of Rs. 12.50 per m2. –NCERTSolution. Diameter of cylindrical pillar = 50 cm

radius (r) m0.25cm25cm2

50

also, height (h) = 3.5 m

Now, curved surface = 2rh 2m3.50.257222 = 5.5 m2

Cost of painting 1 m2 = Rs. 12.50 Cost of painting 5.5 m2 = Rs. 12.50 × 5.5 = Rs. 68.75 Ans.

Example 8. In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth.The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given forfolding it over the top and bottom of the frame. Find how much cloth is required for covering thelampshade. –NCERT


Solution. Here, radius cm10cm2

20)( r

and, height (h) = 30 cm + 2 × 2.5 cm = 35 cm(for margin)

Cloth required for covering the lampshade = Its curved surface area = 2 rh

2cm35107222 = 2200 cm2 Ans.

Example 9. It is required to make a closed cylindrical tank of height 120 cm and base diameter 140 cm from ametal sheet. How many square metres of the sheet is required for the same?

Solution. We have, diameter of base = 140 cm.

radius of base cm70cm2

140

and, height of cylinder = 120 cm. Total surface area of required tank

= 2 r (r + h)

2cm120)(70707222

2m8.36 222 m1000083600cm83600cm190440 Ans.

Example 10. The students of a Vidyalaya were asked to participate in a competition for making and decoratingpenholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be ofradius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. Ifthere were 35 competitors, how much cardboard was required to be bought for the competition?

–NCERTSolution. Cardboard required by each competitor

= curved surface area of one penholder + base area= 2rh + r2, where r = 3 cm, h = 10.5 cm

22 cm(3)72210.53

7222

= (198 + 28.28) cm2 = 226.28 cm2 (approx) Cardboard required for 35 competitors

= 35 × 226.28 cm2 = 7920 cm2 (approx) Ans.Example 11. The diameter of a roller is 84 cm and its length is 100 cm. It takes 300 complete revolution to move

once over to level a playground. Find the area of the playground in m2.

Solution. Radius of roller cm42cm2

84)( r

Length of the cylindrical roller (h) = 100 cm.Area moved by the roller in one revolution

2cm1004272222 rh


Area moved in 300 revolutions

2cm300100427222

= 44 × 180000 cm2 = 44 × 18 m2 = 792 m2 Ans.Example 12. Find (i) the curved surface area of a cylindrical petrol storage tank that is 4.2 m in diameter and

4.5 m high. (ii) how much steel was actually used if 121

of the steel actually used was wasted in

making the closed tank? —NCERT

Solution. Here, m54m12m224 .h,..r

(i) Curved surface area = 2 r h

2m54127222 ..

= 59.4 m2

(ii) Total surface area of closed tank

)(222 2 hrrrhr

2)5412(127222 m...

= 87.12 m2

Let the total sheet used for making the cylindrical tank be x m2. Given, wastage .mx 2

12

according to given question, 128712

.xx

2m049511

12128712871211 ..x.x

Steel used for making closed tank including wastage = 95.04 m2 Ans.Example 13. 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. —NCERTSolution. (i) Here, r l = 308 cm2, l = 14 cm.

cm.7

222

30830814722 rr

and, (ii) Total surface area = r (r + l)

2cm14)(77722

= 22 × 21 cm2 = 462 cm2

Example 14. How many meters of cloth, 5 m wide, will be required to make a conical tent, the radius of whosebase is 7 m and height is 24 m?

Solution. Radius of the tent, r = 7 mheight of the tent, h = 24 m

slant height, m25m625m247 2222 hrl


curved surface area = r l

22 m550m257722

i.e., area of the cloth = 550 m2

Now, length of cloth required m110m5

550widtharea

length of cloth required = 110 m.Example 15. 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 Rs. 70. –NCERT

Solution. (i) here, r = 24 m, h = 10 mLet l be the slant height of the cone. then,

l2 = h2 + r2 2222 1024 rhl

m.26 676100576

(ii) Canvas required to make the conical tent = curved surface of the cone

2cm2624722

rl

Now, Rate of canvas for 1 m2 = Rs. 70

Total cost of canvas 702624722Rs. = Rs. 137280 Ans.

Example 16. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find thearea of the sheet required to make 10 such caps. –NCERT

Solution. here, radius of cap (r) = 7 cm height of cap (h) = 24 cm

Let l be the slant height. then,

cm2562549576724 2222 rhl

Sheet required for one cap = curved surface of the cone = rl

22 cm550cm257722

Sheet required for 10 such caps = 10 × 550 cm2 = 5500 cm2. Ans.Example 17. A bus stop in 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 ofeach of the cones is to be painted and the cost of painting is Rs. 12 per m2, what will be the cost

of painting all these cones? (Use = 3.14 and 021041 .. ). –NCERT

Solution. here, radius m0.2cm20cm240)( r and height (h) = 1 m

slant height m0210411040)( 22 ...hrl


Now, curved surface of 1 cone = rl 2m2012072 ..

Curved surface of 50 such cones 2m0212072250 ..

Now, cost of painting 1 m2 = Rs. 12

Total cost of painting 1.020.27225012Rs. = Rs. 384.68 (approx) Ans.

Example 18. A corn cob (see figure), shaped somewhat like a cone, has the radius of its broadest end as 2.1 cmand length as 20 cm. If each 1 cm2 of the surface of the cob carries an average of four grains, findhow many grains you would find on entire cob?

—(NCERT)

Solution. We have, r = 2.1 cm, h = 20 cmlet, slant height be l cm.

then, cm4004.41cm(20)(2.1) 2222 hrl

cm20.11cm404.41 curved surface area of corn cob = r l

2cm20.112.1722

= 132.726 cm2

Now, number of grains on 1 cm2 = 4 number of grains on 132.726 cm2 = 4 × 132.726

= 530.904 531

Hence, total number of grains on the corn cob = 531 Ans.Example 19. The surface area of a sphere is 154 cm2. Find its radius. —NCERTSolution. Let the radius of the sphere be r cm.

then, 4 r2 = 154 (given)

449

2247154

41542

r

cm27cm

449

r

radius of the sphere cm.27


Example 20. The radius of a spherical baloon increases from 7 cm to 14 cm as air is being pumped into it. Findthe ratio of surface areas of the balloon in the two cases. —NCERT

Solution. Let S1 and S2 be the total surface area in two cases of r = 7 cm and R = 14 cm. S1 = 4 r2 = 4 (7)2 cm2

and S2 = 4 R2 = 4 (14)2 cm2

Required ratio 4:1i.e.41

14144π774π

SS

2

1

Ans.

Example 21. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratioof their surface areas. –NCERT

Solution. Let the diameter of earth be R and that of the moon will be 4R

The radii of moon and earth are 2and

8RR

respectively..

Ratio of their surface area

41641

24

84

2

2

R

R

Ans.16:1.e.i161

14

641

Example 22. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find theouter curved surface area of bowl. –NCERT

Solution. Inner radius, r = 5 cmThickness of Steel = 0.25 cm Outer radius, R = (r + 0.25) cm = (5 + 0.25) cm = 5.25 cm

Outer curved surface 22 cm25525572222

..r

= 173.25 cm2

Example 23. The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respec-tively. Find the cost of painting the vessel all over at 15 paisa per cm2.

Solution. Outer radius of vessel, R = 14 cmInner radius of vessel, r = 10 cmArea of the outer surface = 2 R2

= 2 × (14)2 cm2 = 392 cm2

Area of the inner surface = 2 r2

= 2 × (10)2 cm2 = 200 cm2

Area of the ring at the top = (R2 – r2) = (142 – 102) cm2

= (14 – 10) (14 + 10) cm2

= 96 cm2

Total area to be painted =3 92 cm2 + 200 cm2 + 96 cm2

= 688 cm2

Now, cost of painting 1 cm2 = 15 paisa Rs.10015


Cost of painting 688 Rsπ68810015cm2

Rs722688

10015

= 324.34 Rs. Ans.Example 24. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold?

