RS Aggarwal Solutions: Volumes and Surface Area of Solids- 1 | Mathematics (Maths) Class 9 PDF Download

 Page 1


Question:1
Find the volume, the lateral surface are and the total surface area of the cuboid whose dimensions are:
i length = 12 cm, breadth = 8 cm and height = 4.5 cm
ii length = 26 m, breadth = 14 m and height = 6.5 m
iii length = 15 m, breadth = 6 m and height = 5 dm
iv length = 24 m, breadth = 25 cm and height = 6 m
Solution:
i
Here, l = 12 cm, b = 8 cm, h = 4.5 cm
Volume of the cuboid = l ×b ×h
                                  = 12 ×8 ×4. 5 cm
3
= 432 cm
3
Total Surface area = 2(lb + lh+ bh)
                              = 2 12 ×8 + 12 ×4. 5 +8 ×4. 5 cm
2
= 2 96 +54 + 36 cm
2
= 2 × 186  cm
2
= 372  cm 
2
Lateral surface area = 2(l +b)×h
                              = [2(12 +8)×4. 5] cm
2
= [2(20)×4. 5]  cm
2
= 40 ×4. 5  cm
2
= 180 cm
2
ii
Here, l = 26 m; b = 14 m; h =6.5 m
Volume of the cuboid =  l ×b ×h
                                = (26 ×14 ×6. 5)m
3
= 2366 m
3
Total surface area = 2(lb + lh+ bh)
                             = 2 26 ×14 +26 ×6. 5 +6. 5 ×14 m
2
= 2 364 +169 +91 m
2
= 2 ×624 m
2
= 1248 m
2
Lateral surface area = 2(l +b)×h 
                              = [2(26 +14)×6. 5] m
2
= [2 ×40 ×6. 5] m
2
= 520 m
2
iii
Here, l = 15 m; b = 6 m; h = 5 dm = 0.5 m
Volume of the cuboid =  l ×b ×h
                                = 15 ×6 ×0. 5 m
3
= 45 m
3
Total surface area = 2(lb + lh+ bh) 
                          = 2 15 ×6 +15 ×0. 5 +6 ×0. 5 m
2
= 2 90 +7. 5 +3 m
2
= 2 ×100. 5 m
2
= 201m
2
Lateral surface area = 2(l +b)×h 
                               = [2(15 +6)×0. 5] m
2
= [2 ×21 ×0. 5] m
2
= 21 m
2
iv
Here, l = 24 m; b = 25 cm = 0.25 m; h =6 m
Volume of the cuboid =  l ×b ×h 
                                = (24 ×0. 25 ×6) m
3
= 36 m
3
Total Surface area = 2(lb + lh+ bh)
                            = 2 24 ×0. 25 +24 ×6 +0. 25 ×6 m
2
= 2 6 +144 + 1. 5 m
2
= 2 ×151. 5 m
2
= 303 m
2
Lateral surface area = 2(l +b)×h 
                             = [2(24 +0. 25)×6] m
2
= [2 ×24. 25 ×6] m
2
= 291 m
2
Question:2
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What is the volume of a packet containing 12 such matchboxes?
Solution:
Volume of each matchbox = 4 × 2.5 × 1.5 = 15 cm
3
? Volume of 12 matchboxes = 12 × 15 = 180 cm
3
Thus, the volume of a packet containing 12 such matchboxes is 180 cm
3
.
Question:3
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (Given, 1 m
3
 = 1000 litres.)
Solution:
Volume of water in the tank = Length × Breadth × Height = 6 × 5 × 4.5 = 135 m
3
? Volume of water in litres = 135 × 1000 = 135000 L        (1 m
3
 = 1000 L)
Thus, the water tank can hold 135000 L of water.
Question:4
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank if its length and depth are respectively 10 m and 2.5 m. (Given, 1000 litres = 1 m
3
.)
Solution:
Capacity of the tank = 50000 L = 
50000
1000
 = 50 m
3
           (1000 L = 1 m
3
)
Length of the tank = 10 m
Height ordepth of the tank = 2.5 m
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Page 2


Question:1
Find the volume, the lateral surface are and the total surface area of the cuboid whose dimensions are:
i length = 12 cm, breadth = 8 cm and height = 4.5 cm
ii length = 26 m, breadth = 14 m and height = 6.5 m
iii length = 15 m, breadth = 6 m and height = 5 dm
iv length = 24 m, breadth = 25 cm and height = 6 m
Solution:
i
Here, l = 12 cm, b = 8 cm, h = 4.5 cm
Volume of the cuboid = l ×b ×h
                                  = 12 ×8 ×4. 5 cm
3
= 432 cm
3
Total Surface area = 2(lb + lh+ bh)
                              = 2 12 ×8 + 12 ×4. 5 +8 ×4. 5 cm
2
= 2 96 +54 + 36 cm
2
= 2 × 186  cm
2
= 372  cm 
2
Lateral surface area = 2(l +b)×h
                              = [2(12 +8)×4. 5] cm
2
= [2(20)×4. 5]  cm
2
= 40 ×4. 5  cm
2
= 180 cm
2
ii
Here, l = 26 m; b = 14 m; h =6.5 m
Volume of the cuboid =  l ×b ×h
                                = (26 ×14 ×6. 5)m
3
= 2366 m
3
Total surface area = 2(lb + lh+ bh)
                             = 2 26 ×14 +26 ×6. 5 +6. 5 ×14 m
2
= 2 364 +169 +91 m
2
= 2 ×624 m
2
= 1248 m
2
Lateral surface area = 2(l +b)×h 
                              = [2(26 +14)×6. 5] m
2
= [2 ×40 ×6. 5] m
2
= 520 m
2
iii
Here, l = 15 m; b = 6 m; h = 5 dm = 0.5 m
Volume of the cuboid =  l ×b ×h
                                = 15 ×6 ×0. 5 m
3
= 45 m
3
Total surface area = 2(lb + lh+ bh) 
                          = 2 15 ×6 +15 ×0. 5 +6 ×0. 5 m
2
= 2 90 +7. 5 +3 m
2
= 2 ×100. 5 m
2
= 201m
2
Lateral surface area = 2(l +b)×h 
                               = [2(15 +6)×0. 5] m
2
= [2 ×21 ×0. 5] m
2
= 21 m
2
iv
Here, l = 24 m; b = 25 cm = 0.25 m; h =6 m
Volume of the cuboid =  l ×b ×h 
                                = (24 ×0. 25 ×6) m
3
= 36 m
3
Total Surface area = 2(lb + lh+ bh)
                            = 2 24 ×0. 25 +24 ×6 +0. 25 ×6 m
2
= 2 6 +144 + 1. 5 m
2
= 2 ×151. 5 m
2
= 303 m
2
Lateral surface area = 2(l +b)×h 
                             = [2(24 +0. 25)×6] m
2
= [2 ×24. 25 ×6] m
2
= 291 m
2
Question:2
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What is the volume of a packet containing 12 such matchboxes?
