Brilliant’s Composite Mathematics Class 8 Solutions Chapter 17

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Brilliant’s Composite Mathematics Class 8 Solutions Chapter 17 Volumes and Surface Areas

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Brilliant’s Composite Mathematics Class 8 Math Book, Chapter 17, Volumes and Surface Areas. Here students can easily find step by step solutions of all the problems for Volumes and Surface Areas, Exercise 17.1, 17.2 and 17.3 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.

Volumes and Surface Areas Exercise 17.1 Solution

Question no – (1)

Solution :

Volume = base area × height

= 165 × 17

= 2805 cm³

Question no – (2)

Solution :

(i) Volume,

= base area × height

= 22/7 × (9)² × 14

= 3564 cm².

Question no – (3)

Solution :

Curved Surface area,

= 2 πrh = 22/7 × 21 × 10

= 1320 cm²

Question no – (4)

Solution :

Length of the sheet = 16 cm,

Now, breadth of the sheet,

= 176/16

= 11 cm.

When we roll the sheet along its breadth, then 2πr = 11

Or, r = 11 × 7/2 × 22

= 7/4 cm.

And h = 16 cm.

∴ Volume = πr²h

= 22/7 × (7/4)² × 16

= 154 cm³.

Question no – (5)

Solution :

We have, πr²h = 69300

Or, r² = 69300 × 7/22 × 50

Or, r² = 3 × 3 × 7 × 7

Or, r = 3 × 7

= 21 cm.

Lateral surface area = 2πrh

= 2 × 22/7 × 21 × 50

= 6600 cm².

Question no – (6)

Solution :

Volume of the cylinder/volume of the reduced cylinder

= πr²h/r (r/2)² h

= 4 : 1

Question no – (7)

Solution :

Volume of the one coin,

= πr²h = r × (3/4)² × 1/5

= 9π/80 cm³.

Volume of cylinder,

= π × (3)² × 8

= 72 π cm³.

∴ Number of coins,

= 72π / 9π / 80

= 72π × 80/9π

= 640

Question no – (8)

Solution :

Volume = πr² = 6358

Or, r² = 6358 × 7/22 × 28

Or, r² = 17 × 17/2 × 2

Or, r = 17/2 cm.

∴ Lateral surface area

= 2πrh = 2 × 22/7 × 17/2 × 28

= 1496 cm².

Question no – (9)

Solution :

Volume at the 1^st tin = (8)² × 13

= 832 cm³.

Volume of the 2^nd tin

= 22/7 × (7)² × 15

= 2310 cm³.

∴ 2310 – 832

= 1478 cm³.

Question no – (10)

Solution :

Weight = volume × weigh per cm³.

= π {(4.5)² – (4)²} × 77 × 8

= 22/7 × 8.5 × 0.5 × 77 × 8

= 2057 gm

= 2.057 kg.

Question no – (11)

Solution :

Capacity of tank,

= πr²h = 22/7 × (14)² × 7

= 4312 m³

= 4312 KL.

Thus, it can hold 4312 kiloliter of water.

Question no – (22)

Solution :

Total cost = area × number of pillars × cost per square metre

= 2 πrh × 15 × 1.25

= 2 × 22/7 × 25 × 350 × 15 × 1.25

= 103.20 rupees.

Hence, the cost of the white whishing will be 103.20 rupees.

Question no – (23)

Solution :

Required quantity of sheet metal = 2πr² + 2πrh

= 2πr (r + h) = 2 × 22/7 × 70 × (70 + 100)

= 74800 cm²

= 74800/10000 m²

= 7.48m²

Thus, 7.48 m² sheet metal will be required.

Question no – (24)

Solution :

Quantity of soil = πr²h

= 22/7 × (7)² × 12

=  1848 m³.

Now, volume of the embankment,

= π (R² – r²)h = 18.48

Or, 22/7 × (142 – 72) × h = 1848

Or,  22 × 21 × h = 1848

Or, h = 1848/22 × 21

= 4 m.

Therefore, the height of the embankment will be 4 m.

Question no – (25)

Solution :

Quantity of soil,

= πr²h = 22/7 × (7/2)² × 16

= 616 m³.

Now, 25 × 16 × h = 616

Or, h = 616/400

= 1.54 m.

Hence, the height of the platform will be 1.54 m

Question no – (26)

Solution :

Quantity of soil,

= 22/7 × (14)² × 8

= 448 m³

Now, The soil was spread  over the area of,

= (120 × 200) – 22/7 × (14)²

= 24000 – 616

= 23384 m²

∴ The level at earth will be raised by

= 448/23384 m

= 211 m

= 21.1 cm

Therefore, Earth’s level will rise 21.1 cm.

