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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 cm^{3}

**Question no – (2)**

**Solution :**

**(i)** Volume,

= base area × height

= 22/7 × (9)^{2} × 14

= 3564 cm^{2}.

**Question no – (3)**

**Solution :**

Curved Surface area,

= 2 πrh = 22/7 × 21 × 10

= 1320 cm^{2}

**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 = πr^{2}h

= 22/7 × (7/4)^{2} × 16

= 154 cm^{3}.

**Question no – (5)**

**Solution :**

We have, πr^{2}h = 69300

Or, r^{2} = 69300 × 7/22 × 50

Or, r^{2 }= 3 × 3 × 7 × 7

Or, r = 3 × 7

= 21 cm.

Lateral surface area = 2πrh

= 2 × 22/7 × 21 × 50

= 6600 cm^{2}.

**Question no – (6)**

**Solution :**

Volume of the cylinder/volume of the reduced cylinder

= πr^{2}h/r (r/2)^{2} h

= 4 : 1

**Question no – (7)**

**Solution :**

Volume of the one coin,

= πr^{2}h = r × (3/4)^{2} × 1/5

= 9π/80 cm^{3}.

Volume of cylinder,

= π × (3)^{2} × 8

= 72 π cm^{3}.

**∴** Number of coins,

= 72π / 9π / 80

= 72π × 80/9π

= 640

**Question no – (8)**

**Solution :**

Volume = πr^{2} = 6358

Or, r^{2} = 6358 × 7/22 × 28

Or, r^{2} = 17 × 17/2 × 2

Or, r = 17/2 cm.

**∴** Lateral surface area

= 2πrh = 2 × 22/7 × 17/2 × 28

= 1496 cm^{2}.

**Question no – (9)**

**Solution :**

Volume at the 1^{st} tin = (8)^{2} × 13

= 832 cm^{3}.

Volume of the 2^{nd} tin

= 22/7 × (7)^{2} × 15

= 2310 cm^{3}.

**∴** 2310 – 832

= 1478 cm^{3}.

**Question no – (10)**

**Solution :**

Weight = volume × weigh per cm^{3}.

= 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,

= πr^{2}h = 22/7 × (14)^{2} × 7

= 4312 m^{3}

= 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 = 2πr^{2} + 2πrh

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

= 74800 cm^{2}

= 74800/10000 m^{2}

= 7.48m^{2}

Thus, 7.48 m^{2 }sheet metal will be required.

**Question no – (24)**

**Solution :**

Quantity of soil = πr^{2}h

= 22/7 × (7)^{2} × 12

= 1848 m^{3}.

Now, volume of the embankment,

= π (R^{2} – r^{2})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,

= πr^{2}h = 22/7 × (7/2)^{2} × 16

= 616 m^{3}.

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 m^{3}

Now, The soil was spread over the area of,

= (120 × 200) – 22/7 × (14)^{2}

= 24000 – 616

= 23384 m^{2}

**∴** 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 = 1/3πr^{2}h

= 1/3 × 22/7 × (28)^{2 }× 56

= 41888 cm^{2}

**Question no – (2)**

**Solution :**

**(i)** slant height = πrl

= √(14)^{2} + (48)^{2}

= √2500

= 50 cm.

Lateral surface = πrl

= 22/7 × 14 × 50

= 2200 cm^{2}

**Question no – (3)**

**Solution : **

Curved surface area = πrl

= 22/7 × 35 ×√(35)^{2} + (84)^{2}

= 22 × 5 × 91

= 10010 cm^{2}.

Volume = 1/3πr^{2}h

= 1/3 × 22/7 × (35)^{2}× 84

= 197800 cm^{3}.

**Question no – (5)**

**Solution :**

Lateral surface,

= πrl = 22/7 × 7 × 9

= 198 cm^{2}

**Question no – (6)**

**Solution : **

Area of curved surface = πrl

= 22/7 × 12 ×√(16)^{2} + (12)^{2}

= 22/7 × 12 × 20

= 753.6 m^{2}

**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,

= √8^{2} + 15^{2 }= 17 cm

Now, 2πr = 17

Or, π = 17 × 7/2 × 22

= 119/44 cm

**∴ **Total area,

= πr^{2} + 1/2 × 8 × 15

= 22/7 × 119 × 119/44 × 44 × 60

= 628.57 cm^{2}

**Question no – (10)**

**Solution :**

Measure of canvas = πrl

= 22/7 × 5 × 6.3

= 99 m^{2}.