Solution:
Volume of each matchbox = 4 × 2.5 × 1.5 = 15 cm
3
? Volume of 12 matchboxes = 12 × 15 = 180 cm
3
Thus, the volume of a packet containing 12 such matchboxes is 180 cm
3
.
Question:3
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (Given, 1 m
3
 = 1000 litres.)
Solution:
Volume of water in the tank = Length × Breadth × Height = 6 × 5 × 4.5 = 135 m
3
? Volume of water in litres = 135 × 1000 = 135000 L        (1 m
3
 = 1000 L)
Thus, the water tank can hold 135000 L of water.
Question:4
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank if its length and depth are respectively 10 m and 2.5 m. (Given, 1000 litres = 1 m
3
.)
Solution:
Capacity of the tank = 50000 L = 
50000
1000
 = 50 m
3
           (1000 L = 1 m
3
)
Length of the tank = 10 m
Height ordepth of the tank = 2.5 m
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Now,
Volume of the cuboidal tank = Length × Breadth × Height
? Breadth of the tank = 
Volume of the tank
Length × Height
=
50
10×2.5
=
50
25
 = 2 m
Thus, the breadth of the tank is 2 m.
Question:5
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates, each measuring 1.5 m × 1.25 m × 0.5 m, that can be stored in the godown.
Solution:
Volume of the godown = 40 m × 25 m × 15 m = 15000 m
3
Volume of each wooden crate = 1.5 m × 1.25 m × 0.5 m = 0.9375 m
3
? Maximum number of wooden crates that can be stored in the godown
=
Volume of the godown
Volume of each wooden crate
=
15000
0.9375
= 16000
Question:6
How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 80 cm deep?
Solution:
Number of planks = 
volume of the pit in cm
3
volume of 1 plank in cm
3
Volume of one plank = l ×b ×h cm
3
                                  = 500 ×25 ×10  cm
3
= 125000 cm
3
Volume of the pit = l ×b ×h cm
3
Here, l = 20 m = 2000 cm; b = 6 m = 600 cm; h = 80 cmi. e. , volume of the pit = 2000 ×600 ×80 cm
3
                                  = 96000000 cm
3
? Number of planks = 
96000000
125000
 = 
96000
125
= 768
Question:7
How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?
Solution:
Length of the wall = 8 m = 800 cm
Breadth of the wall = 22.5 cm
Height of the wall = 6 m = 600 cm
i.e., volume of wall = 800 ×22. 5 ×600 cm
3
= 10800000 cm
3
Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm
i.e., volume of one brick = 25 ×11. 25 ×6 = 1687. 5 cm
3
? Number of bricks = 
volume of the wall
volume of one brick
 = 
10800000
1687.5
= 6400
Question:8
Find the capacity of a closed rectangular cistern whose length is 8 m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet required to make the cistern.
Solution:
Length of the cistern, l = 8 m
Breadth of the cistern, b = 6 m
Height ordepth of the cistern, h = 2.5 m
? Capacity of the cistern
= Volume of the cistern
= l × b × h
= 8 × 6 × 2.5
= 120 m
3
Also,
Area of the iron sheet required to make the cistern
= Total surface area of the cistern
= 2(lb + bh + hl)
= 28 ×6 +6 ×2.5 +2.5 ×8
= 2 × 83
= 166 m
2
Question:9
The dimensions of a room are 9m ×8m ×6.5m. It has one door of dimensions 2m ×1.5m and two windows, each of dimensions 1.5m ×1m. Find the cost of whitewashing the walls at Rs
25 per square metre.
Solution:
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Page 3


Question:1
Find the volume, the lateral surface are and the total surface area of the cuboid whose dimensions are:
i length = 12 cm, breadth = 8 cm and height = 4.5 cm
ii length = 26 m, breadth = 14 m and height = 6.5 m
iii length = 15 m, breadth = 6 m and height = 5 dm
iv length = 24 m, breadth = 25 cm and height = 6 m
Solution:
i
Here, l = 12 cm, b = 8 cm, h = 4.5 cm
Volume of the cuboid = l ×b ×h
                                  = 12 ×8 ×4. 5 cm
3
= 432 cm
3
Total Surface area = 2(lb + lh+ bh)
                              = 2 12 ×8 + 12 ×4. 5 +8 ×4. 5 cm
2
= 2 96 +54 + 36 cm
2
= 2 × 186  cm
2
= 372  cm 
2
Lateral surface area = 2(l +b)×h
                              = [2(12 +8)×4. 5] cm
2
= [2(20)×4. 5]  cm
2
= 40 ×4. 5  cm
2
= 180 cm
2
ii
Here, l = 26 m; b = 14 m; h =6.5 m
Volume of the cuboid =  l ×b ×h
                                = (26 ×14 ×6. 5)m
3
= 2366 m
3
Total surface area = 2(lb + lh+ bh)
                             = 2 26 ×14 +26 ×6. 5 +6. 5 ×14 m
2
= 2 364 +169 +91 m
2
= 2 ×624 m
2
= 1248 m
2
Lateral surface area = 2(l +b)×h 
                              = [2(26 +14)×6. 5] m
2
= [2 ×40 ×6. 5] m
2
= 520 m
2
iii
Here, l = 15 m; b = 6 m; h = 5 dm = 0.5 m
Volume of the cuboid =  l ×b ×h
                                = 15 ×6 ×0. 5 m
3
= 45 m
3
Total surface area = 2(lb + lh+ bh) 
                          = 2 15 ×6 +15 ×0. 5 +6 ×0. 5 m
2
= 2 90 +7. 5 +3 m
2
= 2 ×100. 5 m
2
= 201m
2
Lateral surface area = 2(l +b)×h 
                               = [2(15 +6)×0. 5] m
2
= [2 ×21 ×0. 5] m
2
= 21 m
2
iv
Here, l = 24 m; b = 25 cm = 0.25 m; h =6 m
Volume of the cuboid =  l ×b ×h 
                                = (24 ×0. 25 ×6) m
3
= 36 m
3
Total Surface area = 2(lb + lh+ bh)
                            = 2 24 ×0. 25 +24 ×6 +0. 25 ×6 m
2
= 2 6 +144 + 1. 5 m
2
= 2 ×151. 5 m
2
= 303 m
2
Lateral surface area = 2(l +b)×h 
                             = [2(24 +0. 25)×6] m
2
= [2 ×24. 25 ×6] m
2
= 291 m
2
Question:2
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What is the volume of a packet containing 12 such matchboxes?