Volumes and Surface Areas Exercise 17.2 Solution

Question no – (1)

Solution :

(i) Volume = 1/3πr²h

= 1/3 × 22/7 × (28)² × 56

= 41888 cm²

Question no – (2)

Solution :

(i) slant height = πrl

= √(14)² + (48)²

= √2500

= 50 cm.

Lateral surface = πrl

= 22/7 × 14 × 50

= 2200 cm²

Question no – (3)

Solution : 

Curved surface area = πrl

= 22/7 × 35 ×√(35)² + (84)²

= 22 × 5 × 91

= 10010 cm².

Volume = 1/3πr²h

= 1/3 × 22/7 × (35)²× 84

= 197800 cm³.

Question no – (5)

Solution :

Lateral surface,

= πrl = 22/7 × 7 × 9

= 198 cm²

Question no – (6)

Solution : 

Area of curved surface = πrl

= 22/7 × 12 ×√(16)² + (12)²

= 22/7 × 12 × 20

= 753.6 m²

Question no – (8)

Solution : 

Let the height of the cone be h cm.

∴ π (2)² (60 = 1/3π (2)² h

Or, h = 6 × 3

= 18/ cm.

Question no – (9)

Solution : 

The longer side,

= √8² + 15² = 17 cm

Now, 2πr = 17

Or, π = 17 × 7/2 × 22

= 119/44 cm

∴ Total area,

= πr² + 1/2 × 8 × 15

= 22/7 × 119 × 119/44 × 44 × 60

= 628.57 cm²

Question no – (10)

Solution :

Measure of canvas = πrl

= 22/7 × 5 × 6.3

= 99 m².

∴ Length of the canvas,

= 99/1.5

= 66m

Cost = 66 × 22.50

= 1485 rupees

Question no – (11)

Solution :

Capacity = 1/3πr²h

= 1/3 × 22/7 × (2)²× 6

= 8 × 3.14

= 25.12 KL.

Question no – (12)

Solution :

Let their height be x, 3x units and let their radii be 3y, y units

∴ Ratio of their volume

= 1/3 × 22/7 × (3y)²× x/1/3 × 22/7 × (y)²× 3x

= 9/3

= 3/1

Question no – (13)

Solution :

Total volume = volume of cylinder + volume of cone

= πr²h₁ + 1/3πr²h₂

= πr² (h₁ + h₂/3)

= 22/7 × (8)²× (240 + 36/3)

= 22/7 × 64 × 252

= 50688 cm³.

∴ Now the Weight,

= 50688 × 7.8/1000

= 395.688 kg.

Question no – (14)

Solution :

Let the height be h cm.

∴ 1/3π (1.6)2 (3.6)

= 1/3π (1.2)2 h

Or, h = 1.6 × 1.6 × 3.63/1.2 × 1.2

= 6.4 cm

Question no – (16)

Solution :

Slant height of the tent

= √122 + 162

= 20 cm

∴ required cloth,

= πrl = 22/7 × 12 × 20

= 3.19 × 12 × 20

Required length of cloth,

= 3.14 × 12 × 20/3

= 251.2 m

Question no – (18)

Solution :

Canvas required = canvas required for cylindrical part + canvas required for conical part

= 2πrh + πrl

= 2 × 22/7 × 52.5 × 3 + 22/7 × 52.5 ×√(18)² + (52.5)²

= 330 + 165 × 55.5

= 9467.5 m²

Question no – (19)

Solution : 

Volume of the cone,

= 1/3πr²h

= 1/3 × 22/7 × (3)²× 7

= 66 cm³.

∴ Side of the height cube will be 3√66 cm.

Question no – (20)

Solution :

Let the height be h cm.

∴ 1/3π (1.6)² (3.6)

= 1/3π(1.2)² h

Or, h = 1.6 × 1.6 × 3.0/1.2 × 1.2

= 6.4 cm

Question no – (21)

Solution :

Let the height be h cm.

∴ 40 × 7 × 5

= 1/3 × 22/7 × (25)²× h

Or, h = 8 × 7 × 3 × 7/25 × 22

Or, h = 2.14 cm.

Question no – (22)

Solution :

Let the height be h cm. and radius be π cm.

∴ 1/3πr² (h) = πr² (5)

Or, h = 5 × 3

= 15 cm.

