Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.4
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Page 1
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
EXERCISE 21.4 PAGE NO: 21.30
1. Find the length of the longest rod that can be placed in a room 12 m long, 9 m
broad and 8 m high.
Solution:
Given details are,
Length of room = 12 m
Breadth of room = 9m
Height of room = 8m
So,
Length of longest rod that can be placed in room = diagonal of room (cuboid)
= v(l
2
+ b
2
+ h
2
)
= v(12
2
+ 9
2
+ 8
2
)
= v(144+81+64)
= v(289)
= 17m
2. If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then
prove that 1/V = 2/S (1/a + 1/b + 1/c)
Solution:
Let us consider,
V = volume of cuboid
S = surface area of cuboid
Dimensions of cuboid = a, b, c
So,
S = 2 (ab + bc + ca)
V = abc
S/V = 2 (ab + bc + ca) / abc
= 2[(ab/abc) + (bc/abc) + (ca/abc)]
= 2 (1/a + 1/b + 1/c)
1/V = 2/S (1/a + 1/b + 1/c)
Hence proved.
3. The areas of three adjacent faces of a cuboid are x, y, and z. If the volume is V,
prove that V
2
= xyz.
Solution:
Let us consider,
Areas of three faces of cuboid as x,y,z
So, Let length of cuboid be = l
Page 2
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
EXERCISE 21.4 PAGE NO: 21.30
1. Find the length of the longest rod that can be placed in a room 12 m long, 9 m
broad and 8 m high.
Solution:
Given details are,
Length of room = 12 m
Breadth of room = 9m
Height of room = 8m
So,
Length of longest rod that can be placed in room = diagonal of room (cuboid)
= v(l
2
+ b
2
+ h
2
)
= v(12
2
+ 9
2
+ 8
2
)
= v(144+81+64)
= v(289)
= 17m
2. If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then
prove that 1/V = 2/S (1/a + 1/b + 1/c)
Solution:
Let us consider,
V = volume of cuboid
S = surface area of cuboid
Dimensions of cuboid = a, b, c
So,
S = 2 (ab + bc + ca)
V = abc
S/V = 2 (ab + bc + ca) / abc
= 2[(ab/abc) + (bc/abc) + (ca/abc)]
= 2 (1/a + 1/b + 1/c)
1/V = 2/S (1/a + 1/b + 1/c)
Hence proved.
3. The areas of three adjacent faces of a cuboid are x, y, and z. If the volume is V,
prove that V
2
= xyz.
Solution:
Let us consider,
Areas of three faces of cuboid as x,y,z
So, Let length of cuboid be = l
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Breadth of cuboid be = b
Height of cuboid be = h
Let, x = l×b
y = b×h
z = h×l
Else we can write as
xyz = l
2
b
2
h
2
….. (i)
If ‘V’ is volume of cuboid = V = lbh
V
2
= l
2
b
2
h
2
= xyz …… from (i)
? V
2
= xyz
Hence proved.
4. A rectangular water reservoir contains 105 m
3
of water. Find the depth of the
water in the reservoir if its base measures 12 m by 3.5 m.
Solution:
Given details are,
Capacity of water reservoir = 105 m
3
Length of base of reservoir = 12 m
Width of base = 3.5 m
Let the depth of reservoir be ‘h’ m
l × b × h = 105
h = 105 / (l × b)
= 105 / (12×3.5)
= 105/42
= 2.5m
? Depth of reservoir is 2.5 m
5. Cubes A, B, C having edges 18 cm, 24 cm and 30 cm respectively are melted and
moulded into a new cube D. Find the edge of the bigger cube D.
Solution:
Given details are,
Edge length of cube A = 18 cm
Edge length of cube B = 24 cm
Edge length of cube C = 30 cm
Then,
Volume of cube A = v
1
= 18
3
= 5832cm
3
Volume of cube B = v
2
= 24
3
= 13824cm
3
Volume of cube C = v
3
= 30
3
= 27000cm
3
Page 3
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
EXERCISE 21.4 PAGE NO: 21.30
1. Find the length of the longest rod that can be placed in a room 12 m long, 9 m
broad and 8 m high.