(1 m3 = 1000 l). –NCERTSolution. Here, l = 6m, b = 5m and h = 4.5 m

Volume of the tank = lbh = (6 × 5 × 4.5) m3 = 135 m3

The tank can hold = 135 × 1000 litres = 135000 litres of water. ( 1 m3 = 1000 litres)Example 25. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length

and depth are respectively 2.5 m and 10 m.—NCERT

Solution. Given capacity of a cuboidal tank33 m50m

10005000050000 l

Let the breadth of cuboidal tank be b m.according to given question, we have

2.5 × b × 10 = 50

225505025 bbb

breadth of the tank is 2m. Ans.Example 26. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into

the sea in a minute?—NCERT

Solution. Volume of water that flows in 1 hour (60 minutes)= volume of water of a cuboid whose dimensions are 3 m, 40 m and 2000 m.

( 2 km = 2000 m)= 3 × 40 × 2000 m3

Volume of water that flows in 1 minute

Ans.m4000 3

3m60

2000403

Example 27. Three cubes whose edges are 3 cm, 4 cm and 5 cm respectively are melted and recast into a singlecube. find the surface area of the new cube.

Solution. Let x cm be the edge of new cube. Then, volume of the new cube = sum of the volumes of threecubes. x3 = 33 + 43 + 53 = 27 + 64 + 125 = 216 = (6)3

x = 6 cm. Edge of the new cube is 6 cm.and, surface area of the new cube = 6 (6)2 cm2 = 216 cm2 Ans.

Example 28. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tankmeasuring 20 m × 15 m × 6 m. For how many days will the water of this tank last? –NCERT

Solution. Here l = 20 m, b = 15 m, and h = 6 m Capacity of the tank = lbh = (20 × 15 × 6) m3 = 1800 m3

Water requirement per person per day = 150 litres


Water required for 4000 person per day = (4000 × 150) l

33 m600m1000

1504000

Number of days the water will last

dayperrequiredwaterTotaltankofCapacity

30600

1800

Thus, the water will last for 30 days Ans.Example 29. A godown measures 60 m × 25 m × 10 m. Find the maximum number of wooden crates each

measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown. –NCERTSolution. Volume of the godown = (60 × 25 × 10) m3 = 15000 m3

Volume of 1 crate = (1.5 × 1.25 × 0.5) m3 = 0.9375 m3

Number of crate that can be stored in the godown

crate1ofVolumegodowntheofVolume

Ans.1600093750

15000.

Example 30. If the lateral surface of cylinder is 94.2 cm2 and its height is 5 cm, then find (i) radius of its base (ii)its volume (use = 3.14) . –NCERT

Solution. (i) Let r be the radius of the base and h be the height of the cylinder. Then, Lateral surface = 94.2 cm2

2rh = 94.2 2 × 3.14 × r × 5 = 94.2

351432

294

..r

Thus, the radius of its base = 3 cm.(ii) Volume of the cylinder = r2h = (3.14 × 32 × 5) cm3

= 141.3 cm3 Ans.Example 31. It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of

painting is at the rate of Rs. 20 per m2, Find : (i) inner curved surface area of the vessel (ii) radius of the base(iii) capacity of the vessel –NCERT

Solution. (i) inner curved surface area of the vessel paintingofRate

paintingofcostTotal

2m110

2m

202200


(ii) Let r be the radius of the base and h be the height of the cylindrical vessel. 2rh = 110

110107222 r

75147

102227110 .r

Thus, the radius of the base = 1.75 m

(iii) Capacity of the vessel = r2h 31047

47

722 m

= 96.25 m3

Example 32. The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres ofmetal sheet would be needed to make it? –NCERT

Solution. Capacity of a closed cylindrical vessel = 15.4 litres

33 m0.0154m1000

1415

.

Let r be the radius of the base and h be the height of the vessel. Then,Volume = r2h = r2 × 1 = r2 ( h = 1m) r2 = 0.0154

01540722 2 .r

0049022

7015402 ..r

07000490 ..r

Thus, the radius of the base of vessel = 0.07 m.Metal sheet needed to make the vessel

= Total surface area of the vessel= 2rh + 2r2 = 2r (h + r)

2m0.07)(10.077222

= 44 × 0.01 × 1.07 m2 = 0.4708 m2

Example 33. A patient in a hospital in given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filledwith soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250patients? –NCERT

Solution. Diameter of the cylindrical bowl = 7 cm

Radius cm27

Height of serving bowl = 4 cm.


Soup served to 1 patient = Volume of the bowl = r2h

3cm427

27

722

= 154 cm3

Soup served to 250 patients = (250 × 154) cm3 = 38500 cm3 = 38.5 l. Ans.Example 34. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length

of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 gm.—NCERT

Solution. here, inner radius cm12cm2

24)( r

and outer radius (R) = cm14cm228

h = length of the pipe = 35 cm. volume of wood used in making the pipe

hrhR 22

)( 22 rRh

322 cm]12[1435722

3cm12)(1412)(1435722

33 cm5720cm26235722

Now, 1 cm3 of wood = 0.6 g 5720 cm3 of wood = 0.6 × 5720 g = 3432.0 g

= 3.432 kg. Ans.Example 35. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior.

The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of thepencil is 10 cm, find the weight of the whole pencil if the specific gravity of the wood is 0.7 gm/cm3

and that of the graphite is 2.1 gm/cm3. —NCERTSolution. For graphite cylindrical rod :

radius (r) of graphite cylinder cm201cm

101

21

and, length of graphite rod (h) = 10 cm. volume of graphite cylindrical rod = r2 h

32

cm10201

722

Weight of graphite used for pencil = volume × specific gravity

gm1210201

201

722 . gm)2.1cm1( 3

= 0.165 gm.


Again, for pencil including graphite rod, we have,

radius of pencil cm207mm

27(R)

and, length of pencil (h) = 10 cm

volume of pencil = hR2

32

cm10207

722

volume of wood used for pencil

hrhR 22

33 cm101

201

201

722cm10

207

207

722

3cm48201

711

weight of wood gm0.748201

711

gm)0.7cm1( 3

= 2.64 gm. Total weight = 2.64 gm + 0.165 gm

= 2.805 gm Ans.Example 36. A well of diameter 3m is dug 14 m deep. The earth taken out of it has been spread evenly all around

it to a width of 4 m to form an embankment. Find the height of the embankment.

Solution. Radius of the well ,mr23)( height (h) = 14 m

Volume of the earth taken out of the well

332 m99m1423

23

722

hr

Outer radius of the embankment

m2

11m4m23

R

Area of embankment = outer area – inner area22 rR

23

211

23

211

722

23

211

722 22

2m8847722

Height of the embankment

Ans.m1.125 m89m

8899

AreaVolume


Example 37. The height of a cone is 15 cm. If its volume is 1570 cm3. Find the radius of the base (Use = 3.14). –NCERT

Solution. Here h = 15 cm and volume = 1570 cm3

Let the radius of the base of the cone be r cm.Now, Volume = 1570 cm3

157031 2 hr

15701514331 2 r.

1005143

15702

.

r

10100 r

Thus, the radius of the base of cone is 10 cm Ans.Example 38. The radius and height of a right circular cone are in the ratio of 5 : 12. If its volume is 2512 cm3, find

the slant height and radius of the base of the cone. (use = 3.14)Solution. Let the radius of cone be 5x and height be 12 x.

Volume of the cone hr 2

31

according to question, we get

2512)12()5(14331 2 xx.

25121225100314

31 2 xx

2831425123 xx

radius of base = 5x = 5 × 2 cm = 10 cmand, height = 12 x = 12 × 2 cm = 24 cm

slant height 266765761002410 2222 hr radius of cone = 10 cm and slant height = 26 cm Ans.

