ML Aggarwal: Mensuration - 4

 Page 1


 
1. The surface area of a cube is 384 cm
2
. Find  
(i) the length of an edge  
(ii) volume of the cube. 
Solution: 
Given, 
Surface area of a cube = 384 cm
2
 
(i) Surface area of cube = 6(side)
2
 
Hence, edge (side) = 

 


 
= 




 
= v64 
= 8 cm 
(ii) Volume = (Edge)
 3
 = (8)
3 
= 8 × 8 × 8 cm
3
 = 512 cm
3
 
  
2. Find the total surface area of a solid cylinder of radius 5 cm and 
height 10 cm. Leave your answer in terms of n.  
Solution: 
Given, 
Radius of a solid cylinder (r) = 5 cm 
 
Height (h) = 10 cm 
 
Hence, 
Page 2


 
1. The surface area of a cube is 384 cm
2
. Find  
(i) the length of an edge  
(ii) volume of the cube. 
Solution: 
Given, 
Surface area of a cube = 384 cm
2
 
(i) Surface area of cube = 6(side)
2
 
Hence, edge (side) = 

 


 
= 




 
= v64 
= 8 cm 
(ii) Volume = (Edge)
 3
 = (8)
3 
= 8 × 8 × 8 cm
3
 = 512 cm
3
 
  
2. Find the total surface area of a solid cylinder of radius 5 cm and 
height 10 cm. Leave your answer in terms of n.  
Solution: 
Given, 
Radius of a solid cylinder (r) = 5 cm 
 
Height (h) = 10 cm 
 
Hence, 
Total surface area = 2prh + 2pr
2
 
 
= 2rp (h + r)  
 
= 2p × 5(10 + 5)  
 
= p × 10 × 15 
= 150p cm
2 
 
 
3. An aquarium is in the form of a cuboid whose external measures 
are 70 cm × 28 cm × 35 cm. The base, side faces and back face are to 
be covered with coloured paper. Find the area of the paper needed. 
Solution: 
Given, a cuboid shaped aquarium 
 
Length (l) = 70 cm 
 
Breadth (b) = 28 cm 
 
and height (h) = 35 cm 
Page 3


 
1. The surface area of a cube is 384 cm
2
. Find  
(i) the length of an edge  
(ii) volume of the cube. 
Solution: 
Given, 
Surface area of a cube = 384 cm
2
 
(i) Surface area of cube = 6(side)
2
 
Hence, edge (side) = 

 


 
= 




 
= v64 
= 8 cm 
(ii) Volume = (Edge)
 3
 = (8)
3 
= 8 × 8 × 8 cm
3
 = 512 cm
3
 
  
2. Find the total surface area of a solid cylinder of radius 5 cm and 
height 10 cm. Leave your answer in terms of n.  
Solution: 
Given, 
Radius of a solid cylinder (r) = 5 cm 
 
Height (h) = 10 cm 
 
Hence, 
Total surface area = 2prh + 2pr
2
 
 
= 2rp (h + r)  
 
= 2p × 5(10 + 5)  
 
= p × 10 × 15 
= 150p cm
2 
 
 
3. An aquarium is in the form of a cuboid whose external measures 
are 70 cm × 28 cm × 35 cm. The base, side faces and back face are to 
be covered with coloured paper. Find the area of the paper needed. 
Solution: 
Given, a cuboid shaped aquarium 
 
Length (l) = 70 cm 
 
Breadth (b) = 28 cm 
 
and height (h) = 35 cm 
Now, 
 
Area of base = 70 × 28 cm
3
  
= 1960 cm
3
 
 
Area of side face = (28 × 35) × 2 cm
2 
 
= 1960 cm
2
 
Area of back face = 70 × 35 cm
2
  
= 2450 cm 
 
Thus, the total area = 1960 + 1960 + 2450 = 6370 cm
2
 
 
Hence, area of paper required is 6370 cm
2
. 
 