Solution:
Given details are,
Length of room = 12 m
Breadth of room = 9m
Height of room = 8m
So,
Length of longest rod that can be placed in room = diagonal of room (cuboid)
= v(l
2
+ b
2
+ h
2
)
= v(12
2
+ 9
2
+ 8
2
)
= v(144+81+64)
= v(289)
= 17m
2. If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then
prove that 1/V = 2/S (1/a + 1/b + 1/c)
Solution:
Let us consider,
V = volume of cuboid
S = surface area of cuboid
Dimensions of cuboid = a, b, c
So,
S = 2 (ab + bc + ca)
V = abc
S/V = 2 (ab + bc + ca) / abc
= 2[(ab/abc) + (bc/abc) + (ca/abc)]
= 2 (1/a + 1/b + 1/c)
1/V = 2/S (1/a + 1/b + 1/c)
Hence proved.
3. The areas of three adjacent faces of a cuboid are x, y, and z. If the volume is V,
prove that V
2
= xyz.
Solution:
Let us consider,
Areas of three faces of cuboid as x,y,z
So, Let length of cuboid be = l
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Breadth of cuboid be = b
Height of cuboid be = h
Let, x = l×b
y = b×h
z = h×l
Else we can write as
xyz = l
2
b
2
h
2
….. (i)
If ‘V’ is volume of cuboid = V = lbh
V
2
= l
2
b
2
h
2
= xyz …… from (i)
? V
2
= xyz
Hence proved.
4. A rectangular water reservoir contains 105 m
3
of water. Find the depth of the
water in the reservoir if its base measures 12 m by 3.5 m.
Solution:
Given details are,
Capacity of water reservoir = 105 m
3
Length of base of reservoir = 12 m
Width of base = 3.5 m
Let the depth of reservoir be ‘h’ m
l × b × h = 105
h = 105 / (l × b)
= 105 / (12×3.5)
= 105/42
= 2.5m
? Depth of reservoir is 2.5 m
5. Cubes A, B, C having edges 18 cm, 24 cm and 30 cm respectively are melted and
moulded into a new cube D. Find the edge of the bigger cube D.
Solution:
Given details are,
Edge length of cube A = 18 cm
Edge length of cube B = 24 cm
Edge length of cube C = 30 cm
Then,
Volume of cube A = v
1
= 18
3
= 5832cm
3
Volume of cube B = v
2
= 24
3
= 13824cm
3
Volume of cube C = v
3
= 30
3
= 27000cm
3
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Total volume of cube A,B,C = 5832 + 13824 + 27000 = 46656 cm
3
Let ‘a’ be the length of edge of newly formed cube.
a
3
= 46656
a =
3
v(46656)
= 36
? Edge of bigger cube is 36cm
6. The breadth of a room is twice its height, one half of its length and the volume of
the room is 512 cu. Dm. Find its dimensions.
Solution:
Given,
Breadth of room is twice of its height, b = 2h or h = b/2 … (i)
Breadth is one half of length, b = l/2 or l = 2b … (ii)
Volume of the room = lbh = 512 dm
3
… (iii)
By substituting (i) and (ii) in (iii)
2b × b × b/2 = 512
b
3
= 512
b =
3
v(512)
= 8
? Breadth of cube = b = 8 dm
Length of cube = 2b = 2×8 = 16 dm
Height of cube = b/2 = 8/2 = 4 dm
7. A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the
cost of iron sheet used at the rate of Rs. 5 per metre sheet, sheet being 2 m wide.
Solution:
Given,
Length of tank, l = 12 m
Width of tank, b = 9m
Depth of tank, h = 4m
Area of sheet required = total surface area of tank
= 2 (lb×bh×hl)
= 2 (12×9 + 9×4 + 4×12)
= 2 (108 + 36 + 48)
= 2 (192)
= 384 m
2
Let length be l
1
Page 4
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
EXERCISE 21.4 PAGE NO: 21.30
1. Find the length of the longest rod that can be placed in a room 12 m long, 9 m
broad and 8 m high.