Example 39. 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

—NCERTSolution. Given, volume of cone = 9856 cm3

and, radius of base (r) cm14cm228

(i) we know, volume of cone hr 2

31

98561414722

31

h


cm.48

cm

1422298563h

(ii) Slant height of a cone (l) 22 rh

cm(14)(48) 22

cm1962304

cm50 cm2500(iii) Curved surface area of a cone

= r l

.cm2200 2 2cm5014722

Example 40. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find itsvolume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvasrequired. –NCERT

Solution. Diameter of the base of the cone = 10.5 m

radius m5.25m2

510

.r

Height of the cone = 3m

Volume of the cone hr 2

31 3m3255255

722

31

.. = 86.625 m3

To find the slant height l

We have, 22222 (5.25)3 rhl = 9 + 27.5625 = 36.5625

(approx.)m04676562536 ..l

Canvas required to protect wheat from rain = Curved surface area

2m04676255722

..rl = 99.77 m2 (approx)

Example 41. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find thevolume of the solid so obtained?

Solution. The solid obtained is a cone with r = 5 cm and h = 12 cm.

Volume hr 2

31

3cm12553.1431

3cm3.14100

= 314 cm3 Ans.


Example 42. If the surface area of a sphere is 616 cm2, find its volume.Solution. Let r be the radius of sphere.

then, 4 r2 = 616

49224

761646162

r

cm7cm49 r

volume of sphere 33 cm777722

34

34

r

= 1437.3 cm3 Ans.Example 43. The diameter of the moon is approximately one-fourth the diameter of the earth. What fraction of

the volume of the earth is the volume of the moon?–NCERT

Solution. Let the radius of earth be R.

then, radius of moon .R4

volume of earth 3

34 R

and, volume of moon .R 3

434

164

434

34

moonofvolumeearthofvolume

3

3

R

R

i.e., volume of earth is 64 times the volume of the moon.

i.e., volume of moon is 641

times that of earth.

Example 44. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is8.9 g per cm3? –NCERT

Solution. Diameter of the ball = 4.2 cm

Radius cm2.1cm2

4.2

Volume of the ball 3

34 r

33 cm38.808cm2.12.12.1722

34

Now, Density of metal = 8.9 gm per cm3

Mass of the ball = 38.808 × 8.9 g = 345.3912 g = 345.4 g (approx) Ans.


Example 45. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m, then find thevolume of the iron used to make the tank. –NCERT

Solution. Let R cm and r cm be respectively the external and internal radii of the hemispherical vessel, thenR = 1.01 m and r = 1 m ( as thickness = 1cm = 0.01 m)

Volume of iron used= External volume – Internal volume

)(32

32

32 3333 rRrR

333 m](1)–[(1.01)722

32

33 m0.0303012144m1)(1.030301

2144

= 0.06348 m3 (approx)Example 46. A dome of a building is in the form of hemi-sphere. From inside, it was white-washed at the cost

of Rs. 498.96. If the cost of white-washing is Rs. 2 per square meter, find: (i) the inside surface area of the dome(ii) the volume of the air inside the dome.

–NCERTSolution. Let r be the inner radius of the hemispherical dome. Then, inside surface area of the hemisphere

= 2 r2.Since, at the rate of Rs. 2 per square metre, the total cost of white-wash is Rs. 498.96,

surface area of hemisphere 22 m249.48m2

498.96

according to question, 482492 2 .r

6939222

7482492 ..r

m6.3m6939 .r(i) Inside surface are of the dome

= 2 r2 = 249.48 m2 (calculated above)(ii) Inside volume of the dome

333 m(6.3)722

32

32

r

= 523.908 m3 Ans.Example 47. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere

with surface are S. Find the:(i) radius r of the new sphere.(ii) ratio of S and S. –NCERT

Solution. (i) Volume of 27 solid spheres of radius 3

3427 rr ...(1)

Volume of the new sphere of radius 3

34

rr ...(2)


According to the problem, we have,

33

3427

34 rr

rr 3 333 )3(27 rrr

(ii) Required ratio 9:1

91

9)3(44

2

2

2

2

2

2

rr

rr

rr

SS

Example 48. A wooden bookshelf has external dimensions as follows : Height = 110 cm, Depth = 25 cm, Breadth= 85 cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to bepolished and the inner faces are to be painted. If the rate of polishing is 20 paisa per cm2 and therate of painting is 10 paisa per cm2, find the total expenses required for polishing and painting thesurface of the bookshelf. –NCERT

Solution. Area to be polished= (110 × 85 + 2 × 85 × 25 + 2 × 25 × 110 + 4 × 75 × 5 + 2 × 110 × 5) cm2

= (9350 + 4250 + 5500 + 1500 + 1100) cm2 = 21700 cm2

cost of polishing @ 20 paisa per cm2 = 4340Rs.1002021700

Also, Area to be painted = (6 × 75 × 20 + 2 × 90 × 20 + 75 × 90) cm2

= (9000 + 3600 + 6750) cm2 = 19350 cm2

Cost of painting @ 10 paisa per cm2

1935Rs.Rs.1001019350

Total expense = Rs. 4340 + Rs. 1935 = Rs. 6275 Ans.Example 49. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on

small supports as shown in figure. Eight such spheres are used for this purpose, and are to bepainted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be paintedblack. Find the cost of paint required if silver paint costs 25 paisa per cm2 and black paint costs 5paisa per cm2. –NCERT


Solution. Clearly, we have to subtract the area of the circle on which sphere is resting while calculating thecost of silver paint.Surface area to be painted silver

= 8 (curved surface area of the sphere – area of circle on which sphere is resting)

= 8 (4 R2 – r2) where cm51cm221 .r,R

222 cm(438.75)8πcm2.25)(4418πcm2.254

44148π

Cost of silver paint @ 25 paisa per cm2

(approx)2757.86Rs.10025438.75

7228Rs.

Again, surface area to be painted black= 8 × curved surface area of cylinder

= 8 × 2rh = 22 cm528cm71.572228

Cost of black paint @ 5 paisa per cm2

26.40Rs.100

5528Rs.

Total cost of painting = Rs. 2757.86 + Rs. 26.40 = Rs. 2784.26 (approx) Ans.

Example 50. The diameter of a sphere is decreased by 25%. By what percent does its curved surface decrease? –NCERT

Solution. Let d be the diameter of the sphere. Then, its surface area 22

24 dd

On decreasing its diameter by 25%,

New diameter ddddd%dd43

10075

10025of251


New surface area 22

1

43

214

24

dd

22

169

6494 d.d.

Decrease in surface area

2222

167

1691

169 dddd

Percentage decrease in surface area

100%areasurfaceoriginal

areasurfaceindecrease

%%d

d100

16710016

7

2

2

= 43.75% Ans.

PRACTICE EXERCISEQuestions based on Surface area of cuboid and cube

1. Find the surface area of a cuboid 16 m long, 14 m broad and 7 m high.2. Find the length of the longest pole that can be placed in a room 12 m long, 8 m broad and 9 m high.3. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white-

washing the walls of the room and the ceiling at the rate of Rs. 8.50 per m2.4. Find the percentage increase in the surface area of a cube when each side is doubled.5. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of

dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container? —NCERT6. A small indoor green house (herbarium) is made entirely of glass panes (including base) held together

with tape. It is 30 cm long, 25 cm wide and 25 cm high.(i) What is the area of the glass?(ii) How much of tape is needed for all the 12 edges? —NCERT

7. Three cubes each of side 6 cm are joined end to end. Find the surface of the resulting cuboid.8. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at top. Ignoring the

thickness of the plastic sheet, find:(i) area of the sheet required to make the box.(ii) the cost of the sheet for it, if a sheet measuring 1 m2 cost Rs. 22.

9. If the surface area of the cube is 96 cm2, find its edge and length of its diagonal.10. The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of

covering it with sheet of paper at the rate of Rs. 4 and Rs. 4.50 per m2 is Rs. 650. Find the dimensions ofthe box.

11. Mary wants to decorate her christmas tree. She wants to place the tree on a wooden box covered withcoloured paper with picture of Santaclaus on it (see figure). She must know the exact quantity of paperto buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively,how many square sheets of paper of side 40 cm would she require? —NCERT


12. The length and breadth of a hall are in the ratio 4 : 3 and its height is 5.5 meters. The cost of decoratingits walls (including doors and windows) at Rs. 6.60 per m2 is Rs. 5082. Find the length and breadth of theroom.

Questions based on Surface area of a cylinder

13. The curved surface area of a right circular cylinder of height 14 cm is 176 cm2. Find the diameter of thebase of the cylinder.

14. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm.(see figure). Find its : —NCERT

(i) inner curved surface area(ii) outer curved surface area(iii) Total surface area

15. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find: (i) its inner curved surface area,(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2.

16. An iron pipe 20 cm long has exterior diameter 25 cm. If the thickness of the pipe is 1 cm, find the totalsurface area of the pipe.

17. A rectangular sheet of paper 88 cm × 50 cm is rolled along its length and a cylinder is formed. Find curvedsurface area of the cylinder formed.

18. A solid cylinder has total surface area of 462 cm2. Its curved surface area is one-third of its total surfacearea. Find the radius and height of the cylinder.

19. The total surface area of a hollow metal cylinder, open at both the ends of external radius 8 cm and height10 cm is 338 cm2. Find thickness of the metal in the cylinder.


20. In the given figure, you see the frame of a lamp shade. It is to be covered with a decorative cloth. Theframe has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it overthe top and bottom of the frame. Find how much cloth is required for covering the lampshade.

—NCERT

Question based on surface area of a Cone

21. Find the curved surface area of a cone, if its slant height is 50 cm and the diameter of its base is 28 cm.22. Find the total surface area of a cone, if its slant height is 21 cm and diameter of its base is 24 cm.

—NCERT

23. The curved surface area of a cone is 4070 cm2 and its radius is 35 cm. What is its slant height?

722πuse

24. The radius and slant height of a cone are in the ratio 4 : 7. If its curved surface area is 792 cm2, find its

radius.

722πuse

25. The circumference of the base of a 10 m high conical tent is 44 m. Calculate the length of canvas used in

making the tent if width of canvas is 2 m.

722πuse

26. 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 cuttingis approximately 20 cm. (use = 3.14) —NCERT

27. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost ofwhite-washing its curved surface at the rate of Rs. 210 per 100 m2? —NCERT

28. The curved surface area of a cone of radius 6 cm is 188.4 cm2. Find its height.

Questions based on surface area of sphere and hemi-sphere

29. The surface area of a sphere is 5544 cm2. Find its radius.30. A hemi-spherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the

inside at the rate of Rs. 16 per 100 cm2. —NCERT


31. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of theirsurface areas. —NCERT

32. In the given figure, a right cylinder just encloses a sphere of radius r. Find

(i) Surface area of the sphere (ii) curved surface area of the cylinder(iii) ratio of the areas obtained in (i) and (ii).

33. The internal and external diameters of a hollow hemi-spherical vessel are 20 cm and 28 cm respectively.Find the cost of painting the vessel all over at 15 paisa per cm2.

34. A toy is in the form of a cone mounted on a hemi-sphere. The diameter of the base and the height of thecone are 6 cm and 4 cm respectively. Find the surface area of the toy. (use = 3.14)

Question based on Volume of Cuboid and Cube

35. The total surface area of a cube is 1350 cm2. Find its volume.36. 500 persons took dip in a rectangular tank which is 80 m long and 50 m broad. What is the rise in the level

of water in the tank, if the average displacement of water by a person is 4 m3?37. Three cubes of a metal with edges 6 cm, 8 cm and 10 cm respectively are melted and formed into a single

cube. Find the edge of the new cube formed. Also, find its volume.38. The volume of a cuboid is 1536 m3. Its length is 16 m, and its breadth and height are in the ratio 3 : 2. Find

surface area of the cuboid.39. A field is 70 m long and 40 m broad. In one corner of the field, a pit which is 10 m long, 8 m broad and 5

m deep, has been dug out. The earth taken out of it is evenly spread over the remaining part of the field.Find the rise in the level of the field.

40. The areas of three adjacent faces of a cuboid are 15 cm2, 40 cm2 and 24 cm2. Find the volume of thecuboid.

41. How many bricks, each measuring 25 cm × 15 cm × 8 cm will be required to build a wall 10 m × 4 dm × 5 mwhen one-tenth of its volume is occupied by mortar?

42. A rectangular reservoir is 120 m long and 75 m wide. At what speed per hour must water flow into itthrough a square pipe of 20 cm wide so that the water rises by 2.4 m in 18 hours.

Question based on Volume of a Cylinder

43. The radius of a cylinder is 14 cm and its height is 40 cm. Find (i) curved surface area (ii) the total surfacearea (iii) volume of the cylinder.

44. The total surface area of a cylinder is 462 cm2. Its curved surface is one-third of its total surface area. Findthe volume of the cylinder.

45. The curved surface area and the volume of a pillar are 264 m2 and 396 m3 respectively. Find the diameterand the height of the pillar.


46. The sum of the height and radius of the base of a solid cylinder is 37 m. If the total surface area of thecylinder is 1628 m2, find its volume.

47. A cylindrical tube, open at both ends, is made up of metal. The internal radius of the tube is 5.2 cm andits length is 25 cm. The thickness of the metal is 8 mm. Calculate the volume of the metal.

48. A soft drink is available in two packs : (i) a tin can with a rectangular base of length 5 cm and width 4 cm,having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.Which container has greater capacity and by how much? —NCERT

49. If the diameter of the cross-section of a wire is decreased by 5%, how much percent will the length beincreased so that the volume remains the same? —NCERT

50. Water flows out through a circular pipe, whose internal diameter is 2 cm, at the rate of 70 cm per secondinto a cylindrical tank, the radius of whose base is 40 cm. By how much time will the level of water rise inhalf an hour?

51. A rectangular piece of paper is 22 cm long and 12 cm wide. A cylinder is formed by rolling the paper alongits length. Find the volume of the cylinder.

52. A well, with inner radius 4m, is dug 14 m deep. The earth taken out of it has been spread evenly all round

it to a width of 3m to form an embankement. Find the height of this embankement.

722πuse

Question based on Volume of a Cone

53. The base radii of two cones of the same height are in the ratio 3 : 4. Find the ratio of their volumes.

54. A cone of height 24 cm has curved surface area 550 cm2. Find its volume.

722πuse

55. The radius and height of a right circular cone are in the ratio of 5 : 12. If its volume is 314 cm3, find the slantheight and radius of the base of the cone. (use = 31.4)

56. Find the slant height and curved surface area of a cone whose volume is 12935 cm3 and the radius of thebase is 21 cm.

57. A semi-circular thin sheet of metal of diameter 28 cm is bent and an open conical cup is made. Find thecapacity of the cup.

58. Monica has a piece of canvas whose area is 551 m2. She uses it to have a conical tent made with a baseradius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amountsto approximately 1 m2, find the volume of the tent that can be made with it. —NCERT

59. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm.60. A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is

made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curvedsurface area.

Question based on Volume of a Sphere and Hemi-Sphere

61. Find the surface area of a sphere whose volume is 606.375 m3.62. A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameter 0.6 cm.

Find the number of small balls thus obtained.63. The surface areas of two spheres are in the ratio 1 : 4. Find the ratio of their volumes.64. The diameter of a metallic sphere is 6 cm. It is melted and drawn into a wire having diameter of the cross-

section as 2 mm. Find the length of the wire.


65. The radii of the internal and external surfaces of a metallic spherical shell are 3 cm and 5 cm respectively.

It is melted and recast into a solid cylinder of height cm.3210 Find the diameter of the base of the

cylinder.66. The largest sphere is carved out of a cube of side 7 cm. Find the volume of the sphere. (use = 3.14)67. Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.68. A hemi-spherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. Find the volume

of the steel used in making it.69. A hemi-spherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical

shaped small bottles of diameter 3 cm and height 4 cm. How many bottles are required to empty the bowl?70. A hemi-sphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Find the height

of the cone.

Miscellaneous Questions

71. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume

be 271

of the volume of the given cone, at what height above the base is the section made?

72. A circus tent consists of a cylindrical base surmounted by a conical roof. The radius of the cylinder is 20m. The height of the tent is 63 m and that of the cylindrical base is 42 m. Find the volume of air containedin the tent and the area of canvas used for making it.

73. How many litres of water flows out of pipe having an area of cross-section of 5 cm2 in one minute, if thespeed of water in the pipe is 30 cm/sec ?