Question no – (25)

Solution : 

Area of base in the tent = πr²

= 22/7 × 12 × 12

= 452.16 m²

So, the base has the place for,

= 452.10/2

= 226.8 ≈ 226 people.

Air in the tent = 1/3πr²h

= 1/3 × 22/7 × 144 × 9

= 1356.48 m³.

So, the tent has enough air for 1356.48/15 ≈ 90  people

∴ We can see that the base of the tent has enough place for 226 people, but the air in it is enough for only just 90 people.

∴ We can conclude that only 90 persons can be accompanied in that tent

Question no – (26)

Solution :

Let the initial and reduced heights be h, 2h cm

∴ Initial volume = 1/3πr²h,

Reduced volume

= 1/3πr² (2h)

= 2/3 πr²h

∴ Clearly volume doubles

Volumes and Surface Areas Exercise 17.3 Solution

Question no – (1)

Solution :

(i) Volume,

= 4/3πr³ = 4/3 × 22/7 × 6³

= 905.14 cm³

Curved surface = 4πr²

= 4 × 3.14 × 6²

= 452.57 cm²

Question no – (2)

Solution :

(i) Radius = 8/2 = 4 cm.

∴ Volume = 4/3πr³

= 4/3 × 3.14 × 4³

= 267.95 cm³.

Curved surface,

= 4πr² = 4 × 3.14 × 4²

= 318.86 m²

Question no – (8)

Solution :

Let the initial and final radius of the balloon is r cm

∴ Initial volume = 4/3πr³,

Final volume,

= 4/3π (2π)³

= 32/3πr³

Now, final volume/initial volume

= 32/3πr³/34/3πr³

= 8

∴ The volume becomes 8 times

Question no – (9)

Solution :

The external and internal radil are,

= 30/2 = 15 cm, 15 – 2 = 13 cm

∴ Volume = 4/3π (R³ –r³)

= 4/3 × 22/7 × (15³ – 13³)

= 4/3 × 22/7 × (3375 – 2197)

= 4/3 × 22/7 × 1178

Question no – (10)

Solution :

Number of bullets = volume of cube/volume of each bullet

= 22 × 22 × 22/22/7 × (1)³

= 7 × 22 × 22

= 3388

Hence, 3388 special bullet can be made.

Question no – (11)

Solution :

Volume of the shot put,

= 4/3πr³

= 4/3 × 22/7 × (7)³

= 4 × 22 × 49

= 4312 cm³.

∴ Weight of the shot put,

= 4312 × 8

= 34496 gm

= 34.496 kg.

Question no – (13)

Solution :

Given, 4πr² = 154

Or, r² = 154 × 7/4 × 22

Or, r² = 7 × 7/2 × 2

Or, r = 7/2 cm

∴ r = 3.5 cm.

∴ Volume = 4/3πr³

= 4/3 × 22/7 × 7 × 7 × 7/2 × 2 × 2

= 539/3

= 179.67 cm³

Question no – (15)

Solution :

Let the radius of the sphere be r

∴ Surface area = 4πr²

Lateral surface area of the cylinder,

= 2πr(h) = 2πr(2r)

= 4πr²

Therefore, Surface area of the sphere.

Question no – (17)

Solution :

Let the radius of the base be r.

Now, 2/3πr³ = 1/3πr²h

Or, h = 2r

Or, h : r = 2 : 1

∴ The ratio of their ratio = 2 : 1

Question no – (18)

Solution :

Radius of the sphere,

= 18/2

= 9 cm.

∴ Let the length of the wire be h cm.

∴ 4/3πr³ = πr²₂h

Or, 4/3 (9)³ = (0.02)2h

Or, h = 243 × 4/0.0004

= 2430000 cm

= 24300 m

Question no – (19)

Solution :

Number of cubes = volume of the ball/volume of each cube

= 4/3 × 22/7 × 21/2 × 21/2 × 21/2/(1)³

= 21 × 21 × 11

= 4851.

Question no – (21)

Solution :

Volume of the steel,

= 2/3 π (R³ – r³)

= 2/3 × 314 {(5.25)³ – 5³}

= 4.25 cm³.

Question no – (23)

Solution :

Area of surface = area of the cylindrical part + area of the spherical part.

= 2πrh + 4πr²

= 2 × 22/7 × 0.1 × 3.6 + 4 × 22/7 × 0.1 × 0.7

= 44 (36 + 1.4)

= 77.44 m².

∴ Total expense,

= 7744 × 10

= 774.4 rupees

Previous Chapter Solution : 

👉 Chapter 16