Solution:
Volume of each matchbox = 4 × 2.5 × 1.5 = 15 cm
3
? Volume of 12 matchboxes = 12 × 15 = 180 cm
3
Thus, the volume of a packet containing 12 such matchboxes is 180 cm
3
.
Question:3
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (Given, 1 m
3
 = 1000 litres.)
Solution:
Volume of water in the tank = Length × Breadth × Height = 6 × 5 × 4.5 = 135 m
3
? Volume of water in litres = 135 × 1000 = 135000 L        (1 m
3
 = 1000 L)
Thus, the water tank can hold 135000 L of water.
Question:4
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank if its length and depth are respectively 10 m and 2.5 m. (Given, 1000 litres = 1 m
3
.)
Solution:
Capacity of the tank = 50000 L = 
50000
1000
 = 50 m
3
           (1000 L = 1 m
3
)
Length of the tank = 10 m
Height ordepth of the tank = 2.5 m
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Now,
Volume of the cuboidal tank = Length × Breadth × Height
? Breadth of the tank = 
Volume of the tank
Length × Height
=
50
10×2.5
=
50
25
 = 2 m
Thus, the breadth of the tank is 2 m.
Question:5
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates, each measuring 1.5 m × 1.25 m × 0.5 m, that can be stored in the godown.
Solution:
Volume of the godown = 40 m × 25 m × 15 m = 15000 m
3
Volume of each wooden crate = 1.5 m × 1.25 m × 0.5 m = 0.9375 m
3
? Maximum number of wooden crates that can be stored in the godown
=
Volume of the godown
Volume of each wooden crate
=
15000
0.9375
= 16000
Question:6
How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 80 cm deep?
Solution:
Number of planks = 
volume of the pit in cm
3
volume of 1 plank in cm
3
Volume of one plank = l ×b ×h cm
3
                                  = 500 ×25 ×10  cm
3
= 125000 cm
3
Volume of the pit = l ×b ×h cm
3
Here, l = 20 m = 2000 cm; b = 6 m = 600 cm; h = 80 cmi. e. , volume of the pit = 2000 ×600 ×80 cm
3
                                  = 96000000 cm
3
? Number of planks = 
96000000
125000
 = 
96000
125
= 768
Question:7
How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?
Solution:
Length of the wall = 8 m = 800 cm
Breadth of the wall = 22.5 cm
Height of the wall = 6 m = 600 cm
i.e., volume of wall = 800 ×22. 5 ×600 cm
3
= 10800000 cm
3
Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm
i.e., volume of one brick = 25 ×11. 25 ×6 = 1687. 5 cm
3
? Number of bricks = 
volume of the wall
volume of one brick
 = 
10800000
1687.5
= 6400
Question:8
Find the capacity of a closed rectangular cistern whose length is 8 m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet required to make the cistern.
Solution:
Length of the cistern, l = 8 m
Breadth of the cistern, b = 6 m
Height ordepth of the cistern, h = 2.5 m
? Capacity of the cistern
= Volume of the cistern
= l × b × h
= 8 × 6 × 2.5
= 120 m
3
Also,
Area of the iron sheet required to make the cistern
= Total surface area of the cistern
= 2(lb + bh + hl)
= 28 ×6 +6 ×2.5 +2.5 ×8
= 2 × 83
= 166 m
2
Question:9
The dimensions of a room are 9m ×8m ×6.5m. It has one door of dimensions 2m ×1.5m and two windows, each of dimensions 1.5m ×1m. Find the cost of whitewashing the walls at Rs
25 per square metre.
Solution:
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Length of the room, l = 9 m
Breadth of the room, b = 8 m
Height of the room, h = 6.5 m
Now,
Area of the walls to be whitewashed
= Curved surface area of the room - Area of the door - 2 × Area of each window
= 2h(l + b) - 2 m × 1.5 m - 2 × 1.5 m × 1 m
= 2 × 6.5 × 9 +8 - 3 - 3
= 221 - 6
= 215 m
2
? Cost of whitewashing the walls at Rs 25 per square metre
= Area of the walls to be whitewashed × Rs 25 per square metre
= 215 × 25
= Rs 5,375
Question:10
A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring (22 cm × 12.5 cm × 7.5 cm). If 
1
12
 of the total volume of the wall consists of mortar, how many bricks are
there in the wall?
Solution:
Length of the wall = 15 m = 1500 cm
Breadth of the wall = 30 cm
Height of the wall = 4 m = 400 cm
Volume of wall =  = 1500 ×30 ×400 cm
3
= 18000000 cm
3
Now, volume of each brick = 22 ×12. 5 ×7. 5 cm
3
                                         = 2062. 5 cm
3
Also, volume of the mortar = 
1
12
×volume of the wall
                                        =
18000000
12
= 1500000 cm
3
Total volume of the bricks in the wall = volume of the wall - volume of the mortar
                                                           = 18000000 -1500000 cm
3
= 16500000 cm
3
? Number of bricks = 
volume of bricks
volume of one brick
=
16500000
2062.5
= 8000 bricks
Question:11
How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cm
3
 of iron weighs
15 g, find the weight of the empty box in kilograms.