74. A sphere of diameter 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of thevessel is 6 cm. If the sphere is completely submerged in water, find by how much will the surface level ofwater be raised.

75. An ice-cream cone has a hemispherical top. If the height of the conical portion is 9 cm and base radius

2.5 cm, find the volume of ice-cream in the ice-cream cone.

722πuse

76. In the given figure, a solid is made of a cylinder with hemispherical ends. If the entire length of the solidis 108 cm and the diameter of the hemispherical ends is 36 cm, find the cost of polishing the surface of thesolid at the rate of 7 paisa per cm2.

77. A spherical copper ball of diameter 9 cm is melted and drawn into a wire, the diameter of whose thicknessis 2 mm. Find the length of the wire in meters.

78. The difference between the inside and outside surfaces of a cylindrical water pipe 14 m long is 88 m2. Ifthe volume of pipe be 176 m3. Find the inner and outer radii of the water pipe.


79. Water is flowing at the rate of 2.5 km/hr through a circular pipe 20 cm internal diameter, into a circularcistern of diameter 20 m and depth 2.5 m. In how much time will the cistern be filled?

80. A conical vessel of radius 6 cm and height 8 cm is filled with water. A sphere is lowered into the water (seefigure), and its size is such that when it touches the sides of the conical vessel, it is just immersed. Howmuch water will remain in the cone after the overflow?

PRACTICE TESTM.M. : 30 Time : 1 hour


Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.

1. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate ofRs. 10 per m2 is Rs. 15000, find the height of the hall.

2. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surfaceof the pillar at the rate of Rs. 1250 per m2.

3. A right triangle PQR with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volumeof the solid so obtained.

4. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?5. A joker’s cap is in the form of a right circular cone of base 7 cm and height 24 cm. Find the area of the

sheet required to make 10 such caps.6. 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.7. A village, having a population of 4000, requires 150 l of water per head per day. It has a tank measuring

20 m × 15 m × 6 m. For how many days will the water of this tank last ?8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with

soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?9. The radius and height of a cone are in the ratio 4 : 3. The area of the base is 154 cm2. Find the area of the

curved surface.10. The diameter of a sphere is decreased by 25%. By what percent its curved surface area decrease?

ANSWERS OF PRACTICE EXERCISE1. 868 cm2 2. 17 m 3. Rs. 629 4. 300%

5. 100 6. (i) 4250 cm2 (ii) 320 cm 7. 360 cm2 8. (i) 5.45 m2 (ii) 119.90 Rs.

9. 4 cm, cm34 10. 10 m, 15 m, 20 m 11. 7 12. 40 m, 30 m

13. 4 cm 14. (i) 968 cm2 (ii) 1064.80 cm2 (iii) 2038.08 cm2 15. (i) 110 m2 (ii) Rs. 4400


16. 3168 cm2 17. 440 cm2 18. r = 7 cm, h = 14 cm 19. 3 cm

20. 2200 cm2 21. 2200 cm2 22. 1244.57 cm2 23. 37 cm

24. 12 cm 25. 134.2 m 26. 63 m 27. Rs. 1155

28. 8 cm 29. 21 cm 30. Rs. 27.72 31. 1 : 16

32. (i) 4 r2 (ii) 4r2 (iii) 1 : 1 33. 324.34 Rs 34. 103.62 cm2

35. 3375 cm3 36. 50 cm 37. 12 cm, 1728 cm3 38. 832 cm2

39. 14.7 cm 40. 120 cm3 41. 6000 42. 30 km/hr

43. (i) 3520 cm2 (ii) 4752 cm2 (iii) 24640 cm3 44. 539 cm3 45. 6 m, 14 m

46. 4620 m3 47. 704 cm3 48. cylindrical tin, 85 cm2 49. 10.8%

50. 78.75 cm 51. 462 m3 52. 6.8 m (approx) 53. 9 : 16

54. 1232 cm3 55. 13 cm, 5 cm 56. 35 cm, 2310 cm2 57. 622.3 cm2

58. 1232 m3 59. 190.93 cm3 60. 415.8 cm3, 233.9 cm2 61. 346.5 m2

62. 1000 63. 1 : 8 64. 36 m 65. 7 cm

66. 179.5 cm3 67. 6 : 68. 56.83 cm3 69. 54

70. 28.44 cm 71. 20 cm 72. 61600 m3, 7102.85 m2 73. 9 l

74. 1 cm 75. 91.66 cm3 76. Rs. 855.36 77. 121.5 m

78. 1.5 m, 2.5 m 79. 10 hours 80. 188.57 cm3

ANSWERS OF PRACTICE TEST1. 6 m 2. Rs. 68.75 3. 100 cm3 4. 0.303 l (approx.)

5. 5500 cm2 6. 173.25 cm2 7. 3 days 8. 38.5 l

9. 192.5 cm2 10. 43.75%

MATHEMATICS–IX STATISTICS 173

CHAPTER 14STATISTICS

Points to Remember :1. Facts or figures, collected with a definite pupose, are called Data.2. Statistics is the area of study dealing with the collection, presentantion, analysis and interpretation of

data.3. The data collected by the investigator himself with a definite objective in mind are known as Primary

data.4. The data collected by somone else, other than the investigator, are known as Secondary data.5. Any character which is capable of taking reversal different values is called a variable.6. Each group into which the raw data are condensed is known as class-interval. Each class is bounded by

two figures known as its limits. The figure on the left is lower limit and figure on the right is upper limit.7. The difference between true upper limit and true lower limit of a class is known as its class-size.

8. Mid-value of a class (or class mark) = 2

limitlowerlimitupper

9. Class size is the difference between any two successive class marks (mid-values).10. The difference between the maximum value and the minimum value of the variable is known as Range.11. The count of number of observations in a particular class is known as its Frequency.12. The data can be presented graphically in the form of bar graphs, histograms and frequency polygons.13. The three measures of central tendency for an ungrouped data are :

(i) Mean : It is found by adding all the values of the observations and dividing it by the total number ofobservations. It is denoted by x .

So,n

x

nx........xxx

n

ii

n

121 .

For an ungrouped frequency distribution,

n

ii

n

iii

n

nn

f

xf

f........ffxf......xfxfx

1

1

21

2211

(ii) Median : It is the value of the middle-most observation(s).

If n is an odd number, then median = value of the thn

21

observation.

and, if n is an even number, then median = mean of the values of thth

12

and2

nn

observations.

(iii) Mode : The mode is the most frequently occurring observation.Empirical formula for calculating mode is given by, Mode = 3 (Median) – 2 (Mean)

174 STATISTICS MATHEMATICS–IX


Example 1. The relative humidity (in %) of a certain city for a month of 30 days was as follows :98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.189.2 92.3 97.1 93.5 82.7 95.1 97.2 93.3 95.2 97.396.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89.0 (i) Construct a grouped frequency distribution table with classes 84-86, 86-88 etc. (ii) Which month or season do you think this data is about ?(iii) What is the range of this data? —NCERT

Solution. (i) Frequency distribution table

30Total41009879896696947949229290290881888618684

FrequencyMarksTally%)(inhumidityRelative

|||||||||||||||||||||

||||||

(ii) Month-June or season-Monsoon(iii) Range = Maximum observation – minimum observation

= 99.2 – 84.9 = 14.3Example 2. The value of upto 50 decimal places is given below :

3.14159265358979323846264338327950288419716939937510(i) List the digits from 0 to 9 and make a frequency distribution of the digits after the decimal point.(ii) What are the most and the least frequency occurring digits? —NCERT

Solution. (i) Frequency distribution table

5089584746554483525120

Total|||||||

||||||||||||||||||||

|||||||||||||||||

FrequencyMarksTallyDigit

(ii) Most frequency occuring digits are 3 and 9, and least occurring digit is 0.


Example 3. The following table gives the life times of 400 neon lamps:

481000-90062900-80074800-70086700-60060600-50056500-40014400-300

)( lampsofNo.hoursintimeLife

(i) Represent the given information with the help of a histogram.(ii) How many lamps have a life time of more than 700 hours?

