ML Aggarwal Solutions for Class 9 Maths Chapter 16: Mensuration

ML Aggarwal Solutions for Class 9 Maths Chapter 16 Mensuration is an important study material for the students from the exam perspective. Students can refer to the solutions designed by the subject matter experts at BYJU’S having vast conceptual knowledge. The main aim of providing ML Aggarwal Solutions for Class 9 Maths Chapter 16 Mensuration PDF is to boost the exam preparation of students.

Chapter 16 consists of problems on finding the area of various figures like triangles, polygons, etc. The ML Aggarwal Solutions help students understand the concepts and their applications in a wide range. Each problem is solved after conducting wide research on the concepts to help students, irrespective of their intelligence quotient.

ML Aggarwal Solutions for Class 9 Maths Chapter 16: Mensuration Download PDF

 

ml aggarwal solutions for class 9 maths chapter 16 001
ml aggarwal solutions for class 9 maths chapter 16 002
ml aggarwal solutions for class 9 maths chapter 16 003
ml aggarwal solutions for class 9 maths chapter 16 004
ml aggarwal solutions for class 9 maths chapter 16 005
ml aggarwal solutions for class 9 maths chapter 16 006
ml aggarwal solutions for class 9 maths chapter 16 007
ml aggarwal solutions for class 9 maths chapter 16 008
ml aggarwal solutions for class 9 maths chapter 16 009
ml aggarwal solutions for class 9 maths chapter 16 010
ml aggarwal solutions for class 9 maths chapter 16 011
ml aggarwal solutions for class 9 maths chapter 16 012
ml aggarwal solutions for class 9 maths chapter 16 013
ml aggarwal solutions for class 9 maths chapter 16 014
ml aggarwal solutions for class 9 maths chapter 16 015
ml aggarwal solutions for class 9 maths chapter 16 016
ml aggarwal solutions for class 9 maths chapter 16 017
ml aggarwal solutions for class 9 maths chapter 16 018
ml aggarwal solutions for class 9 maths chapter 16 019
ml aggarwal solutions for class 9 maths chapter 16 020
ml aggarwal solutions for class 9 maths chapter 16 021
ml aggarwal solutions for class 9 maths chapter 16 022
ml aggarwal solutions for class 9 maths chapter 16 023
ml aggarwal solutions for class 9 maths chapter 16 024
ml aggarwal solutions for class 9 maths chapter 16 025
ml aggarwal solutions for class 9 maths chapter 16 026
ml aggarwal solutions for class 9 maths chapter 16 027
ml aggarwal solutions for class 9 maths chapter 16 028
ml aggarwal solutions for class 9 maths chapter 16 029
ml aggarwal solutions for class 9 maths chapter 16 030
ml aggarwal solutions for class 9 maths chapter 16 031
ml aggarwal solutions for class 9 maths chapter 16 032
ml aggarwal solutions for class 9 maths chapter 16 033
ml aggarwal solutions for class 9 maths chapter 16 034
ml aggarwal solutions for class 9 maths chapter 16 035
ml aggarwal solutions for class 9 maths chapter 16 036
ml aggarwal solutions for class 9 maths chapter 16 037
ml aggarwal solutions for class 9 maths chapter 16 038
ml aggarwal solutions for class 9 maths chapter 16 039
ml aggarwal solutions for class 9 maths chapter 16 040
ml aggarwal solutions for class 9 maths chapter 16 041
ml aggarwal solutions for class 9 maths chapter 16 042
ml aggarwal solutions for class 9 maths chapter 16 043
ml aggarwal solutions for class 9 maths chapter 16 044
ml aggarwal solutions for class 9 maths chapter 16 045
ml aggarwal solutions for class 9 maths chapter 16 046
ml aggarwal solutions for class 9 maths chapter 16 047
ml aggarwal solutions for class 9 maths chapter 16 048
ml aggarwal solutions for class 9 maths chapter 16 049
ml aggarwal solutions for class 9 maths chapter 16 050
ml aggarwal solutions for class 9 maths chapter 16 051
ml aggarwal solutions for class 9 maths chapter 16 052
ml aggarwal solutions for class 9 maths chapter 16 053
ml aggarwal solutions for class 9 maths chapter 16 054
ml aggarwal solutions for class 9 maths chapter 16 055
ml aggarwal solutions for class 9 maths chapter 16 056
ml aggarwal solutions for class 9 maths chapter 16 057
ml aggarwal solutions for class 9 maths chapter 16 058
ml aggarwal solutions for class 9 maths chapter 16 059
ml aggarwal solutions for class 9 maths chapter 16 060
ml aggarwal solutions for class 9 maths chapter 16 061
ml aggarwal solutions for class 9 maths chapter 16 062
ml aggarwal solutions for class 9 maths chapter 16 063
ml aggarwal solutions for class 9 maths chapter 16 064
ml aggarwal solutions for class 9 maths chapter 16 065
ml aggarwal solutions for class 9 maths chapter 16 066
ml aggarwal solutions for class 9 maths chapter 16 067
ml aggarwal solutions for class 9 maths chapter 16 068
ml aggarwal solutions for class 9 maths chapter 16 069
ml aggarwal solutions for class 9 maths chapter 16 070
ml aggarwal solutions for class 9 maths chapter 16 071
ml aggarwal solutions for class 9 maths chapter 16 072
ml aggarwal solutions for class 9 maths chapter 16 073
ml aggarwal solutions for class 9 maths chapter 16 074
ml aggarwal solutions for class 9 maths chapter 16 075
ml aggarwal solutions for class 9 maths chapter 16 076
ml aggarwal solutions for class 9 maths chapter 16 077
ml aggarwal solutions for class 9 maths chapter 16 078
ml aggarwal solutions for class 9 maths chapter 16 079
ml aggarwal solutions for class 9 maths chapter 16 080
ml aggarwal solutions for class 9 maths chapter 16 081
ml aggarwal solutions for class 9 maths chapter 16 082
ml aggarwal solutions for class 9 maths chapter 16 083
ml aggarwal solutions for class 9 maths chapter 16 084
ml aggarwal solutions for class 9 maths chapter 16 085
ml aggarwal solutions for class 9 maths chapter 16 086
ml aggarwal solutions for class 9 maths chapter 16 087
ml aggarwal solutions for class 9 maths chapter 16 088
ml aggarwal solutions for class 9 maths chapter 16 089
ml aggarwal solutions for class 9 maths chapter 16 090
ml aggarwal solutions for class 9 maths chapter 16 091
ml aggarwal solutions for class 9 maths chapter 16 092
ml aggarwal solutions for class 9 maths chapter 16 093
ml aggarwal solutions for class 9 maths chapter 16 094
ml aggarwal solutions for class 9 maths chapter 16 095
ml aggarwal solutions for class 9 maths chapter 16 096
ml aggarwal solutions for class 9 maths chapter 16 097
ml aggarwal solutions for class 9 maths chapter 16 098
ml aggarwal solutions for class 9 maths chapter 16 099
ml aggarwal solutions for class 9 maths chapter 16 100
ml aggarwal solutions for class 9 maths chapter 16 101
ml aggarwal solutions for class 9 maths chapter 16 102
ml aggarwal solutions for class 9 maths chapter 16 103
ml aggarwal solutions for class 9 maths chapter 16 104
ml aggarwal solutions for class 9 maths chapter 16 105
ml aggarwal solutions for class 9 maths chapter 16 106
ml aggarwal solutions for class 9 maths chapter 16 107
ml aggarwal solutions for class 9 maths chapter 16 108
ml aggarwal solutions for class 9 maths chapter 16 109
ml aggarwal solutions for class 9 maths chapter 16 110
ml aggarwal solutions for class 9 maths chapter 16 111
ml aggarwal solutions for class 9 maths chapter 16 112
ml aggarwal solutions for class 9 maths chapter 16 113
ml aggarwal solutions for class 9 maths chapter 16 114
ml aggarwal solutions for class 9 maths chapter 16 115
ml aggarwal solutions for class 9 maths chapter 16 116
ml aggarwal solutions for class 9 maths chapter 16 117
ml aggarwal solutions for class 9 maths chapter 16 118
ml aggarwal solutions for class 9 maths chapter 16 119
ml aggarwal solutions for class 9 maths chapter 16 120
ml aggarwal solutions for class 9 maths chapter 16 121
ml aggarwal solutions for class 9 maths chapter 16 122
ml aggarwal solutions for class 9 maths chapter 16 123
ml aggarwal solutions for class 9 maths chapter 16 124
ml aggarwal solutions for class 9 maths chapter 16 125
ml aggarwal solutions for class 9 maths chapter 16 126
ml aggarwal solutions for class 9 maths chapter 16 127
ml aggarwal solutions for class 9 maths chapter 16 128
ml aggarwal solutions for class 9 maths chapter 16 129
ml aggarwal solutions for class 9 maths chapter 16 130
ml aggarwal solutions for class 9 maths chapter 16 131

 

Access ML Aggarwal Solutions for Class 9 Maths Chapter 16: Mensuration

Exercise 16.1

1. Find the area of a triangle whose base is 6 cm and corresponding height is 4 cm.

Solution:

It is given that

Base of triangle = 6 cm

Height of triangle = 4 cm

We know that

Area of triangle = ½ × base × height

Substituting the values

= ½ × 6 × 4

By further calculation

= 6 × 2

= 12 cm2

2. Find the area of a triangle whose sides are

(i) 3 cm, 4 cm and 5 cm

(ii) 29 cm, 20 cm and 21 cm

(iii) 12 cm, 9.6 cm and 7.2 cm

Solution:

(i) Consider a = 3 cm, b = 4 cm and c = 5 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (3 + 4 + 5)/ 2

= 12/2

= 6 cm

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 1

= 6 cm2

(ii) Consider a = 29 cm, b = 20 cm and c = 21 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (29 + 20 + 21)/ 2

= 70/2

= 35 cm

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 2

So we get

= 7 × 5 × 3 × 2

= 210 cm2

(iii) Consider a = 12 cm, b = 9.6 cm and c = 7.2 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (12 + 9.6 + 7.2)/ 2

= 28.8/2

= 14.4 cm

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 3

ML Aggarwal Solutions for Class 9 Chapter 16 Image 4

So we get

= 2.4 × 2.4 × 6

= 34.56 cm2

3. Find the area of a triangle whose sides are 34 cm, 20 cm and 42 cm. hence, find the length of the altitude corresponding to the shortest side.

Solution:

Consider 34 cm, 20 cm and 42 cm as the sides of triangle

a = 34 cm, b = 20 cm and c = 42 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (34 + 20 + 42)/ 2

= 96/2

= 48 cm

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 5

So we get

= 14 × 6 × 4

= 336 cm2

Here the shortest side of the triangle is 20 cm

Consider h cm as the corresponding altitude

Area of triangle = ½ × base × height

Substituting the values

336 = ½ × 20 × h

By further calculation

h = (336 × 2)/ 20

So we get

h = 336/10

h = 33.6 cm

Hence, the required altitude of the triangle is 33.6 cm.

4. The sides of a triangular field are 975 m, 1050 m and 1125 m. If this field is sold at the rate of Rs 1000 per hectare, find its selling price. (1 hectare = 10000 m2)

Solution:

It is given that

a = 975 m, b = 1050 m and c = 1125 m

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (975 + 1050 + 1125)/ 2

= 3150/2

= 1575 cm

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 6

So we get

= 525 × 450 × 2

It is given that

1 hectare = 10000 m2

= (525 × 900)/ 10000

By further calculation

= (525 × 9)/ 100

= 4725/100

= 47.25 hectares

We know that

Selling price of 1 hectare field = Rs 1000

Selling price of 47.25 hectare field = 1000 × 47.25 = Rs 47250

5. The base of a right angled triangle is 12 cm and its hypotenuse is 13 cm long. Find its area and the perimeter.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 7

Solution:

It is given that

ABC is a right angled triangle

BC = 12 cm and AB = 13 cm

Using the Pythagoras theorem

AB2 = AC2 + BC2

Substituting the values

132 = AC2 + 122

By further calculation

AC2 = 132 – 122

So we get

AC2 = 169 – 144 = 25

AC = √25 = 5 cm

We know that

Area of triangle ABC = ½ × base × height

Substituting the values

= ½ × 12 × 5

= 30 cm2

Similarly

Perimeter of triangle ABC = AB + BC + CA

Substituting the values

= 13 + 12 + 5

= 30 cm

6. Find the area of an equilateral triangle whose side is 8 m. Give your answer correct to two decimal places.

Solution:

It is given that

Side of equilateral triangle = 8 m

We know that

Area of equilateral triangle = √3/4 (side)2

Substituting the values

= √3/4 × 8 × 8

By further calculation

= √3 × 2 × 8

= 1.73 × 16

= 27.71 m2

7. If the area of an equilateral triangle is 81√3 cm2 find its perimeter.

Solution:

We know that

Area of equilateral triangle = √3/4 (side)2

Substituting the values

81 √3 = √3/4 (side)2

By further calculation

Side2 = (81 √3 × 4)/ √3

So we get

(side)2 = 81 × 4

side = √(81 × 4)

side = 9 × 2 = 18 cm

So the perimeter of equilateral triangle = 3 × side

= 3 × 18

= 54 cm

8. If the perimeter of an equilateral triangle is 36 cm, calculate its area and height.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 16 Image 8

We know that

Perimeter of an equilateral triangle = 3 × side

Substituting the values

36 = 3 × side

By further calculation

side = 36/3 = 12 cm

So AB = BC = CA = 12 cm

Here

Area of equilateral triangle = √3/4 (side)2

Substituting the values

= √3/4 (12)2

By further calculation

= √3/4 × 12 × 12

So we get

= √3 × 3 × 12

= 1.73 × 36

= 62.4 cm2

In triangle ABD

Using Pythagoras Theorem

AB2 = AD2 + BD2

Here BD = 12/2 = 6 cm

Substituting the values

122 = AD2 + 62

By further calculation

144 =AD2 + 36

AD2 = 144 – 36 = 108

So we get

AD = √108 = 10.4

Therefore, the required height is 10.4 cm.

9. (i) If the length of the sides of a triangle are in the ratio 3: 4: 5 and its perimeter is 48 cm, find its area.

(ii) The sides of a triangular plot are in the ratio 3: 5: 7 and its perimeter is 300 m. Find its area.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 16 Image 9

(i) Consider ABC as the triangle

Ratio of the sides are 3x, 4x and 5x

Take a = 3x cm, b = 4x cm and c = 5x cm

We know that

a + b + c = 48

Substituting the values

3x + 4x + 5x = 48

12x = 48

So we get

x = 48/12 = 4

Here

a = 3x = 3 × 4 = 12 cm

b = 4x = 4 × 4 = 16 cm

c = 5x = 5 × 4 = 20 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (12 + 16 + 20)/ 2

= 48/2

= 24 cm

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 10

ML Aggarwal Solutions for Class 9 Chapter 16 Image 11

So we get

= 12 × 4 × 2

= 96 cm2

(ii) It is given that

Sides of a triangle are in the ratio = 3: 5: 7

Perimeter = 300 m

We know that

First side = (300 × 3)/ sum of ration

Substituting the values

= (300 × 3)/ (3 + 5 + 7)

By further calculation

= (300 × 3)/ 15

= 60 m

Second side = (300 × 5)/ 15 = 100 m

Third side = (300 × 7)/ 15 = 140 m

Here

S = perimeter/2 = 300/2 = 150 m

So we get

ML Aggarwal Solutions for Class 9 Chapter 16 Image 12

ML Aggarwal Solutions for Class 9 Chapter 16 Image 13

We get

= 1500 × 1.732

= 2598 m2

10. ABC is a triangle in which AB = AC = 4 cm and ∠A = 900. Calculate the area of △ABC. Also find the length of perpendicular from A to BC.

Solution:

It is given that

AB = AC = 4 cm

Using the Pythagoras theorem

BC2 = AB2 + AC2

Substituting the values

BC2 = 42 + 42

By further calculation

BC2 = 16 + 16 = 32

BC = √32 = 4√2 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 14

We know that

Area of △ABC = ½ × BC × h

Substituting the values

8 = ½ × 4√2 × h

By further calculation

h = (8 × 2)/ 4√2

We can write it as

h = (2 × 2)/ √2 × √2/√2

So we get

h = 4√2/2 = 2 × √2

h = 2 × 1.41 = 2.82 cm

11. Find the area of an isosceles triangle whose equal sides are 12 cm each and the perimeter is 30 cm.

Solution:

Consider ABC as the isosceles triangles

ML Aggarwal Solutions for Class 9 Chapter 16 Image 15

Here AB = AC = 12cm

Perimeter = 30 cm

So BC = 30 – (12 + 12) = 30 – 24 = 6 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= 30/ 2

= 15 cm

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 16

ML Aggarwal Solutions for Class 9 Chapter 16 Image 17

We can write it as

= 9 × 3.873

= 34.857

= 34.86 cm2

12. Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm.

Solution:

It is given that

Base = 6 cm

Perimeter = 16 cm

Consider ABC as an isosceles triangle in which

AB = AC = x

So BC = 6 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 18

We know that

Perimeter of △ABC = AB + BC + AC

Substituting the values

16 = x + 6 + x

By further calculation

16 = 2x + 6

16 – 6 = 2x

10 = 2x

So we get

x = 10/2 = 5

Here AB = AC = 5 cm

BC = ½ × 6 = 3 cm

In △ABD

AB2 = AD2 + BD2

Substituting the values

52 = AD2 + 32

25 = AD2 + 9

By further calculation

AD2 = 25 – 9 = 16

So we get

AD = 4 cm

Here

Area of △ABC = ½ × base × height

Substituting the values

= ½ × 6 × 4

= 3 × 4

= 12 cm2

13. The sides of a right angled triangle containing the right angle are 5x cm and (3x – 1) cm. Calculate the length of the hypotenuse of the triangle if its area is 60 cm2.

