# Rd Sharma Solutions Class 8 Chapter 21 Mensuration – II Volumes and Surface Areas of a Cuboid and a Cube

Welcome to NCTB Solution. Here with this post we are going to help 8th Class students for the Solutions of RD Sharma Class 8 Mathematics, Chapter 21, Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube). Here students can easily find Exercise wise solution for chapter 21, Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube). Students will find proper solutions for Exercise 21.1, 21.2, 21.3 and 21.4 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.

Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube) Exercise 21.1 Solution :

Question no – (1)

Solution :

(i) Given in the question,

Length = 12 cm,

height = 6 cm

∴ Volume is,

= (12 × 8 × 6) …(we know Volume = l × b × h)

= 576 cm3

Therefore, the volume of a cuboid is 576 cm3

(ii) In the given question,

Length = 1.2 m,

Height = 15 cm

∴ Volume is,

= (1200 × 80 × 15) (Volume = l × b × h)

= 540000 cm3

Therefore, the volume of a cuboid is 540000 cm3

(iii) In the question we get,

Length = 15 cm,

Height = 8 cm

∴ Volume is,

= (15 × 2500 × 8) (∵ Volume = l × b × h)

= 3000 cm3

Therefore, the volume of a cuboid is 3000 cm3

Question no – (2)

Solution :

(i) Cube whose side is = 4 cm

Volume is,

4 = 4 × 4 × 4

= 64 cm3

Therefore, the volume of a cube is 64 cm3

(ii) Cube whose side is = 8 cm

Volume is,

8 = 8 × 8 × 8

= 512 cm3

So, the volume of a cube is 512 cm3

(iii) Cube whose side is = 1.5 dm

Volume is,

= 1.5 dm = 1.5 × 1.5 × 1.5

= 3375 cm3

Thus, the volume of a cube is 3375 cm3

(iv) Cube whose side is = 1.2 m

Volume is,

= 1.2 m = 1.2 × 1.2 × 1.2

= 1.728 m3

Hence, the volume of a cube is 1.728 m3

(v) Cube whose side is = 25 mm

Volume is,

= 25 m = 25 × 25 × 25

= 15625 mm3

Therefore, the volume of a cube is 15625 mm3

Question no – (3)

Solution :

Given in the question,

Volume 100 cm3,

Length = 5 cm

Let, height = h

According to questions,

h × l × b = 100

or, h = 100/l × b = 100/5 × 4 [∴ l = 5 cm b = 4 cm]

or, h = 5 cm

Therefore, the height of a cuboid will be 5 cm.

Question no – (4)

Solution :

In the question we get,

Length is = 10 cm

Let, height = h

According to questions,

= h × l × b = 300

or, h = 300/l × b = 300/10 × 5 [∴ l = 10, b = 5]

or, h = 6 cm

Therefore, the height of the vessel is 6 cm.

Question no – (5)

Solution :

As per the question,

Length is = 8 cm

Can hold = 4 litres of milk

4 litres = 4 × 1000 = 4000 cm3

Let, h × 8 × 50 = 400

or, h = 4000/8 × 50

= 10 cm

Hence, its height should be 10 cm.

Question no – (6)

Solution :

Given in the question,

Length is = 4 cm

Contain wood = 36 cm3

Let, height = h

According to questions,

h × 4 × 3 = 36

or, h = 36/4 × 3

h = 3 cm

Therefore, the height cuboidal wooden block will be 3 cm.

Question no – (7)

Solution :

(i) The volume of a cube, if its edge is halved will be 1/8 times.

(ii) The volume of a cube, if its edge is trebled will be 27 times.

Question no – (8)

Solution :

(i) If Length is doubled, height is same and breadth is halved than the volume of a cuboid will be Same.

(ii) If length is doubled, height is doubled and breadth is same then the the volume will be 4 times.

