# Joy of Mathematics Class 8 Solutions Chapter 21

## Joy of Mathematics Class 8 Solutions Chapter 21 Surface Area and Volume of Solids

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 21 Surface Area and Volume of Solids. Here students can easily find step by step solutions of all the problems for Surface Area and Volume of Solids. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 21.1 and 21.2

Surface Area and Volume of Solids Exercise 21.1 Solution

Question no – (1)

Solution :

Here, l = long = 6 dm

B = wide = 3.4 dm

H = high = 3 dm

Surface area of card board = 2(lb + bh + hl) dm²

= 2 (6 × 3.4 + 3.4 × 3 × 6)

= 2 (20.4 + 10.2 + 18)

= 2 (48.6)

= 97.2 dm²

= 97.2 × 100

= 9720 cm²

Surface area of card board in dm units 9.72 dm and in cm unit’s 5720 cm²

Let, the x card board used

Now, x × 5/6 = 9720

=> 5x = 58320

=> x = 11664

11664 cm² card board is used .

Question no – (2)

Solution :

1st packages = 42 dm × 10 dm × 9 m

Surface area of 1st packages

= 2 (42 × 15 + 15 × 9 × 9 × 42) dm²

= 2 × 1143 dm²

= 2286 dm²

2nd packages = 18 dm × 15 dm × 21 dm

Surface area of 2nd packages

= 2 (18 × 15 + 15 × 18 + 21 × 18) dm²

= 2 × 963 dm²

= 1926 dm²

2nd packages make less quantity.

2nd packages make less quantity by,

= (2286 – 1926) dm²

= 360 dm²

Question no – (3)

Solution :

Here, long (i) = 1.8 m = 180 cm

Wide (w) = 1.5 m = 150 cm

High (h) = 1.2 m = 120 cm

Volume of the wooden card

= (180 × 150 × 120 cm³

= 3240000 cm³

And the volume of the cube which wide is 6 cm

= (6)³

= 216

Number of iron cubes that can be fitted in the wooden create

= 3240000/216

= 15000

Question no – (4)

Solution :

Here, long (i) = 5.6 m

Wide (b) = 4.3 m

High (h) = 4 m

Area of the wall = 2h (l + b) cm²

= 2 × 4 × (5.6 + 4.3)

= 8 × 9.9

= 79.2 cm²

the cost price of painting the wall at 65/- per sq. meter

= 79.2 × 65

= 5148 Rs

Question no – (5)

Solution :

Here, length (l) = 8 m

Wide (b) = 6 m

High (h0 = 4 m

Area of the room

= 2h (l + b) m²

= 2 × 4 (8 + 6) m²

= 8 × 14 m²

= 112 m²

Area of two windows

= 1.2 × 1 × 2

= 3 m²

And, area of two doors

= 2.2 × 1.2 × 2

= 5.28 m²

Now, cost of white washing the walls at 3 /- per sq. meter

= (112 × 3) – (3 × 3 + 5.28 × 3)

= 336 – (9 + 15.84)

= 336 – 24.84

= 311.16 Rs

Question no – (6)

Solution :

Let, the height of the room = x m

Length (l) = 4.2 m

Now, area of the room

= 2h (l + b) m²

= 2x (4.2 + 3) m²

According to the question,

2x (7.2) = 51.2

=> x = 51.2/2 × 7.2

= 512/144

= 3.56

So, the height of the room = 3.56 m

Question no – (7)

Solution :

The flat roof dimensions = 4 × 6 m

Rainfall height = 4.2 cm = 0.042 m

Volume of the rain water collected at roof

= 4 × 6 × 0.042 m³

Let, the depth of the water tank be = x m

4 × 6 × 0.042 = 3.2 × 1.5 × x

=> x = 4 × 6 × 0.042/3.2 × 1.5

=> 1008/4800

= 0.21 m

= 210 mm

So, the depth of the tank = 210 mm

Question no – (8)

Solution :

Here, volume of the lamp = 136 cm³

Volume of the cube which side is 2 cm

= (2 × 2 × 2) cm²

Number of cube = volume of lamp/volume of cube

= 136/2 × 2 × 2

= 17

Weight of one cube

= 207/17

= 12.17 g

Question no – (9)

Solution :

Let , the length of the square based cube = a and breadth of the square based cube = a (since the cuboid has a square base )

Height (h) = 10 cm

Side of the cube = 8 cm

Now, volume of the cube = 8 × 8 × 8 cm³

Volume of 5 cube = 5 × 8 × 8 × 8 cm³

Now, according to the question,

5 × 8 × 8 × 8 = a × a × 10

=> a² = 5 × 8 × 8 × 8/10

=> a = √4 × 8 × 8

=> √266

=> a = 16

side of the square based of the cuboid 16 cm

Question no – (10)