Solution:

Consider ABC as a right angled triangle

AB = 5x cm and BC = (3x – 1) cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 19

We know that

Area of △ABC = ½ × AB × BC

Substituting the values

60 = ½ × 5x (3x – 1)

By further calculation

120 = 5x (3x – 1)

120 = 15x2 – 5x

It can be written as

15x2 – 5x – 120 = 0

Taking out the common terms

5 (3x2 – x – 24) = 0

3x2 – x – 24 = 0

3x2 – 9x + 8x – 24 = 0

Taking out the common terms

3x (x – 3) + 8 (x – 3) = 0

(3x + 8) (x – 3) = 0

Here

3x + 8 = 0 or x – 3 = 0

We can write it as

3x = -8 or x = 3

x = -8/3 or x = 3

x = -8/3 is not possible

So x = 3

AB = 5 × 3 = 15 cm

BC = (3 × 3 – 1) = 9 – 1 = 8 cm

In right angled △ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = 152 + 82

By further calculation

AC2 = 225 + 64 = 289

AC2 = 172

So AC = 17 cm

Therefore, the hypotenuse of the right angled triangle is 17 cm.

14. In △ABC, ∠B = 900, AB = (2A + 1) cm and BC = (A + 1) cm. If the area of the △ABC is 60 cm2, find its perimeter.

Solution:

It is given that

AB = (2x + 1) cm

BC = (x + 1) cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 20

We know that

Area of △ABC = ½ × AB × BC

Substituting the values

60 = ½ × (2x + 1) (x + 1)

By cross multiplication

60 × 2 = (2x + 1) (x + 1)

By further calculation

120 = 2x2 + 3x + 1

We can write it as

0 = 2x2 + 3x + 1 – 120

0 = 2x2 + 3x – 119

So we get

2x2 + 3x – 119 = 0

2x2 + 17x – 14x – 119 = 0

Taking out the common terms

x (2x + 17) – 7 (2x + 17) = 0

(x – 7) (2x + 17) = 0

Here

x – 7 = 0 or 2x + 17 = 0

x = 7 or 2x = – 17

x = 7 or x = -17/2

AB = (2x + 1) = 2 × 7 + 1

AB = 14 + 1 = 15 cm

BC = (x + 1) = 7 + 1 = 8 cm

In right angled △ABC

Using Pythagoras Theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = 152 + 82

AC2 = 225 + 64

AC2 = 289

So we get

AC = 17 cm

So the perimeter = AB + BC + AC

Substituting the values

= 15 + 8 + 17

= 40 cm

15. If the perimeter of a right angled triangle is 60 cm and its hypotenuse is 25 cm, find its area.

Solution:

We know that

Perimeter of a right angled triangle = 60 cm

Hypotenuse = 25 cm

Here the sum of two sides = 60 – 25 = 35 cm

Consider base = x cm

Altitude = (35 – x) cm

Using the Pythagoras theorem

x2 + (35 – x)2 = 252

By further calculation

x2 + 1225 + x2 – 70x = 625

2x2 – 70x + 1225 – 625 = 0

2x2 – 70x + 600 = 0

Dividing by 2

x2 – 35x + 300 = 0

x2 – 15x – 20x + 300 = 0

Taking out the common terms

x (x – 15) – 20 (x – 15) = 0

(x – 15) (x – 20) = 0

Here

x – 15 = 0

So we get

x = 15

Similarly

x – 20 = 0

So we get

x = 20 cm

So 15 cm and 20 cm are the sides of the triangle

Area = ½ × base × altitude

Substituting the values

= ½ × 15 × 20

= 150 cm2

16. The perimeter of an isosceles triangle is 40 cm. The base is two third of the sum of equal sides. Find the length of each side.

Solution:

It is given that

Perimeter of an isosceles triangle = 40 cm

Consider x cm as each equal side

We know that

Base = 2/3 (2x) = 4/3 x

So according to the sum

2x + 4/3 x = 40

By further calculation

6x + 4x = 120

10x = 120

By division

x = 120/10 = 12

Therefore, the length of each equal side is 12 cm.

17. If the area of an isosceles triangle is 60 cm2 and the length of each of its equal sides is 13 cm, find its base.

Solution:

It is given that

Area of isosceles triangle = 60 cm2

Length of each equal side = 13 cm

Consider base BC = x cm

Construct AD perpendicular to BC which bisects BC at D

So BD = DC = x/2 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 21

In right △ABD

AB2 = BD2 + AD2

Substituting the values

132 = (x/2)2 + AD2

By further calculation

169 = x2/4 + AD2

AD2 = 169 – x2/4 …… (1)

We know that

Area = 60 cm2

AD = (area × 2)/ base

Substituting the values

AD = (60 × 2)/ x = 120/x …… (2)

Using both the equations

169 – x2 /4 = (120/x)2

By further calculation

(676 – x2)/ 4 = 14400/x2

By cross multiplication

676x2 – x4 = 57600

We can write it as

x4 – 676x2 + 57600 = 0

x4 – 576x2 – 100x2 + 57600 = 0

Taking out the common terms

x2 (x2 – 576) – 100 (x2 – 576) = 0

(x2 – 576) (x2 – 100) = 0

Here

x2 – 576 = 0 where x2 = 576

So x = 24

Similarly

x2 – 100 = 0 where x2 = 100

So x = 10

Hence, the base is 10 cm or 24 cm.

18. The base of a triangular field is 3 times its height if the cost of cultivating the field at the rate of Rs 25 per 100 m2 is Rs 60000; find its base and height.

Solution:

It is given that

Cost of cultivating the field at the rate of Rs 25 per 100 m2 = Rs 60000

Here the cost of cultivating the field of Rs 25 for 100 m2

So the cost of cultivating the field of Rs 1 = 100/25 m2

Cost of cultivating the field of Rs 60000 = 100/25 × 60000

= 4 × 60000

= 240000 m2

So the area of field = 240000 m2

½ × base × height = 240000 ….. (1)

Consider

Height of triangular field = h m2

Base of triangular field = 3h m2

Substituting the values in equation (1)

½ × 3h × h = 240000

By further calculation

½ × 3h2 = 240000

h2 = (240000 × 2)/ 3

h2 = 80000 × 2

h2 = 160000

So we get

h = √160000 = 400

Here the height of triangular field = 400 m

Base of triangular field = 3 × 400 = 1200 m2

19. A triangular park ABC has sides 120 m, 80 m and 50 m (as shown in the given figure). A gardener Dhania has to put a fence around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of Rs 20 per metre leaving a space 3 m wide for a gate on one side.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 22

Solution:

It is given that

ABC is a triangular park with sides 120 m, 80 m and 50 m.

Here the perimeter of triangle ABC = 120 + 80 + 50 = 250 m

ML Aggarwal Solutions for Class 9 Chapter 16 Image 23

We know that

Portion at which a gate is build = 3m

Remaining perimeter = 250 – 3 = 247 m

So the length of fence around it = 247 m

Rate of fencing = Rs 20 per m

Total cost of fencing = 20 × 247 = Rs 4940

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= 250/2

= 125 cm

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 24

20. An umbrella is made by stitching 10 triangular pieces of cloth of two different colors (shown in the given figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each color is required for the umbrella?

ML Aggarwal Solutions for Class 9 Chapter 16 Image 25

Solution:

It is given that

An umbrella is made by stitching 10 triangular pieces of cloth of two different colors

20 cm, 50 cm, 50 cm are the measurement of each triangle

So we get

s/2 = (20 + 50 + 50)/ 2

s/2 = 120/2 = 60

We know that

ML Aggarwal Solutions for Class 9 Chapter 16 Image 26

ML Aggarwal Solutions for Class 9 Chapter 16 Image 27

Here the area of 5 triangular piece of first color = 5 × 200 √6 = 1000 √6 cm2

Area of triangular piece of second color = 1000 √6 cm2

21. (a) In the figure (1) given below, ABC is an equilateral triangle with each side of length 10 cm. In △BCD, ∠D = 900 and CD = 6 cm.

Find the area of the shaded region. Give your answer correct to one decimal place.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 28

(b) In the figure (ii) given, ABC is an isosceles right angled triangle and DEFG is a rectangle. If AD = AE = 3 cm and DB = EC = 4 cm, find the area of the shaded region.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 29

Solution:

(a) It is given that

ABC is an equilateral triangle of side = 10 cm

We know that

Area of equilateral triangle ABC = √3/4 × (side)2

Substituting the values

= √3/4 × 102

= √3/4 × 100

So we get

= √3× 25

= 1.73 × 25

= 43.3 cm2

In right angled triangle BDC

∠D = 900

BC = 10 cm

CD = 6 cm

Using Pythagoras Theorem

BD2 + DC2 = BC2

Substituting the values

BD2 + 62 = 102

BD2 + 36 = 100

So we get

BD2 = 100 – 36 = 64

BD = √64 = 8 cm

We know that

Area of triangle BDC = ½ × base × height

So we get

= ½ × BD × DC

Substituting the values

= ½ × 8 × 6

= 4 × 6

= 24 cm2

Here the area of shaded portion = Area of triangle ABC – Area of triangle BDC

Substituting the values

= 43.3 – 24

= 19.3 cm2

(b) It is given that

AD = AE = 3 cm

DB = EC = 4 cm

By addition we get

AD + DB = AE + EC = (3 + 4) cm

AB = AC = 7 cm

∠A = 900

We know that

Area of right triangle ADE = ½ × AD × AE

Substituting the values

= ½ × 3 × 3

= 9/2 cm2

So triangle BDG is an isosceles right triangle

Similarly

DG2 + BG2 = BD2

DG2 + DG2 = 42

By further calculation

2DG2 = 16

DG2 = 16/2 = 8

DG = √8 cm

Area of triangle BDG = ½ × BG × DG

We can write it as

= ½ × DG × DG

Substituting the values

= ½ (√8)2

= ½ × 8

= 4 cm2

Area of isosceles right triangle EFC = 4 cm2

So the area of shaded portion = 9/2 + 4 + 4

Taking LCM

= (9 + 8 + 8)/ 2

= 25/2

= 12.5 cm2

Exercise 16.2

1. (i) Find the area of quadrilateral whose one diagonal is 20 cm long and the perpendiculars to this diagonal from other vertices are of length 9 cm and 15 cm.

(ii) Find the area of a quadrilateral whose diagonals are of length 18 cm and 12 cm and they intersect each other at right angles.

Solution:

(i) Consider ABCD as a quadrilateral in which AC = 20 cm

We know that

BY = 9 cm and DY = 15 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 30

Here

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

We can write it as

= ½ × base × height + ½ × base × height

So we get

= ½ × AC × BX + ½ × AC × DY

Substituting the values

= (½ × 20 × 9) + (½ × 20 × 15)

By further calculation

= (10 × 9 + 10 × 15)

= 90 + 150

= 240 cm2

(ii) Consider ABCD as a quadrilateral in which the diagonals AC and BD intersect each other at M at right angles

AC = 18 cm and BD = 12 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 31

We know that

Area of quadrilateral ABCD = ½ × diagonal AC × diagonal BD

Substituting the values

= ½ × 18 × 12

By further calculation

= 9 × 12

= 108 cm2

2. Find the area of the quadrilateral field ABCD whose sides AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and ∠A = 900.

Solution:

It is given that

ABCD is a quadrilateral field

AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and ∠A = 900

ML Aggarwal Solutions for Class 9 Chapter 16 Image 32

In triangle BAD

∠A = 900

Using the Pythagoras Theorem

BD2 = BA2 + AD2

Substituting the values

BD2 = 402 + 92

By further calculation

BD2 = 1600 + 81 = 1681

So we get

BD = 41

We know that

Area of quadrilateral ABCD = Area of △BAD + Area of △BDC

It can be written as

= ½ × base × height + Area of △BDC

Substituting the values

= ½ × 40 × 9 + Area of △BDC

By further calculation

= 180 m2 + Area of △BDC

Determining the area of △BDC

Consider a = BD = 41 m, b = CD = 15 m, c = BC = 28 m

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (41 + 15 + 28)/ 2

= 42 cm

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 33

So we get

= 2 × 7 × 3 × 3

= 126 m2

So the area of quadrilateral ABCD = 180 m2 + Area of △BDC

Substituting the values

= 180 + 126

= 306 m2

3. Find the area of the quadrilateral ABCD in which ∠BCA = 900, AB = 13 cm and ACD is an equilateral triangle of side 12 cm.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 34

Solution:

It is given that

ABCD is a quadrilateral in which ∠BCA = 900 and AB = 13 cm

ABCD is an equilateral triangle in which AC = CD = AD = 12 cm

In right angled △ABC

Using Pythagoras theorem,

AB2 = AC2 + BC2

Substituting the values

132 = 122 + BC2

By further calculation

BC2 = 132 – 122

BC2 = 169 – 144 = 25

So we get

BC = √25 = 5 cm

We know that

Area of quadrilateral ABCD = Area of △ABC + Area of △ACD

It can be written as

= ½ × base × height + √3/4 × side2

= ½ × AC × BC + √3/4 × 122

Substituting the values

= ½ × 12 × 5 + √3/4 × 12 × 12

So we get

= 6 × 5 + √3 × 3 × 12

= 30 + 36√3

Substituting the value of √3

= 30 + 36 × 1.732

= 30 + 62.28

= 92.28 cm2

4. Find the area of quadrilateral ABCD in which ∠B = 900, AB = 6 cm, BC = 8 cm and CD = AD = 13 cm.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 35

Solution:

It is given that

ABCD is a quadrilateral in which ∠B = 900, AB = 6 cm, BC = 8 cm and CD = AD = 13 cm

In △ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = 62 + 82

By further calculation

AC2 = 36 + 64 = 100

So we get

AC2 = 102

AC = 10 cm

We know that

Area of quadrilateral ABCD = Area of △ABC + Area of △ACD

It can be written as

= ½ × base × height + Area of △ACD

= ½ × AB × BC + Area of △ACD

Substituting the values

= ½ × 6 × 8 + Area of △ACD

By further calculation

= 24 cm2 + Area of △ACD …… (1)

Finding the area of △ACD

Consider a = AC = 10 cm, b = CD = 13 cm, c = AD = 13 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (10 + 13 + 13)/ 2

= (10 + 26)/2

= 36/2

= 18 cm

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 36

So we get

= 3 × 2 × 2 × 5

= 60 cm2

Using equation (1)

Area of quadrilateral ABCD = 24 cm2 + Area of △ACD

Substituting the values

= 24 + 60

= 84 cm2

5. The perimeter of a rectangular cardboard is 96 cm; if its breadth is 18 cm, find the length and the area of the cardboard.

Solution:

Consider ABCD as a rectangle

Take length = l cm

Breadth = 18 cm

Perimeter = 96 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 37

We know that

2 × (l + b) = 96 cm

Substituting the values

2 × (l + 18) = 96 cm

By further calculation

(l + 18) = 96/2

l + 18 = 48

So we get

l = 48 – 18 = 30 cm

Here

Area of rectangular cardboard = l × b

Substituting the values

= 30 × 18

= 540 cm2

6. The length of a rectangular hall is 5 cm more than its breadth, if the area of the hall is 594 m2, find its perimeter.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 38

Solution:

Consider ABCD is a rectangular hall

Take Breadth = x m

Length = (x + 5) m

We know that

Area of rectangular field = l × b

Substituting the values

594 = x (x + 5)

By further calculation

594 = x2 + 5x

0 = x2 + 5x – 594

x2 + 5x – 594 = 0

It can be written as

x2 + 27x – 22x – 594 = 0

Taking out the common terms

x (x + 27) – 22 (x + 27) = 0

So we get

(x – 22) (x + 27) = 0

Here

x – 22 = 0 or x + 27 = 0

We get

x = 22 m or x = -27 which is not possible

We know that

Breadth = 22 m

Length = (x + 5) = 22 + 5 = 27 m

Perimeter = 2 (l + b)

Substituting the values

= 2 (27 + 22)

By further calculation

= 2 × 49

= 98 m

7. (a) The diagram (i) given below shows two paths drawn inside a rectangular field 50 m long and 35 m wide. The width of each path is 5 metres. Find the area of the shaded portion.

(b) In the diagram (ii) given below, calculate the area of the shaded portion. All measurements are in centimetres.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 39

Solution:

(a) We know that

Area of shaded portion = Area of rectangle ABCD + Area of rectangle PQRS – Area of square LMNO

ML Aggarwal Solutions for Class 9 Chapter 16 Image 40

Substituting the values

= 50 × 5 + 5 × 35 – 5 × 5

By further calculation

= 250 + 175 – 25

So we get

= 250 + 150

= 400 m2

(b) We know that

Area of shaded portion = Area of ABCD – 5 × Area of any small square

ML Aggarwal Solutions for Class 9 Chapter 16 Image 41

It can be written as

= l × b – 5 × side × side

Substituting the values

= 8 × 6 – 5 × 2 × 2

By further calculation

= 48 – 20

= 28 cm2

8. A rectangular plot 20 m long and 14 m wide is to be covered with grass leaving 2 m all around. Find the area to be laid with grass.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 16 Image 42

Consider ABCD as a plot

Length of plot = 20 m

Breadth of plot = 14 m

Take PQRS as the grassy plot

Here

Length of grassy lawn = 20 – 2 × 2

By further calculation

= 20 – 4

= 16 m

Breadth of grassy lawn = 14 – 2 × 2

By further calculation

= 14 – 4

= 10 m

Area of grassy lawn = length × breadth

Substituting the values

= 16 × 10

= 160 m2

9. The shaded region of the given diagram represents the lawn in front of a house. On three sides of the lawn there are flower beds of width 2 m.

(i) Find the length and the breadth of the lawn.

