# Frank ICSE Class 8 Solutions Chapter 18

## Frank ICSE Mathematics Class 8 Solutions Chapter 18 Mensuration

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Frank ICSE Mathematics Class 8 Math Book, Chapter 18, Mensuration. Here students can easily find step by step solutions of all the problems for Mensuration, Exercise 18.1, 18.2, 18.3, 18.4 and 18.5 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 18 solutions. Here in this post all the solutions are based on latest Syllabus.

Mensuration Exercise 18.1 Solution :

Question no – (1)

Solution :

= r = 7cm

Area = πr2 = 22/7 × 7

= 154 cm2

Perimeter = 2π =2 × 22/7 × 7

= 44 cm

Question no – (2)

Solution :

PR = √(16)2 + (12)2

= √256 + 133

= √400

= 20 cm

Radius = 20/2 = 10 cm

Area = πr2 = (3.14) × 10 × 10

= 314 cm2

Question number – (3)

Solution :

Area of the large semicircle

= 1/2πr2 = 22/7 × (21/2) × (21/2) × 1/2

= 346.5/2 cm2

= 173.25cm2

Area of the small semicircles

= 3πr2 = 3/2 × 22/7 × 7/2

= 57.75 cm2

Total area = 173.25 + 57.75

= 231 cm2

Question no – (4)

Solution :

Length of the wire = (6.6 × 3) cm = 19.8 cm

= 2πr = 19.8

or, 2 × 22/7 × r = 19.8

or, r = 19.8 × 7/2 × 22

= 0.45 × 7

= 3.15 cm

Question no – (5)

Solution :

Perimeter = 2πr = 440

or, 2r = 440 × 7/22 = 140 cm

Area = πr2 = 22/7 × 70 × 70

= 15400 m2

Total cost 15400 × 225

= 3465000 rupees

Question no – (6)

Solution :

The diameter of the circle will be 14 cm

= 14/2

= 7 cm

Area of the circle

= πr2 = 22/7 × 7 × 7

= 154 cm2

Remaining Area

= (21 × 14) – 154

= 294 – 154

= 140 cm2

Question no – (7)

Solution :

Breadth of the rectangle = diameter of circle = 140 cm

Radius of the circle = 40 cm

Total Area = Area of rectangle + 2 identical semicircles

Area of rectangle + area of circle

= {250 – 80) × (140)} + {22/7 × 40 × 40}

= (170 × 140) + (3.14 ×40 × 40)

= 238000 + 5024

= 18824 cm2

Question no – (8)

Solution :

(i) Diameter = 42m, radius = 21m

Perimeter = 2r + πr = 21 (2 + 22/7)

= 21 × 36/7

= 108 m

Area = πr2/2 = 22/7 × 1/2 × 21

= 693m2

Cost of fencing = (108 × 7)

= 756 rupees

(ii)

Perimeter = 21 + 28 + (22/7 × 35/2)

= 21 + 28 + 55 = 104 m

Area = (1/2 × 21 × 28) + (1/2 × 22/7 × 35/2 × 35/2)

= (14 × 21) + (1925/4)

= 294 + 481.25

= 775.25 m2

∴ Cost of fencing = (104 × 7) = 728 rupees

(iii)

Perimeter = 20 + 20 + 21 + (21/2 × 22/7)

= 20 + 20 + 21 + 33 = 94 m

Area = (21 × 20) + (1/2 × 22/7 × 21/2 × 21/2)

= 420 + 165.75

= 585.75 m2′

Question no – (9)

Solution :

Shaded area – Area of large semicircle –

= 2 (area of smaller semicircle)

= πR2 – 2πr2

= 22/7 [914)2 – 2 (7)2]

= 22/7 × (7)2 × [22 – 2]

= 22 × 7 × 2

= 308 cm2

Question no – (10)

Solution :

Diameter of each circle = 14/2 = 7 cm

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

(b) Remaining area = Area of rectangle – 6 (area of circle)

= (21 × 20) – 6 [22/7 × (3.5) × (3.5)]

= 420 – (38.5 × 6)

= 420 – 231

= 189 cm2

Mensuration Exercise 18.2 Solution :

Question no – (2)

Solution :

Area = (12 × 15) + (1/2 × 6 × 15)

= 180 + 45

= 225 cm2

Question no – 3)

Solution :

Sum of the parallel sides = 108/12 cm

= 9 cm

Question no – (4)

Solution :

Suppose initial height = h

Suppose final height = 3h

initial base (sum) = b

initial final base = 3b

initial area = bh/2 final area = 9bh/2

Area will be 9 times of the initial area

Question no – (5)

Solution :

Area = 16/2 × (22.3 + 28.7)

= 16/2 × 51 = 816/2 cm2

= 408cm2

Question no – (6)

Solution :

Distance between parallel sides = Area/(sum of then/2)

= 984/(31.5 + 50.5)/2= 984/82/2

= 24 cm

Question no – (7)