Solution:
The external dimensions of the box are 36 cm, 25 cm and 16.5 cm.
Thickness of the iron = 1.5 cm
? Inner length of the box = 36 - 1.5 -1.5 = 33 cm
Inner breadth of the box = 25 - 1.5 -1.5 = 22 cm
Inner height of the box = 16.5 - 1.5 = 15 cm
Now,
Volume of iron in the open box
= Volume of the outer box - Volume of the inner box
= 36 × 25 × 16.5 - 33 × 22 × 15
= 14850 - 10890
= 3960 cm
3
It is given that 1 cm
3
 of iron weighs 15 g.
? Weight of the empty box = 3960 × 15 = 59400 g = 
59400
1000
 = 59.4 kg         1kg = 1000g
Question:12
A box made of sheet metal costs  6480 at  120 per square metre. If the box is 5 m long and 3 m wide, find its height.
Solution:
Length of the box =5 m
Breadth of the box =3 m
Area of the sheet required = 
total cost
cost per metre square 
Let h m be the height of the box.
Then area of the sheet = total surface area of the box    
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Page 4


Question:1
Find the volume, the lateral surface are and the total surface area of the cuboid whose dimensions are:
i length = 12 cm, breadth = 8 cm and height = 4.5 cm
ii length = 26 m, breadth = 14 m and height = 6.5 m
iii length = 15 m, breadth = 6 m and height = 5 dm
iv length = 24 m, breadth = 25 cm and height = 6 m
Solution:
i
Here, l = 12 cm, b = 8 cm, h = 4.5 cm
Volume of the cuboid = l ×b ×h
                                  = 12 ×8 ×4. 5 cm
3
= 432 cm
3
Total Surface area = 2(lb + lh+ bh)
                              = 2 12 ×8 + 12 ×4. 5 +8 ×4. 5 cm
2
= 2 96 +54 + 36 cm
2
= 2 × 186  cm
2
= 372  cm 
2
Lateral surface area = 2(l +b)×h
                              = [2(12 +8)×4. 5] cm
2
= [2(20)×4. 5]  cm
2
= 40 ×4. 5  cm
2
= 180 cm
2
ii
Here, l = 26 m; b = 14 m; h =6.5 m
Volume of the cuboid =  l ×b ×h
                                = (26 ×14 ×6. 5)m
3
= 2366 m
3
Total surface area = 2(lb + lh+ bh)
                             = 2 26 ×14 +26 ×6. 5 +6. 5 ×14 m
2
= 2 364 +169 +91 m
2
= 2 ×624 m
2
= 1248 m
2
Lateral surface area = 2(l +b)×h 
                              = [2(26 +14)×6. 5] m
2
= [2 ×40 ×6. 5] m
2
= 520 m
2
iii
Here, l = 15 m; b = 6 m; h = 5 dm = 0.5 m
Volume of the cuboid =  l ×b ×h
                                = 15 ×6 ×0. 5 m
3
= 45 m
3
Total surface area = 2(lb + lh+ bh) 
                          = 2 15 ×6 +15 ×0. 5 +6 ×0. 5 m
2
= 2 90 +7. 5 +3 m
2
= 2 ×100. 5 m
2
= 201m
2
Lateral surface area = 2(l +b)×h 
                               = [2(15 +6)×0. 5] m
2
= [2 ×21 ×0. 5] m
2
= 21 m
2
iv
Here, l = 24 m; b = 25 cm = 0.25 m; h =6 m
Volume of the cuboid =  l ×b ×h 
                                = (24 ×0. 25 ×6) m
3
= 36 m
3
Total Surface area = 2(lb + lh+ bh)
                            = 2 24 ×0. 25 +24 ×6 +0. 25 ×6 m
2
= 2 6 +144 + 1. 5 m
2
= 2 ×151. 5 m
2
= 303 m
2
Lateral surface area = 2(l +b)×h 
                             = [2(24 +0. 25)×6] m
2
= [2 ×24. 25 ×6] m
2
= 291 m
2
Question:2
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What is the volume of a packet containing 12 such matchboxes?
Solution:
Volume of each matchbox = 4 × 2.5 × 1.5 = 15 cm
3
? Volume of 12 matchboxes = 12 × 15 = 180 cm
3
Thus, the volume of a packet containing 12 such matchboxes is 180 cm
3
.
Question:3
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (Given, 1 m
3
 = 1000 litres.)
Solution:
Volume of water in the tank = Length × Breadth × Height = 6 × 5 × 4.5 = 135 m
3
? Volume of water in litres = 135 × 1000 = 135000 L        (1 m
3
 = 1000 L)
Thus, the water tank can hold 135000 L of water.
Question:4
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank if its length and depth are respectively 10 m and 2.5 m. (Given, 1000 litres = 1 m
3
.)
Solution:
Capacity of the tank = 50000 L = 
50000
1000
 = 50 m
3
           (1000 L = 1 m
3
)
Length of the tank = 10 m
Height ordepth of the tank = 2.5 m
( )
( ) ( )
( ) ( )
( )
( ) ( )
( ) ( )
Now,
Volume of the cuboidal tank = Length × Breadth × Height
? Breadth of the tank = 
Volume of the tank
Length × Height
=
50
10×2.5
=
50
25
 = 2 m
Thus, the breadth of the tank is 2 m.
Question:5
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates, each measuring 1.5 m × 1.25 m × 0.5 m, that can be stored in the godown.
Solution:
Volume of the godown = 40 m × 25 m × 15 m = 15000 m
3
Volume of each wooden crate = 1.5 m × 1.25 m × 0.5 m = 0.9375 m
3
? Maximum number of wooden crates that can be stored in the godown
=
Volume of the godown
Volume of each wooden crate
=
15000
0.9375
= 16000
Question:6
How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 80 cm deep?
