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,
breadth = 8 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,
Breadth = 30 cm,
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,
Breadth = 2.5 dm,
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
Breadth = 4 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
Breadth is = 5 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
Breadth is = 50 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
Breadth is = 3 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
Breadth = 4 cm
Height = 3.5 cm
Volume = 1.5 cm3
∴ Block’s volume,
= 9 × 4 × 3.5
= 126 cm3
∴ 1 beads volume = 1.5 cm3
∴ No of beads,
= 126/1.5
= 84
Thus, 84 beads can be made from the block.
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 :
Let, breadth = b
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,
Breadth = 10 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,
Breadth = 2.5 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
breadth 25 dm
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 m3 clay 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 :
Let, breadth = B
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 m3 diesel.
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
Breadth is = 0.4 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)3 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 dm3 = 106 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,
breadth = 12 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,
Breadth = 8 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,
Breadth = 4 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,
Breadth = 30 dm,
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 cm2
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 m2
Therefore, the sum of the areas will be 190 m2
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
Breadth = b
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
Breadth = 3m
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
Breadth = b
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
Let, breadth = b
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 :
👉 Chapter 22 👈