—NCERTSolution. (i)

(ii) No. of lamps having life time more than 700 hours= 74 + 62 + 48 = 184

Example 4. The following two tables give the distribution of students of two sections according to the marksobtained by them :

Section A Section BMarks Frequency Marks Frequency

0-10 3 0-10 510-20 9 10-20 1920-30 17 20-30 1530-40 12 30-40 1040-50 9 40-50 1

Represent the marks of the students of both the sections on the same graph by frequencypolygon.


Solution. Required frquency polygon is as follows :

Example 5. Draw histogram of the weekly pocket expenses of 125 students of a school given below :

05100-903090-602560-504050-301530-201020-10

)( studentsofNo.Rs.inensesexppocketWeekly

Solution. Here, we observe that class intervals are unequal, so we will first adjust the frequencies of eachclass interval. Here, the minimum class size is 10.We know, Adjusted frequency of a class interval

classtheoffrequencysizeclass

sizeclassMinimum

The adjusted frequency of each class interval is given below :

055101005100-90

103030103090-60

252510102560-50

204020104050-30

151510101530-20

101010101020-10

frequencyAdjustedFrequencyexpensespocketWeekly


So, required histogram is given below.

Example 6. In a mathematics test given to 15 students, the following marks (out of 100) are recorded :41, 48, 39, 46, 52, 54, 62, 40, 96, 52, 98, 40, 42, 52, 60.Find the mean, median and mode of the above marks.

Solution. (i) Mean 15

)(

15

1 i

ixx

1560524240985296406256545246394841

85415822 .

(ii) Arranging the data in the ascending order :

39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98

here, n = 15, which is odd.

median = value of thn

21 observation

th

2

115 observation = 8th observation = 52

(iii) Since, 52 occurs most frequently i.e. 3 times, so mode is 52.


Example 7. Find the mean salary of 60 workers of a factory from the following table: —NCERT

11000039000480006700086000

105000124000163000

)( workersofNo.RsinSalary

Solution.

30500060Total100001100002700039000320004800042000670004800086000500001050004800012400048000163000

)(

iii

iiii

xff

xffworkersofNo.x.RsinSalary

Mean Ans.5083.33Rs.

60

305000)(i

ii

fxf

x

Example 8. The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excludednumber.

Solution. Here, n = 5, .x 18

Now, 90185

ii x

nxx

So, total of 5 numbers is 90.Let the excluded number be a. Then, total of 4 numbers is 90 – a.

Mean of 4 numbers 4

90 a

164

90

a( Given, new mean = 16)

90 – a = 64 a = 26 the excluded number is 26. Ans.


Example 9. The median of the observations 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending orderis 24. Find x.

Solution. Here, n = 10. Since n is even :

Median 2

nobservatio12

nobservatio2

thth nn

2nobservatio6thnobservatio5th24

324242

622

)4()2(24

xxxx

x = 21 Ans.Example 10. Find the mode for the following data :

14, 25, 28, 14, 18, 17, 18, 14, 23, 22, 14, 18.Solution. Arranging the given data in ascending order:

14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28Since, 14 occurs maximum number of times (4 times), 14 is the required mode.

PRACTICE EXERCISE1. Construct a frequency table for the following ages (in years) of 30 students using equal class intervals,

one of them being 9-12, where 12 is not included.18, 12, 7, 6, 11, 15, 21, 9, 13, 8, 15, 17, 19, 22, 14, 21, 8, 23, 12, 17, 6, 18, 15, 23, 16,22, 9, 21, 16, 11

2. The electricity bills (in Rs.) of 40 houses in a locality are given below :116, 127, 100, 107, 80, 82, 65, 91, 101, 95, 87, 105, 81, 129, 92, 75, 78, 89, 61, 121, 128, 63, 76, 84, 62, 98, 65, 95,108, 115, 101, 65, 52, 59, 81, 87, 130, 118, 108, 116Construct a grouped frequency table.

3. For the following data of weekly wages (in Rs.) received by 30 workers in a factory, construct a groupedfrequency distribution table.258, 215, 320, 300, 290, 311, 242, 272, 268, 210, 242, 258, 268, 220, 210, 240, 280, 316, 306,215, 236, 319, 304, 278, 254, 292, 306, 332, 318, 300

4. Construct a frequency table, with equal class-intervals from the following data on the weekly wages(in Rs.) of 25 labourers working in a factory, taking one of the class-intervals as 460-500 (500 notincluded).580, 625, 485, 537, 540, 425, 637, 605, 607, 430, 611, 632, 600, 640, 638, 612, 584, 440, 536,515, 449, 480, 556, 561, 508

5. Given below are two cumulative frequency distribution tables. Form a frequency distribution table foreach of these.

(i) (ii)

8060Below7350Below6040Below3930Below2820Below1510Belowpersonsof.No)yearsin(Ages

6060thanMore5250thanMore3940thanMore2730thanMore1720thanMore010thanMoreStudentsof.NoobtainedMarks


6. On a certain day, the temperature in a city was recorded as under :

1822262420)(p.m.6p.m.3a.m.11a.m.8a.m.5

CineTemperaturTime

Draw a bar graph to represent the above data.7. Read the bar graph given below and answer the questions given below :

(i) What information is given by the bar graph? (ii) In which year was the production maximum?(iii) After which year was there a sudden fall in the production?(iv) Find the ratio between the maximum and minimum production during the given period.

8. The table given below shows the number of blinds in a village :

30Total21008048060

106040940205200

blindsof.NogroupAge

Draw a histogram to represent the above data.9. The following table shows the average daily earnings of 40 general stores in a market, during a certain

week.

40Total4900-850

12850-8007800-7502750-700

10700-6505650-600

)( storesof.No.RsinearningDaily


Draw a histogram to represent the above data.10. The following table gives the heights of 50 students of a class. Draw a frequency polygon to represent

this.

50Total1591-1807801-16523651-15018501-1351351-120

)( Studentsof.NocminHeight

11. In a study of diabetic patients in a village, the following observations were noted. Represent the givendata by a frequency polygon.

53Total470-60960-502050-401240-30630-20220-10

)( Patientsof.NoYearsinAge

12. Draw a histogram and a frequency polygon on the same graph to represent the following data :

135Total1009-803008-704007-602506-503050-40

)( Personsof.NocminWeight

13. Draw a histogram to represent the following frequency distribution.

1808-50805-30

1003-20902-15615-10

FrequencyIntervalClass


14. Draw a histogram for the marks of students given below :

74Total606-50

1005-451854-303203-10801-0Studentsof.NoMarks

15. The runs scored by two teams A and B on the first 120 balls in a cricket match are given below :

420120-1081216108-9620896-8412684-72101272-6081060-48

182048-3616436-2421224-1241012-0

BTeamATeamballsofNumber

Represent the data of both the teams on the same graph with the help of frequency polygons.16. Find the mean for each of the following sets of numbers :

(i) 25, 12, 37, 19, 43, 40, 11 (ii) 6.2, 4.9, 7.1, 2.9, 5.7, 8.3(iii) 12, 22, 32, 42, 52, 62 (iv) 13, 23, 33, 43, 53

17. Calculate the mean ( x ) for each of the following distribution :

(i) 62531108642

fx

(ii) 31523630252015105

fx

18. The following table shows the number of accidents met by 120 workers in a factory during a month :

1325213436543210

workersofNo.accidentsofNo.

Find the average number of accidents per workers.19. The marks obtained out of 50 by 80 students in a test are given below :

437181216785353833302524222015

StudentsofNo.Marks

Calculate the average marks.20. If the mean of the following data is 18.75, find the value of p.

28710530251510

i

i

fpx


21. The average of height of 30 boys out of a class of 50, is 160 cm. If the average height of the remainingboys is 165 cm, find the average height of the whole class.

22. The average of six numbers is 30. If the average of first four is 25 and that of last three is 35, find thefourth number.

23. The mean of 100 observations was calculated as 40. It was found later on that one of the observationswas misread as 83 instead of 53. Find the corrected mean.