**∴** Length of the canvas,

= 99/1.5

= 66m

Cost = 66 × 22.50

= 1485 rupees

**Question no – (11)**

**Solution :**

Capacity = 1/3πr^{2}h

= 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

= πr^{2}h_{1} + 1/3πr^{2}h_{2}

= πr^{2} (h_{1} + h_{2}/3)

= 22/7 × (8)^{2}× (240 + 36/3)

= 22/7 × 64 × 252

= 50688 cm^{3}.

**∴** 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 m^{2}

**Question no – (19)**

**Solution : **

Volume of the cone,

= 1/3πr^{2}h

= 1/3 × 22/7 × (3)^{2}× 7

= 66 cm^{3}.

**∴** 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.

**∴** 1/3πr^{2} (h) = πr^{2} (5)

Or, h = 5 × 3

= 15 cm.

**Question no – (25)**

**Solution : **

Area of base in the tent = πr^{2}

= 22/7 × 12 × 12

= 452.16 m^{2}

So, the base has the place for,

= 452.10/2

= 226.8 ≈ 226 people.

Air in the tent = 1/3πr^{2}h

= 1/3 × 22/7 × 144 × 9

= 1356.48 m^{3}.

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 = 1/3πr^{2}h,

Reduced volume

= 1/3πr^{2} (2h)

= 2/3 πr^{2}h

**∴** Clearly volume doubles

**Volumes and Surface Areas Exercise 17.3 Solution**

**Question no – (1)**

**Solution :**

**(i)** Volume,

= 4/3πr^{3} = 4/3 × 22/7 × 6^{3}

= 905.14 cm^{3}

Curved surface = 4πr^{2}

= 4 × 3.14 × 6^{2}

= 452.57 cm^{2}

**Question no – (2)**

**Solution :**

**(i)** Radius = 8/2 = 4 cm.

**∴** Volume = 4/3πr^{3}

= 4/3 × 3.14 × 4^{3}

= 267.95 cm^{3}.

**Curved surface,**

= 4πr^{2} = 4 × 3.14 × 4^{2}

= 318.86 m^{2}

**Question no – (8)**

**Solution :**

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

**∴** Initial volume = 4/3πr^{3},

Final volume,

= 4/3π (2π)^{3}

= 32/3πr^{3}

Now, final volume/initial volume

= 32/3πr^{3}/34/3πr^{3}

= 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 (R^{3} –r^{3})

= 4/3 × 22/7 × (15^{3} – 13^{3})

= 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,

= 4/3πr^{3}

= 4/3 × 22/7 × (7)^{3}

= 4 × 22 × 49

= 4312 cm^{3}.

**∴** Weight of the shot put,

= 4312 × 8

= 34496 gm

= 34.496 kg.

**Question no – (13)**

**Solution :**

Given, 4πr^{2} = 154

Or, r^{2} = 154 × 7/4 × 22

Or, r^{2} = 7 × 7/2 × 2

Or, r = 7/2 cm

**∴** r = 3.5 cm.

**∴** Volume = 4/3πr^{3}

= 4/3 × 22/7 × 7 × 7 × 7/2 × 2 × 2

= 539/3

= 179.67 cm^{3}

**Question no – (15)**

**Solution :**

Let the radius of the sphere be r

**∴** Surface area = 4πr^{2}

Lateral surface area of the cylinder,

= 2πr(h) = 2πr(2r)

= 4πr^{2}

Therefore, Surface area of the sphere.

**Question no – (17)**

**Solution :**

Let the radius of the base be r.

Now, 2/3πr^{3} = 1/3πr^{2}h

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.

**∴** 4/3πr^{3} = πr^{2}_{2}h

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 (R^{3} – r^{3})

= 2/3 × 314 {(5.25)^{3} – 5^{3}}

= 4.25 cm^{3}.

**Question no – (23)**

**Solution :**

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

= 2πrh + 4πr^{2}

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

= 44 (36 + 1.4)

= 77.44 m^{2}.

**∴ **Total expense,

= 7744 × 10

= 774.4 rupees

**Previous Chapter Solution : **