(ii) Hence, or otherwise, find the area of the flower – beds.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 43

Solution:

Consider BCDE as the lawn

ML Aggarwal Solutions for Class 9 Chapter 16 Image 44

(i) We know that

Length of lawn BCDE = BC

It can be written as

= AD – AB – CD

Substituting the values

= 30 – 2 – 2

By further calculation

= 30 – 4

= 26 m

Breadth of lawn BDCE = BE

It can be written as

= AG – GH

Substituting the values

= 12 – 2

= 10 m

(ii) We know that

Area of flower beds = Area of rectangle ADFG – Area of lawn BCDE

It can be written as

= AD × AG – BC × BE

Substituting the values

= 30 × 12 – 26 × 10

By further calculation

= 360 – 260

= 100 m2

10. A foot path of uniform width runs all around the inside of a rectangular field 50 m long and 38 m wide. If the area of the path is 492 m2, find its width.

Solution:

Consider ABCD as a rectangular field having

Length = 50 m

Breadth = 38 m

ML Aggarwal Solutions for Class 9 Chapter 16 Image 45

We know that

Area of rectangular field ABCD = l × b

Substituting the values

= 50 × 38

= 1900 m2

Let x m as the width of foot path all around the inside of a rectangular field

Length of rectangular field PQRS = (50 – x – x) = (50 – 2x) m

Breadth of rectangular field PQRS = (38 – x – x) = (38 – 2x) m

Here

Area of foot path = Area of rectangular field ABCD – Area of rectangular field PQRS

Substituting the values

492 = 1900 – (50 – 2x) (38 – 2x)

It can be written as

492 = 1900 – [50 (38 – 2x) – 2x (38 – 2x)]

By further calculation

492 = 1900 – (1900 – 100x – 76x + 4x2)

492 = 1900 – 1900 + 100x + 76x – 4x2

On further simplification

492 = 176x – 4x2

Taking out 4 as common

492 = 4 (44x – x2)

44x – x2 = 492/4 = 123

We get

x2 – 44x + 123 = 0

It can be written as

x2 – 41x – 3x + 123 = 0

Taking out the common terms

x (x – 41) – 3 (x – 41) = 0

(x – 3) (x – 41) = 0

Here

x – 3 = 0 or x – 41 = 0

So x = 3 m or x = 41 m which is not possible

Therefore, width is 3 m.

11. The cost of enclosing a rectangular garden with a fence all around at the rate of Rs 15 per metre is Rs 5400. If the length of the garden is 100 m, find the area of the garden.

Solution:

Consider ABCD as a rectangular garden

Length = 100 m

Take breadth = x m

ML Aggarwal Solutions for Class 9 Chapter 16 Image 46

We know that

Perimeter of the garden = 2 (l + b)

Substituting the values

= 2 (100 + x)

= (200 + 2x) m

We know that

Cost of 1 m to enclosing a rectangular garden = Rs 15

So the cost of (200 + 2x) m to enclosing a rectangular garden = 15 (200 + 2x)

= 3000 + 30x

Given cost = Rs 5400

We get

3000 + 30x = 5400

It can be written as

30x = 5400 – 3000

x = 2400/30 = 80 m

Breadth of garden = 80 m

So the area of rectangular field = l × b

Substituting the values

= 100 × 80

= 8000 m2

12. A rectangular floor which measures 15 m × 8 m is to be laid with tiles measuring 50 cm × 25 cm find the number of tiles required further, if a carpet is laid on the floor so that a space of 1 m exists between its edges and the edges of the floor, what fraction of the floor is uncovered?

Solution:

Consider ABCD as a rectangular field of measurement 15m × 8m

Length = 15 m

Breadth = 8 m

ML Aggarwal Solutions for Class 9 Chapter 16 Image 47

Here the area = l × b = 15 × 8 = 120 m2

Measurement of tiles = 50 cm × 25 cm

Length = 50 cm = 50/100 = ½ m

Breadth = 25 cm = 25/100 = ¼ m

So the area of one tile = ½ × ¼ = 1/8 m2

No. of required tiles = Area of rectangular field/Area of one tile

Substituting the values

= 120/ (1/8)

By further calculation

= (120 × 8)/ 1

= 960 tiles

Length of carpet = 15 – 1 – 1

= 15 – 2

= 13 m

Breadth of carpet = 8 – 1 – 1

= 8 – 2

= 6 m

Area of carpet = l × b

= 13 × 6

= 78 m2

We know that

Area of floor which is uncovered by carpet = Area of floor – Area of carpet

Substituting the values

= 120 – 78

= 42 m2

Fraction = Area of floor which is uncovered by carpet/ Area of floor

Substituting the values

= 42/120

= 7/20

13. The width of a rectangular room is 3/5 of its length x metres. If its perimeter is y metres, write an equation connecting x and y. Find the floor area of the room if its perimeter is 32 m.

Solution:

It is given that

Length of rectangular room = x m

Width of rectangular room = 3/5 of its length

= 3x/5 m

Perimeter = y m

We know that

Perimeter = 2 (l + b)

Substituting the values

y = 2 [(5x + 3x)/ 5]

By further calculation

y = 2 × 8x/5

y = 16x/5

We get

5y = 16x

16x = 5y ….. (1)

Equation (1) is the required relation between x and y

Given perimeter = 32 m

So y = 32 m

Now substituting the value of y in equation (1)

16x = 5 × 32

By further calculation

x = (5 × 32)/ 16

x = (5 × 2)/ 1

x = 10 m

Breadth = 3/5 × x

Substituting the value of x

= 3/5 × 10

= 3 × 2

= 6 m

Here the floor area of the room = l × b

Substituting the values

= 10 × 6

= 60 m2

14. A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.

Solution:

Consider ABCD as a rectangular garden

ML Aggarwal Solutions for Class 9 Chapter 16 Image 48

Length = 10 m

Breadth = 16 m

So the area of ABCD = l × b

Substituting the values

= 10 × 16

= 160 m2

Consider x m as the width of the walk

Length of rectangular garden PQRS = 10 – x – x = (10 – 2x) m

Breadth of rectangular garden PQRS = 16 – x – x = (16 – 2x) m

15. A rectangular room is 6 m long, 4.8 m wide and 3.5 m high. Find the inner surface of the four walls.

Solution:

It is given that

Length of rectangular room = 6 m

Breadth of rectangular room = 4.8 m

Height of rectangular room = 3.5 m

Here

Inner surface area of four wall = 2 (l + b) × h

Substituting the values

= 2 (6 + 4.8) × 3.5

By further calculation

= 2 × 10.8 × 3.5

= 21.6 × 3.5

= 75.6 m2

16. A rectangular plot of land measures 41 metres in length and 22.5 metres in width. A boundary wall 2 metres high is built all around the plot at a distance of 1.5 m from the plot. Find the inner surface area of the boundary wall.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 16 Image 49

It is given that

Length of rectangular plot = 41 metres

Breadth of rectangular plot = 22.5 metres

Height of boundary wall = 2 metre

Here

Boundary wall is built at a distance of 1.5 m

New length = 41 + 1.5 + 1.5

= 41 + 3

= 44 m

New breadth = 22.5 + 1.5 + 1.5

= 22.5 + 3

= 25.5 m

We know that

Inner surface area of the boundary wall = 2 (l + b) × h

Substituting the values

= 2 (44 + 25.5) × 2

By further calculation

= 2 × 69.5 × 2

= 2 × 139

= 278 m2

17. (a) Find the perimeter and area of the figure

(i) given below in which all corners are right angled.

(b) Find the perimeter and area of the figure

(ii) given below in which all corners are right angles.

(c) Find the area and perimeter of the figure

(iii) given below in which all corners are right angled and all measurement in centimetres.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 50

Solution:

(a) It is given that

ML Aggarwal Solutions for Class 9 Chapter 16 Image 51

AB = 2m, BE = 4m, FE = 4m and FG = 1.5 m

So BD = 4 + 1.5 = 5.5 m

AC = BD = 5.5 m

CG = (4 + 2) = 6 m

We know that

Perimeter of figure (i) = AC + CG + GF + FE + EB + BA

Substituting the values

= 5.5 + 6 + 1.5 + 4 + 4 + 2

= 23 m

Here

Area of given figure = Area of ABEDC + Area of FEDG

It can be written as

= length × breadth + length × breadth

Substituting the values

= 2 × 5.5 + 4 × 1.5

= 11 + 6

= 17 m2

(b) It is given that

AB = CD = 3m

HI = AC = 7m

JE = BE = 5m

GF = DE = 2m

DG = EF = 8m

GH = JI = 2m

ML Aggarwal Solutions for Class 9 Chapter 16 Image 52

We know that

CH = CD + DG + GH

Substituting the values

= 3 + 8 + 2

= 13 m

Perimeter of the given figure = AB + AC + CH + HI + IJ + JF + FE + BE

Substituting the values

= 3 + 7 + 13 + 7 + 2 + 5 + 8 + 5

= 50 m

Here

Area of given figure = Area of first figure + Area of second figure + Area of third figure

Substituting the values

= 7 × 3 + 8 × 2 + 7 × 2

By further calculation

= 21 + 16 + 14

= 51 m2

(c) It is given that

AB = 12 cm

AL = BC = 7 cm

JK = DE = 5 cm

HJ = GF = 3 cm

LK = HG = CD = 2 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 53

We know that

Perimeter of given figure = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA

Substituting the values

= 12 + 7 + 2 + 5 + 3 + 3 + 2 + 3 + 3 + 5 + 2 + 7

= 54 cm

Here

Area of given figure =Area of first part + Area of second part + Area of third part + Area of fourth part + Area of fifth part

Substituting the values

= 7 × 2 + 2 × 3 + (2 + 3) × 2 + 2 × 3 + 7 × 2

By further calculation

= 14 + 6 + 10 + 6 + 14

= 50 cm2

18. The length and the breadth of a rectangle are 12 cm and 9 cm respectively. Find the height of a triangle whose base is 9 cm and whose area is one third that of rectangle.

Solution:

It is given that

Length of a rectangle = 12 cm

Breadth of a rectangle = 9 cm

So the area = l × b

Substituting the values

= 12 × 9

= 108 cm2

ML Aggarwal Solutions for Class 9 Chapter 16 Image 54

Using the condition

Area of triangle ABC = 1/3 × area of rectangle

Substituting the values

= 1/3 × 108

= 36 cm2

Consider h cm as the height of triangle ABC

Area of triangle ABC = ½ × base × height

Substituting the values

36 = ½ × 9 × h

By further calculation

36 × 2 = 9 × h

h = (36 × 2)/ 9

So we get

h = 4 × 2

h = 8 cm

Therefore, height of triangle ABC is 8 cm.

19. The area of a square plot is 484 m2. Find the length of its one side and the length of its one diagonal.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 55

Solution:

It is given that

ABCD is a square plot having area = 484 m2

Sides of square are AB, BC, CD and AD

We know that

Area of square = side × side

Substituting the values

484 = (side)2

So we get

Side = √484 = 22 m

AB = BC = 22 m

In triangle ABC

Using Pythagoras Theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = 222 + 222

AC2 = 484 + 484 = 968

By further calculation

AC = √968 = √ (484 × 2)

AC = 22 × √2

So we get

AC = 22 × 1.414 = 31.11 m

Therefore, length of side is 22 m and length of diagonal is 31.11 m.

20. A square has the perimeter 56 m. Find its area and the length of one diagonal correct up to two decimal places.

Solution:

Consider ABCD as a square with side x m

Perimeter of square = 4 × side

Substituting the values

56 = 4x

By further calculation

x = 56/4 = 14 m

ML Aggarwal Solutions for Class 9 Chapter 16 Image 56

In triangle ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = 142 + 142

By further calculation

AC2 = 196 + 196 = 392

So we get

AC = √392

AC = √ (196 × 2)

AC = 14 √2

Substituting the value of √2

AC = 14 × 1.414

AC = 19.80 m

Therefore, the side of square is 14 m and diagonal is 19.80 m.

21. A wire when bent in the form of an equilateral triangle encloses an area of 36 √3 cm2. Find the area enclosed by the same wire when bent to form:

(i) a square, and

(ii) a rectangle whose length is 2 cm more than its width.

Solution:

It is given that

Area of equilateral triangle = 36 √3 cm2

Consider x cm as the side of equilateral triangle

We know that

Area = √3/4 (side)2

Substituting the values

36 √3 = √3/4 × (x)2

By further calculation

x2 = (36 √3 × 4)/ √3

x2 = 36 × 4

So we get

x = √ (36 × 4)

x = 6 × 2

x = 12 cm

Here

Perimeter of equilateral triangle = 3 × side

= 3 × 12

= 36 cm

(i) We know that

Perimeter of equilateral triangle = Perimeter of square

It can be written as

36 = 4 × side

So we get

Side = 36/4 = 9 cm

Area of square = side × side

= 9 × 9

= 81 cm2

(ii) We know that

Perimeter of triangle = Perimeter of rectangle …. (1)

According to the condition of rectangle

Length is 2 cm more than its width

Width of rectangle = x cm

Length of rectangle = (x + 2) cm

Perimeter of rectangle = 2 (l + b)

Substituting the values

= 2 [(x + 2) + x]

By further calculation

= 2 (2x + 2)

= 4x + 4

Using equation (1)

4x + 4 = Perimeter of triangle

4x + 4 = 36

By further calculation

4x = 36 – 4 = 32 cm

So we get

x = 32/4 = 8 cm

Here

Length of rectangle = 8 + 2 = 10 cm

Breadth of rectangle = 8 cm

Area of rectangle = length × breadth

= 10 × 8

= 80 cm2

22. Two adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 8 cm, find the area of the parallelogram. Also find the distance between shorter sides.

Solution:

Consider ABCD as a parallelogram

Longer side AB = 15 cm

Shorter side = 10 cm

Distance between longer side DM = 8 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 57

Consider DN as the distance between the shorter side

Area of parallelogram ABCD = base × height

We can write it as

= AB × DM

Substituting the values

= 15 × 8

= 120 cm2

If base is AD

Area of parallelogram = AD × DN

Substituting the values

120 = 10 × DN

So we get

DN = 120/10 = 12 cm

Therefore, the area of parallelogram is 120 cm2 and the distance between shorter side is 12 cm.

23. ABCD is a parallelogram with sides AB = 12 cm, BC = 10 cm and diagonal AC = 16 cm. Find the area of the parallelogram. Also find the distance between its shorter sides.

Solution:

It is given that

ABCD is a parallelogram

AB = 12 cm, BC = 10 cm and AC = 16 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 58

Area of triangle ABC

BC = a = 10 cm

AC = b = 16 cm

AB = c = 12 cm

We know that

s = (a + b + c)/ 2

Substituting the values

s = (10 + 16 + 12)/ 2

By further calculation

s = 38/2 = 19 cm

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 59

So we get

= 3 √399 cm2

We know that

Area of parallelogram = 2 × Area of triangle ABC

Substituting the values

= 2 × 3 √399

= 6 √399

So we get

= 6 × 19.96

= 119.8 cm2

Consider DM as the distance between the shorter lines

Base = AD = BC = 10 cm

Area of parallelogram = AD × DM

Substituting the values

119.8 = 10 × DM

By further calculation

DM = 119.8/ 10

DM = 11.98 cm

Therefore, the distance between shorter lines is 11.98 cm.

24. Diagonals AC and BD of a parallelogram ABCD intersect at O. Given that AB = 12 cm and perpendicular distance between AB and DC is 6 cm. Calculate the area of the triangle AOD.

Solution:

It is given that

ABCD is a parallelogram

AC and BD are the diagonal which intersect at O

AB = 12 cm and DM = 6 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 60

We know that

Area of parallelogram ABCD = AB × DM

Substituting the values

= 12 × 6

= 72 cm2

Similarly

Area of triangle AOD = ¼ × Area of parallelogram

Substituting the values

= ¼ × 72

= 18 cm2

25. ABCD is a parallelogram with side AB = 10 cm. Its diagonals AC and BD are of length 12 cm and 16 cm respectively. Find the area of the parallelogram ABCD.

Solution:

It is given that

ABCD is a parallelogram

ML Aggarwal Solutions for Class 9 Chapter 16 Image 61

AB = 10 cm, AC = 12 cm

AO = CO = 12/2 = 6 cm

BD = 16 cm

BO = OD = 16/2 = 8 cm

In triangle AOB

a = 10 cm, b = AO = 6 cm, c = BO = 8 cm

We know that

s = (a + b + c)/ 2

Substituting the values

s = (10 + 6 + 8)/ 2

By further calculation

s = 24/2 = 12 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 62

So we get

= 12 × 2

= 24 cm2

We know that

Area of parallelogram ABCD = 4 × Area of triangle AOB

Substituting the values

= 4 × 24

= 96 cm2

26. The area of a parallelogram is p cm2 and its height is q cm. A second parallelogram has equal area but its base is r cm more than that of the first. Obtain an expression in terms of p, q and r for the height h of the second parallelogram.

Solution:

It is given that

Area of a parallelogram = p cm2

Height of first parallelogram = q cm

We know that

Area of parallelogram = base × height

Substituting the values

p = base × q

Base = p/q

Here

Base of second parallelogram = (p/q + r)

Taking LCM

= (p + qr)/ q cm

Area of second parallelogram = Area of first parallelogram = p cm2

It can be written as

Base × height = p cm2

Substituting the values

[(p + qr)/q] × h = p

So we get

h = pq/ (p + qr) cm

Therefore, the height of second parallelogram is h = pq/ (p + qr) cm.

27. What is the area of a rhombus whose diagonals are 12 cm and 16 cm?

Solution:

It is given that

ABCD is a rhombus

BD = 12 cm and AC = 16 cm are diagonals

ML Aggarwal Solutions for Class 9 Chapter 16 Image 63

We know that

Area of rhombus ABCD = ½ × AC × BD

Substituting the values

= ½ × 16 × 12

By further calculation

= 8 × 12

= 96 cm2

28. The area of a rhombus is 98 cm2. If one of its diagonal is 14 cm, what is the length of the other diagonal?

Solution:

It is given that

Area of rhombus = 98 cm2

One of its diagonal = 14 cm

We know that

Area of rhombus = ½ × product of diagonals

Substituting the values

98 = ½ × one diagonal × other diagonal

98 = ½ × 14 × other diagonal

By further calculation

Other diagonal = (98 × 2)/ 14

= 7 × 2

= 14 cm

Therefore, the other diagonal is 14 cm.