4. The internal dimensions of rectangular hall are 15 m × 12 m × 4 m. 
There are 4 windows each of dimension 2 m × 1.5 m and 2 doors each 
of dimension 1.5 m × 2.5 m. find the cost of white washing all four 
walls of the hall, if the cost of white washing is ?5 per m
2
. What will 
be the cost of white washing if the ceiling of the hall is also white 
washed? 
Solution: 
Given, 
Internal dimension of rectangular hall = 15m × 12 m × 4 m 
Now, 
 
Area of 4-walls = 2(l + b) × h 
 
= 2(15 + 12) × 4 
 
Page 4


 
1. The surface area of a cube is 384 cm
2
. Find  
(i) the length of an edge  
(ii) volume of the cube. 
Solution: 
Given, 
Surface area of a cube = 384 cm
2
 
(i) Surface area of cube = 6(side)
2
 
Hence, edge (side) = 

 


 
= 




 
= v64 
= 8 cm 
(ii) Volume = (Edge)
 3
 = (8)
3 
= 8 × 8 × 8 cm
3
 = 512 cm
3
 
  
2. Find the total surface area of a solid cylinder of radius 5 cm and 
height 10 cm. Leave your answer in terms of n.  
Solution: 
Given, 
Radius of a solid cylinder (r) = 5 cm 
 
Height (h) = 10 cm 
 
Hence, 
Total surface area = 2prh + 2pr
2
 
 
= 2rp (h + r)  
 
= 2p × 5(10 + 5)  
 
= p × 10 × 15 
= 150p cm
2 
 
 
3. An aquarium is in the form of a cuboid whose external measures 
are 70 cm × 28 cm × 35 cm. The base, side faces and back face are to 
be covered with coloured paper. Find the area of the paper needed. 
Solution: 
Given, a cuboid shaped aquarium 
 
Length (l) = 70 cm 
 
Breadth (b) = 28 cm 
 
and height (h) = 35 cm 
Now, 
 
Area of base = 70 × 28 cm
3
  
= 1960 cm
3
 
 
Area of side face = (28 × 35) × 2 cm
2 
 
= 1960 cm
2
 
Area of back face = 70 × 35 cm
2
  
= 2450 cm 
 
Thus, the total area = 1960 + 1960 + 2450 = 6370 cm
2
 
 
Hence, area of paper required is 6370 cm
2
. 
 
4. The internal dimensions of rectangular hall are 15 m × 12 m × 4 m. 
There are 4 windows each of dimension 2 m × 1.5 m and 2 doors each 
of dimension 1.5 m × 2.5 m. find the cost of white washing all four 
walls of the hall, if the cost of white washing is ?5 per m
2
. What will 
be the cost of white washing if the ceiling of the hall is also white 
washed? 
Solution: 
Given, 
Internal dimension of rectangular hall = 15m × 12 m × 4 m 
Now, 
 
Area of 4-walls = 2(l + b) × h 
 
= 2(15 + 12) × 4 
 
= 2 × 27 × 4 m
2
 
 
= 216 m
2
 
 
Area of 4 windows of size = (2 × 1.5) × 4 = 12 m
2
 
 
Area of 2 door of size = 2 × (1.5 × 2.5) = 7.5 m
2
 
 
So, area of remaining hall = 216 – (12 + 7.5) = 216 – 19.5 m
2
 = 196.5 m
2
 
 
And, 
Cost of white washing the walls all four halls of the house is at the rate 
of ?5 per m
2 
 
= 196.5 × 5 = ?982.50 
 
Area of ceiling = l × b = 15 × 12 = 180 m
2
 
 
Cost of white washing = 180 × 5 = ?900 
 
Therefore, the total cost for white washing = ?982.50 + 900.00 
= ?1882.50 
  
5. A swimming pool is 50 m in length, 30 m in breadth and 2.5 m in 
depth. Find the cost of cementing its floor and walls at the rate of 
?27 per square metre. 
Solution: 
Given, 
Length of swimming pool = 50 m 
 
Breadth of swimming pool = 30 m 
Page 5


 
1. The surface area of a cube is 384 cm
2
. Find  
(i) the length of an edge  
(ii) volume of the cube. 
Solution: 
Given, 
Surface area of a cube = 384 cm
2
 
(i) Surface area of cube = 6(side)
2
 
Hence, edge (side) = 

 