Solution:
Number of planks = 
volume of the pit in cm
3
volume of 1 plank in cm
3
Volume of one plank = l ×b ×h cm
3
                                  = 500 ×25 ×10  cm
3
= 125000 cm
3
Volume of the pit = l ×b ×h cm
3
Here, l = 20 m = 2000 cm; b = 6 m = 600 cm; h = 80 cmi. e. , volume of the pit = 2000 ×600 ×80 cm
3
                                  = 96000000 cm
3
? Number of planks = 
96000000
125000
 = 
96000
125
= 768
Question:7
How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?
Solution:
Length of the wall = 8 m = 800 cm
Breadth of the wall = 22.5 cm
Height of the wall = 6 m = 600 cm
i.e., volume of wall = 800 ×22. 5 ×600 cm
3
= 10800000 cm
3
Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm
i.e., volume of one brick = 25 ×11. 25 ×6 = 1687. 5 cm
3
? Number of bricks = 
volume of the wall
volume of one brick
 = 
10800000
1687.5
= 6400
Question:8
Find the capacity of a closed rectangular cistern whose length is 8 m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet required to make the cistern.
Solution:
Length of the cistern, l = 8 m
Breadth of the cistern, b = 6 m
Height ordepth of the cistern, h = 2.5 m
? Capacity of the cistern
= Volume of the cistern
= l × b × h
= 8 × 6 × 2.5
= 120 m
3
Also,
Area of the iron sheet required to make the cistern
= Total surface area of the cistern
= 2(lb + bh + hl)
= 28 ×6 +6 ×2.5 +2.5 ×8
= 2 × 83
= 166 m
2
Question:9
The dimensions of a room are 9m ×8m ×6.5m. It has one door of dimensions 2m ×1.5m and two windows, each of dimensions 1.5m ×1m. Find the cost of whitewashing the walls at Rs
25 per square metre.
Solution:
( )
( )
Length of the room, l = 9 m
Breadth of the room, b = 8 m
Height of the room, h = 6.5 m
Now,
Area of the walls to be whitewashed
= Curved surface area of the room - Area of the door - 2 × Area of each window
= 2h(l + b) - 2 m × 1.5 m - 2 × 1.5 m × 1 m
= 2 × 6.5 × 9 +8 - 3 - 3
= 221 - 6
= 215 m
2
? Cost of whitewashing the walls at Rs 25 per square metre
= Area of the walls to be whitewashed × Rs 25 per square metre
= 215 × 25
= Rs 5,375
Question:10
A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring (22 cm × 12.5 cm × 7.5 cm). If 
1
12
 of the total volume of the wall consists of mortar, how many bricks are
there in the wall?
Solution:
Length of the wall = 15 m = 1500 cm
Breadth of the wall = 30 cm
Height of the wall = 4 m = 400 cm
Volume of wall =  = 1500 ×30 ×400 cm
3
= 18000000 cm
3
Now, volume of each brick = 22 ×12. 5 ×7. 5 cm
3
                                         = 2062. 5 cm
3
Also, volume of the mortar = 
1
12
×volume of the wall
                                        =
18000000
12
= 1500000 cm
3
Total volume of the bricks in the wall = volume of the wall - volume of the mortar
                                                           = 18000000 -1500000 cm
3
= 16500000 cm
3
? Number of bricks = 
volume of bricks
volume of one brick
=
16500000
2062.5
= 8000 bricks
Question:11
How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cm
3
 of iron weighs
15 g, find the weight of the empty box in kilograms.
Solution:
The external dimensions of the box are 36 cm, 25 cm and 16.5 cm.
Thickness of the iron = 1.5 cm
? Inner length of the box = 36 - 1.5 -1.5 = 33 cm
Inner breadth of the box = 25 - 1.5 -1.5 = 22 cm
Inner height of the box = 16.5 - 1.5 = 15 cm
Now,
Volume of iron in the open box
= Volume of the outer box - Volume of the inner box
= 36 × 25 × 16.5 - 33 × 22 × 15
= 14850 - 10890
= 3960 cm
3
It is given that 1 cm
3
 of iron weighs 15 g.
? Weight of the empty box = 3960 × 15 = 59400 g = 
59400
1000
 = 59.4 kg         1kg = 1000g
Question:12
A box made of sheet metal costs  6480 at  120 per square metre. If the box is 5 m long and 3 m wide, find its height.
Solution:
Length of the box =5 m
Breadth of the box =3 m
Area of the sheet required = 
total cost
cost per metre square 
Let h m be the height of the box.
Then area of the sheet = total surface area of the box    
( ) ( ) ( ) ( )
                                 = 2 lb +lh +bh m
2
= 2 5 ×3 +5 ×h +3 ×h m
2
= 2 15 +8h = 30 +16h m
2
Now, 30 +16h =
6480
120
? 30 +16h = 54 ? 16h = 24 ? h = 1. 5 m
? The height of the box is 1.5 m.
Question:13
The volume of a cuboid is 1536 m
3
. Its length is 16 m, and its breadth and height are in the ratio 3 : 2. Find the breadth and height of the cuboid.
Solution:
Length of the cuboid = 16 m
Suppose that the breadth and height of the cuboid are 3x m and 2x m, respectively.
Then 1536 = 16 ×3x ×2x ? 1536 = 16 ×6x
2
? x
2
=
1536
96
= 16 ? x = v 16 = 4
? The breadth and height of the cuboid are 12 m and 8 m, respectively.
Question:14
How many persons can be accommodated in a dining hall of dimensions (20 m × 16 m × 4.5 m), assuming that each person requires 5 cubic metres of air?
Solution:
Volume of the dining hall = 20 ×16 ×4. 5 m
3
= 1440 m
3
Volume of air required by each person = 5 m
3
? Capacity of the dining hall = 
volume of dining hall
volume of air required by each person
=
1440
5
= 288 persons
Question:15
A classroom is 10 m long, 6.4 m wide and 5 m high. If each student be given 1.6 m
2
 of the floor area, how many students can be accommodated in the room? How many cubic metres of
air would each student get?
Solution:
Length of the classroom = 10 m
Breadth of the classroom = 6.4 m
Height of the classroom = 5 m
Area of the floor = length × breadth = 10 × 6.4 m
2
No. of students =
area of the floor
area given to one student on the floor
 =
10×6.4
1.6
=
640
16
= 40 students
Now, volume of the classroom=10 ×6. 4 ×5 m
3
? Air required by each student =
volume of the room
number of students
=
10×6.4×5
40
 m
3
= 8 m
3
Question:16
The surface area of a cuboid is 758 cm
2
. Its length and breadth are 14 cm and 11 cm respectively. Find its height.