24. The mean of 10 numbers is 18. If 3 is subtracted from every number, what will be the new mean?

25. If x is the mean of n observations x1, x2, ...., xn, then prove that

n

ii xx

10)( i.e. the algebraic sum of

deviations from mean is zero.26. Find the median of following data :

(i) 14, 6, 18, 9, 23, 22, 10, 19, 24(ii) 17, 13, 28, 19, 23, 22, 12, 32

27. The numbers 5, 7, 10, 12, 2x – 8, 2x + 10, 35, 41, 42, 50 are arranged in ascending order. If their median is25, find the value of x.

28. Find the median of the following observations :46, 64, 58, 87, 41, 77, 35, 55, 90, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the newmedian.

29. Find out the mode of the following data :(i) 14, 28, 19, 25, 14, 31, 17, 14, 12, 27(ii) 8.3, 8.9, 8.1, 8.7, 8.9, 7.9, 8.7, 8.9, 8.1

30. Given below is the number of pairs of shoes of different sizes sold in a day by the owner of the shop.

196554322987654321

soldpairsof.NoshoeofSize

What is the modal shoe size?

PRACTICE TEST


General Instructions :Each question carry 3 marks.

1. Three coins were tossed 30 times simultaneously. Each time the number of heads occuring was noteddown as follows:0 1 2 2 1 2 3 1 3 01 3 1 1 2 2 0 1 2 13 0 0 1 1 2 3 2 2 0Prepare a frequency distribution table for this data.

2. A random survey of the number of children of various age groups playing in a park was found as follows:

41091263517-1515-1010-77-55-33-22-1

childrenof.No)yearsin(Age

Draw a histogram to represent the data above.


3. Find mean ( x ) for the following distribution:

137423353025201510

fx

4. The following observations have been arranged in ascending order. If the median of the data is 63, findthe value of x.29, 32, 48, 50, x, x + 2, 72, 78, 84, 95

5. The marks obtained by 80 students in a test are given. Find the modal marks.

815241510844362820124

studentsof.NoMarks


1.737445

24-2121-1818-1515-1212-99-6Frequency

Class

2.144758362140-130130-120120-110110-100100-9090-8080-7070-6060-50

FrequencyClass

3.2934624

340-320320-300300-280280-260260-240240-220220-200

workersofNo.

.)Rsin(WagesWeekly

4.573424

660-620620-580580-540540-500500-460460-420

workersofNo.

.)Rsin(WagesWeekly

5. (i) 7132111131560-5050-4040-3030-2020-1010-0)(

personsofNo.yearinAge

(ii) 171012138100-8080-6060-4040-2020-0

StudentsofNo.obtainedMarks


6.

7. (i) The given bar graph shows the production (in million tonnes) of food grains during the period from2000 to 2004.

(ii) 2002 (iii) 2000 (iv) 5 : 2

8.

9.

10. 11.


12. 13.

14.

15.

0

6

9

5

2

3


16. (i) 26.71 (ii) 5.85 (iii) 15.16 (iv) 4517. (i) 7.05 (ii) 15.25 18. 1.4 19. 26.7 (approx) 20. 2021. 162 cm 22. 25 23. 39.7 24. 1526. (i) 18 (ii) 20.5 27. 12 28. 58, 5829. (i) 14 (ii) 8.9 30. 8

ANSWERS OF PRACTICE TEST3. 22 4. 62 5. 28

188 PROBABILITY MATHEMATICS–IX

CHAPTER 15PROBABILITY

Points to Remember :1. An activity which gives a result is called an experiment.2. An experiment which can be repeated a number of times under the same set of conditions, and the

outcomes are not predictable is called a Random Experiment.3. Performing an experiment is called a trial.4. Any outcome of an experiment is known as an event.5. In n trials of a random experiments, if an event E happens m times, then the probability of happening of

E is given by,

nmEEP

outcomespossibleofnumberTotaltofavouroutcomesofNumber)(

6. For any event E, which is associated to an experiment, we have .EP 1)(0

7. If E1, E2, E3, ....., En are n elemantary events associated to a random experiment, thenP(E1) + P(E2) + P(E3) + ........... + P(En) = 1


Example 1. In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find theprobability that she did not hit the boundary.

—NCERTSolution. Since, she hits a boundary 6 times, it means she has missed 30 – 6 = 24 times.

Required probability= P (she did not hit the boundary)

playssheballsofnumberTotalboundarythehitnotdidshetimeofNumber

Ans.0.854

3024

Example 2. Three coins are tossed simultaneously 200 times with the following frequencies of differentoutcomes:

28777223Frequencyheadsnoheads1heads2heads3Outcome

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.—NCERT

Solution. Out of total of 200 outcomes, 2 heads come for 72 times.

P (2 heads coming up) Ans.259

20072

MATHEMATICS–IX PROBABILITY 189

Example 3. An organisation selected 2400 families at random and surveyed them to determine a relationshipbetween income level and the number of vehicles in a family. The information gathered is listed inthe table below:

Monthly income Vehicles per family (in Rs.) 0 1 2 Above 2

Less than 7000 10 160 25 0 7000-10000 0 305 27 210000-13000 1 535 29 113000-16000 2 469 59 2516000 or more 1 579 82 88

Suppose a family is chosen. Find the probability that the family chosen is : (i) earning Rs. 10000-13000 per month and owning exactly 2 vehicles. (ii) earning Rs. 16000 or more per month and owning exactly 1 vehicle.(iii) earning less than Rs. 7000 per month and does not own any vehicle. (iv) earning Rs. 13000-16000 per month and owning more than 2 vehicles. (v) owning not more than 1 vehicle. —NCERT

Solution. The total number of families = 2400(i) Number of families earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles = 29

Required probability 2400

29

(ii) Number of families earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579


2400579

(iii) Number of families earning less than Rs. 7000 per month and does not own any vehicle = 10


240010

(iv) Number of families earning Rs. 13000 – 16000 per month are owning more than 2 vehicles = 25.


2400

25

(v) Number of families owning not more than 1 vehicle= families having no vehicle + families having 1 vehicle= (10 + 0 + 1 + 2 + 1) + (160 + 305 + 535 + 469 + 579)= 2162


24002162

Example 4. A die is throw 300 times and the outcomes are noted in as given below :

483530474570Frequency654321Outcome

If a die is thrown at random, find the probability of getting.(i) 1 (ii) 2 (iii) 3 (iv) 4 (v) 5 (vi) 6


Solution. Total number of trials = 300

(i) P (getting 1) 0.23307

30070

(ii) P (getting 2) 0.1530045

(iii) P (getting 3) 0.16 156030047 .

(iv) P (getting 4) 0.130030

(v) P (getting 5) 0.11630035

(iv) P (getting 6) 0.1630048

Example 5. The table given below shows the marks obtained by 90 students of a class in a test with maximummarks 100.

9172518147studentsofNo.75Above7560604545303015150Marks

A student of the class is selected at random. Find the probability that student gets : (i) less than 30% marks (ii) 60 or more marks(iii) marks between 45 and 75. (iv) distinction

Solution. Total number of students = 90

(i) P (getting less than 30% marks) 0.233

9021

90147

(ii) P (getting 60 or more marks) 0.288

9026

90917

(iii) P (getting marks between 45 and 75) 0.466

9042

901725

(iv) P (getting distinction) 0.1909

Example 6. A tyre manufacturing company kept a record of the distance covered before a type to be replaced.Following table shows the result of 1000 cases.

90375335200tyresofNo.1400thanmore1400900900400400thanlesskm)(inDistance

If you buy a tyre of this company, what is the probability that : (i) it will need to be replaced before it has covered 400 km.(ii) it will last more than 900 km?


(iii) it will need to be replaced? (iv) it will not need to be replaced at all? (v) it will need to be replaced after it has covered somewhere between 400 km and 1400 km?

Solution. We have, total number of trials = 1000 (i) The number of tyres that needs to be replaced before it has covered 400 km = 200.

Required probability 0.21000200

(ii) The number of tyres that last more than 900 km = 375 + 90 = 465


(iii) Since, all the tyres we have considered to be replaced.


(iv) The number of tyres that do not need to be replaced at all = 0


0

(v) The number of tyres which require replacement after covering somewhere between 400 kmand 1400 km is 335 + 375 = 710.