Question no – (9)

Solution :

Volume1 = 5 × 6 × 7 = 210 cm3

Volume 2 = 4 × 7 × 8 = 224 cm3

Volumes3 = 2 × 3 × 13 = 78 cm3

Total volume =   512 cm3

Method cubes volume = 512 cm3

∴ method cubes h × l × b = 512 = × 8 × 8 × 8

Height = h = 8 cm

Question no – (10)

Solution :

Volume of iron piece,

= 50 × 40 × 10

= 20000 cm3

1 cm3 of iron weights = 8 gm.

20000 cm3 of iron weights,

=  8 × 20000

= 160000 gm.

= 160 kg

Therefore, the weight of the solid iron piece will be 160 kg.

Question no – (11)

Solution :

We know that,

3 m = 300 cm

Volume of wooden log,

= 300 × 75 × 50

= 1,125,000 cm3

1 cubical volume,

= 25 × 25 × 25

= 15,625 cm3

No of blocks volume,

= 1,125,000/15,625

= 72

Hence, 72 cubical block can be cut.

Question no – (12)

Solution :

Given in the question,

Length = 9 cm

Height = 3.5 cm

Volume = 1.5 cm3

Block’s volume,

= 9 × 4 × 3.5

= 126 cm3

1 beads volume = 1.5 cm3

= 126/1.5

= 84

Question no – (13)

Solution :

Box’s volume,

= 2 × 3 × 10

= 60 cm3

Cartons volume,

= 40 × 36 × 24

= 34560 cm3

∴ No of box,

= 24560/60

= 576

Question no – (14)

Solution :

Cuboidal block’s volume,

= 50 × 45 × 34

= 76500 cm³

Cuboids volume,

= 5 × 3 × 2

= 30 cm²

Number of cuboids,

= 76500/30

= 2550

Thus, 2550 cuboids can be obtained from this block.

Question no – (15)

Solution :

Length of cube = l

Length of cube A = LA = 3lB

= CubeA/CubeB = (lA)3/(lB)3 = (3lB × lA × lA)/(lB)3

or, CubeA/CubeB = 27lB/lB3 = 27/1

∴ Cube A : Cube B = 27 : 1

Therefore, the ratio of the volume is 27 : 1

Question no – (16)

Solution :

Ice-cream brick’s volume,

= 20 × 10 × 7

= 1400 cm3

Inner dimension of bricks stood volume,

= 100 × 50 × 42

= 210000 cm3

No of bricks stood,

= 210000/1400

= 150

Hence, 150 such bricks can be stored in deep fridge.

Question no – (17)

Solution :

According to the question,

= V1 = (2)3 = 8 cm3

= V2 = (4)3 = 64 cm3

= 8 × V1

Question no – (18)

Solution :

Tea packet volume,

= 10 × 6 × 4

= 240 cm3

Cardboard volume,

= 50 × 30 × 20

= 30000 cm3

Number of tea packets,

= 30000/240

= 125

Therefore, 125 such tea-packets can be placed in a cardboard box.

Question no – (19)

Solution :

Metal block’s volume,

= 5 × 4 × 3

= 60 cm3

Another block’s volume,

= 15 × 8 × 3

= 360 cm3

∴ 60 cm3 blocks weight = 1 kg

∴ 1 cm3 block’s weight = 1/60 kg

∴ 360 cm3 block’s weight

= 360/60 kg

= 6 kg

Hence, the weight of a block of the same metal is 6 kg.

Question no – (20)

Solution :

Box’s  volume,

= 56 × 40 × 25

= 56000 cm3

Soap cake’s volume,

= 7 × 5 × 2.5

= 87.5 cm3

No of cake’s volume,

= 56000/87.5

= 640

Thus, 640 soap cakes can be placed in the box.

Question no – (21)

Solution :

According to questions,

= 3 × 4 × b = 48

or, b = 48/3 × 4

= 48/12

= 4 cm

Therefore, the breadth of cuboidal box will be 4 cm.

Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube) Exercise 21.2 Solution :

Question no – (1)

Solution :

(i) In the question we get,

Length = 12 m,

Height = 4.5 m

Volume is,

= 12 × 10 × 4.5

= 540 m3

Therefore, the volume in cube meter will be 540 m3

(ii) Given in the question,

Length = 4 m,

Height = 50 cm

Volume is,

= 4 × 2.5 × 0.05

= 0.5 m3

Thus, the volume in cubic metre will be 0.5 m3

(iii) In the question we get,

length = 10 m

height = 25 cm

Volume is,

= 10 × 2.5 × 0.25

= 6.25 m3

Therefore, the volume in cubic metre is 6.25 m3

Question no – (2)

Solution :

(i) Cubes whose side is = 1.5 m.

Volume in cubic decimetre,

= 1.5 m = 15 dm

= (15)3

= 3375 dm3

Thus, volume of the cube is 3375 dm3

(ii) Cubes whose side is = 75 cm

Volume in cubic decimetre,

= 75 cm = 7.5 dm

= (7.5)3

= 421.875 dm3

Hence, volume of the cube is 421.875 dm3

(iii) Cubes whose side is = 2 dm 5 cm

Volume in cubic decimetre,

= 2 dm 5 cm = {(2 × 10) + 5}

= (20 + 5) = 25 cm

= 2.5 dm = (2.5)3

= 15.625 dm3

Therefore, volume of the cube is = 15.625 dm3

Question no – (3)

Solution :

As per the given question,

Well measuring = 3 m by 2 m by 5 m

Clay dug out,

= 3 × 2 × 5

= 30 m3

Therefore, 30 mclay will dug out.

Question no – (4)

Solution :

Let, height = h

According to questions,

Area × h = 106

or, h = 168/Area

= h = 168/28

= h = 6 cm

Thus, the height of cuboid will be 6 cm.

Question no – (5)

Solution :

As per the question,

Length of Tank = 8 m

Breadth of Tank =  6 m

Height of Tank =  2 m

Water contain,

= 8 × 6 × 2

= 96 m3

Therefore, tank can contain 96 m3 water.

Question no – (6)

Solution :

According to questions,

= B × 10 × 2.5 = 50 …(∴ 50000 litres = 50 litres)

or, B = 50/10 × 2.5

= B = 2 m

Hence, the breadth of the tank will be 2 m

Question no – (7)

Solution :

According to the question,

L = 2 m

b = 2 m

h = 40 cm = 0.4 m

Diesel hold,

= 2 × 2 × 0.4

= 1.6 m3

Thus, it can hold 1.6 mdiesel.

Question no – (8)

Solution :

Given in the question,

Length of  room = 5 m,

Breadth of  room = 4.5 m

Height of  room = 3 m

Volume of Air contains,

= 5 × 4.5 × 3

= 67.5 m3

Thus, the volume of the air in the room will be 67.5 m3

Question no – (9)

Solution :

As per the given question,

length of tank = 3 m

Breadth of tank = 2 m

Depth of tank = 1 m .

Water hold,

= 3 × 2 × 1

= 6 m3

= 6000 litres

Therefore, the tank can hold 6000 litres water.

Question no – (10)

Solution :

We know that,

3 m = 300 cm

Planks volume,

= 300 × 15 × 5

= 22500 cm3

Block’s volume,

= 600 × 75 × 45

= 20,475000 cm3

Number of planks,

= 20,475000/22500

= 910

Therefore, 910 planks cab be prepared.

Question no – (11)

Solution :

Brick’s volume,

= 25 × 10 × 8

= 2000 cm3

Wall volume,

= 500 × 300 × 16

= 2400000 cm3

Number of bricks,

= 2400000/2000

= 1200

Therefore, 1200 bricks will be required.

Question no – (12)

Solution :

Tank volume,

= 20 × 15 × 6

= 1800 m3

= 1800000 litres

Water consumption,

= 4000 × 150

= 6000000 litres

Water of this tank will left,

= 1800000/600000

= 3 days

Therefore, water of this tank will last 3 days.