Solution :

Here, the dimensions = 19 m × 16 m × 10 m

Volume of the outer dimension wood

= (13 × 16 × 10) m²

= 2080 m²

Thickness

= 50 cm

= 0.5 m

Volume of the inner wood,

= (13 – 2 × 0.5) × (16 – 2 × 0.5) × (10 – 2 × 0.5) m²

= 1620 m²

Volume of the wood

= (2080-1620) m²

= 460 m²

Question no – (11)

Solution :

Here, the dimension of cuboid = 63 cm × 35 cm × 21 cm

Volume of cuboid,

= (63 × 35 × 21) cm³

= 46,305 cm³

Cube of length 7 cm and its volume

= 7 × 7 × 7 cm³

= 343 cm³

Number of cube,

= 46,305/343

= 135

So, the number of cube 135

Question no – (12)

Solution :

Here, shape of dimension = 80 cm × 50 cm

Area of the sheet

= (80 × 50) cm

= 4000 cm²

Now, new length = 80 – 10 – 10

= 60 cm

And new breadth = 50 – 10 – 10

= 50 – 20

= 30

New area of sheet = 60 × 30 cm²

= 1800 cm²

Surface area of the cuboid

= 2h (l + b)

According to the question,

2 h (l + b) = 1800

=> 2h (60 + 30) = 1800

=>2h = 1800/90

=> h = 10

Now, volume of the cuboid = l × b × h

= 60 × 30 × 10

= 18000 cm³

Question no –(13)

Solution :

Here, length of cuboid (i) = 15 + 15 = 30 cm

Height (h) = 15 cm

Surface area of cuboid = 2 (lb + bh + lh)

= 2 (30 × 15 + 15 × 15 + 15 × 15)

= 2 × 1125 cm²

= 2250 cm²

Volume of the cuboid = l × b × h

= 30 × 15 × 15

= 30 × 225

= 6750 cm³

So, the surface area of the cuboid 2250 cm² and volume of the cuboid = 6750 cm³

Question no – (14)

Solution :

Dimension of cuboid = 36 cm × 75 cm × 80 cm

∴ Volume of the cuboid = 36 × 75 × 80 cm³

Let, the sides of the cube be x

Volume of the cube = x³cm³

According of the question,

x³ = 36 × 75 × 80

=> x³ = 21600

=> x = 60

The surface area of the cube

= 6 (60)²

= 6 × 360 cm²

= 2160 cm²

Question no (15)

Solution :

Let, the length, breadth and height of the cuboid are 1 cm, b cm and h cm.

then area of the adjacent rectangular faces are lb, bh, hl, cm²

Now, lb = 80 → (1)

bh = 48 → (2)

hl = 60 → (3)

Multiplying (1), (2) and (3) we get,

lb × bh × hl = 230400

=> (lbh)² = 230400

=> lbh = √230400

=> 480

Area of the cuboid,

= 480/80, 480/48, 480/60,

= 6, 10, 8 cm

Surface Area and Volume of Solids Exercise 21.2 Solution :

Question no – (1)

Solution :

Here, radius of the cylinder = (r) = 70cm

Height of the cylinder (h)

= 7 m

= 700 cm

(a) Curved surface area of the cylinder = 2πrh

= 2 × 22/7 × 70 × 700 (∵ π = 22/7)

= 44 × 7000)

= 308000 cm2

(b) Total surface area of the cylinder

= 2πr (h + r)

= 2 × 22/7 × 70 (700 + 70)

= 2 × 22/7 × 70 × 770

= 44 × 770

= 311080 cm2

(c) Volume of the cylinder = πr2h

= 22/7 × 70 × 70 × 700

= 220 × 4900

= 1078000 cm2

Question no – (2)

Solution :

Let, the radius of the right circular cylinder = r

∴ Height of the cylinder (h) = 20cm

Now, volume of the cylinder = πr2h

According to the question,

πr2h = 770

=> r2 = 770/πh

= 770 × 7/22 × 20

=> r = √7 × 11 × 7/2 × 11 × 2

=> r = 7/2

= 3.5

Radius of the cylinder = 3.5 cm

Curved surface area = 2πrh

= 2 × 22/7 × 35/10 × 20

= 440 cm2

Question no – (3)

Solution :

(a) Let, the radius of the base = r

Circumference of the base = 2πr

2πr = 88

=> r = 88/2π

= 88 × 7/2 × 22 (∵ π = 22/7)

= 14

Height (h) = 26 cm

Now,

Total surface area of the cylinder = 2πr = (h + r)

= 2 × 22/7 × 14 (26 + 14)