29. The perimeter of a rhombus is 45 cm. If its height is 8 cm, calculate its area.

Solution:

It is given that

ABCD is a rhombus

Consider x cm as each side

ML Aggarwal Solutions for Class 9 Chapter 16 Image 64

Perimeter = 45 cm

AB + BC + CD + AD = 45 cm

Substituting the values

x + x + x + x = 45

4x = 45

By division

x = 45/4 cm

We know that

Height = 8 cm

Area of rhombus = base × height

Substituting the values

= 45/4 × 8

= 45 × 2

= 90 cm2

30. PQRS is a rhombus. If it is given that PQ = 3 cm and the height of the rhombus is 2.5 cm, calculate its area.

Solution:

It is given that

PQRS is a rhombus

ML Aggarwal Solutions for Class 9 Chapter 16 Image 65

PQ = 3 cm

Height = 2.5 cm

Consider PQ as the base of rhombus PQRS.

SM = 2.5 cm is the height of rhombus

We know that

Area of rhombus PQRS = base × height

Substituting the values

= 3 × 2.5

= 7.5 cm2

31. If the diagonals of a rhombus are 8 cm and 6 cm, find its perimeter.

Solution:

Consider ABCD as a rhombus with AC and BD as two diagonals

Here

AC = 8 cm and BD = 6 cm

AO = 4 cm and BO = 3 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 66

In triangle ABC

Using Pythagoras theorem

AB2 = AO2 + BO 2

Substituting the values

AB2 = 42 + 32

By further calculation

AB2 = 16 + 9 = 25

So we get

AB = √25 = 5 cm

Side of rhombus ABCD = 5 cm

Here

Perimeter of rhombus = 4 × side

Substituting the values

= 4 × 5

= 20 cm

32. If the sides of a rhombus are 5 cm each and one diagonal is 8 cm, calculate

(i) the length of the other diagonal, and

(ii) the area of the rhombus.

Solution:

It is given that

ABCD is a rhombus with side AB, BC, CD and AD

AB = BC = CD = AD = 5 cm

AC = 8 cm and AO = 4 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 67

In triangle AOB

Using Pythagoras theorem

AB2 = AO2 + BO2

Substituting the values

52 = 42 + BO2

By further calculation

25 = 16 + BO2

BO2 = 25 – 16 = 9

So we get

BO = √9 = 3 cm

BD = 2 × BO = 2 × 3 = 6 cm

Length of other diagonal = 6 cm

Area of rhombus = ½ × product of diagonals

Substituting the values

= ½ × 8 × 6

By further calculation

= 4 × 6

= 24 cm2

33. (a) The diagram (i) given below is a trapezium. Find the length of BC and the area of the trapezium. Assume AB = 5 cm, AD = 4 cm, CD = 8 cm.

(b) The diagram (ii) given below is a trapezium. Find

(i) AB

(ii) area of trapezium ABCD

(c) The cross-section of a canal is shown in figure (iii) given below. If the canal is 8 m wide at the top and 6 m wide at the bottom and the area of the cross-section is 16.8 m2, calculate its depth.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 68

Solution:

(a) It is given that

ABCD is a trapezium

AB = 5 cm, AD = 4 cm and CD = 8 cm

Construct BN perpendicular to CD

ML Aggarwal Solutions for Class 9 Chapter 16 Image 69

Here

BN = 4 cm

CN = CD – ND

CN = CD – AO

CN = 8 – 5 = 3 cm

In triangle BCN

Using Pythagoras theorem

BC2 = BN2 + CN2

Substituting the values

BC2 = 42 + 32

By further calculation

BC2 = 16 + 9 = 25

BC = √25 = 5 cm

Length of BC = 5 cm

Area of trapezium = ½ × sum of parallel sides × height

It can be written as

= ½ × (AB + CD) × AD

Substituting the values

= ½ × (5 + 8) × 4

By further calculation

= ½ × 13 × 4

So we get

= 13 × 2

= 26 cm2

Area of trapezium = 26 cm2

(b) From the figure (ii)

AD = 8 units

BC = 2 units

CD = 10 units

Construct CN perpendicular to AD

AN = 2 units

ML Aggarwal Solutions for Class 9 Chapter 16 Image 70

We know that

DN = AD – DN

Substituting the values

= 8 – 6

= 2 units

In triangle CDN

Using Pythagoras theorem

CD2 = DN2 + NC2

Substituting the values

102 = 62 + NC2

By further calculation

NC2 = 102 – 62

NC2 = 100 – 36 = 64

So we get

NC = √64 = 8 units

From the figure NC = AB = 8 units

We know that

Area of trapezium = ½ × sum of parallel sides × height

It can be written as

= ½ × (BC + AD) × AB

Substituting the values

= ½ × (2 + 8) × 8

By further calculation

= ½ × 10 × 8

= 5 × 8

= 40 sq. units

(c) Consider ABCD as the cross section of canal in the shape of trapezium.

AB = 6 m, DC = 8 m

Take AL as the depth of canal

ML Aggarwal Solutions for Class 9 Chapter 16 Image 71

So the area of cross-section = 16.8 m2

It can be written as

½ × sum of parallel sides × depth = 16.8

½ × (AB + DC) × AL = 16.8

Substituting the values

½ × (6 + 8) × AL = 16.8

By further calculation

½ × 14 × AL = 16.8

AL = (16.8 × 2)/ 14

So we get

AL = (16.8 × 1)/7

AL = 2.4 m

34. The distance between parallel sides of a trapezium is 12 cm and the distance between mid-points of other sides is 18 cm. Find the area of the trapezium.

Solution:

Consider ABCD as a trapezium in which AB || DC

Height CL = 12 cm

E and F are the mid-points of sides AD and BC

EF = 18 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 72

We know that

EF = ½ (AB + DC) = 18 cm

Here

Area of trapezium ABCD = ½ (AB + DC) × height

Substituting the values

= 18 × 12

= 216 cm2

35. The area of a trapezium is 540 cm2. If the ratio of parallel sides is 7: 5 and the distance between them is 18 cm, find the length of parallel sides.

Solution:

It is given that

Area of trapezium = 540 cm2

Ratio of parallel sides = 7: 5

Consider 7x cm as one parallel side

Other parallel side = 5x cm

Distance between the parallel sides = height = 18 cm

We know that

Area of trapezium = ½ × sum of parallel sides × height

Substituting the values

540 = ½ × (7x + 5x) × 18

By further calculation

540 = ½ × 12x × 18

540 = 6x × 18

540 = 108x

x = 540/108 = 5

Here

First parallel side = 7x = 7 × 5 = 35 cm

Second parallel side = 5x = 5 × 5 = 25 cm

36. The parallel sides of an isosceles trapezium are in the ratio 2: 3. If its height is 4 cm and area is 60 cn2, find the perimeter.

Solution:

It is given that

ABCD is an isosceles trapezium

BC = AD

Height = 4 cm

Consider CD = 2x and AB = 3x

ML Aggarwal Solutions for Class 9 Chapter 16 Image 73

We know that

Area of trapezium = ½ (sum of parallel sides) × height

Substituting the values

60 = ½ × (2x + 3x) × 4

By further calculation

60 = ½ × 5x × 4

60 = 5x × 2

60 = 10x

So we get

x = 60/10 = 6

Here

CD = 2x = 2 × 6 = 12 cm

AB = 3x = 3 × 6 = 18 cm

AN = BM

We can write it as

AN = AB – BN

AN = AB – (MN + BM)

We know that

MN = CD

AN = AB – (CD + BM)

Similarly BM = AN

AN = AB – (CD + AN)

Substituting the values

AN = 18 – (12 + AN)

AN = 18 – 12 – AN

AN + AN = 6

2AN = 6

By division

AN = 6/2 = 3

In triangle AND

Using Pythagoras theorem

AD2 = DN2 + AN2

Here DN = 4 cm

AD2 = 42 + 32

By further calculation

AD2 = 16 + 9 = 25

So we get

AD= √25 = 5 cm

Here AD = BC = 5 cm

Perimeter of trapezium = AB + BC + CD + AD

Substituting the values

= 18 + 5 + 12 + 5

= 40 cm

37. The area of a parallelogram is 98 cm2. If one altitude is half the corresponding base, determine the base and the altitude of the parallelogram.

Solution:

It is given that

Area of parallelogram = 98 cm2

Condition – If one altitude is half the corresponding base

Take base = x cm

Corresponding altitude = x/2 cm

We know that

Area of parallelogram = base × altitude

Substituting the values

98 = x × x/2

98 = x2/2

By cross multiplication

x2 = 98 × 2 = 196

So we get

x = √196 = 14 cm

Base = 14 cm

Here altitude = 14/2 = 7 cm

38. The length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.

Solution:

The dimensions of rectangular garden are

Breadth = x m

Length = (x + 12) m

We know that

Area = l × b

Substituting the values

Area = (x + 12) × x

Area = (x2 + 12x) m2

Perimeter = 2 (l + b)

Substituting the values

= 2 [(x + 12) + x]

= 2 [ x + 12 + x]

= 2 [2x + 12]

= (4x + 24) m

ML Aggarwal Solutions for Class 9 Chapter 16 Image 74

Based on the question

Numerical value of area = 4 × numerical value of perimeter

x 2 + 12x = 4 × (4x + 24)

By further calculation

x2 + 12x = 16x + 96

x2 + 12x – 16x – 96 = 0

x2 – 4x – 96 = 0

It can be written as

x2 – 12x + 8x – 96 = 0

Taking out the common terms

x (x – 12) + 8 (x – 12) = 0

(x + 8) (x – 12) = 0

Here

x + 8 = 0 or x – 12 = 0

So we get

x = -8 (not possible) or x = 12

Breadth of rectangular garden = 12 m

Length of rectangular garden = 12 + 12 = 24 m

39. If the perimeter of a rectangular plot is 68 m and length of its diagonal is 26 m, find its area.

Solution:

It is given that

Perimeter of a rectangular plot = 68 m

Length of its diagonal = 26 m

ABCD is a rectangular plot of length x m and breath y m

ML Aggarwal Solutions for Class 9 Chapter 16 Image 75

Perimeter = 2 (length + breadth)

Substituting the values

68 = 2 (x + y)

By further calculation

68/2 = x + y

34 = x + y

So we get

x = 34 – y …… (1)

In triangle ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

262 = x2 + y2

x2 + y2 = 676

Now substituting the value of x in equation (1)

(34 – y)2 + y2 = 676

1156 + y2 – 68y + y2 = 676

By further calculation

2y2 – 68y + 1156 – 676 = 0

2y2 – 68y – 480 = 0

Taking 2 as common

2 (y2 – 34y – 240) = 0

y2 – 34y – 240 = 0

It can be written as

y2 – 24y – 10y – 240 = 0

Taking out the common terms

y (y – 24) – 10 (y – 24) = 0

(y – 10) (y – 24) = 0

Here

y – 10 = 0 or y – 24 = 0

y = 10 m or y = 24 m

Now substituting the value of y in equation (1)

y = 10 m, x = 34 – 10 = 24 m

y = 24 m, x = 34 – 24 = 10 m

Area in both cases = xy

= 24 × 10 or 10 × 24

= 240 m2

Therefore, the area of the rectangular block is 240 m2.

40. A rectangle has twice the area of a square. The length of the rectangle is 12 cm greater and the width is 8 cm greater than 2 side of a square. Find the perimeter of the square.

Solution:

Consider

Side of a square = x cm

Length of rectangle = (x + 12) cm

Breadth of rectangle = (x + 8) cm

We know that

Area of square = side × side = x × x = x2 cm2

Area of rectangle = l × b

Substituting the values

= (x + 12) (x + 8) cm2

Based on the question

Area of rectangle = 2 × area of square

Substituting the values

(x + 12) (x + 8) = 2 × x2

It can be written as

x (x + 8) + 12 (x + 8) = 2x2

x2 + 8x + 12x + 96 = 2x2

x2 – 2x2 + 8x + 12x + 96 = 0

By further calculation

-x2 + 20x + 96 = 0

– (x2 – 20x – 96) = 0

x2 – 20x – 96 = 0

We can write it as

x2 – 24x + 4x – 96 = 0

x (x – 24) + 4 (x – 24) = 0

(x + 4) (x – 24) = 0

Here

x + 4 = 0 or x – 24 = 0

x = -4 or x = 24 cm

Side of square = 24 cm

Perimeter of square = 4 × side

Substituting the values

= 4 × 24

= 96 cm

41. The perimeter of a square is 48 cm. The area of a rectangle is 4 cm2 less than the area of the square. If the length of the rectangle is 4 cm greater than its breadth, find the perimeter of the rectangle.

Solution:

It is given that

Perimeter of a square = 48 cm

Side = perimeter/4 = 48/4 = 12 cm

We know that

Area = side2 = 122 = 144 cm2

Area of rectangle = 144 – 4 = 140 cm2

Take breadth of rectangle = x cm

Length of rectangle = x + 4 cm

So the area = (x + 4) × x cm2

Substituting the values

(x + 4) x = 140

By further calculation

x2 + 4x – 140 = 0

x2 + 14x – 10x – 140 = 0

Taking out the common terms

x (x + 14) – 10 (x + 14) = 0

(x + 14) (x – 10) = 0

Here

x + 14 = 0 where x = – 14

x – 10 = 0 where x = 10

Breadth = 10 cm

Length = 10 + 4 = 14 cm

Perimeter = 2 (l + b)

Substituting the values

= 2 (14 + 10)

= 2 × 24

= 48 cm

42. In the adjoining figure, ABCD is a rectangle with sides AB = 10 cm and BC = 8 cm. HAD and BFC are equilateral triangle; AEB and DCG are right angled isosceles triangles. Find the area of the shaded region and the perimeter of the figure.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 16 Image 76

It is given that

ABCD is a rectangle with sides AB = 10 cm and BC = 8 cm

HAD and BFC are equilateral triangle with each side = 8 cm

AEB and DCG are right angled isosceles triangles with hypotenuse = 10 cm

Consider AE = EB = x cm

In triangle ABE

AE2 + EB2 = AB2

Substituting the values

x2 + x2 = 102

2x2 = 100

By further calculation

x2 = 100/2 = 50

x = √50 = √ (25 × 2) = 5√2 cm

We know that

Area of triangle AEB = Area of triangle GCD

It can be written as

= ½ × x × x

= ½ x2 cm2

Substituting the value of x

= ½ × 50

= 25 cm2

Area of triangle HAD = Area of BFC

It can be written as

= √3/4 × 82

= √3/4 × 64

= 16√3 cm2

Area of shaded portion = Area of rectangle ABCD + 2 area of triangle AEB + 2 area of triangle BFC

Substituting the values

= (10 × 8 + 2 × 25 + 2 × 16√3)

By further calculation

= 80 + 50 + 32√3

So we get

= (130 + 32√3) cm2

Here

Perimeter of the figure = AE + EB + BF + FC + CD + GD + DH + HA

It can be written as

= 4AE + 4BF

Substituting the values

= 4 × 5√2 + 4 × 8

So we get

= 20√2 + 32

= (32 + 20√2) cm

43. (a) Find the area enclosed by the figure (i) given below, where ABC is an equilateral triangle and DEFG is an isosceles trapezium. All measurements are in centimetres.

(b) Find the area enclosed by the figure (ii) given below. AH measurements are in centimetres.

(c) In the figure (iii) given below, from a 24 cm × 24 cm piece of cardboard, a block in the shape of letter M is cut off. Find the area of the cardboard left over, all measurements are in centimetres.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 77

Solution:

(a) It is given that

ABC is an equilateral triangle and DEFG is an isosceles trapezium

EF = GD = 5 cm

DE = 6 cm

GF = GB + BC + CF

Substituting the values

= 3 + 6 + 3

= 12 cm

AB = AC = BC = 6 cm

Join BD and CE

In right triangle CEF

CE2 = EF2 – CF2

Substituting the values

= 52 – 32

= 25 – 9

= 16

So we get

CE = √16 = 4 cm

Area of triangle ABC = √3/4 × 62

By further calculation

= √3/4 × 36

= 9√3 cm2

Area of trapezium DEFG = ½ (DE + GF) × CE

Substituting the values

= ½ × (6 + 12) × 4

By further calculation

= ½ × 18 × 4

= 36 cm2

So the area of figure = 9√3 + 36

Substituting the values

= 9 × 1.732 + 36

= 15.59 + 36

= 51.59 cm2

(b) We know that

Length of rectangle = 2 + 2 + 2 + 2 = 8 cm

Width of rectangle = 2 cm

Area of rectangle = l × b

Substituting the values

= 8 × 2

= 16 cm2

Here

Area of each trap = ½ (2 + 2) × (6 – 2)

By further calculation

= ½ × 4 × 4

= 8 cm2

So the total area = area of rectangle + area of 2 trapezium

Substituting the values

= 16 + 8 + 8

= 32 cm2

(c) We know that

Length of each rectangle = 24 cm

Width of each rectangle = 6 cm

Area of each rectangle = l × b

Substituting the values

= 24 × 6

= 144 cm2

Base of each parallelogram = 8 cm

Height of each parallelogram = 6 cm

So the area of each parallelogram = 8 × 6 = 48 cm2

Here

Area of the M-shaped figure = 2 × 144 + 2 × 48

So we get

= 288 + 96

= 384 cm2

Area of the square cardboard = 24 × 24 = 576 cm2

Area of the removing cardboard = 576 – 384 = 192 cm2

44. (a) The figure (i) given below shows the cross-section of the concrete structure with the measurements as given. Calculate the area of cross-section.

(b) The figure (ii) given below shows a field with the measurements given in metres. Find the area of the field.