Solution:
Length of the cuboid = 14 cm
Breadth of the cuboid = 11 cm
Let the height of the cuboid be x cm.
Surface area of the cuboid = 758 cm
2
Then 758 = 2 14 ×11 +14 ×x +11 ×x ? 758 = 2 154 +14x +11x ? 758  = 2 154 +25x ? 758  = 308 +50x ? 50x = 758 -308 = 450 ? x =
450
50
= 9
? The height of the cuboid is 9 cm.
Question:17
In a shower, 5 cm of rain falls. Find the volume of water that falls on 2 hectares of ground.
Solution:
Volume of the water that falls on the ground = area of ground ×depth
                                                                 = 20000 ×0. 05   m
3
= 1000 m
3
Question:18
Find the volume, the lateral surface area, the total surface area and the diagonal of a cube, each of whose edges measures 9 m. (Take v 3 = 1. 73.)
Solution:
Here, a = 9 m
Volume of the cube = a
3
= 9
3
 m
3
= 729 m
3
Lateral surface area of the cube = 4a
2
= 4 ×9
2
m
2
= 4 ×81 m
2
= 324 m
2
Total surface area of the cube = 6a
2
= 6 ×9
2
m
2
= 6 ×81 m
2
= 486 m
2
? Diagonal of the cube = v 3a = v 3 ×9 = 15. 57 m
 
Question:19
The total surface area of a cube is 1176 cm
2
. Find its volume.
Solution:
Suppose that the side of cube is x cm.
Total surface area of cube = 1176 sq cm
Then 1176 = 6x
2
? x
2
=
1176
6
= 196 ? x = v 196 = 14 
i.e., the side of the cube is 14 cm.
? Volume of the cube = x
3
 = 14
3
 cm
3
= 2744 cm
3
Question:20
The lateral surface area of a cube is 900 cm
2
. Find its volume.
( ) ( ) ( ) ( )
( )
( ) ( ) ( )
Page 5


Question:1
Find the volume, the lateral surface are and the total surface area of the cuboid whose dimensions are:
i length = 12 cm, breadth = 8 cm and height = 4.5 cm
ii length = 26 m, breadth = 14 m and height = 6.5 m
iii length = 15 m, breadth = 6 m and height = 5 dm
iv length = 24 m, breadth = 25 cm and height = 6 m
Solution:
i
Here, l = 12 cm, b = 8 cm, h = 4.5 cm
Volume of the cuboid = l ×b ×h
                                  = 12 ×8 ×4. 5 cm
3
= 432 cm
3
Total Surface area = 2(lb + lh+ bh)
                              = 2 12 ×8 + 12 ×4. 5 +8 ×4. 5 cm
2
= 2 96 +54 + 36 cm
2
= 2 × 186  cm
2
= 372  cm 
2
Lateral surface area = 2(l +b)×h
                              = [2(12 +8)×4. 5] cm
2
= [2(20)×4. 5]  cm
2
= 40 ×4. 5  cm
2
= 180 cm
2
ii
Here, l = 26 m; b = 14 m; h =6.5 m
Volume of the cuboid =  l ×b ×h
                                = (26 ×14 ×6. 5)m
3
= 2366 m
3
Total surface area = 2(lb + lh+ bh)
                             = 2 26 ×14 +26 ×6. 5 +6. 5 ×14 m
2
= 2 364 +169 +91 m
2
= 2 ×624 m
2
= 1248 m
2
Lateral surface area = 2(l +b)×h 
                              = [2(26 +14)×6. 5] m
2
= [2 ×40 ×6. 5] m
2
= 520 m
2
iii
Here, l = 15 m; b = 6 m; h = 5 dm = 0.5 m
Volume of the cuboid =  l ×b ×h
                                = 15 ×6 ×0. 5 m
3
= 45 m
3
Total surface area = 2(lb + lh+ bh) 
                          = 2 15 ×6 +15 ×0. 5 +6 ×0. 5 m
2
= 2 90 +7. 5 +3 m
2
= 2 ×100. 5 m
2
= 201m
2
Lateral surface area = 2(l +b)×h 
                               = [2(15 +6)×0. 5] m
2
= [2 ×21 ×0. 5] m
2
= 21 m
2
iv
Here, l = 24 m; b = 25 cm = 0.25 m; h =6 m
Volume of the cuboid =  l ×b ×h 
                                = (24 ×0. 25 ×6) m
3
= 36 m
3
Total Surface area = 2(lb + lh+ bh)
                            = 2 24 ×0. 25 +24 ×6 +0. 25 ×6 m
2
= 2 6 +144 + 1. 5 m
2
= 2 ×151. 5 m
2
= 303 m
2
Lateral surface area = 2(l +b)×h 
                             = [2(24 +0. 25)×6] m
2
= [2 ×24. 25 ×6] m
2
= 291 m
2
Question:2
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What is the volume of a packet containing 12 such matchboxes?
Solution:
Volume of each matchbox = 4 × 2.5 × 1.5 = 15 cm
3
? Volume of 12 matchboxes = 12 × 15 = 180 cm
3
Thus, the volume of a packet containing 12 such matchboxes is 180 cm
3
.
Question:3
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (Given, 1 m
3
 = 1000 litres.)
Solution:
Volume of water in the tank = Length × Breadth × Height = 6 × 5 × 4.5 = 135 m
3
? Volume of water in litres = 135 × 1000 = 135000 L        (1 m
3
 = 1000 L)
Thus, the water tank can hold 135000 L of water.
Question:4
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank if its length and depth are respectively 10 m and 2.5 m. (Given, 1000 litres = 1 m
3
.)
Solution:
Capacity of the tank = 50000 L = 
50000
1000
 = 50 m
3
           (1000 L = 1 m
3
)
Length of the tank = 10 m
Height ordepth of the tank = 2.5 m
( )
( ) ( )
( ) ( )
( )
( ) ( )
( ) ( )
Now,
Volume of the cuboidal tank = Length × Breadth × Height
? Breadth of the tank = 
Volume of the tank
Length × Height
=
50
10×2.5
=
50
25
 = 2 m
Thus, the breadth of the tank is 2 m.