Solution:
Given details are,
Length of room = 12 m
Breadth of room = 9m
Height of room = 8m
So,
Length of longest rod that can be placed in room = diagonal of room (cuboid)
= v(l
2
+ b
2
+ h
2
)
= v(12
2
+ 9
2
+ 8
2
)
= v(144+81+64)
= v(289)
= 17m
2. If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then
prove that 1/V = 2/S (1/a + 1/b + 1/c)
Solution:
Let us consider,
V = volume of cuboid
S = surface area of cuboid
Dimensions of cuboid = a, b, c
So,
S = 2 (ab + bc + ca)
V = abc
S/V = 2 (ab + bc + ca) / abc
= 2[(ab/abc) + (bc/abc) + (ca/abc)]
= 2 (1/a + 1/b + 1/c)
1/V = 2/S (1/a + 1/b + 1/c)
Hence proved.
3. The areas of three adjacent faces of a cuboid are x, y, and z. If the volume is V,
prove that V
2
= xyz.
Solution:
Let us consider,
Areas of three faces of cuboid as x,y,z
So, Let length of cuboid be = l
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Breadth of cuboid be = b
Height of cuboid be = h
Let, x = l×b
y = b×h
z = h×l
Else we can write as
xyz = l
2
b
2
h
2
….. (i)
If ‘V’ is volume of cuboid = V = lbh
V
2
= l
2
b
2
h
2
= xyz …… from (i)
? V
2
= xyz
Hence proved.
4. A rectangular water reservoir contains 105 m
3
of water. Find the depth of the
water in the reservoir if its base measures 12 m by 3.5 m.
Solution:
Given details are,
Capacity of water reservoir = 105 m
3
Length of base of reservoir = 12 m
Width of base = 3.5 m
Let the depth of reservoir be ‘h’ m
l × b × h = 105
h = 105 / (l × b)
= 105 / (12×3.5)
= 105/42
= 2.5m
? Depth of reservoir is 2.5 m
5. Cubes A, B, C having edges 18 cm, 24 cm and 30 cm respectively are melted and
moulded into a new cube D. Find the edge of the bigger cube D.
Solution:
Given details are,
Edge length of cube A = 18 cm
Edge length of cube B = 24 cm
Edge length of cube C = 30 cm
Then,
Volume of cube A = v
1
= 18
3
= 5832cm
3
Volume of cube B = v
2
= 24
3
= 13824cm
3
Volume of cube C = v
3
= 30
3
= 27000cm
3
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Total volume of cube A,B,C = 5832 + 13824 + 27000 = 46656 cm
3
Let ‘a’ be the length of edge of newly formed cube.
a
3
= 46656
a =
3
v(46656)
= 36
? Edge of bigger cube is 36cm
6. The breadth of a room is twice its height, one half of its length and the volume of
the room is 512 cu. Dm. Find its dimensions.
Solution:
Given,
Breadth of room is twice of its height, b = 2h or h = b/2 … (i)
Breadth is one half of length, b = l/2 or l = 2b … (ii)
Volume of the room = lbh = 512 dm
3
… (iii)
By substituting (i) and (ii) in (iii)
2b × b × b/2 = 512
b
3
= 512
b =
3
v(512)
= 8
? Breadth of cube = b = 8 dm
Length of cube = 2b = 2×8 = 16 dm
Height of cube = b/2 = 8/2 = 4 dm
7. A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the
cost of iron sheet used at the rate of Rs. 5 per metre sheet, sheet being 2 m wide.
Solution:
Given,
Length of tank, l = 12 m
Width of tank, b = 9m
Depth of tank, h = 4m
Area of sheet required = total surface area of tank
= 2 (lb×bh×hl)
= 2 (12×9 + 9×4 + 4×12)
= 2 (108 + 36 + 48)
= 2 (192)
= 384 m
2
Let length be l
1
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Breadth be b
1
Given, b
1
= 2m
l
1
× b
1
= 384
l
1
= 384/b
1
= 384/2
= 192m
? Cost of iron sheet at the rate of Rs 5 per metre = 5 × 192 = Rs 960
8. A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the
tank are 12m×8m×6m, find the cost of iron sheet at Rs. 17.50 per metre.