 
= 




 
= v64 
= 8 cm 
(ii) Volume = (Edge)
 3
 = (8)
3 
= 8 × 8 × 8 cm
3
 = 512 cm
3
 
  
2. Find the total surface area of a solid cylinder of radius 5 cm and 
height 10 cm. Leave your answer in terms of n.  
Solution: 
Given, 
Radius of a solid cylinder (r) = 5 cm 
 
Height (h) = 10 cm 
 
Hence, 
Total surface area = 2prh + 2pr
2
 
 
= 2rp (h + r)  
 
= 2p × 5(10 + 5)  
 
= p × 10 × 15 
= 150p cm
2 
 
 
3. An aquarium is in the form of a cuboid whose external measures 
are 70 cm × 28 cm × 35 cm. The base, side faces and back face are to 
be covered with coloured paper. Find the area of the paper needed. 
Solution: 
Given, a cuboid shaped aquarium 
 
Length (l) = 70 cm 
 
Breadth (b) = 28 cm 
 
and height (h) = 35 cm 
Now, 
 
Area of base = 70 × 28 cm
3
  
= 1960 cm
3
 
 
Area of side face = (28 × 35) × 2 cm
2 
 
= 1960 cm
2
 
Area of back face = 70 × 35 cm
2
  
= 2450 cm 
 
Thus, the total area = 1960 + 1960 + 2450 = 6370 cm
2
 
 
Hence, area of paper required is 6370 cm
2
. 
 
4. The internal dimensions of rectangular hall are 15 m × 12 m × 4 m. 
There are 4 windows each of dimension 2 m × 1.5 m and 2 doors each 
of dimension 1.5 m × 2.5 m. find the cost of white washing all four 
walls of the hall, if the cost of white washing is ?5 per m
2
. What will 
be the cost of white washing if the ceiling of the hall is also white 
washed? 
Solution: 
Given, 
Internal dimension of rectangular hall = 15m × 12 m × 4 m 
Now, 
 
Area of 4-walls = 2(l + b) × h 
 
= 2(15 + 12) × 4 
 
= 2 × 27 × 4 m
2
 
 
= 216 m
2
 
 
Area of 4 windows of size = (2 × 1.5) × 4 = 12 m
2
 
 
Area of 2 door of size = 2 × (1.5 × 2.5) = 7.5 m
2
 
 
So, area of remaining hall = 216 – (12 + 7.5) = 216 – 19.5 m
2
 = 196.5 m
2
 
 
And, 
Cost of white washing the walls all four halls of the house is at the rate 
of ?5 per m
2 
 
= 196.5 × 5 = ?982.50 
 
Area of ceiling = l × b = 15 × 12 = 180 m
2
 
 
Cost of white washing = 180 × 5 = ?900 
 
Therefore, the total cost for white washing = ?982.50 + 900.00 
= ?1882.50 
  
5. A swimming pool is 50 m in length, 30 m in breadth and 2.5 m in 
depth. Find the cost of cementing its floor and walls at the rate of 
?27 per square metre. 
Solution: 
Given, 
Length of swimming pool = 50 m 
 
Breadth of swimming pool = 30 m 
 
Depth (Height) of swimming pool = 2·5 m 
 
Now, 
Area of floor = 50 × 30 = 1500 m
2
 
 
Area of four walls = 2 (50 + 30) × 2.5 = 160 × 2.5 = 400 m
2
 
 
so, the area to be cemented = 1500 m
2
 + 400 m
2
 = 1900 m
2
 
 
Cost of cementing 1m
2
 = ?27 
 
Hence, 
Cost of cementing 1900m
2 
= ?27 × 1900 = ?51300 
 
6. The floor of a rectangular hall has a perimeter 236 m. Its height is 
4·5 m. Find the cost of painting its four walls (doors and windows be 
ignored) at the rate of Rs. 8.40 per square metre. 
Solution: 
Given, 
Perimeter of Hall = 236 m. 
 
Height = 4.5 m 
 
Perimeter = 2 (l + b) = 236 m 
 
Area of four walls = 2 (l + b) × h 
= 236 × 4.5