Question:5
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates, each measuring 1.5 m × 1.25 m × 0.5 m, that can be stored in the godown.
Solution:
Volume of the godown = 40 m × 25 m × 15 m = 15000 m
3
Volume of each wooden crate = 1.5 m × 1.25 m × 0.5 m = 0.9375 m
3
? Maximum number of wooden crates that can be stored in the godown
=
Volume of the godown
Volume of each wooden crate
=
15000
0.9375
= 16000
Question:6
How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 80 cm deep?
Solution:
Number of planks = 
volume of the pit in cm
3
volume of 1 plank in cm
3
Volume of one plank = l ×b ×h cm
3
                                  = 500 ×25 ×10  cm
3
= 125000 cm
3
Volume of the pit = l ×b ×h cm
3
Here, l = 20 m = 2000 cm; b = 6 m = 600 cm; h = 80 cmi. e. , volume of the pit = 2000 ×600 ×80 cm
3
                                  = 96000000 cm
3
? Number of planks = 
96000000
125000
 = 
96000
125
= 768
Question:7
How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?
Solution:
Length of the wall = 8 m = 800 cm
Breadth of the wall = 22.5 cm
Height of the wall = 6 m = 600 cm
i.e., volume of wall = 800 ×22. 5 ×600 cm
3
= 10800000 cm
3
Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm
i.e., volume of one brick = 25 ×11. 25 ×6 = 1687. 5 cm
3
? Number of bricks = 
volume of the wall
volume of one brick
 = 
10800000
1687.5
= 6400
Question:8
Find the capacity of a closed rectangular cistern whose length is 8 m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet required to make the cistern.
Solution:
Length of the cistern, l = 8 m
Breadth of the cistern, b = 6 m
Height ordepth of the cistern, h = 2.5 m
? Capacity of the cistern
= Volume of the cistern
= l × b × h
= 8 × 6 × 2.5
= 120 m
3
Also,
Area of the iron sheet required to make the cistern
= Total surface area of the cistern
= 2(lb + bh + hl)
= 28 ×6 +6 ×2.5 +2.5 ×8
= 2 × 83
= 166 m
2
Question:9
The dimensions of a room are 9m ×8m ×6.5m. It has one door of dimensions 2m ×1.5m and two windows, each of dimensions 1.5m ×1m. Find the cost of whitewashing the walls at Rs
25 per square metre.
Solution:
( )
( )
Length of the room, l = 9 m
Breadth of the room, b = 8 m
Height of the room, h = 6.5 m
Now,
Area of the walls to be whitewashed
= Curved surface area of the room - Area of the door - 2 × Area of each window
= 2h(l + b) - 2 m × 1.5 m - 2 × 1.5 m × 1 m
= 2 × 6.5 × 9 +8 - 3 - 3
= 221 - 6
= 215 m
2
? Cost of whitewashing the walls at Rs 25 per square metre
= Area of the walls to be whitewashed × Rs 25 per square metre
= 215 × 25
= Rs 5,375
Question:10
A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring (22 cm × 12.5 cm × 7.5 cm). If 
1
12
 of the total volume of the wall consists of mortar, how many bricks are
there in the wall?
Solution:
Length of the wall = 15 m = 1500 cm
Breadth of the wall = 30 cm
Height of the wall = 4 m = 400 cm
Volume of wall =  = 1500 ×30 ×400 cm
3
= 18000000 cm
3
Now, volume of each brick = 22 ×12. 5 ×7. 5 cm
3
                                         = 2062. 5 cm
3
Also, volume of the mortar = 
1
12
×volume of the wall
                                        =
18000000
12
= 1500000 cm
3
Total volume of the bricks in the wall = volume of the wall - volume of the mortar
                                                           = 18000000 -1500000 cm
3
= 16500000 cm
3
? Number of bricks = 
volume of bricks
volume of one brick
=
16500000
2062.5
= 8000 bricks
Question:11
How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cm
3
 of iron weighs
15 g, find the weight of the empty box in kilograms.
Solution:
The external dimensions of the box are 36 cm, 25 cm and 16.5 cm.
Thickness of the iron = 1.5 cm
? Inner length of the box = 36 - 1.5 -1.5 = 33 cm
Inner breadth of the box = 25 - 1.5 -1.5 = 22 cm
Inner height of the box = 16.5 - 1.5 = 15 cm
Now,
Volume of iron in the open box
= Volume of the outer box - Volume of the inner box
= 36 × 25 × 16.5 - 33 × 22 × 15
= 14850 - 10890
= 3960 cm
3
It is given that 1 cm
3
 of iron weighs 15 g.
? Weight of the empty box = 3960 × 15 = 59400 g = 
59400
1000
 = 59.4 kg         1kg = 1000g
Question:12
A box made of sheet metal costs  6480 at  120 per square metre. If the box is 5 m long and 3 m wide, find its height.
Solution:
Length of the box =5 m
Breadth of the box =3 m
Area of the sheet required = 
total cost
cost per metre square 
Let h m be the height of the box.
Then area of the sheet = total surface area of the box    
( ) ( ) ( ) ( )
                                 = 2 lb +lh +bh m
2
= 2 5 ×3 +5 ×h +3 ×h m
2
= 2 15 +8h = 30 +16h m
2
Now, 30 +16h =
6480
120
? 30 +16h = 54 ? 16h = 24 ? h = 1. 5 m
? The height of the box is 1.5 m.
Question:13
The volume of a cuboid is 1536 m
3
. Its length is 16 m, and its breadth and height are in the ratio 3 : 2. Find the breadth and height of the cuboid.
Solution:
Length of the cuboid = 16 m
Suppose that the breadth and height of the cuboid are 3x m and 2x m, respectively.
Then 1536 = 16 ×3x ×2x ? 1536 = 16 ×6x
2
? x
2
=
1536
96
= 16 ? x = v 16 = 4
? The breadth and height of the cuboid are 12 m and 8 m, respectively.