Question no – (13)

Solution :

Wells volume,

= 14 × 8 × 6

= 672 m3

Let, the earth level rise = x m

According to questions,

= x × 70 × 60 = 672

or, x = 672/70 × 60 = 0.16

or, x = 0.16 m

= x = 16 cm

Therefore, earth level will rise 16 m.

Question no – (14)

Solution :

Let, rise in level of water = x m

According to questions,

= x × 250 × 130 = 3250

Or, x = 3250/250 × 130

= 32250/32500 = 0.1m

or, x = 0.1 m

Hence, the rise in the level of water will be 0.1 m.

Question no – (15)

Solution :

Let, thickness = h cm

Length is = 5 m

According to questions,

= H × 5 × 0.4 = 0.6

or, H = 0.6/5 × 0.4

= 0.6/2

= 0.3 m

Therefore, the beam will be 0.3 m thick.

Question no – (16)

Solution :

Height = 6 cm 0.06 m

Area of field,

= 3 hectares = 3 × 10000

= 300000 m2

Volume of water = area × height

= 30000 × 0.6

= 1800 m3

= 1800000 litres.

Therefore, 1800000 litres will fall on there.

Question no – (17)

Solution :

Volume of 1 cube = 1 cm

= 0.1 m = (0.01)m3

Volume of 4000 cube,

= 400 × (0.01)3 m3

Let, 3rd edge of cuboidal beam = h

According to questions,

= h × 0.5 × 8 = 4000 × (0.01)3

or, h = 4000 × 0.01 × 0.01 × 0.01/0.5 × 8 = 0.001

h = 0.001 m

Therefore, the third edge will be 0.001 m

Question no – (18)

Solution :

Volume of metal block,

= 2.25 × 1.5 × 0.27

= 0.911125 m3

Volume of cube,

= 45 cm

= 0.45 m

=  (0.45)3

Number of cube,

= 2.25 × 1.5 × 0.27/0.45 × 0.45 × 0.45

= 0.91125/0.091125

= 10

Hence, 10 cubes will formed.

Question no – (19)

Solution :

Volume of iron,

= 600 cm × 6 cm × 2 cm

= 7200 cm3

1 cm3 of iron weight = 8 gm

7200 if iron height,

= 8 × 7200

= 57600

= 57.6 kg

Therefore, the weight of this piece is 57.6 kg.

Question no – (20)

Solution :

(i) 1 m3 = 106 cm3

(ii) 1 litre = 1 cubic decimetre

(iii) 1 kl = 1 m3

(iv) The volume of cube of side 8 cm is 512 cm3

(v) The volume of a wooden cuboid of length 10 cm and breadth 8 cm is 4000 cm3. The height of the cuboid is 50 cm.

= h 4000/8 × 0.10

= 50 cm

(vi) 1 cu.dm = 106 cu.mm

(vii) 1 cu. km = 109 cu. m

(viii) 1 litre = 103 cu.cm

(ix) 1 ml = 1 cu.cm

(x) 1 kl = 1000 dm106 cm3

Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube) Exercise 21.3 Solution :

Question no – (1)

Solution :

(i) In the question we get,

length = 10 cm,

height = 14 cm

Surface Area = 2 [(10 × 12)] + (10 × 14) + (12 × 14))]

= 2 [120 + 140 + 16]

= 2 × 428

= 856 cm2

Therefore, the surface area of a cuboid is 856 cm2

(ii) In the question we get,

Length = 6 dm,

Height = 10 dm

Surface Area,

= 2 [(6 × 8) + (6 × 10) + (8 × 10)]

= 2 [84 + 60 + 80]

= 2 × 188

= 376 dm2

Hence, the surface area of a cuboid will be 376 dm2

(iii) From the question we get,

Length = 2 m,

Height = 5 m

Surface Area,

= 2 [(2 × 4) + (2 × 5) + (4 × 5)]