(c) Calculate the area of the pentagon ABCDE shown in fig (iii) below, given that AX = BX = 6 cm, EY = CY = 4 cm, DE = DC = 5 cm, DX = 9 cm and DX is perpendicular to EC and AB.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 78

Solution:

(a) From the figure (i)

AB = 1.8 m, CD = 0.6 m, DE = 1.2 m

EF = 0.3 m, AF = 2.4 m

Construct DE to meet AB in G

∠FEG = ∠GAF = 900

So, AGEF is a rectangle

ML Aggarwal Solutions for Class 9 Chapter 16 Image 79

We know that

Area of given figure = Area of rectangle AGEF + Area of trapezium GBCD

It can be written as

= l × b + ½ (sum of parallel sides × height)

= AF × AG + ½ (GB + CD) × DG

Substituting the values

= 2.4 × 0.3 + ½ [(AB – AG) + CD] × (DE + EG)

Here AG = FE and using EG = AF

= 0.72 + ½ [(1.8 – 0.3) + 0.6] × (1.2 + 2.4)

By further calculation

= 0.72 + ½ [1.5 + 0.6] × 3.6

= 0.72 + ½ × 2.1 × 3.6

So we get

= 0.72 + 2.1 × 1.8

= 0.72 + 3.78

= 4.5 m2

(b) It is given that

ABCD is a pentagonal field

AX = 12 m, BX = 30 m, XZ = 15 m, CZ = 25 m,

DZ = 10 m, AD = 12 + 15 + 10 = 37 m, EY = 20 m

ML Aggarwal Solutions for Class 9 Chapter 16 Image 80

We know that

Area of pentagonal field ABCDE = Area of triangle ABX + Area of trapezium BCZX + Area of triangle CDZ + Area of triangle AED

It can be written as

= ½ × base × height + ½ (sum of parallel sides) × height + ½ × base × height + ½ × base × height

= ½ × BX × AX + ½ (BX + CZ) × XZ + ½ × CZ × DZ + ½ × AD × EY

Substituting the values

= ½ × 30 × 12 + ½ (30 + 25) × 15 + ½ × 25 × 10 + ½ × 37 × 20

By further calculation

= 15 × 12 + 7.5 × 55 + 25 × 5 + 37 × 10

So we get

= 180 + 412.5 + 125 + 370

= 1087.5 m2

(c) It is given that

ABCDE is a pentagon

AX = BX = 6 cm, EY = CY = 4 cm

DE = DC = 5 cm, DX = 9 cm

Construct DX perpendicular to EC and AB

ML Aggarwal Solutions for Class 9 Chapter 16 Image 81

In triangle DEY

Using Pythagoras Theorem

DE2 = DY2 + EY2

Substituting the values

52 = DY2 + 42

25 = DY2 + 16

DY2 = 25 – 16 = 9

So we get

DY = √9 = 3 cm

Here

Area of pentagonal field ABCDE = Area of triangle DEY + Area of triangle DCY + Area of trapezium EYXA + Area of trapezium CYXB

It can be written as

= ½ × base × height + ½ × base × height + ½ × (sum of parallel sides) × height + ½ × (sum of parallel sides) × height

= ½ × EY × DY + ½ × CY × DY + ½ × (EY + AX) × XY + ½ × (CY + BX) × XY

Substituting the values

= ½ × 4 × 3 + ½ × 4 × 3 + ½ × (4 + 6) × (DX – DY) + ½ (4 + 6) × (DX – DY)

By further calculation

= 2 × 3 + 2 × 2 + ½ × 10 × (9 – 3) + ½ × 10 × (9 – 3)

So we get

= 6 + 6 + 5 × 6 + 5 × 6

= 6 + 6 + 30 + 30

= 72 cm2

45. If the length and the breadth of a room are increased by 1 metre the area is increased by 21 square metres. If the length is increased by 1 metre and breadth is decreased by 1 metre the area is decreased by 5 square metres. Find the perimeter of the room.

Solution:

Take length of room = x m

Breadth of room = y m

Here

Area of room = l × b = xy m2

We know that

Length is increased by 1 m then new length = (x + 1) m

Breadth is increased by 1 m then new breadth = (y + 1) m

So the new area = new length × new breadth

Substituting the values

= (x + 1) (y + 1) m2

Based on the question

xy = (x + 1) (y + 1) – 21

By further calculation

xy = x (y + 1) + 1 (y + 1) – 21

xy = xy + x + y + 1 – 21

So we get

0 = x + y + 1 – 21

0 = x + y – 20

x + y – 20 = 0

x + y = 20 ….. (1)

Similarly

Length is increased by 1 metre then new length = (x + 1) metre

Breadth is decreased by 1 metre than new breadth = (y – 1) metre

So the new area = new length × new breadth

= (x + 1) (y – 1) m2

Based on the question

xy = (x + 1) (y – 1) + 5

By further calculation

xy = x (y – 1) + 1 (y – 1) + 5

xy = xy – x + y – 1 + 5

0 = -x + y + 4

So we get

x – y = 4 …… (2)

By adding equations (1) and (2)

2x = 24

x = 24/2 = 12 m

Now substituting the value of x in equation (1)

12 + y = 20

y = 20 – 12 = 8 m

Here

Length of room = 12 m

Breadth of room = 8 m

So the perimeter = 2 (l + b)

Substituting the values

= 2 (12 + 8)

= 2 × 20

= 40 m

46. A triangle and a parallelogram have the same base and 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.

Solution:

It is given that

Sides of a triangle = 26 cm, 28 cm and 30 cm

We know that

s = (a + b + c)/ 2

Substituting the values

s = (26 + 28 + 30)/ 2

s = 84/2

s = 42 cm

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 82

So we get

= 2 × 4 × 6 × 7

= 336 cm2

We know that

Base = 28 cm

Height = Area/base

Substituting the values

= 336/28

= 12 cm

47. A rectangle of area 105 cm2 has its length equal to x cm. Write down its breadth in terms of x. Given that its perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of the rectangle.

Solution:

It is given that

Area of rectangle = 105 cm2

Length of rectangle = x cm

We know that

Area = length × breadth

Substituting the values

105 = x × breadth

x = 105/ x cm

Perimeter of rectangle = 44 cm

So we get

2 (l + b) = 44

2 (x + 105/x) = 44

By further calculation

(x2 + 105)/ x = 22

By cross multiplication

x2 + 105 = 22x

x2 – 22x + 105 = 0

We can write it as

x2 – 15x – 7x + 105 = 0

x (x – 15) – 7 (x – 15) = 0

(x – 7) (x – 15) = 0

Here

x – 7 = 0 or x – 15 = 0

x = 7 cm or x = 15 cm

If x = 7 cm,

Breadth = 105/7 = 15 cm

If x = 15 cm,

Breadth = 105/15 = 7 cm

Therefore, the required dimensions of rectangle are 15 cm and 7 cm.

48. The perimeter of a rectangular plot is 180 m and its area is 1800 m2. Take the length of plot as x m. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the value of the length, breadth and the area to write an equation in x. Solve the equation to calculate the length and breadth of the plot.

Solution:

It is given that

Perimeter of a rectangular plot = 180 m

Area of a rectangular plot = 1800 m2

Take length of rectangle = x m

Here

Perimeter = 2 (length + breadth)

Substituting the values

180 = 2 (x + breadth)

180/2 = x + breadth

x + breadth = 90

Breadth = 90 – x m

We know that

Area of rectangle = l × b

Substituting the values

1800 = x × (90 – x)

90x – x2 = 1800

It can be written as

– (x2 – 90x) = 1800

x2 – 90x = – 1800

By further calculation

x2 – 90x + 1800 = 0

x2 – 60x – 30x + 1800 = 0

x (x – 60) – 30 (x – 60) = 0

(x – 30) (x – 60) = 0

Here

x – 30 = 0 or x – 60 = 0

x = 30 or x = 60

If x = 30 m

Breadth = 90 – 30 = 60 m

If x = 60 m

Breadth = 90 – 60 = 30 m

Therefore, the required length of rectangle is 60 m and the breadth of rectangle is 30 m.

Exercise 16.3

1. Find the length of the diameter of a circle whose circumference is 44 cm.

Solution:

Consider radius of the circle = r cm

Circumference = 2 πr

We know that

2 πr = 44

So we get

(2 × 22)/ 7 r = 44

By further calculation

r = (44 × 7)/ (2 × 22) = 7 cm

Diameter = 2r = 2 × 7 = 14 cm

2. Find the radius and area of a circle if its circumference is 18π cm.

Solution:

Consider the radius of the circle = r

Circumference = 2 πr

We know that

2 πr = 18π

So we get

2r = 18

r = 18/2 = 9 cm

Here

Area = πr2

Substituting the value of r

= π × 9 × 9

= 81π cm2

3. Find the perimeter of a semicircular plate of radius 3.85 cm.

Solution:

It is given that

Radius of semicircular plate = 3.85 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 83\

We know that

Length of semicircular plat = πr

Perimeter = πr + 2r = r (π + 2)

Substituting the values

= 3.85 (22/7 + 2)

By further calculation

= 3.85 × 36/7

= 0.55 × 36

= 19.8 cm

4. Find the radius and circumference of a circle whose area is 144 π cm2.

Solution:

It is given that

Area of a circle = 144 π cm2

Consider radius = r

π r2 = 144 π

r2 = 144

So we get

r = √144 = 12 cm

Here

Circumference = 2 πr

So we get

= 2 × 12 × π

= 24π cm

5. A sheet is 11 cm long and 2 cm wide. Circular pieces 0.5 cm in diameter are cut from it to prepare discs. Calculate the number of discs that can be prepared.

Solution:

It is given that

Length of sheet = 11 cm

Width of sheet = 2 cm

We have to cut the sheet to a square of side 0.5 cm

Here

Number of squares = 11/0.5 × 2/0.5

Multiply and divide by 10

= (11 × 10)/5 × (2 × 10)/5

By further calculation

= 22 × 4

= 88

Hence, the number of discs will be equal to number of squares cut out = 88.

6. If the area of a semicircular region is 77 cm2, find its perimeter.

Solution:

It is given that

Area of a semicircular region = 77 cm2

Consider r as the radius of the region

ML Aggarwal Solutions for Class 9 Chapter 16 Image 84

We know that

Area = ½ πr2

½ πr2 = 77

By further calculation

½ × 22/7 r2 = 77

So we get

r2 = (77 × 2 × 7)/ 22 = 49 = 72

r = 7 cm

Here

Perimeter of the region = πr + 2r

By further calculation

= 22/7 × 7 + 2 × 7

So we get

= 22 + 14

= 36 cm

7. (a) In the figure (i) given below, AC and BD are two perpendicular diameters of a circle ABCD. Given that the area of shaded portion is 308 cm2, calculate

(i) the length of AC and

(ii) the circumference of the circle

(b) In the figure (ii) given below, AC and BD are two perpendicular diameters of a circle with centre O. If AC = 16 cm, calculate the area and perimeter of the shaded part. (Take π = 3.14)

ML Aggarwal Solutions for Class 9 Chapter 16 Image 85

Solution:

(a) It is given that

Area of shaded portion = Area of semicircle = 308 cm2

Consider r as the radius of circle

½ πr2 = 308

By further calculation

½ × 22/7 r2 = 308

r2 = (308 × 2 × 7)/ 22

r2 = 196 = 142

So r = 14 cm

(i) AC = 2r = 2 × 14 = 28 cm

(ii) We know that

Circumference of the circle = 2πr

Substituting the values

= 28 × 22/7

So we get

= 4 × 22

= 88 cm

(b) We know that

Diameter of circle = 16 cm

Radius of circle = 16/2 = 8 cm

Here

Area of shaded part = 2 × area of one quadrant

So we get

= ½ πr2

Substituting the values

= ½ × 3.14 × 8 × 8

= 100.48 cm2

We know that

Perimeter of shaded part = ½ of circumference + 4r

It can be written as

= ½ × 2πr + 4r

= πr + 4r

Taking r as common

= r (π + 4)

Substituting the values

= 8 (3.14 + 4)

= 8 × 7.14

= 57.12 cm

8. A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77 cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/sec, calculate the number of complete revolutions the wheel makes in raising the bucket.

Solution:

It is given that

Diameter of wheel = 77 cm

So radius of wheel = 77/2 cm

We know that

Circumference of wheel = 2πr

Substituting the values

= 2 × 22/7 × 77/2

= 242 cm

9. The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm, find its speed in km/hr. Give your answer correct to the nearest km.

Solution:

It is given that

Diameter of wheel = 84 cm

Radius of wheel = 84/2 = 42 cm

Here

Circumference of wheel = 2πr

Substituting the values

= 2 × 22/7 × 42

= 264 cm

So the distance covered in 5 reductions = 264 × 5 = 1320 cm

Time = 1 second

We know that

Speed of wheel = 1320/1 × (60 × 60)/ (100 × 1000)

= 47.25 km/hr

= 48 km/hr

10. The circumference of a circle is 123.2 cm. Calculate:

(i) the radius of the circle in cm.

(ii) the area of the circle in cm2, correct to the nearest cm2.

(iii) the effect on the area of the circle if the radius is doubled.

Solution:

It is given that

Circumference of a circle = 123.2 cm

Consider radius = r cm

(i) We know that

2πr = 123.2

By further calculation

(2 × 22)/ 7 r = 1232/10

So we get

r = (1232 × 7)/ (10 × 2 × 22)

r = 19.6 cm

Therefore, radius of the circle is 19.6 cm

(ii) Here

Area of the circle = πr2

Substituting the values

= 22/7 × 19.6 × 19.6

= 1207.36

= 1207 cm2

(iii) We know that

If radius is doubled = 19.6 × 2 = 39.2 cm

So the area of circle = πr2

Substituting the values

= 22/7 × 39.2 × 39.2

= 4829.44 cm2

Effect on area = 4829.44/1207 = 4 times

11. (a) In the figure (i) given below, the area enclosed between the concentric circles is 770 cm2. Given that the radius of the outer circle is 21 cm, calculate the radius of the inner circle.

(b) In the figure (ii) given below, the area enclosed between the circumferences of two concentric circles is 346.5 cm2. The circumference of the inner circle is 88 cm. Calculate the radius of the outer circle.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 86

Solution:

(a) It is given that

Radius of outer circle (R) = 21 cm

Consider r cm as the radius of inner circle

We know that

Area of the ring = π (R2 – r2)

Substituting the values

= 22/7 (212 – r2)

= 22/7 (441 – r2)

Area of the ring = 770 cm2

So we get

22/7 (441 – r2) = 770

By further calculation

441 – r2 = (770 × 7)/ 22 = 245

r2 = 441 – 245 = 196

r = √196 = 14

Hence, the radius of inner circle is 14 cm.

(b) It is given that

Area of ring = 346.5 cm2

Circumference of inner circle = 88 cm

We know that

Radius = (88 × 7)/ (2 × 22) = 14 cm

Consider R cm as the radius of outer circle

Area of ring = π (R2 – r2)

Substituting the values

= 22/7 (R2 – 142)

= 22/7 (R2 – 196) cm2

Area of ring = 346.5 cm2

By equating we get

22/7 (R2 – 196) = 346.5

By further calculation

R2 – 196 = (346.5 × 7)/ 22 = 110.25

R2 = 110.25 + 196 = 306.25

So we get

R = √306.25 = 17.5

Hence, the radius of outer circle is 17.5 cm.

12. A road 3.5 m wide surrounds a circular plot whose circumference is 44m. Find the cost of paving the road at Rs 50 per m2.

Solution:

It is given that

Circumference of circular plot = 44 m

Radius of circular plot = (44 × 7)/ (22 × 2) = 7 m

Width of the road = 3.5 m

So the radius of outer circle = 7 + 3.5 = 10.5 m

We know that

Area of road = π (R2 – r2)

Substituting the values

= 22/7 (10.52 – 72)

We can write it as

= 22/7 (10.5 + 7) (10 – 7)

= 22/7 × 17.5 × 3.5

= 192.5 m2

Here

Rate of paving the road = Rs 50 per m2

Total cost = 192.5 × 50 = Rs 9625

13. The sum of diameters of two circles is 14 cm and the difference of their circumferences is 8 cm. Find the circumference of the two circles.

Solution:

It is given that

Sum of the diameters of two circles = 14 cm

Consider R and r as the radii of two circles

2R + 2r = 14

Dividing by 2

R + r = 7 …… (1)

We know that

Difference of their circumferences = 8 cm

2 πR – 2 πr = 8

Taking out the common terms

2 π (R – r) = 8

(2 × 22)/ 7 (R – r) = 8

By further calculation

R – r = (8 × 7)/ (2 × 22) = 14/11 ….. (2)

By adding both the equations

2R = 7 + 14/11

By taking LCM

2R = (77 + 14)/ 11 = 91/11

By further calculation

R = 91/ (11 × 2) = 91/22

From equation (1)

R + r = 7

Substituting the value of R

91/22 + r = 7

r = 7 – 91/22

Taking LCM

r = (154 – 91)/ 22 = 63/22

We know that

Circumference of first circle = 2 πr

Substituting the values

= 2 × 22/7 × 91/22

= 26 cm

Circumference of second circle = 2 π R

Substituting the values

= 2 × 22/7 × 63/22

= 18 cm

14. Find the circumference of the circle whose area is equal to the sum of the areas of three circles with radius 2 cm, 3 cm and 6 cm.

Solution:

It is given that

Radius of first circle = 2 cm

Area of first circle = πr2

= π (2)2

= 4 π cm2

Radius of second circle = 3 cm

Area of first circle = πr2

= π (3)2

= 9 π cm2

Radius of second circle = 6 cm

Area of first circle = πr2

= π (6)2

= 36 π cm2

So the total area of the three circles = 4 π + 9 π + 36 π = 49 π cm2

Area of the given circle = 49 π cm2

We know that

Radius = √ (49 π/ π) = √49 = 7 cm

Circumference = 2 πr = 2 × 22/7 × 7 = 44 cm

15. A copper wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent into the form of a circle, find the area of the circle.