Solution:
Given details are,
Dimensions of tank = 12m × 8m × 6m
Where, length = 12m
Breadth = 8m
Height = 6m
Area of sheet required = total surface area of tank with one top open
= l × b + 2 (l×h + b×h)
= 12 × 8 + 2 (12×6 + 8×6)
= 96 + 240
= 336 m
2
Let length be l
1
Breadth be b
1
Given, b
1
= 4m
l
1
× b
1
= 336
l
1
= 336/b
1
= 336/4
= 84m
? Cost of iron sheet at the rate of Rs 17.50 per metre = 17.50 × 84 = Rs 1470
9. Three equal cubes are placed adjacently in a row. Find the ratio of total surface
area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution:
Given details are,
Let edge length of three equal cubes = a
Then,
Sum of surface area of 3 cubes = 3 × 6a
2
= 18a
2
When these cubes are placed in a row adjacently they form a cuboid.
Length of new cuboid formed = a + a + a = 3a
Page 5
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
EXERCISE 21.4 PAGE NO: 21.30
1. Find the length of the longest rod that can be placed in a room 12 m long, 9 m
broad and 8 m high.
Solution:
Given details are,
Length of room = 12 m
Breadth of room = 9m
Height of room = 8m
So,
Length of longest rod that can be placed in room = diagonal of room (cuboid)
= v(l
2
+ b
2
+ h
2
)
= v(12
2
+ 9
2
+ 8
2
)
= v(144+81+64)
= v(289)
= 17m
2. If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then
prove that 1/V = 2/S (1/a + 1/b + 1/c)
Solution:
Let us consider,
V = volume of cuboid
S = surface area of cuboid
Dimensions of cuboid = a, b, c
So,
S = 2 (ab + bc + ca)
V = abc
S/V = 2 (ab + bc + ca) / abc
= 2[(ab/abc) + (bc/abc) + (ca/abc)]
= 2 (1/a + 1/b + 1/c)
1/V = 2/S (1/a + 1/b + 1/c)
Hence proved.
3. The areas of three adjacent faces of a cuboid are x, y, and z. If the volume is V,
prove that V
2
= xyz.
Solution:
Let us consider,
Areas of three faces of cuboid as x,y,z
So, Let length of cuboid be = l
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Breadth of cuboid be = b
Height of cuboid be = h
Let, x = l×b
y = b×h
z = h×l
Else we can write as
xyz = l
2
b
2
h
2
….. (i)
If ‘V’ is volume of cuboid = V = lbh
V
2
= l
2
b
2
h
2
= xyz …… from (i)
? V
2
= xyz
Hence proved.
4. A rectangular water reservoir contains 105 m
3
of water. Find the depth of the
water in the reservoir if its base measures 12 m by 3.5 m.
Solution:
Given details are,
Capacity of water reservoir = 105 m
3
Length of base of reservoir = 12 m
Width of base = 3.5 m
Let the depth of reservoir be ‘h’ m
l × b × h = 105
h = 105 / (l × b)
= 105 / (12×3.5)
= 105/42
= 2.5m
? Depth of reservoir is 2.5 m
5. Cubes A, B, C having edges 18 cm, 24 cm and 30 cm respectively are melted and
moulded into a new cube D. Find the edge of the bigger cube D.
Solution:
Given details are,
Edge length of cube A = 18 cm
Edge length of cube B = 24 cm
Edge length of cube C = 30 cm
Then,
Volume of cube A = v
1
= 18
3
= 5832cm
3
Volume of cube B = v
2
= 24
3
= 13824cm
3
Volume of cube C = v
3
= 30
3
= 27000cm
3
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Total volume of cube A,B,C = 5832 + 13824 + 27000 = 46656 cm
3
Let ‘a’ be the length of edge of newly formed cube.
a
3
= 46656
a =
3
v(46656)
= 36
? Edge of bigger cube is 36cm
6. The breadth of a room is twice its height, one half of its length and the volume of
the room is 512 cu. Dm. Find its dimensions.