= 2 [8 + 10 + 20]

= 2 × 38

= 76 m2

Thus, the surface area of a cuboid is 76 m2

(iv) From the question we get,

Length = 3.2 m,

Height = 250 cm

Surface Area,

= 2 [(3.2 × 3) + (3.2 × 2.5)]

= 2 [9.6 + 8 + 7.5]

= 2 × 25.1

= 50.2 m2

Therefore, the surface area of a cuboid is 50.2 m2

Question no – (2)

Solution :

(i) Cube whose volume is = 1.2 m

Surface area of cube,

∴ 1.2 m = 6 (1.2)2

= 6 × 1.44

= 8.64 m2

Thus, the surface area of cube is 8.64 m2

(ii) Cube whose volume is = 27 cm

Surface area of cube,

∴ 27 cm =  6(27)2

= 6 × 729

= 4374 cm2

Hence, the surface area of cube will be 4374 cm2

(iii) Cube whose volume is = 3 cm

Surface area of cube,

∴ 3 cm = 6 (6)2

= 6 × 9

= 54 cm2

Thus, the surface area of cube will be 54 cm2

(iv) Cube whose volume is = 6 m

Surface area of cube,

∴ 6 m = 6 (6)2

= 6 × 36

= 216 m2

Hence, the surface area of cube will be 216 m2

(v) Cube whose volume is = 2.1 m

Surface area of cube,

∴ 2.1 m = 6 (2.1)2

= 6 × 4.41

= 26.46 m2

Therefore, the surface area of cube is 26.46 m2

Question no – (3)

Solution :

Given, Cuboidal box is 5 cm by 5 cm by 4 cm.

Surface Area,

= 2[5 × 5) + (5 × 4) + (5 × 4)

= 2 [25 + 20 + 20]

= 2 × 65

= 130 cm2

Therefore, the surface area cuboidal box 130 cm

Question no – (4)

Solution :

(i) Let, one side of cube = l

According to questions,
= l3 = 343

or, l = 3√343

or, l = 7

Surface Area,

= 6 (7)2 = 6 × 72

= 294 cm2

Thus, the surface area of cube will be 294 cm2

(ii) Let, one side cube = l

According to questions,

= l3 = 216

or, = 3√216 = 6

Surface Area,

= 6 × (6)2

= 216 dm2

Hence, the surface area of cube will be 216 dm2

Question no – (5)

Solution :

(i) Let, are edge is = l

According to questions,

= 6l2 = 96

or, l2 = 96/6 = 16

or, l = √16

= l = 4

∴ Volume,

= l3 = 43

= 64 cm2

Thus, the volume of the cube will be 64 cm2

(ii) Let, one edge = l

According to questions,

= 6l2 = 150

or, l2 = 150/6 = 25

or, l = √25 = 5

Volume,

= 6 (5)2 = 6 × 25

= 150 m3

Therefore, the volume of a cube will be 150 m3

Question no – (6)

Solution :

Let, common factor = x

Ratio of dimension = 5 : 3 : 1

Length = l = 5x

Breadth = b = 3x height = x

Now, according to the questions,

= 2 [(5x × 3x) + (5x × x) + (3x ×x) = 414

or, 15 x2 + 5x2 + 3x2 = 414/2 = 207

or, 23x2 = 207

or, x2 = 207/23 = 9

or, x = √9 = 3

Length = 5x = 5 × 3 = 15 m

Breadth = 3.x = 3 × 3 = 9 m

Height = x = 3 m

Therefore, the dimensions will be length = 15 m, breadth = 9 m and height = 3 m.