Solution:

It is given that

Area of square = 121 cm2

So side = √121 = 11 cm

Perimeter = 4a = 4 × 11 = 44 cm

Circumference of circle = 44 cm

Radius of circle = (44 × 7)/ (2 × 22) = 7 cm

We know that

Area of the circle = πr2

Substituting the values

= 22/7 (7)2

So we get

= 22/7 × 7× 7

= 154 cm2

16. A copper wire when bent into an equilateral triangle has area 121 √3 cm2. If the same wire is bent into the form of a circle, find the area enclosed by the wire.

Solution:

It is given that

Area of equilateral triangle = 121 √3 cm2

Consider a as the side of triangle

Area = √3/4 a2

It can be written as

√3/4 a2 = 121√3

By further calculation

a2 = (121 × √3 × 4)/ √3

a2 = 484

So we get

a = √484 = 22 cm

Here

Length of the wire = 66 cm

Radius of the circle = 66/2π

By further calculation

= (66 × 7)/ (2 × 22)

= 21/2 cm

We know that

Area of the circle = πr2

By further calculation

= 22/7 × (21/2)2

= 22/7 × 21/2 × 21/2

So we get

= 693/2

= 346.5 cm2

17. (a) Find the circumference of the circle whose area is 16 times the area of the circle with diameter 7 cm.

(b) In the given figure, find the area of the unshaded portion within the rectangle. (Take π = 3.14)

ML Aggarwal Solutions for Class 9 Chapter 16 Image 87

Solution:

(a) It is given that

Diameter of the circle = 7 cm

Radius of the circle = 7/2 cm

We know that

Area of the circle = πr2

Substituting the values

= 22/7 × 7/2 × 7/2

= 77/2 cm2

Here

Area of bigger circle = 77/2 × 16 = 616 cm2

Consider r as the radius

πr2 = 616

We can write it as

22/7 r2 = 616

By further calculation

r2 = (616 × 7)/ 22

r2 = 196 cm2

So we get

r = √196 = 14 cm

Circumference = 2πr

Substituting the values

= 2 × 22/7 × 14

= 88 cm

(b) It is given that

Radius of each circle = 3 cm

Diameter of each circle = 2 × 3 = 6 cm

Here

Length of rectangle (l) = 6 + 6 + 3 = 15 cm

Breadth of rectangle (b) = 6 cm

So the area of rectangle = l × b

Substituting the values

= 15 × 6

= 90 cm2

We know that

Area of 2 ½ circles = 5/2 πr2

Substituting the values

= 5/2 × 3.14 × 3 × 3

By further calculation

= 5 × 1.57 × 9

= 70.65 cm2

So the area of unshaded portion = 90 – 70.65

= 19.35 cm2

18. In the adjoining figure, ABCD is a square of side 21 cm. AC and BD are two diagonals of the square. Two semi-circles are drawn with AD and BC as diameters. Find the area of the shaded region. Take π = 22/7.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 88

Solution:

It is given that

Side of square = 21 cm

So the area of square = side2 = 212 = 441 cm2

We know that

∠AOD + ∠COD + ∠AOB + ∠BOC = 441

Substituting the values

x + x + x + x = 441

4x = 441

So we get

x = 441/4 = 110.25 cm2

Based on the question

We should find the area of shaded portion in square ABCD which is ∠AOD and ∠BOC

∠AOD + ∠BOC = 110.25 + 110.25 = 220.5 cm2

Here

Area of two semicircle = πr2

Substituting the values

= 22/7 × 10.5 × 10.5

= 346.50 cm2

So the area of shaded portion = 220.5 + 346.5 = 567 cm2

19. (a) In the figure (i) given below, ABCD is a square of side 14 cm and APD and BPC are semicircles. Find the area and the perimeter of the shaded region.

(b) In the figure (ii) given below, ABCD is a square of side 14 cm. Find the area of the shaded region.

(c) In the figure (iii) given below, the diameter of the semicircle is equal to 14 cm. Calculate the area of the shaded region. Take π = 22/7.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 89

Solution:

(a) It is given that

ABCD is a square of each side (a) = 14 cm

APD and BPC are semi-circle with diameter = 14 cm

Radius of each semi-circle (a) = 14/2 = 7 cm

(i) We know that

Area of square = a2 = 142 = 196 cm2

Area of two semicircles = 2 × ½ πr2

= πr2

Substituting the values

= 22/7 × 7 × 7

= 154 cm2

So the area of shaded portion = 196 – 154 = 42 cm2

(ii) Here

Length of arcs of two semicircles = 2πr

Substituting the values

= 2 × 22/7 × 7

= 44 cm

So the perimeter of shaded portion = 44 + 14 + 14 = 72 cm

(b) It is given that

ABCD is a square whose each side (a) = 14 cm

4 circles are drawn which touch each other and the sides of squares

Radius of each circle (r) = 7/2 = 3.5 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 90

(i) We know that

Area of square ABCD = a2 = 142 = 196 cm2

Area of 4 circles = 4 × πr2

Substituting the values

= 4 × 22/7 × 7/2 × 7/2

= 154 cm2

So the area of shaded portion = 196 – 154 = 42 cm2

(ii) Here

Perimeter of 4 circles = 4 × 2 πr

Substituting the values

= 4 × 2 × 22/7 × 7/2

= 88 cm

So the perimeter of shaded portion = perimeter of 4 circles + perimeter of square

Substituting the values

= 88 + 4 × 14

So we get

= 88 + 56

= 144 cm

(c) We know that

Area of rectangle ACDE = ED × AE

Substituting the values

= 14 × 7

= 98 cm2

ML Aggarwal Solutions for Class 9 Chapter 16 Image 91

Here

Area of semicircle DEF = πr2/2

Substituting the values

= (22 × 7 × 7)/ (7 × 2)

= 77 cm2

So the area of shaded region = 77 + (98 – 2 × ¼ × 22/7 × 7 × 7)

= 77 + 21

= 98 cm2

20. (a) Find the area and the perimeter of the shaded region in figure (i) given below. The dimensions are in centimetres.

(b) In the figure (ii) given below, area of △ABC = 35 cm2. Find the area of the shaded region.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 92

Solution:

(a) It is given that

There are 2 semicircles where the smaller is inside the larger

Radius of larger semicircles (R) = 14 cm

Radius of smaller circle (r) = 14/2 = 7 cm

(i) We know that

Area of shaded portion = Area of larger semicircle – Area of smaller circle

= ½ πR2 – ½ πr2

We can write it as

= ½ π (R2 – r2)

Substituting the values

= ½ × 22/7 (142 – 72)

By further calculation

= 11/7 (14 + 7) (14 – 7)

So we get

= 11/7 × 21 × 7

= 231 cm2

(ii) Here

Perimeter of shaded portion = circumference of larger semicircle + circumference of smaller semicircle + radius of larger semicircle

= πR + πr + R

Substituting the values

= 22/7 × 14 + 22/7 × 7 + 14

By further calculation

= 44 + 22 + 14

= 80 cm

(b) We know that

Area of △ABC formed in a semicircle = 3.5 cm

Altitude CD = 5 cm

So the base AB = (area × 2)/ altitude

Substituting the values

= (35 × 2)/ 5

= 14 cm

Here

Diameter of semicircle = 14 cm

Radius of semicircle (R) = 14/2 = 7 cm

So the area of semicircle = ½ πR2

Substituting the values

= ½ × 22/7 × 7 × 7

= 77 cm2

Area of shaded portion = Area of semicircle – Area of triangle

= 77 – 35

= 42 cm2

21. (a) In the figure (i) given below, AOBC is a quadrant of a circle of radius 10m. Calculate the area of the shaded portion. Take π = 3.14 and give your answer correct to two significant figures.

(b) In the figure (ii) given below, OAB is a quadrant of a circle. The radius OA = 3.5 cm and OD = 2 cm. Calculate the area of the shaded portion.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 93

Solution:

(a) In the figure

Shaded portion = Quadrant – △AOB

Radius of the quadrant = 10 m

Here

Area of quadrant = ¼ πr2

Substituting the values

= ¼ × 3.14 × 10 × 10

By further calculation

= (3.14 × 100)/ 4

= 314/4

= 78.5 m2

We know that

Area of △AOB = ½ × AO × OB

Substituting the values

= ½ × 10 × 10

= 50 m2

So the area of shaded portion = 78.5 – 50 = 28.5 m2

(b) In the figure

Radius of quadrant = 3.5 cm

(i) We know that

Area of quadrant = ¼ πr2

Substituting the values

= ¼ × 22/7 × 3.5 × 3.5

= 9.625 cm2

(ii) Here

Area of △AOD = ½ × AO × OD

Substituting the values

= ½ × 3.5 × 2

= 3.5 cm2

So the area of shaded portion = Area of quadrant – Area of △AOD

Substituting the values

= 9.625 – 3.6

= 6.125 cm2

22. A student takes a rectangular piece of paper 30 cm long and 21 cm wide. Find the area of the biggest circle that can be cut out from the paper. Also find the area of the paper left after cutting out the circle. (Take π = 22/7)

Solution:

It is given that

Length of rectangle = 30 cm

Width of rectangle = 21 cm

We know that

Area of rectangle = l × b

= 30 × 21

= 630 cm2

So the radius of the biggest circle = 21/2 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 94

Here

Area of the circle = πr2

Substituting the values

= 22/7 × 21/2 × 21/2

So we get

= 693/2

= 346.5 cm2

So the area of remaining part = 630 – 346.5 = 283.5 cm2

23. A rectangle with one side 4 cm is inscribed in a circle of radius 2.5 cm. Find the area of the rectangle.

Solution:

Consider ABCD as a rectangle

AB = 4 cm

Diameter of circle AC = 2.5 × 2 = 5 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 95

Here

ML Aggarwal Solutions for Class 9 Chapter 16 Image 96

So we get

= 3 cm

We know that

Area of rectangle = AB × BC

Substituting the values

= 4 × 3

= 12 cm2

24. (a) In the figure (i) given below, calculate the area of the shaded region correct to two decimal places. (Take π = 3.142)

(b) In the figure (ii) given below, ABC is an isosceles right angled triangle with ∠ABC = 900. A semicircle is drawn with AC as diameter. If AB = BC = 7 cm, find the area of the shaded region. Take π = 22/7.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 97

Solution:

(a) We know that

ABCD is a rectangle which is inscribed in a circle of length = 12 cm

Width = 5 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 98

= 13 cm

Diameter of circle = AC = 13 cm

Radius of circle = 13/2 = 6.5 cm

Here

Area of circle = πr2

Substituting the values

= 3.142 × (6.5)2

By further calculation

= 3.142 × 42.25

= 132.75 cm2

Area of rectangle = l × b

Substituting the values

= 12 × 5

= 60 cm2

So the area of the shaded portion = 132.75 – 60 = 72.75 cm2

(b) We know that

Area of △ABC = ½ × AB × BC

Substituting the values

= ½ × 7 × 7

= 49/2 cm2

Here

AC2 = AB2 + BC2

Substituting

= 49 + 49

So we get

AC = 7√2

Radius of semi-circle = 7√2/ 2 cm

Area of semi-circle = π/2 × (7√2/ 2)2

By further calculation

= ½ × 22/7 × 98/4

= 77/2 cm2

Area of the shaded region = Area of the semi-circle – Area of △ABC

Substituting the values

= 77/2 – 49/2

= 28/2

= 14 cm2

25. A circular field has perimeter 660 m. A plot in the shape of a square having its vertices on the circumference is marked in the field. Calculate the area of the square field.

Solution:

It is given that

Perimeter of circular field = 660 m

Radius of the field = 660/2 π

Substituting the values

= (660 × 7)/ (2 × 22)

= 105 m

Here

ABCD is a square which is inscribed in the circle where AC is the diagonal which is the diameter of the circular field

ML Aggarwal Solutions for Class 9 Chapter 16 Image 99

Consider a as the side of the square

AC = √2 a

a = AC/√2

Substituting the values

a = (105 × 2)/ √2

Multiply and divide by √2

a = (105 × 2 × √2)/ (√2 × √2)

By further calculation

a = (105 × 2 × √2)/ 2

a = 105 √2 m

We know that

Area of the square = a2

It can be written as

= (105√2)2

= 105√2 × 105√2

= 22050 m2

26. In the adjoining figure, ABCD is a square. Find the ratio between

(i) the circumferences

(ii) the areas of the incircle and the circumcircle of the square.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 100

Solution:

Consider side of the square = 2a

Area = (2a)2 = 4a2

Diagonal of AC = √2AB

ML Aggarwal Solutions for Class 9 Chapter 16 Image 101

(i) We know that

Radius of the circumcircle = ½ AC

It can be written as

= ½ (√2 × AB)

Substituting the values

= √2/2 × 2a

= √2a

Circumference = 2 πr = 2 × π × √2a = 2√2 πa

Radius of incircle = AB = ½ × 2a = a

Circumference = 2 πr = 2 πa

Here

Ratio between the circumference incircle and circumcircle = 2 πa: 2√2 πa

= 1: √2

(ii) We know that

Area of incircle = πr2 = πa2

Area of circumcircle = πR2

= π (√2a)2

= π2a2

= 2 πa2

So the ratio = πa2: 2 πa2 = 1: 2

27. (a) The figure (i) given below shows a running track surrounding a grassed enclosure PQRSTU. The enclosure consists of a rectangle PQST with a semicircular region at each end.

PQ = 200 m, PT = 70 m

ML Aggarwal Solutions for Class 9 Chapter 16 Image 102

(i) Calculate the area of the grassed enclosure in m2.

(ii) Given that the track is of constant width 7 m, calculate the outer perimeter ABCDEF of the track.

(b) In the figure (ii) given below, the inside perimeter of a practice running track with semi-circular ends and straight parallel sides is 312 m. The length of the straight portion of the track is 90 m. If the track has a uniform width of 2m throughout, find its area.

Solution:

(a) It is given that

Length of PQ = 200 m

Width PT = 70 m

(i) We know that

Area of rectangle PQST = l × b

= 200 × 70

= 14000 m2

Radius of each semi-circular part on either side of rectangle = 70/2 = 35 m

Area of both semi-circular parts = 2 × ½ πr2

Substituting the values

= 22/7 × 35 × 35

= 3850 m2

So the total area of grassed enclosure = 1400 + 3850 = 17850 m2

(ii) We know that

Width of track around the enclosure = 7 m

Outer length = 200 m

So the width = 70 + 7 × 2

= 70 + 14

= 84 m

Outer radius = 84/2 = 42 m

Here

Circumference of both semi-circular part = 2πr

Substituting the values

= 2 × 22/7 × 42

= 264 m

Outer perimeter = 264 + 200 × 2

= 264 + 400

= 664 m

(b) It is given that

Inside perimeter = 312 m

Total length of the parallel sides = 90 + 90 = 180 m

Circumference of two semi-circles = 312 – 180 = 132 m

Here

Radius of each semi-circle = 132/2π

So we get

= 66/3.14

= 21.02 m

Diameter of each semi-circle = 66/π × 2

So we get

= 132/ π

= 132/3.14

Multiply and divide by 100

= (132 × 100)/ 314

= 42.04 m

Width of track = 2 m

Outer diameter = 42.04 + 4 = 46.04 m

Radius = 46.04/2 = 23.02 m

We know that

Area of two semi-circles = 2 × ½ × πR2

= πR2

Substituting the values

= 3.14 × (23.02)2

= 3.14 × 23.02 × 23.02

= 1663.95 m2

Area of rectangle = 90 × 46.04

= 4143.6 m2

Total area = 1663.95 + 4143.60 = 5807.55 m2

Area of two inner circles = 2 × ½ πr2

Substituting the values

= 3.14 × 21.02 × 21.02

= 1387.38 m2

Area of inner rectangle = 90 × 42.04

= 3783.6 m2

Total inner area = 3783.60 + 1387.38

= 5170.98 m2

Here

Area of path = 5807.55 – 5170.98

= 636.57 m2

28. (a) In the figure (i) given below, two circles with centres A and B touch each other at the point C. If AC = 8 cm and AB = 3 cm, find the area of the shaded region.

