# Brilliant’s Composite Mathematics Class 8 Solutions Chapter 17

## 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,

= 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, r= 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 1st tin = (8)2 × 13

= 832 cm3.

Volume of the 2nd 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 msheet 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)× 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 + 15= 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 :

= 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 :

Updated: May 29, 2023 — 10:11 am