= 2 × 22/7 × 14 × 40

= 3520 cm2

Volume of the cylinder = πr2h

= 22/7 × 14 × 14 × 26

= 16016 cm3

(b) Let, the radius of the base = r cm

Area of the base = πr2 cm2

πr2 = 346.5

=> r= 346.5 × 7/22 (∵ π = 22/7)

=> r2 = 110.25

=> r = 10.5

height (h) = 24.5 cm

Total surface area of the cylinder

= 2πr (h + r)

= 2 × 22/7 × 10.5 (24.5 + 10.5)

= 2 × 22/7 × 105/10 × 35

= 2310 cm2

Volume of the cylinder = πr2h

= 22/7 × 10.5 × 10.5 × 24.5

= 8489.25 cm3

Question no – (4)

Solution :

Let, the height of the cylinder = h

diameter f the cylinder = 7 cm

= 7/2 cm

= 3.5 cm

Volume of the cylinder = πr2h

= 22/7 × 3.5 × 3.5 × h

According to the question,

22/7 × 3.5 × 3.5 × h = 462

=> h = 462 × 7/22 × 3.5 × 3.5

= 12

Thus the height of the cylinder 12 cm.

Question no – (5)

Solution :

Diameter of the cylinder milk tank = 3.5m

Radius of the cylinder milk tank

= 3.5/2 m

= 1.75 m

Height (h) = 7m

Volume of the cylinder milk tank

= π × (1.75)2 × 7 m3

= 22/7 × (1.75)2 × 7 m3

= 67.375m3

= 67375 liter

Question no – (6)

Solution :

Here, diameter of the road roller = 1.26m

Height of the cylinder (h) = 1 m

Area leveled by road roller in one revolution = 2πrh

= 2 × 22/7 × 1.26/100 × 1

= 36 × 22/100 = 792/100

= 7.92

Area leveled in 500 complete revolution

= 7.92 × 500

= 3960

Question no – (7)

Solution :

Here, radius of the cylinder (r) = 35/2 cm

= 17.5 cm

= 0.175 m

Height of the cylinder (h) = 3m

Surface area of the cylinder lamp

= 2πrh

= 2 × 22/7 × 0.175 × 3

= 3.3 m2

Surface area of 1000 lamp

= 3.3 × 1000 m2

= 3300 m2

The total cost price of painting

= 15 × 3300m2

= 49500m2

Question no – (8)

Solution :

Here, length (l) = 22cm

Wide (b) = 12cm

Height of the cylinder (h) = 12cm

Circumference of base of cylinder = 22

2πr = 22

=> r = 22 × 7/2 × 22

=> r = 3.5cm

Volume of the cylinder = πr2h

= 22/7 × 3.5 × 3.5 × 12

= 462 cm3

Now, circumference of base of cylinder = 12 cm

2πr = 12

=> r = 12 × 7/2 × 22

= 42/22

= 1.909 cm

and the volume of the cylinder = πr2h

= 22/7 × (1.909)2 × 12

Question  no – (9)

Solution :

Here,

Side of the cube = 3.08 m

= 308 cm

= 308000km

Radius of the wire = 1.4cm

Let, the height of the wire = h

Now,

πr2h = 308000

=> h = 308000/π.r2

=> 308000/3.14 × (1.4)2 (π = 3.14)

= 50045 cm

= 5km

Question no – (10)

Solution :

Let, the radius of the cylinder = r and height of the cylinder = h

Curved surface area = 2πrh

Total surface area = 2πr (h + r)

Now, according to the question,

2πrh/2πr(h + r) = 3/5

=> h/h + r = 3/5

=> 5h = 3h + 3r

=> 5h – 3h = 3r

=> 2h = 3r

=> h/r = 3/2

Ratio of the height to the radius of the cylinder = 3 : 2

Question no – (11)

Solution :

Let, the radius of the cylinder 2x and 3x and height of the cylinder 9h and 4h

Volume of the first cylinder (v1) = πr2h

= π × (2x)2 × 9h

Volume of the second cylinder (v2) = π(3x)2 × 4h

v1/v2 = π × 4x × 9h/ π × 9x2 × 4h

v1/v2 = 1/1

Ratio of their volume 1 : 1

Question no – (12)

Solution :

Here, the length of the pipe (h) = 35m

Outer radius of the pipe (R) = 16/2 = 8cm

Inner radius of pipe (r) = 14/2 = 7cm

Volume of the metal = πh (R– r2) cm3

= 22/7 × 35 × (64 – 49)cm3

= 22 × 5 × 15 cm3

= 1650 cm3

= 0.165 m3

Volume of the metal 0.165 m3

Question no – (13)

Solution :

Depth of well = 12 m

Volume of earth dug out = πr2h

= 22/7 × 7 × 7 × 12

= 1848 m3

Let, height of the embankment = h m

wide = 7 m

Previous Chapter Solution :

Updated: May 30, 2023 — 3:44 pm