(b) The quadrants shown in the figure (ii) given below are each of radius 7 cm. Calculate the area of the shaded portion.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 103

Solution:

(a) It is given that

AC = 8 cm

BC = AC – AB = 8 – 3 = 5 cm

We know that

Area of big circle of radius AC = πR2

Substituting the values

= 22/7 × 8 × 8

By further calculation

= 64 × 22/7 cm2

Area of smaller circle = πr2

Substituting the values

= 22/7 × 5 × 5

By further calculation

= (25 × 22)/ 7 cm2

Here

Area of shaded portion = (64 × 22)/ 7 – (25 × 22)/ 7

Taking out the common terms

= 22/7 (64 – 25)

By further calculation

= 22/7 × 39

= 122.57 cm2

(b) We know that

Radius of each quadrant = 7 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 104

Here

Area of shaded region = Area of square – 4 area of the quadrant

It can be written as

= (side)2 – 4 × ¼ πr2

Substituting the values

= 142 – 22/7 × 7 × 7

By further calculation

= 196 – 154

= 42 cm2

29. (a) In the figure (i) given below, two circular flower beds have been shown on the two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 105

(b) In the figure (ii) given below, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. (π = 3.14)

ML Aggarwal Solutions for Class 9 Chapter 16 Image 106

Solution:

(a) It is given that

Side of square lawn ABCD (a) = 56 cm

Area = a2 = 52 = 3136 cm2

We know that

Length of the diagonal of the square = √2 a

= √2 × 56 cm

Radius of each quadrant = (√2 × 56)/ 2

= 28 √2 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 107

So the area of each segment = ¼ πr2 – area of △OBC

Substituting the values

= ¼ × 22/7 × 28√2 × 28√2 – ½ × 28√2 × 28√2

Taking out the common terms

= 28√2 × 28√2 (1/4 × 22/7 – 1/2)

By further calculation

= 784 × 2 (11/14 – ½)

So we get

= 784 × 2 × 4/14

= 448 cm2

Here

Area of two segments = 448 × 2 = 896 cm2

So the total area of the lawn and beds = 3136 + 896 = 4032 cm2

(b) In the figure

OPBQ is a quadrant and OABC is a square which is inscribed in a side of square = 20 cm

OB is joined

Here

OB = √2 a = √2 ×20 cm

Radius of quadrant = OB = 20√2cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 108

We know that

Area of quadrant = ¼ πr2

Substituting the values

= ¼ × 3.14 × (20√2)2

By further calculation

= ¼ × 3.14 × 800

So we get

= 314 × 2

= 628 cm2

Area of square = a2 = 202 = 400 cm2

So the area of shaded portion = 628 – 400 = 228 cm2

30. (a) In the figure (i) given below, ABCD is a rectangle, AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area if the shaded portion.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 109

(b) In the figure (ii) given below, O is the centre of a circle with AC = 24 cm, AB = 7 cm and ∠BOD = 900. Find the area of the shaded region. (Use π = 3.14)

ML Aggarwal Solutions for Class 9 Chapter 16 Image 110

Solution:

(a) It is given that

ABCD is a rectangle

Three semicircles are drawn with AB = 14 cm and BC = 7 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 111

We know that

Area of rectangle ABCD = l × b

Substituting the values

= 14 × 7

= 98 cm2

Radius of each outer semicircles = 7/2 cm

So the area = 2 × ½ πr2

Substituting the values

= 22/7 × 7/2 × 7/2

= 77/2

= 38.5 cm2

Here

Area of semicircle drawn on CD as diameter = ½ πR2

Substituting the values

= ½ × 22/7 × 72

By further calculation

= 11/7 × 7 × 7

= 77 cm2

So the area of shaded region = 98 + 38.5 – 77

= 59.5 cm2

(b) In the figure

AC = 24 cm

AB = 7 cm

∠BOD = 900

ML Aggarwal Solutions for Class 9 Chapter 16 Image 112

In △ABC

Using Pythagoras theorem

BC2 = AC2 + AB2

Substituting the values

BC2 = 242 + 72

By further calculation

BC = √ (576 + 49) = √625 = 25 cm

So the radius of circle = 25/2 cm

We know that

Area of △ABC = ½ × AB × AC

Substituting the values

= ½ × 7 × 24

= 84 cm2

Area of quadrant COD = ¼ πr2

Substituting the values

= ¼ × 3.14 × 25/2 × 25/2

By further calculation

= 1962.5/16

= 122.66 cm2

Area of circle = πr2

Substituting the values

= 3.14 × 25/2 × 25/2

By further calculation

= 1962.5/4

= 490.63 cm2

Area of shaded portion = Area of circle – (Area of △ABC + Area of quadrilateral COD)

Substituting the values

= 490.63 – (84 + 122.66)

By further calculation

= 490.63 – 206.66

= 283.97 cm2

31. (a) In the figure given below ABCD is a square of side 14 cm. A, B, C and D are centres of the equal circle which touch externally in pairs. Find the area of the shaded region.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 113

(b) In the figure (ii) given below, the boundary of the shaded region in the given diagram consists of three semicircular arcs, the smaller being equal. If the diameter of the larger one is 10 cm, calculate.

(i) the length of the boundary.

(ii) the area of the shaded region. (Take π to be 3.14)

ML Aggarwal Solutions for Class 9 Chapter 16 Image 114

Solution:

(a) It is given that

Side of square ABCD = 14 cm

Radius of each circle drawn from A, B, C and D and touching externally in pairs = 14/2 = 7 cm

We know that

Area of square = a2

Substituting the values

= 14 × 14

= 196 cm2

Area of 4 sectors of 900 each = 4 × π × r2

Substituting the values

= 4 × 22/7 × 7 × 7 × ¼

= 154 cm

Area of each sector of 2700 angle = 3/4 πr2

Substituting the values

= ¾ × 22/7 × 7 × 7

= 231/2

= 115.5 cm2

Here

Area of 4 sectors = 115.5 × 4 = 462 cm2

So the area of shaded portion = area of square + area of 4 bigger sector – area of 4 smaller sector

Substituting the values

= 196 + 462 – 154

= 658 – 154

= 504 cm2

(b) We know that

Radius of big semi-circle = 10/2 = 5 cm

Radius of each smaller circle = 5/2 cm

(i) Length of boundary = circumference of bigger semi-circle + 2 circumference of smaller semi-circles

It can be written as

= πR + πr + πr

= 3.14 (R + 2r)

Substituting the values

= 3.14 (5 + 2 × 5/2)

By further calculation

= 3.14 × 10

= 31.4 cm

(ii) Here

Area of shaded region = area of bigger semi-circle + area of one smaller semi-circle – area of other smaller semi-circle

We know that

Area of bigger semi-circle = ½ πR2

Substituting the values

= 3.14/2 × 5 × 5

= 1.57 × 25

= 39.25 cm

32. (a) In the figure (i) given below, the points A, B and C are centres of arcs of circles of radii 5 cm, 3 cm and 2 cm respectively. Find the perimeter and the area of the shaded region. (Take π = 3.14)

(b) In the figure (ii) given below, ABCD is a square of side 4 cm. At each corner of the square a quarter circle of radius 1 cm, and at the centre a circle of diameter 2 cm are drawn. Find the perimeter and the area of the shaded region. Take π = 3.14

ML Aggarwal Solutions for Class 9 Chapter 16 Image 115

Solution:

(a) It is given that

Radius of bigger circle = 5 cm

Radius of small circle (r1) = 3 cm

Radius of smaller circle (r2) = 2 cm

(i) We know that

Perimeter of shaded region = circumference of bigger semi-circle + circumference of small semi-circle + circumference of smaller semi-circle

It can be written as

= πR + π r1 + π r2

= π (R + r1 + r2)

Substituting the values

= π (5 + 3 + 2)

= 3.14 × 10

= 31.4 cm2

(ii) We know that

Area of shaded region = area of bigger semi-circle + area of smaller semi-circle – area of small semicircle

It can be written as

= ½ πR2 + ½ πr22 – ½ πr12

= ½ π (R2 + r22 – r12)

Substituting the values

= ½ π (5 2 + 22 – 32)

By further calculation

= ½ π (25 + 4 – 9)

= ½ π × 20

So we get

= 10 × 3.14

= 31.4 cm2

(b) We know that

Side of square ABCD = 4 cm

Radius of each quadrant circle = 1 cm

Radius of circle in the square = 2/2 = 1 cm

(i) Here

Perimeter of shaded region = circumference of 4 quadrants + circumference of circle + 4 × ½ side of square

It can be written as

= 4 × ¼ (2 πr) + 2 πr + 4 × 2

By further calculation

= 2 πr + 2 πr + 8

= 4 πr + 8

So we get

= 4 × 3.14 × 1 + 8

= 12.56 + 8

= 20.56 cm

(ii) Area of shaded region = area of square – area of 4 quadrants – area of circle

It can be written as

= side2 – 4 × ¼ πr2 – πr2

= 42 – πr2– πr2

By further calculation

= 16 – 2πr2

So we get

= 16 – 2 × 3.14 × 12

= 16 – 6.28

= 9.72 cm2

33. (a) In the figure given below, ABCD is a rectangle. AB = 14 cm, BC = 7 cm. From the rectangle, a quarter circle BFEC and a semicircle DGE are removed. Calculate the area of the remaining piece of the rectangle. (Take π = 22/7)

(b) The figure (ii) given below shows a kite, in which BCD is in the shape of a quadrant of circle of radius 42 cm. ABCD is a square and △CEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 116

Solution:

(a) Here

Area of remaining piece = area of rectangle ABCD – area of semicircle DGE – area of quarter BFEC

Substituting the values

= 14 × 7 – ½ × π (7/2)2 – ¼ π × 72

By further calculation

= 14 × 7 – ½ × 22/7 × 7/2 × 7/2 – ¼ × 22/7 × 7 × 7

So we get

= 98 – 77/4 – 154/4

= 98 – 19.25 – 38.5

= 98 – 57.75

= 40.25 cm2

(b) In the given figure

ABCD is a square of side = radius of quadrant = 42 cm

△CEF is an isosceles right triangle with each side = 6 cm

Area of shaded portion = area of quadrant + area of isosceles right triangle

It can be written as

= ¼ πr2 + ½ EC × FC

Substituting the values

= ¼ × 22/7 × 42 × 42 + ½ × 6 × 6

By further calculation

= 1386 + 18

= 1404 cm2

34. (a) In the figure (i) given below, the boundary of the shaded region in the given diagram consists of four semicircular arcs, the smallest two being equal. If the diameter of the largest is 14 cm and of the smallest is 3.5 cm, calculate

(i) the length of the boundary.

(ii) the area of the shaded region.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 117

(b) In the figure (ii) given below, a piece of cardboard, in the shape of a trapezium ABCD, AB || CD and ∠BOD = 900, quarter circle BFEC is removed. Given AB = BC = 3.5 cm and DE = 2 cm. Calculate the area of the remaining piece of the cardboard.

Solution:

(a) (i) We know that

Length of boundary = Circumference of bigger semi-circle + Circumference of small semi-circle + 2 × circumference of the smaller semi-circles

ML Aggarwal Solutions for Class 9 Chapter 16 Image 118

It can be written as

= πR + πr1 + 2 × πr2

= π (R + r1) + 2πr2

Substituting the values

= 22/7 (7 + 3.5) + 2 × 22/7 × 3.5/2

By further calculation

= 22/7 × 10.5 + 11

So we get

= 33 + 11

= 44 cm

(ii) We know that

Area of shaded region = Area of bigger semicircle + area of small semicircle – 2 × area of smaller semicircles

It can be written as

= ½ π (7)2 + ½ π (3.5)2 – 2 × ½ π (1.75)2

By further calculation

= ½ × 22/7 × 7 × 7 + ½ × 22/7 × 3.5 × 3.5 – 22/7 × 1.75 × 1.75

So we get

= 77 + 19.25 – 9.625

= 86.625 cm2

(b) Here

ABCD is a trapezium in which AB || DC and ∠C = 900

ML Aggarwal Solutions for Class 9 Chapter 16 Image 119

AB = BC = 3.5 cm and DE = 2 cm

Radius of quadrant = 3.5 cm

We know that

Area of trapezium = ½ (AB + DC) × BC

Substituting the values

= ½ (3.5 + 3.5 + 2) × 3.5

By further calculation

= ½ (9 × 3.5)

= 4.5 × 3.5

= 15.75 cm2

So the area of quadrant = ¼ πr2

Substituting the values

= ¼ × 22/7 × 3.5 × 3.5

= 9.625 cm2

Area of shaded portions = 15.75 – 9.625 = 6.125 cm2

35. (a) In the figure (i) given below, ABC is a right angled triangle, ∠B = 900, AB = 28 cm and BC = 21 cm. With AC as diameter a semi-circle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.

(b) In the figure (ii) given below, ABC is an equilateral triangle of side 8 c. A, B and C are the centers of circular arcs of equal radius. Find the area of the shaded region correct upto 2 decimal places. (Take π = 3.142 and √3 = 1.732)

ML Aggarwal Solutions for Class 9 Chapter 16 Image 120

Solution:

(a) In right △ABC

∠B = 900

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

= 282 + 212

= 784 + 441

= 1225

So we get

AC = √1225 = 35 cm

Here

Radius of semi-circle (R) = 35/2

Radius of quadrant (r) = 21 cm

So the area of shaded region = area of △ABC + area of semi-circle – area of quadrant

= ½ × 28 × 21 + ½ πR2 – ¼ r2

Substituting the values

= 294 + ½ × 22/7 × 35/2 × 35/2 – ¼ × 22/7 × 21 × 21

By further calculation

= 294 + 1925/4 – 693/2

= 294 + 481.25 – 346.5

So we get

= 775.25 – 346.50

= 428.75 cm2

(b) We know that

△ABC is an equilateral triangle of side 8 cm

A, B, C are the centres of three circular arcs of equal radius

Radius = 8/2 = 4 cm

Here

Area of △ABC = √3/4a2

Substituting the values

= √3/4 × 8 × 8

= √3/4 × 64

So we get

= 16√3cm2

Substituting the value of √3

= 16 × 1.732

= 27.712 cm2

So the area of 3 equal sectors of 600 whose radius is 4 cm = 3 × πr2 × 60/360

By further calculation

= 3 × 3.142 × 4 × 4 × 1/6

So we get

= 3.142 × 8

= 25.136 cm2

Area of shaded region = 27.712 – 25.136 = 2.576 = 2.58 cm2

36. A circle is inscribed in a regular hexagon of side 2√3 cm. Find

(i) the circumference of the inscribed circle

(ii) the area of the inscribed circle

Solution:

It is given that

ABCDEF is a regular hexagon of side 2√3 cm and a circle is inscribed in it with O as the centre

ML Aggarwal Solutions for Class 9 Chapter 16 Image 121

Radius of inscribed circle = √3/2 × side of regular hexagon

Substituting the values

= √3/2 × 2√3

= 3 cm

(i) We know that

Circumference of the circle = 2πr

Substituting the values

= 2π × 3

By further calculation

= (6 × 22)/ 7

= 132/7 cm

(ii) Area of the circle = πr2

Substituting the values

= π × 3 × 3

By further calculation

= (9 × 22)/ 7

= 198/7 cm2

37. In the figure (i) given below, a chord AB of a circle of radius 10 cm subtends a right angle at the centre O. Find the area of the sector OACB and of the major segment. Take π = 3.14.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 16 Image 122

It is given that

Radius of the circle = 10 cm

Angle at the centre subtended by a chord AB = 900

We know that

Area of sector OACB = πr2 × 90/360

Substituting the values

= 3.14 × 10 × 10 × 90/360

So we get

= 314 × ¼

= 78.5 cm2

Here

Area of △OAB = ½ × 10 × 10 = 50 cm2

Area of minor segment = Area of sector △ACB – Area of △OAB

Substituting the values

= 78.5 – 50

= 28.5 cm2

Area of circle = πr2

Substituting the values

= 3.14 × 10 × 10

= 314 cm2

Area of major segment = area of circle – area of minor segment

Substituting the values

= 314 – 28.5

= 285.5 cm2

Exercise 16.4

1. Find the surface area and volume of a cube whose one edge is 7 cm.

Solution:

It is given that

One edge of cube a = 7 cm

We know that

Surface area of cube = 6a2

Substituting the values

= 6 (7)2

= 6 × 7 × 7

= 294 cm2

Volume of cube = a3

Substituting the values

= 73

= 7 × 7 × 7

= 343 cm3

2. Find the surface area and the volume of a rectangular solid measuring 5 m by 4 m by 3 m. Also find the length of a diagonal.

Solution:

In a rectangular solid

l = 5 m, b = 4 m and h = 3 m

Here

Surface area of rectangular solid = 2 (lb + bh + lh)

Substituting the values

= 2 (5 × 4 + 4 × 3 + 5 × 3)

By further calculation

= 2 (20 + 12 + 15)

= 2 × 47

= 94 sq. m

Volume of rectangular solid = l × b × h

Substituting the values

= 5 × 4 × 3

= 60 m3

We know that

ML Aggarwal Solutions for Class 9 Chapter 16 Image 123

It can be written as

= 5 √2 m

Substituting the value of √2

= 5 × 1.414

= 7.07 m

Therefore, the length of diagonal is 7.07 m.

3. The length and breadth of a rectangular solid are respectively 25 cm and 20 cm. If the volume is 7000 cm3, find its height.

Solution:

It is given that

Length of rectangular solid = 25 cm

Breadth of rectangular solid = 20 cm

Volume of rectangular solid = 7000 cm3

Consider height of rectangular solid = h cm

Here

Volume = l × b × h

Substituting the values

7000 = 25 × 20 × h

By further calculation

25 × 20 × h = 7000

h = 7000/ (25 × 20)

So we get

h = 700/ (25 × 2)

h = 350/25

By division

h = 70/5

h = 14 cm

Therefore, height of rectangular solid is 14 cm.

4. A class room is 10 m long, 6 m broad and 4 m high. How many students can it accommodate if one student needs 1.5 m2 of floor area? How many cubic metres of air will each student have?

Solution:

The given dimensions of class room are

Length (l) = 10 m

Breadth (b) = 6 m

Height (h) = 4 m

We know that

Floor area of class room = l × b

Substituting the values

= 10 × 6

= 60 m2

Here

One student needs 1.5 m2 floor area

So the number of students = 60/1.5

Multiply and divide by 10

= (60 × 10)/ 15

= 600/15

= 40 students

Volume of class room = l × b × h

Substituting the values

= 10 × 6 × 4

= 240 m3

So the cubic metres of air for each student = volume of classroom/ number of students

Substituting the values

= 240/40

= 6 m3

5. (a) The volume of a cuboid is 1440 cm3. Its height is 10 cm and the cross-section is a square. Find the side of the square.

(b) The perimeter of one face of a cube is 20 cm. Find the surface area and the volume of the cube.

Solution:

(a) It is given that

Volume of cuboid = 1440 cm3

Height of cuboid = 10 cm

We know that

Volume of cuboid = area of square × height

Substituting the values

1440 = area of square × 10

By further calculation

Area of square = 1440/10 = 144 cm2

Here

Side × side = 144

So we get

Side = √144 = 12 cm

Therefore, the side of square is 12 cm.

(b) It is given that

Perimeter of one face of a cube = 20 cm

Perimeter of one face of a cube = 4 × side

We can write it as

20 = 4 × side

By further calculation

Side = 20/4 = 5 cm

Here

Area of one face = side × side

Substituting the values

= 5 × 5

= 25 cm2

Area of 6 faces = 6 × 25 = 150 cm2

So the volume of cube = side × side × side

Substituting the values

= 5 × 5 × 5

= 125 cm3

6. Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden box covered with coloured papers with pictures of Santa Claus. She must know the exact quantity of paper to buy for this purpose. If the box has length 80 cm, breadth 40 cm and height 20 cm respectively, then how many square sheets of paper of side 40 cm would she require?