Solution :

Area = 4/2 × (8 + 6)

[80 dm = 8m, 40 dm = 4 m]

= 28m2

Question no – (8)

Solution :

Suppose the parallel sides are – x cm, (x + 6) cm

9/2 × (x + x + 6) = 270

or, 2x + 6 = 30 × 2

or, x = 27 cm

Length of the

x + 6 = 33 cm

Question no – (9)

Solution :

Suppose the other side = x cm

(x + 42/2) × 26 = 793

or, x + 42 = 30.5 × 200

or, x = 61 – 42 = 19 cm

Question no – (10)

Solution :

Suppose there are x trapeziums

x × 8/2 × (17 + 25)

= 3.78 × 100 × 100

or, x = 225

Question no – (11)

Solution :

Area in m2 = 41 × (67 + 58)/2 × 100 × 100

= 41 × 125/2 × 100 × 100

= 41/160 m2

Question no – (12)

Solution :

Since AB = 57, DC = 27

AE = 57 – 27 = 30 cm

And BC + AD = 134 – (57 + 27) = 50 cm – (i)

And, AD2 – BC2 = 302

or, AD – BC = 900/50

= 18 – (ii)

From (i) and (ii) we get, AD = 34 cm

BC = 16 cm

Mensuration Exercise 18.3 Solution :

Question no – (1)

Solution :

(a) π/2 (10)2 + {(20)2 – (1/2 × 12 × 16)}

= 314/2 + 304

= 157 + 304

= 461 cm2

(b) (14 × 14) + {1/2 × 14 × (19 – 14)}

= 196 + 35

= 231 cm2

(c) (24 × 24) – [2 × 1/2 × 10 × (24 – 0)]

= 567 – 150

= 426 cm2

(d) (6 × 6) + (12 × 6) + {1/2 × 4 × (5 + 12)

= 36 + 72 + 34

= 142 cm2

Question no – (2)

Solution :

(a) 1/2 π(14)2 + 1/2 × 28 × (20 + 30)

= 308 + 700

= 1008 cm2

(b) 1/2 π(21/2)2 + (21 × 21) = 639/2 + 441

= 346.5 + 441

= 787.5 cm2

(c) 1/2 π(7)2 + 1/2 π (14)2 + 1/2 × (4 – 28) × (42 – 7 – 14)

= 77 + 308 + 441

= 826 cm2

(d) π [(11)2 – (4)2] = 22/7 × (121 – 16)

= 22/7 × 105

= 22 × 15

= 330 cm2

Question no – (3)

Solution :

Depth = 112/1/2(16 + 12)

= 112/1/2 × 28 =

112/14

= 8m

Question no – (4)

Solution :

= {2 × 1/2 × (3 + 13) × 5} + (13 × 5)

= 195 + 65

= 260 cm2

Question no – (5)

Solution :

Area of flower beds along = 1/2 × (24 + 24) × 10

= 290 m2

Area of the flower beds along breath = 1/2 × (28 + 18) × 10

= 230 m2

Question no – (6)

Solution :

Perimeter = (6 + 5 + 4 + 5 + 8) cm

= 28 cm

Area = (1/2 × 6 × 8) + [1/2 × 4 × {√62 + 82 + 4}]

= 24 + 28 = 52 cm2

Question no – (7)

Solution :

= [1/2 × 20 × (30 + 10) – [1/2 × π × (10)2]

= 400 – 157

= 243 cm2

Question no – (8)

Solution :

= [1/2 × 7 × 7) + {1/2 × 22/7 × (7√2/2)2}

= 49/2 + 77/2

= 63 cm2

Question no – (9)

Solution :

= (1/2 × 8 × 4) + (1/2 × 8 × 2.5)

= 16 + 10

= 26 cm2

Question no – (10)

Solution :

= (9 × 3) + {1/2 × (9 + 3) × 3}

= 27 + 18

= 45 cm2

Question no – (11)

Solution :

= (1/2 × 12 × 14) + (1/2 × 10 × 12) + {1/2 × (12 + 14) × 8} + (1/2 × 30 × 7}

= 353 cm2

Mensuration Exercise 18.4 Solution :

Question no – (1)

Solution :

Cost = 2 {(30 × 25) + (25 + 18) + (30 + 18)} × 12

= 2 × 170 × 12

= 41760

Question no – (2)

Solution :

Cost of whitewashing walks

= [{2 × (150 + 25)] × 6 + (150 × 25] × 20

= 117000 rupees

Cost of polishing the flower

= (150 × 25 × 40) = 15000 rupees

Question no – (3)

Solution :

Side of the cube = √3750/6

= √625

= 25 m

Question no – (4)

Solution :

Painted are = (8 × 7.5) + (2 × 8 × 6) + (2 × 7.5 × 6)

= 60 + 96 + 90

= 246m2

Question no – (5)

Solution :

