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**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 = πr^{2} = 22/7 × 7

= 154 cm^{2}

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 = πr^{2} = (3.14) × 10 × 10

= 314 cm^{2}

**Question number – (3)**

**Solution :**

Area of the large semicircle

= 1/2πr^{2} = 22/7 × (21/2) × (21/2) × 1/2

= 346.5/2 cm^{2}

= 173.25cm^{2}

Area of the small semicircles

= 3πr^{2} = 3/2 × 22/7 × 7/2

= 57.75 cm^{2}

Total area = 173.25 + 57.75

= 231 cm^{2}

**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 = πr^{2} = 22/7 × 70 × 70

= 15400 m^{2}

**∴** Total cost 15400 × 225

= 3465000 rupees

**Question no – (6)**

**Solution :**

The diameter of the circle will be 14 cm

**∴** Radius

= 14/2

= 7 cm

**∴** Area of the circle

= πr^{2} = 22/7 × 7 × 7

= 154 cm^{2}

**∴** Remaining Area

= (21 × 14) – 154

= 294 – 154

= 140 cm^{2}

**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 cm^{2}

**Question no – (8)**

**Solution :**

**(i)** Diameter = 42m, radius = 21m

**∴** Perimeter = 2r + πr = 21 (2 + 22/7)

= 21 × 36/7

= 108 m

Area = πr^{2}/2 = 22/7 × 1/2 × 21

= 693m^{2}

**∴** 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 m^{2}

∴ 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 m^{2′}

**Question no – (9)**

**Solution :**

Shaded area – Area of large semicircle –

= 2 (area of smaller semicircle)

= πR^{2} – 2πr^{2}

= 22/7 [914)^{2} – 2 (7)^{2}]

= 22/7 × (7)^{2} × [22 – 2]

= 22 × 7 × 2

= 308 cm^{2}

**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 cm^{2}

**Mensuration Exercise 18.2 Solution :**

**Question no – (2)**

**Solution :**

**∴** Area = (12 × 15) + (1/2 × 6 × 15)

= 180 + 45

= 225 cm^{2}

**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 cm^{2}

= 408cm^{2}

**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]

= 28m^{2}

**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 m^{2} = 41 × (67 + 58)/2 × 100 × 100

= 41 × 125/2 × 100 × 100

= 41/160 m^{2}

**Question no – (12) **

**Solution :**

Since AB = 57, DC = 27

**∴** AE = 57 – 27 = 30 cm

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

And, AD^{2} – BC^{2} = 30^{2}

or, (AD + BC) (AD – BC) (AD – BC) = 900

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 cm^{2}

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

= 196 + 35

= 231 cm^{2}

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

= 567 – 150

= 426 cm^{2}

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

= 36 + 72 + 34

= 142 cm^{2}

**Question no – (2) **

**Solution : **

**(a)** 1/2 π(14)^{2} + 1/2 × 28 × (20 + 30)

= 308 + 700

= 1008 cm^{2}

**(b)** 1/2 π(21/2)^{2} + (21 × 21) = 639/2 + 441

= 346.5 + 441

= 787.5 cm^{2}

**(c)** 1/2 π(7)^{2} + 1/2 π (14)^{2} + 1/2 × (4 – 28) × (42 – 7 – 14)

= 77 + 308 + 441

= 826 cm^{2}

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

= 22/7 × 105

= 22 × 15

= 330 cm^{2}

**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 cm^{2}

**Question no – (5)**

**Solution :**

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

= 290 m^{2}

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

= 230 m^{2}

**Question no – (6)**

**Solution :**

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

= 28 cm

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

= 24 + 28 = 52 cm^{2}

**Question no – (7)**

**Solution :**

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

= 400 – 157

= 243 cm^{2}

**Question no – (8)**

**Solution :**

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

= 49/2 + 77/2

= 63 cm^{2}

**Question no – (9)**

**Solution :**

= (1/2 × 8 × 4) + (1/2 × 8 × 2.5)