Solution:

It is given that

Length of box (l) = 80 cm

Breadth (b) = 40 cm

Height (h) = 20 cm

We know that

Surface area of the box = 2 (lh + bh + hl)

Substituting the values

= 2 (80 × 40 + 40 × 20 + 20 × 80)

By further calculation

= 2 (320 + 800 + 1600)

= 2 × 5600

= 11200 cm2

So the area of square sheet = side2

= 402

= 1600 cm2

Here

Number of sheets = area of box/ area of one sheet

Substituting the values

= 11200/1600

= 7

7. The volume of a cuboid is 3600 cm3 and its height is 12 cm. The cross-section is a rectangle whose length and breadth are in the ratio 4: 3. Find the perimeter of the cross-section.

Solution:

It is given that

Volume of a cuboid = 3600 cm3

Height of cuboid = 12 cm

We know that

Volume of cuboid = Area of rectangle × height

Substituting the values

3600 = area of rectangle × 12

By further calculation

Area of rectangle = 3600/ 12

Area of rectangle = 300 cm2 ….. (1)

Here

Ratio of length and breadth of rectangle = 4: 3

Consider

Length of rectangle = 4x

Breadth of rectangle = 3x

Area of rectangle = length × breadth

Substituting the values

Area of rectangle = 4x × 3x

So we get

Area of rectangle = 12x2 cm2 ….. (2)

Using equations (1) and (2)

12x2 = 300

x2 = 300/12

So we get

x2 = 25

x = √25 = 5

Here

Length of rectangle = 4 × 5 = 20 cm

Breadth of rectangle = 3 × 5 = 15 cm

Perimeter of the cross section = 2 (l + b)

Substituting the values

= 2 (20 + 15)

= 2 × 35

= 70 cm

8. The volume of a cube is 729 cm3. Find its surface area and the length of a diagonal.

Solution:

It is given that

Volume of a cube = 729 cm3

We can write it as

side × side × side = 729

(side)3 = 729

So we get

ML Aggarwal Solutions for Class 9 Chapter 16 Image 124

Side = 9 cm

We know that

Surface area of cube = 6 (side)2

Substituting the values

= 6 × (9)2

= 6 × 9 × 9

= 486 cm2

So the length of a diagonal = √3 × side

Substituting the value

= √3 × 9

= 1.73 × 9

= 15.57 cm

9. The length of the longest rod which can be kept inside a rectangular box is 17 cm. If the inner length and breadth of the box are 12 cm and 8 cm respectively, find its inner height.

Solution:

Consider h m as the inner height

It is given that

Length of longest rod inside a rectangular box = 17 cm which is same as diagonal of rectangular box

ML Aggarwal Solutions for Class 9 Chapter 16 Image 125

By squaring on both sides

172 = 122 + 82 + h2

By further calculation

289 = 144 + 64 + h2

289 = 208 + h2

So we get

h2 + 208 = 289

h2 = 289 – 208 = 81

h = √81 = 9 cm

Therefore, the inner height of rectangular box is 9 cm.

10. A closed rectangular box has inner dimensions 90 cm by 80 cm by 70 cm. Calculate its capacity and the area of tin-foil needed to line its inner surface.

Solution:

It is given that

Inner length of rectangular box = 90 cm

Inner breadth of rectangular box = 80 cm

Inner height of rectangular box = 70 cm

We know that

Capacity of rectangular box = volume of rectangular box = l × b × h

Substituting the values

= 90 × 80 × 70

= 504000 cm3

Here

Required area of tin foil = 2 (lb + bh + lh)

Substituting the values

= 2 (90 × 80 + 80 × 70 + 90 × 70)

By further calculation

= 2 (7200 + 5600 + 6300)

So we get

= 2 × 19100

= 38200 cm2

11. The internal measurements of a box are 20 cm long, 16 cm wide and 24 cm high. How many 4 cm cubes could be put into the box?

Solution:

It is given that

Volume of box = 20 cm × 16 cm × 14 cm

Volume of cubes = 4 cm × 4 cm × 4 cm

We know that

Number of cubes put into the box = volume of box/ volume of cubes

Substituting the values

= (20 × 16 × 24)/ (4 × 4 × 4)

= 5 × 4 × 6

= 120

Therefore, 120 cubes can be put into the box.

12. The internal measurements of a box are 10 cm long, 8 cm wide and 7 cm high. How many cubes of side 2 cm can be put into the box?

Solution:

It is given that

Length of box = 10 cm

Breadth of box = 8 cm

Height of box = 7 cm

We know that

3 number of cubes of side 2 cm can be put in box

i.e. height of box is 7 cm so only 3 cubes can be put height wise

13. A certain quantity of wood costs Rs 250 per m3. A solid cubical block of such wood is bought for Rs 182.25. Calculate the volume of the block and use the method of factors to find the length of one edge of the block.

Solution:

It is given that

Cost of Rs 250 for 1 m3 wood

Cost of Rs 1 for 1/250 m3 wood

Cost of Rs 182.25 for 182.25/250 m3 wood

Here

Quantity of wood = 182.25/250 m3

Multiply and divide by 100

= 18225/ (250 × 100)

So we get

= 18225/ (25 × 1000)

= 729/ 1000

= 0.729 m3

We know that

Volume of given block = 0.729 m3

Consider the length of one edge of block = x m

So we get

x3 = 0.729 m3

Now taking cube root on both sides

ML Aggarwal Solutions for Class 9 Chapter 16 Image 126

ML Aggarwal Solutions for Class 9 Chapter 16 Image 127

So we get

= (3 × 3)/ (2 × 5)

= 9/10

= 0.9 m

Therefore, the length of one edge of the block is 0.9 m.

14. A cube of 11 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base of the vessel are 15 cm × 12 cm, find the rise in the water level in centimetres correct to 2 decimal places, assuming that no water over flows.

Solution:

It is given that

Edge of cube = 11 cm

Volume of cube = edge3

Substituting the values

= 113

= 11 × 11 × 11

= 1331 cm3

We know that

The dimensions of the base of the vessel are 15 cm × 12 cm

Consider the rise in the water level = h cm

So the volume of cube = volume of vessel

Substituting the values

1331 = 15 × 12 × h

By further calculation

h = 1331/ (15 × 12)

So we get

h = 1331/ 180 = 7.39 cm

Therefore, the rise in the water level is 7.39 cm.

15. A rectangular container, whose base is a square of side 6 cm, stands on a horizontal table and holds water upto 1 cm from the top. When a cube is placed in the water and is completely submerged, the water rises to the top and 2 cm3 of water over flows, calculate the volume of the cube.

Solution:

If the base of rectangular container is a square

l = 6 cm and b = 6 cm

When a cube is placed in it, water rises to top i.e. through height 1 cm and 2 cm3 of water overflows

ML Aggarwal Solutions for Class 9 Chapter 16 Image 128

We know that

Volume of cube = Volume of water displaced

Substituting the values

= 6 × 6 × 1 + 2

= 36 + 2

= 38 cm3

16. (a) Two cubes, each with 12 cm edge, are joined end to end. Find the surface area of the resulting cuboid.

(b) A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between the surface area of the original cube and the sum of the surface areas of the new cubes.

Solution:

(a) We know that

By joining two cubes end to end a cuboid is formed whose dimensions are

l = 12 + 12 = 24 cm

b = 12 cm

h = 12 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 129

Here

Total surface area of cuboid = 2 (lb + bh + hl)

Substituting the values

= 2 (24 × 12 + 12 × 12 + 12 × 24)

By further calculation

= 2 (288 + 144 + 288)

So we get

= 2 × 720

= 1440 cm2

(b) We know that

Side of a cube = 12 cm

Here

Volume = side3 = 123 = 1728 cm3

ML Aggarwal Solutions for Class 9 Chapter 16 Image 130

If cut into 8 equal cubes

Volume of each cube = 1728/8 = 216 cm3

ML Aggarwal Solutions for Class 9 Chapter 16 Image 131

Surface area of original cube = 6 × side2

Substituting the values

= 6 × 122

= 6 × 144

= 864 cm2

Surface area of one smaller cube = 6 × 62

So we get

= 6 × 36

= 216 cm2

Surface area of 8 cube = 216 × 8

= 1728 cm2

So the ratio between their areas = 864: 1728 = 1: 2

17. A cube of a metal of 6 cm edge is melted and cast into a cuboid whose base is 9 cm × 8 cm. Find the height of the cuboid.

Solution:

It is given that

Edge of melted cube = 6 cm

Volume of melted cube = 6 cm × 6 cm × 6 cm = 216 cm3

Dimensions of cuboid are

Length = 9 cm

Breadth = 8 cm

h cm is the height

We know that

Volume of cuboid = l × b × h

Substituting the values

= 9 × 8 × h

= 72 h cm3

Here

Volume of cuboid = Volume of melted metal cube

Substituting the values

72h = 216

h = 216/72 = 3 cm

Therefore, the height of cuboid is 3 cm.

18. The area of a playground is 4800 m2. Find the cost of covering it with gravel 1 cm deep, if the gravel costs Rs 260 per cubic metre.

Solution:

It is given that

Area of playground = 4800 m2

We can write it as

l × b = 4800

We know that

Depth of level = 1 cm

h = 1 cm = 1/100 m

Here

Volume of gravel = l × b × h

Substituting the values

= 4800 × 1/100

= 48 m3

Cost of gravel = Rs 260 per cubic metre

So the total cost = 260 × 48 = Rs 12480

19. A field is 30 m long and 18 m broad. A pit 6 m long, 4 m wide and 3 m deep is dug out from the middle of the field and the earth removed is evenly spread over the remaining area of the field. Find the rise in the level of the remaining part of the field in centimetres correct to two decimal places.

Solution:

Consider ABCD is a field

Let ABCD be a part of the field where a pit is dug

Here

Volume of the earth dug out = 6 × 4 × 3 = 72 m3

h m is the level raised over the field uniformly

ML Aggarwal Solutions for Class 9 Chapter 16 Image 132

Now divide the raised level of the field into parts I and II

Volume of part I = 14 × 6 × h = 84h m3

Volume of part II = 24 × 18 × h = 432h m3

Total volume of part I and II = [84h + 432h] = 516h m3

ML Aggarwal Solutions for Class 9 Chapter 16 Image 133

We know that

516 = volume of earth dug out

Substituting the values

516 = 72

So we get

h = 72/516 = 0.1395 m

Multiply by 100

h = 0.1395 × 100

h = 13.95 cm

Therefore, the level has been raised by 13.95 cm.

20. A rectangular plot is 24 m long and 20 m wide. A cubical pit of edge 4 m is dug at each of the four corners of the field and the soil removed is evenly spread over the remaining part of the plot. By what height does the remaining plot get raised?

Solution:

It is given that

Length of plot (l) = 24 m

Width of plot (b) = 20 m

So the area of plot = l × b = 24 × 20 = 480 m2

We know that

Side of cubical pit = 4 m

Volume of each pit = 43 = 64 m3

Here

Volume of 4 pits at the corners = 4 × 64 = 256 m3

Area of surface of 4 pits = 4a2

= 4 × 42

= 64 m2

So the area of remaining plot = 480 – 64 = 416 m2

Height of the soil spread over the remaining plot = 256/416 = 8/13 m

21. The inner dimensions of a closed wooden box are 2 m, 1.2 m and 0.75 m. The thickness of the wood is 2.5 cm. Find the cost of wood required to make the box if 1 m3 of wood costs Rs 5400.

Solution:

It is given that

Inner dimensions of wooden box are 2 m, 1.2 m and 0.75 m

Thickness of the wood = 2.5 cm

So we get

= 25/10 cm

It can be written as

= 25/10 × 1/100

= 1/10 × ¼

So we get

= 1/40

= 0.025 m

So the external dimensions of wooden box are

(2 + 2 × 0.025), (1.2 + 2 × 0.025), (0.75 + 2 × 0.025)

By further calculation

= (2 + 0.05), (1.2 + 0.05), (0.75 + 0.5)

= 2.05, 1.25, 0.80

Here

Volume of solid = External volume of box – Internal volume of box

Substituting the values

= 2.05 × 1.25 × 0.80 – 2 × 1.2 × 0.75

By further calculation

= 2.05 – 1.80

= 0.25 m3

Cost = Rs 5400 for 1 m3

So the total cost = 5400 × 0.25

Multiply and divide by 100

= 5400 × 25/100

= 54 × 25

= Rs 1350

22. A cubical wooden box of internal edge 1 m is made of 5 cm thick wood. The box is open at the top. If the wood costs Rs 9600 per cubic metre, find the cost of the wood required to make the box.

Solution:

It is given that

Internal edge of cubical wooden box = 1 m

Thickness of wood = 5 cm

We know that

External length = 1 m + 10 cm = 1.1 m

Breadth = 1 m + 10 m = 1.1 m

Height = 1 m + 5 cm = 1.05 m

Here

Volume of the wood used = Outer volume – Inner volume

Substituting the values

= 1.1 × 1.1 × 1.05 – 1 × 1 × 1

By further calculation

= 1.205 – 1

= 0.2705 m3

Cost of 1 m3 = Rs 9600

So the cost of 0.2705 m3 = 9600 × 0.2705

= Rs 2596.80

23. A square brass plate of side x cm is 1 mm thick and weighs 4725 g. If one cc of brass weights 8.4 gm, find the value of x.

Solution:

It is given that

Side of square brass plate = x cm

Here l = x cm and b = x cm

Thickness of plate = 1 mm = 1/10 cm

We know that

Volume of the plate = l × b × h

Substituting the values

= x × x × 1/10

= x2/ 10 cm3 ….. (1)

Here

8.4 gm weight brass having volume = 1 cc

1 gm weight brass having volume = 1/8.4 cc

So the 4725 gm weight brass having volume = 4725 × 1/8.4 = 562.5 cc

Volume of plate = 562.5 cc = 562.5 cm3 …. (2)

Using both the equations

x2/10 = 562.5

By cross multiplication

x2 = 562.5 × 10 = 5625

Here

x = √5625 = 75 cm

Therefore, the value of x is 75 cm.

24. Three cubes whose edges are x cm, 8 cm and 10 cm respectively are melted and recast into a single cube of edge 12 cm. Find x.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 16 Image 134

It is given that

Edges of three cubes are x cm, 8 cm and 10 cm

So the volumes of these cubes are x3, 83 and 103 i.e. x3, 512 cm3 and 1000 cm3

We know that

Edges of new cube formed = 12 cm

Volume of new cube = 123 = 1728 cm3

Based on the question

x3 + 512 + 1000 = 1728

By further calculation

x3 + 1512 = 1728

x3 = 216

It can be written as

x3 = 6 × 6 × 6

So we get

x = 6 cm

25. The area of cross-section of a pipe is 3.5 cm2 and water is flowing out of pipe at the rate of 40 cm/s. How much water is delivered by the pipe in one minute?

Solution:

It is given that

Area of cross-section of pipe = 3.5 cm2

Speed of water = 40 cm/sec

Length of water column in 1 sec = 40 cm

We know that

Volume of water flowing in 1 second = Area of cross section × length

Substituting the values

= 3.5 × 40

= 35 × 4

= 140 cm3

So the volume of water flowing in 1 minute i.e. 60 sec = 140 × 60 cm3

Here

1 litre = 1000 cm3

So we get

Volume = (140 × 60)/ 100

= (14 × 6)/ 10

= 84/10

= 8.4 litres

26. (a) The figure (i) given below shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in cm and all angles in the figure are right angles.

(b) The figure (ii) given below shows the cross section of a concrete wall to be constructed. It is 2 m wide at the top, 3.5 m wide at the bottom and its height is 6 m and its length is 400 m. Calculate

(i) the cross sectional area and

(ii) volume of concrete in the wall.

(c) The figure (iii) given below show the cross section of a swimming pool 10 m broad, 2 m deep at one end and 3 m deep at the other end. Calculate the volume of water it will hold when full, given that its length is 40 m.

ML Aggarwal Solutions for Class 9 Chapter 16 Image 135

Solution:

(a) We know that

The given figure can be divided into two cuboids of dimensions 4 cm, 4 cm, 2 cm, 4 cm, 2 cm and 6 cm

ML Aggarwal Solutions for Class 9 Chapter 16 Image 136

Volume of solid = 4 × 4 × 2 + 4 × 2 × 6

= 32 + 48

= 80 cm3

(b) We know that

Figure (ii) is a trapezium with parallel sides 2 m and 3.5 m

(i) Area of cross section = ½ (sum of parallel sides) × height

Substituting the values

= ½ (2 + 3.5) × 6

By further calculation

= ½ × 5.5 × 6

So we get

= 5.5 × 3

= 16.5 m2

(ii) Volume of concrete in the wall = Area of cross section × length

Substituting the values

= 16.5 × 400

= 165 × 40

= 6600 m3

(c) We know that

From figure (iii) we know that it is trapezium with parallel sides 2 m and 3m

Here

Area of cross section = ½ (sum of parallel sides) × height

Substituting the values

= ½ (2 + 3) × 10

By further calculation

= ½ × 5 × 10

= 5 × 5

= 25 m2

So the volume of water it will hold when full = area of cross section × height

= 25 × 40

= 1000 m3

27. A swimming pool is 50 metres long and 15 metres wide. Its shallow and deep ends are 1 ½ metres and 14 ½ metres deep respectively. If the bottom of the pool slopes uniformly, find the amount of water required to fill the pool.

Solution:

It is given that

Length of swimming pool = 50 m

Width of swimming pool = 15 m

Its shallow and deep ends are 1 ½ m and 5 ½ m deep

We know that

Area of cross section of swimming pool = ½ (sum of parallel sides) × width

Substituting the values

= ½ (1 ½ + 4 ½) × 15

By further calculation

= ½ (3/2 + 9/2) × 15

So we get

= ½ [(3 + 9)/ 2] × 15

= ½ × 12/2 × 15

= ½ × 6 × 15

= 3 × 15

= 45 m2

Here

Amount of water required to fill pool = Area of cross section × length

= 45 × 50

= 2250 m3

Leave a Comment

Your email address will not be published. Required fields are marked *