Question no – (7)

Solution :

Given, Length 25 cm, 0.5 m and Height 15 cm

Surface Area = 2 [(0.25 × 0.05) + (0.25 × 0.5) + (0.15 × 0.5)

= [0.0375 + 0.125 + 0.075]

= 2 × 2 0.2375

= 0.475 m2

Therefore, the area of cardboard required will be 0.475 m2

Question no – (8)

Solution :

Given, the edge of the box is = 12 cm

Surface Area,

= (6 (12)2 = 6 × 144

= 864 cm2

Hence, the surface area of a wooden box is 864 cm2

Question no – (9)

Solution :

Area of tin required for 1 sheet,

= 2 [(26 × 26) + (26 × 45) + (26 × 45)

= 6032 cm2

Area of tin required for 20 sheet,

= 20 × 6032 = 120660 cm2

= 120640/100 × 100 m2

= 12.064 m2

Cost of tin,

= 12.064 × 10

= 120.64 Rs

Thus, the cost of tin sheet used is Rs. 120.64

Question no – (10)

Solution :

As per the given question,
Classroom is 11 m long, 8 m wide and 5 m high.

Area of the floor,

= (11 × 8)

= 88 m2

Area of floor walls,

= [2 (11 + 8)] × 5

= 2 × 19 × 5

= 190 m

Therefore, the sum of the areas will be 190 m

Question no – (11)

Solution :

Area of floor,

= 200 × 15

= 300 m2

Area of walls,

= 2 (20 + 15) × 3

= (2 × 35) × 3 = 70 × 3

= 210

Total area,

= 300 + 210

= 510 m2

Total cost of repairing,

= 510 × 25

= 12750 Rs.

Therefore, the cost of repairing will be 12750 Rs.

Question no – (12)

Solution :

According to the question,

Perimeter of a floor of a room is = 30 m

Height is = 3 m

Area of wall = surface area × height

= 30 × 3

= 90m2

Hence, the area of four walls will be 90m2

Question no – (13)

Solution :

Let, length = l

Height = h

Area of the floor = l × b

Area of one adjacent wall = l × h

Area of 2nd adjacent wall = b × h

Total area = (l × b) × (l × h) × (b × h)

= 1b × 1h × bh

= l2 b2 h2

Question no – (14)

Solution :

Given, Length = 4.5 m

Height = 350 cm = 3.5 m.

Cost of plastering = 8/m2

Area of walls,

= 2 × 3.5 (4.5 + 3)

= 7 × 7.5

= 52.2m2

The cost of plastering,

= 65.7 × 8

= 528 Rs.

Thus, the cost plastering will be 528 Rs.

Question no – (15)

Solution :

Let, Length = l

Height = h

S.A = 2(lb + bh + lh)

= 50 m2

L.S.A = 2h (l + b) = 30 m2

According to the question,

2 (lb + bh + lh) – 2h (l + b) = 50 – 30 = 20

or, 2 (lb + bh + lh) – 2h (l + b) = 20

or, lb + bh + lh + lh – bh = 20/2 = 10

or, lb = 10

Hence, the  Area of base lb = 10m2

Question no – (16)

Solution :

Area of four wall = 2 h (l + b)

= 2 × 3.5 (7 + 6)

= 2 × 35 × 35

= 91 m2

Area of four wall without Door and window,

= 90 – 17

= 74 m2

Total cost of whit washing

= 74 × 1.50

= 111 Rs.

Question no – (17)

Solution :

Area of hall,

= 2385 – 60/1.20

= 1988

Area of hall including door and window,

= 2 × 8 (80 + b)

= 16 (80 + b)

Area of hall 10 doors,

= 10 × (3 × 1.5)

= 45 m2

Area of hall 10 windows,

= 10 (1.5 × 1m)

= 15 m2

According to questions,

= 16 (80 + b) – (45 + 15) = 198

or, 16 (80 + b) = 1988 + 60 = 2048

or, 80 + b = 2048/16 = 128

or, b = 128 – 80

= b = 48 m

Therefore, the breadth of the hall will be 48 m.

Next Chapter Solution :

Updated: June 14, 2023 — 7:15 am