(a) 2 {70 × 60) + (60 × 50) + (70 × 50)

= 2 × 10700

= 21400 cm2

(b) 6 × 60 × 60

= 21000 cm2

(c) (2 × 22/7 × 7/2) × 7

= 154 cm2

(d) 6 × 7 × 7

= 294 cm2

Question no – (6)

Solution :

Cost = {(2 × 22/7 × 3.5) × 8} × 130

= 176 × 130

= 22880 rupees

Question no – (7)

Solution :

Required wrapper = 2 × {(50 × 35) + (35 × 10) + (50 × 10)} × 60

= 321000 cm2

= 312000 cm2 = (312000/1000) m2

= 312 m2

Question no – (8)

Solution :

= 2πrh = 2 × 22/7 × 35/2 × 25

= 275 m2

Question no – (9)

Solution :

Area = 2πrh × 900

= (2 × 22/7 × 98/2 × 125) × 900

= (77 × 500 × 900) cm2

= (77 × 5 × 9) m2 = 3456 m2

Question no – (10)

Solution :

= 2πrh = 2 × 22/7 × 350/2 × (750 – 50)

= (22 × 50 × 700) cm2

= 77m2

Question no – (11)

Solution :

Cost = {2 × (7 + 6) × 4 – 7 (7 × 6) × 15

= 2085 rupees

Mensuration Exercise 18.5 Solution :

Question number – (2)

Solution :

Curved surface area,

2πrh = 2 × 22/7 × 8 × 21

= 1056 cm2

Question no – (4)

Solution :

Suppose height = h πr2h = 577.5

or, 22/7 × (7/2)2 × h = 577.5

or, h = 577.5 × 2/11 × 7 = 15 cm

Question no – (5)

Solution :

Suppose height = h = 14 × 11 × h πr2h

or, r2 = 14 × 11 × 7/22

or, r2 = 72

r = 7

Question number – (6)

Solution :

New edge = √33 + 43 + 53

= √27 + 64 + 125

= √216

= 6 cm

Question no – (7)

Solution :

= No of cubical blocks

= 300 × 75 × 602/30 × 10 × 30

= 50

Question no – (8)

Solution :

No of small boxes

= 150 × 75 × 24/12 × 10 × 3

= 750

Question no – (9)

Solution :

Suppose the height = h cm

= 22 × 18 × 14

= 22/7 × (7)2 × h

or, h = 22 × 18 × 14/22 × 7

= 36 cm

Question no – (10)

Solution :

= 2πrh = 4400 – (i) 2πr = 220 – (ii)

Dividing (i) by (ii) we get h = 20 cm

And from (ii) r = 220 × 7/2 × 22 = 35 cm

Volume = πr2h

= 22/7 × 35 × 20

= 77000 cm3

Question no – (11)

Solution :

Curved surface area = 44 × 20

= 880 cm2

Now, 2πrh = 880

or, r = 880 × 7/2 × 22 × 20

= 7 cm

Volume = πr2h

= 22/7 × 7 × 7 × 20

= 3080 cm3

Question no – (12)

Solution :

No of person = 30 × 17 × 6/5

= 612

Question no – (13)

Solution :

Quantity of sort = πr2h

= 22/7 × (7/2)2 × 20

= 770m2

Height of the platform,

= 770/14 × 11

= 5m

Question no – (14)

Solution :

Suppose breadth of room = x m

2 (2x + x) × 4 = 192

or, x = 8

length = 16m,

height = 4m

Volume = 8 × 16 × 4

= 512m3

Question no – (15)

Solution :

(i) Volume of can = πr2h

= 22/7 × (14) × 20

= 12320 cm3

(ii) 2πrh = 2 × 22/7 × 14 × 20

= 1760 cm2

Revision Exercise Questions Solution :

Question no – (1)

Solution :

Volume of can,

= 6 × (6)3

= 1296 cm3

Surface area,

= (6)2 × 26

= 936 cm2

Question no – (2)

Solution :

Thickness = 4 × 4 × 4/32 × 8

= 1/4 cm

= 0.25

Question number – (3)

Solution :

= πr2h = 594

or, h = 594 × 7/22 × 1/9

= 21 m

Question no – (4)

Solution :

= No of cubes

= 16 × 8 × 4/2 × 2 × 2

= 64

Question no – (5)

Solution :

Since, r = 3 1/2

or, r = 7/2 πr2

= 22/7 × 7/2

= 77/2 cm2

Area lying outside the circle

= (20 × 15) – 77/2

= 300 – 77/2

= 600 – 77/2

= 523/2

= 261.5 cm2

Question no – (6)

Solution :

Volume of first cake

= πr2h = 22/7 × 7 × 7 × 4

= 616 cm3

Volume of second cake

= 14 × 8 × 5 = 560 cm3

The first cake

= (616 – 560)

= 56 cm3 more volume

Next Chapter Solution :

Updated: June 26, 2023 — 6:08 am