Example 7. A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag.What is the probability that the ball drawn is:(i) White (ii) not black (iii) Red or white

Solution. Total number of balls = 3 + 5 + 4 = 12.

(i) P (white ball) 31

124

(ii) P (not black) = 1 – P (black) 127

12

5121251

(iii) P (red or white) 127

12

43 .

Example 8. Cards marked with the numbers 2, 3,...., 101 are placed in a box and mixed thoroughly. One card isdrawn from this box. Find the probability that the number on the card is: (i) an even number (ii) a number less than 14(iii) a perfect square number (iv) a prime number less than 20.

Solution. Total numbers = 100(i) Total number of even numbers = 50

P (an even number) 21

10050

(ii) P (a number less than 14)

253

10012)1332( ,....,,P


(iii) P (a perfect square number) = P (4, 9, 16, ....., 100) 100

9

(iv) P (a prime number less than 20) = P (2, 3, 5, 7, 11, 13, 17, 19)

252

100

8

Example 9. Find the probability that a leap year, selected at random will have 53 sundays.Solution. A leap year has 366 days.

52 weeks = 52 × 7 = 364 daysDays left = 366 – 364 = 252 weeks means there will be 52 sundays.From remaining 2 days, the possibilities are as follows :(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thurs-day, Friday), (Friday, Saturday) and (Saturday, Sunday).So, from 7 possibilities, we have 2 cases in which we have 53rd sunday.


Ans.

Example 10. A bag contains 10 white balls and x black balls. If the probability of drawing a black ball is doublethat of a white ball, find x.

Solution. Total number of balls = 10 + x.

P (black ball) x

x

10

and, P (white ball) x

10

10

according to given question, P (black ball) = 2 P (white ball)

xxx

10

10210

x = 20 Ans. ( 10 + x 0)

PRACTICE EXERCISE1. A number is chosen from 1 to 20. Find the probability that the number chosen is :

(i) a prime number (ii) a composite number(iii) a square number (iv) an odd number (v) an even number (vi) number between 7 and 14

2. A bag contains 9 red and 6 blue balls. Find the probability that a ball drawn from a bag at random is(i) Red ball (ii) blue ball

3. In a sample of 500 items, 120 are found to be defective. Find the probability that the item selected atrandom is(i) defective (ii) non-defective

4. In a school of 1800 students, there are 875 girls. Find the probability that a student chosen at random is(i) a boy (ii) a girl

5. In a cricket match, a batsman hit a boundary 12 times out 45 balls he plays. Find the probability that hedid not hit a boundary.


6. A coin is tossed 700 times and we get head : 385 times; tail : 315 times. When a coin is tossed at random,what is the probability of getting : (i) a head? (ii) a tail?

7. Two coins are tossed 600 times and we get two heads : 138 times, one head : 192 times ; no head : 270times. When two coins are tossed at random, what is the probability of getting :(i) 2 heads? (ii) 1 head? (iii) no head?

8. Three coins are tossed 250 times and we get:3 heads : 46 times; 2 heads : 56 times; 1 head : 70 times; 0 head : 78 times.When three coins are tossed at random, what is the probability of getting :(i) 3 heads ? (ii) 2 heads? (iii) atleast 2 heads? (iv) atmost 2 heads?

9. A die is thrown 300 times and the outcomes are noted as given below :

344239527558sFrequencie654321Outcomes

When a die is thrown at random, what is the probability of getting a:(i) 4 (ii) 6 (iii) number less than 3 (iv) number which is prime

10. In a survey of 350 ladies, it was found that 235 like coffee, while rest of them dislike it. Find the probabilitythat a lady chosen at random:(i) likes coffee (ii) dislikes coffee.

11. On one page of a telephone directory, there are 200 phone numbers. The frequency distribution of theirunits digit is given below :

16151722252221212021Frequency9876543210digitUnits

One of the numbers is chosen at random from the page. What is the probability that the units digit of thechosen number is :(i) 5 (ii) 8 (iii) an even number (iv) an odd number

12. The blood groups of 30 students of class IX are recorded as follows :A, B, O, AB, O, A, O, O, B, A, O, A, B, O, O,A, O, B, A, B, O, A, B, AB, O, O, A, A, O, ABA student is selected at random from the class for blood donation. Find the probability that the bloodgroup of the student chosen is:(i) A (ii) B (iii) AB (iv) O

13. Following are the ages (in years) of 350 patients, getting medical treatment in a hospital.

305575555085PatientsofNo.706060505040403030202010years)(in Age

One of the patients is selected at random. Find the probability that his age is : (i) 40 years or more but less than 50 years. (ii) 30 years or less than it.(iii) less than 10 years.(iv) 50 years or more.


14. Given below is the frequency distribution of wages (in Rs.) of 40 workers in a certain factory:

5895634workersofNo.240220220200200180180160160140140120120100Rs)(inWages

A workers is selected at random. Find the probability that his wages are: (i) less than Rs. 160. (ii) atleast Rs. 180.(iii) more than or equal to Rs. 140 but less than Rs. 200.(iv) more than Rs. 200.

15. The following table gives the life time of 500 neon lamps:

78728996756624Lamps

ofNo.

1000900900800800700700600600500500400400300hours)(in

timeLife

A lamp is selected at random. Find the probability that the life time of the selected lamp is:(i) less than 400 hours(ii) atleast 800 times(iii) atmost 600 hours(iv) between 500 hours to 900 hours.

PRACTICE TEST



Each Questions carry 3 marks.1. Two coins are tossed simultaneously 350 times with the following frequencies of different outcomes:

Two heads : 105 times; One head : 125 times; no head : 120 times.Find the probability of getting :(i) 2 heads (ii) atleast one head

2. Following table shows the marks scored by 80 students in a mathematics test of 100 marks.

61815181085studentsofNo.1007070606050504040303020200Marks

Find the probability that a student obtained:(i) less than 40% marks.(ii) 60 or more marks.

3. 1200 families with 2 children were selected at random and the following data were recorded;

430515255familiesofNo.210familyaingirlsofNo.


If a family is chosen at random, compute the probability that it has :(i) no girl (ii) 2 girls (iii) at most one girl(iv) at least one girl

4. The table given below shows the ages of 90 teachers in a school.

1040328teachersofNo.5950494039302918years)(in Age

A teacher from this school is chosen at random. What is the probability that the selected teacher is :(i) 40 or more than 40 years old ?(ii) age less than 40 years ?

5. An insurance company selected 2000 drivers at random, in a particular city to find out a relationshipbetween age and accidents. The data obtained are given in the following table : Age of drivers Accidents in one year

(in years) 0 1 2 3 over 318-29 440 160 110 61 3530-50 505 125 60 22 18above 50 360 45 35 15 9

Find the probabilities of the following events for a driven chosen at random from the city:(i) being 18-29 years of age and exactly 2 accidents in one year.(ii) being 30-50 years of age and having one or more accidents in a year.(iii) having no accidents in one year.


1. (i) 52

(ii) 2011

(iii) 51

(iv) 103

2. (i) 53

(ii) 52

3. (i) 256

(ii) 2519

4. (i) 7237

(ii) 7235

5. 1511

6. (i) 2011

(ii) 209

7. (i) 10023

(ii) 10032

(iii) 209

8. (i) 12523

(ii) 12528

(iii) 25051

(iv) 125102

9. (i) 10013

(ii) 15017

(iii) 300133

(iv) 300169

10. (i) 7047

(ii) 7023


11. (i) 81

(ii) 403

(iii) 200101

(iv) 20099

12. (i) 103

(ii) 51

(iii) 101

(iv) 52

13. (i) 143

(ii) 7027

(iii) 0 (iv) 7017

14. (i) 4013

(ii) 2011

(iii) 21

(iv) 4013

15. (i) 1256

(ii) 103

(iii) 10033

(iv) 12583


1. (i) 103

(ii) 3523

2. (i) 8033

(ii) 103

3. (i) 8017

(ii) 12043

(iii) 12077

(iv) 8063

4. (i) 95

(ii) 94

5. (i) 20011

(ii) 809

(iii) 400261

Number Systems

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