Question:14
How many persons can be accommodated in a dining hall of dimensions (20 m × 16 m × 4.5 m), assuming that each person requires 5 cubic metres of air?
Solution:
Volume of the dining hall = 20 ×16 ×4. 5 m
3
= 1440 m
3
Volume of air required by each person = 5 m
3
? Capacity of the dining hall = 
volume of dining hall
volume of air required by each person
=
1440
5
= 288 persons
Question:15
A classroom is 10 m long, 6.4 m wide and 5 m high. If each student be given 1.6 m
2
 of the floor area, how many students can be accommodated in the room? How many cubic metres of
air would each student get?
Solution:
Length of the classroom = 10 m
Breadth of the classroom = 6.4 m
Height of the classroom = 5 m
Area of the floor = length × breadth = 10 × 6.4 m
2
No. of students =
area of the floor
area given to one student on the floor
 =
10×6.4
1.6
=
640
16
= 40 students
Now, volume of the classroom=10 ×6. 4 ×5 m
3
? Air required by each student =
volume of the room
number of students
=
10×6.4×5
40
 m
3
= 8 m
3
Question:16
The surface area of a cuboid is 758 cm
2
. Its length and breadth are 14 cm and 11 cm respectively. Find its height.
Solution:
Length of the cuboid = 14 cm
Breadth of the cuboid = 11 cm
Let the height of the cuboid be x cm.
Surface area of the cuboid = 758 cm
2
Then 758 = 2 14 ×11 +14 ×x +11 ×x ? 758 = 2 154 +14x +11x ? 758  = 2 154 +25x ? 758  = 308 +50x ? 50x = 758 -308 = 450 ? x =
450
50
= 9
? The height of the cuboid is 9 cm.
Question:17
In a shower, 5 cm of rain falls. Find the volume of water that falls on 2 hectares of ground.
Solution:
Volume of the water that falls on the ground = area of ground ×depth
                                                                 = 20000 ×0. 05   m
3
= 1000 m
3
Question:18
Find the volume, the lateral surface area, the total surface area and the diagonal of a cube, each of whose edges measures 9 m. (Take v 3 = 1. 73.)
Solution:
Here, a = 9 m
Volume of the cube = a
3
= 9
3
 m
3
= 729 m
3
Lateral surface area of the cube = 4a
2
= 4 ×9
2
m
2
= 4 ×81 m
2
= 324 m
2
Total surface area of the cube = 6a
2
= 6 ×9
2
m
2
= 6 ×81 m
2
= 486 m
2
? Diagonal of the cube = v 3a = v 3 ×9 = 15. 57 m
 
Question:19
The total surface area of a cube is 1176 cm
2
. Find its volume.
Solution:
Suppose that the side of cube is x cm.
Total surface area of cube = 1176 sq cm
Then 1176 = 6x
2
? x
2
=
1176
6
= 196 ? x = v 196 = 14 
i.e., the side of the cube is 14 cm.
? Volume of the cube = x
3
 = 14
3
 cm
3
= 2744 cm
3
Question:20
The lateral surface area of a cube is 900 cm
2
. Find its volume.
( ) ( ) ( ) ( )
( )
( ) ( ) ( )
Solution:
Suppose that the side of cube is x cm.
Lateral surface area of the cube = 900 cm
2
Then 900 = 4x
2
? x
2
=
900
4
= 225 ? x = v 225 = 15
i.e., the side of the cube is 15 cm.
? Volume of the given cube = x
3
 cm
3
= 15
3 
cm
3
= 3375 cm
3
Question:21
The volume of a cube is 512 cm
3
. Find its surface area.
Solution:
Suppose that the side of the given cube is x cm.
Volume of the cube = 512 cm
3
Then 512 = x
3
? x =
3
v 512 = 8i. e. , the side of the cube is 8 cm.
? Surface area of the cube = 6x
2
 cm
2
= 6 ×8
2
 cm
2
= 384 cm
2
Question:22
Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.
Solution:
Three cubes of metal with edges 3cm, 4cm and 5 cm are melted to form a single cube.
? Volume of the new cube = sum of the volumes the old cubes
                                        = 3
3
+4
3
+5
3
cm
3
= 27 +64 +125 cm
3
= 216 cm
3
Suppose the edge of the new cube = x cm
Then we have:
Then 216 = x
3
? x =
3
v 216 = 6 i. e. , the edge of the new cube is 6 cm.
? Lateral surface area of the new cube = 4x
2
 cm
2
= 4 ×6
2
 cm
2
= 144 cm
2
Question:23
Find the length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5 m).
Solution:
Length of the longest pole = length of the diagonal of the room
                                      =
v
l
2
+b
2
+h
2
 m =
v
10
2
+10
2
+5
2
 m = v 100 +100 +25 = v 225 = 15 m
Question:24
The sum of length, breadth and depth of a cuboid is 19 cm and the length of its diagonal is 11 cm. Find the surface area of the cuboid.
Solution:
Let the length, breadth and height ordepth of the cuboid be l cm, b cm and h cm, respectively.
? l + b + h = 19           .....1
Also,
Length of the diagonal = 11 cm
?
v
l
2
+b
2
+h
2
= 11 ? l
2
+b
2
+h
2
= 121                  . . . . . (2)
Squaring 1, we get
(l + b + h)
2
 = 19
2
? l
2
 + b
2
 + h
2 
+ 2(lb + bh + hl) = 361
? 121 + 2(lb + bh + hl) = 361                      
Using(2)
? 2(lb + bh + hl) = 361 - 121 = 240 cm
2
Thus, the surface area of the cuboid is 240 cm
2
.
Question:25
Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.
Solution:
Let the initial edge of the cube be a units.
? Initial surface area of the cube = 6a
2
 square units
New edge of the cube = a + 50% of a = a +
50
100
a = 1.5a units
? New surface of the cube = 6(1.5a)
2
 = 13.5a
2
 square units
Increase in surface area of the cube = 13.5a
2
 - 6a
2
 = 7.5a
2
 square units
? Percentage increase in the surface area of the cube
=
Increase in surface area of the cube
Initial surface area of the cube
×100 % =
7.5a
2
6a
2
×100 % = 125 %
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