= 16 + 10

= 26 cm^{2}

**Question no – (10)**

**Solution :**

= (9 × 3) + {1/2 × (9 + 3) × 3}

= 27 + 18

= 45 cm^{2}

**Question no – (11)**

**Solution :**

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

= 353 cm^{2}

**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

= 246m^{2}

**Question no – (5)**

**Solution :**

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

= 2 × 10700

= 21400 cm^{2}

**(b)** 6 × 60 × 60

= 21000 cm^{2}

**(c)** (2 × 22/7 × 7/2) × 7

= 154 cm^{2}

**(d)** 6 × 7 × 7

= 294 cm^{2}

**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 cm^{2}

= 312000 cm^{2} = (312000/1000) m^{2}

= 312 m^{2}

**Question no – (8)**

**Solution :**

= 2πrh = 2 × 22/7 × 35/2 × 25

= 275 m^{2}

**Question no – (9)**

**Solution :**

Area = 2πrh × 900

= (2 × 22/7 × 98/2 × 125) × 900

= (77 × 500 × 900) cm^{2}

= (77 × 5 × 9) m^{2} = 3456 m^{2}

**Question no – (10)**

**Solution :**

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

= (22 × 50 × 700) cm^{2}

= 77m^{2}

**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 cm^{2}

**Question no – (4)**

**Solution :**

Suppose height = h πr^{2}h = 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 πr^{2}h

or, r^{2} = 14 × 11 × 7/22

or, r^{2} = 7^{2}

**∴** r = 7

**Question number – (6)**

**Solution :**

New edge = √3^{3} + 4^{3} + 5^{3}

= √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

**∴** radius = 35 cm

**∴** Volume = πr^{2}h

= 22/7 × 35 × 20

= 77000 cm^{3}

**Question no – (11)**

**Solution :**

Curved surface area = 44 × 20

= 880 cm^{2}

Now, 2πrh = 880

or, r = 880 × 7/2 × 22 × 20

= 7 cm

**∴** Volume = πr^{2}h

= 22/7 × 7 × 7 × 20

= 3080 cm^{3}

**Question no – (12)**

**Solution :**

No of person = 30 × 17 × 6/5

= 612

**Question no – (13)**

**Solution :**

Quantity of sort = πr^{2}h

= 22/7 × (7/2)^{2} × 20

= 770m^{2}

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

**∴** Breadth = 8m,

length = 16m,

height = 4m

**∴** Volume = 8 × 16 × 4

= 512m^{3}

**Question no – (15)**

**Solution :**

**(i)** Volume of can = πr^{2}h

= 22/7 × (14) × 20

= 12320 cm^{3}

**(ii)** 2πrh = 2 × 22/7 × 14 × 20

= 1760 cm^{2}

**Revision Exercise Questions Solution : **

**Question no – (1)**

**Solution :**

Volume of can,

= 6 × (6)^{3}

= 1296 cm^{3}

Surface area,

= (6)^{2} × 26

= 936 cm^{2}

**Question no – (2)**

**Solution :**

Thickness = 4 × 4 × 4/32 × 8

= 1/4 cm

= 0.25

**Question number – (3)**

**Solution :**

= πr^{2}h = 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 πr^{2}

= 22/7 × 7/2

= 77/2 cm^{2}

Area lying outside the circle

= (20 × 15) – 77/2

= 300 – 77/2

= 600 – 77/2

= 523/2

= 261.5 cm^{2}

**Question no – (6)**

**Solution :**

Volume of first cake

= πr^{2}h = 22/7 × 7 × 7 × 4

= 616 cm^{3}

Volume of second cake

= 14 × 8 × 5 = 560 cm^{3}

**∴** The first cake

= (616 – 560)

= 56 cm^{3} more volume

**Next Chapter Solution : **

👉 Chapter 19 👈