Solution:
Given,
Breadth of room is twice of its height, b = 2h or h = b/2 … (i)
Breadth is one half of length, b = l/2 or l = 2b … (ii)
Volume of the room = lbh = 512 dm
3
… (iii)
By substituting (i) and (ii) in (iii)
2b × b × b/2 = 512
b
3
= 512
b =
3
v(512)
= 8
? Breadth of cube = b = 8 dm
Length of cube = 2b = 2×8 = 16 dm
Height of cube = b/2 = 8/2 = 4 dm
7. A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the
cost of iron sheet used at the rate of Rs. 5 per metre sheet, sheet being 2 m wide.
Solution:
Given,
Length of tank, l = 12 m
Width of tank, b = 9m
Depth of tank, h = 4m
Area of sheet required = total surface area of tank
= 2 (lb×bh×hl)
= 2 (12×9 + 9×4 + 4×12)
= 2 (108 + 36 + 48)
= 2 (192)
= 384 m
2
Let length be l
1
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Breadth be b
1
Given, b
1
= 2m
l
1
× b
1
= 384
l
1
= 384/b
1
= 384/2
= 192m
? Cost of iron sheet at the rate of Rs 5 per metre = 5 × 192 = Rs 960
8. A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the
tank are 12m×8m×6m, find the cost of iron sheet at Rs. 17.50 per metre.
Solution:
Given details are,
Dimensions of tank = 12m × 8m × 6m
Where, length = 12m
Breadth = 8m
Height = 6m
Area of sheet required = total surface area of tank with one top open
= l × b + 2 (l×h + b×h)
= 12 × 8 + 2 (12×6 + 8×6)
= 96 + 240
= 336 m
2
Let length be l
1
Breadth be b
1
Given, b
1
= 4m
l
1
× b
1
= 336
l
1
= 336/b
1
= 336/4
= 84m
? Cost of iron sheet at the rate of Rs 17.50 per metre = 17.50 × 84 = Rs 1470
9. Three equal cubes are placed adjacently in a row. Find the ratio of total surface
area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution:
Given details are,
Let edge length of three equal cubes = a
Then,
Sum of surface area of 3 cubes = 3 × 6a
2
= 18a
2
When these cubes are placed in a row adjacently they form a cuboid.
Length of new cuboid formed = a + a + a = 3a
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Breadth of cuboid = a
Height of cuboid = a
Total surface area of cuboid = 2 (lb×bh×hl)
= 2 (3a×a + a×a + a×3a)
= 2 (3a
2
+ a
2
+ 3a
2
)
= 2 (7a
2
)
= 14 a
2
Total surface area of new cuboid / sum of surface area of 3 cuboids = 14/18 = 7/9 = 7:9
? The ratio is 7:9
10. The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4
windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by
1 m. Find the cost of painting the walls at Rs. 3.50 per square metre.
Solution:
Given details are,
Dimensions of room = 12.5m × 9m × 7m
Dimensions of each door = 2.5m × 1.2m
Dimensions of each window = 1.5m × 1m
Area of four walls including doors and windows = 2 (l×h + b×h)
= 2 (12.5×7 + 9×7)
= 2 (87.5 + 63)
= 2 (150.5)
= 301 m
2
Area of 2 doors and 4 windows = 2 (2.5×1.2) + 4 (1.5×1)
= 2(3) + 4 (1.5)
= 6 + 6
= 12 m
2
Area of only walls = 301 – 12
= 289 m
2
? Cost of painting the walls at the rate of Rs 3.50 per square metre = Rs (3.50 × 289) =
Rs 1011.50
11. A field is 150m long and 100m wide. A plot (outside the field) 50m long and 30m
wide is dug to a depth of 8m and the earth taken out from the plot is spread evenly
in the field. By how much is the level of field raised?
Solution:
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,Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.4 | Mathematics (Maths) Class 8
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Additional Information about Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.4 for Class 8 Preparation
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