Textbook: Mensuration

 Page 1


140
7
Mensuration
•  Mixed examples on 
 surface area and 
 volume of different 
 solid figures
Let’s recall.
   Last year we have studied surface area and volume of some three dimentional 
figures. Let us recall the formulae to find the surface areas and volumes.
No. Three dimensional 
figure
Formulae
1 . Cuboid
     
Lateral surface area   = 2h ( l + b )
Total surface area  = 2  (lb + bh + hl )
Volume  = lbh
2 .  Cube
           
Lateral surface area =  4l
2
Total surface area = 6l
2
Volume = l
3
3 . Cylinder
                 
Curved surface area =  2prh
Total surface area = 2pr ( r + h )
Volume = pr
2
h
4 . Cone
        
Slant height (l)  = 
hr
22
+
Curved surface area = prl 
Total surface area = pr (r + l)
Volume = 
1
3
 ´ pr
2
h
l
l
h
b
• Arc of circle - length of arc 
• Area of a sector 
• Area of segment of a circle
l
h
h
r
r
Let’s study.
Page 2


140
7
Mensuration
•  Mixed examples on 
 surface area and 
 volume of different 
 solid figures
Let’s recall.
   Last year we have studied surface area and volume of some three dimentional 
figures. Let us recall the formulae to find the surface areas and volumes.
No. Three dimensional 
figure
Formulae
1 . Cuboid
     
Lateral surface area   = 2h ( l + b )
Total surface area  = 2  (lb + bh + hl )
Volume  = lbh
2 .  Cube
           
Lateral surface area =  4l
2
Total surface area = 6l
2
Volume = l
3
3 . Cylinder
                 
Curved surface area =  2prh
Total surface area = 2pr ( r + h )
Volume = pr
2
h
4 . Cone
        
Slant height (l)  = 
hr
22
+
Curved surface area = prl 
Total surface area = pr (r + l)
Volume = 
1
3
 ´ pr
2
h
l
l
h
b
• Arc of circle - length of arc 
• Area of a sector 
• Area of segment of a circle
l
h
h
r
r
Let’s study.
141
Fig 7.1
30 cm
20 cm
20 cm
Fig 7.2
21 cm
10 cm
No.
Three dimensional 
figure
Formulae
5. Sphere
          
Surface area = 4 pr
2
Volume = 
4
3
 pr
3
6.
Hemisphere 
         
Curved surface area = 2pr
2
Total surface area of a solid hemisphere = 3pr
2
Volume = 
2
3
 pr
3
Solve the following examples
As shown in the adjacent figure, a sphere is placed in a 
cylinder. It touches the top, bottom and the curved surface 
of the cylinder. If radius of the base of the cylinder is ‘r’, 
(1) What is the ratio of the radii of the sphere and the  
       cylinder ?
(2) What is the ratio of the curved surface area of the  
       cylinder and the surface area of the sphere ?
(3) What is the ratio of the volumes of the cylinder and the 
      sphere ?
Fig. 7.3
The length, breadth and height of 
an oil can are 20 cm, 20 cm and 30 cm 
respectively as shown in the adjacent 
figure. 
How much oil will it contain ?
   (1 litre = 1000 cm
3
)  
The adjoining figure shows the 
measures of a Joker’s cap. How much 
cloth is needed to make such a cap ?
Let’s think.
r
r
Ex. (1)
Ex. (2)
Page 3


140
7
Mensuration
•  Mixed examples on 
 surface area and 
 volume of different 
 solid figures
Let’s recall.
   Last year we have studied surface area and volume of some three dimentional 
figures. Let us recall the formulae to find the surface areas and volumes.
No. Three dimensional 
figure
Formulae
1 . Cuboid
     
Lateral surface area   = 2h ( l + b )
Total surface area  = 2  (lb + bh + hl )
Volume  = lbh
2 .  Cube
           
Lateral surface area =  4l
2
Total surface area = 6l
2
Volume = l
3
3 . Cylinder
                 
Curved surface area =  2prh
Total surface area = 2pr ( r + h )
Volume = pr
2
h
4 . Cone
        
Slant height (l)  = 
hr
22
+
Curved surface area = prl 
Total surface area = pr (r + l)
Volume = 
1
3
 ´ pr
2
h
l
l
h
b
• Arc of circle - length of arc 
• Area of a sector 
• Area of segment of a circle
l
h
h
r
r
Let’s study.
141
Fig 7.1
30 cm
20 cm
20 cm
Fig 7.2
21 cm
10 cm
No.
Three dimensional 
figure
Formulae
5. Sphere
          
Surface area = 4 pr
2
Volume = 
4
3
 pr
3
6.
Hemisphere 
         
Curved surface area = 2pr
2
Total surface area of a solid hemisphere = 3pr
2
Volume = 
2
3
 pr
3
Solve the following examples
As shown in the adjacent figure, a sphere is placed in a 
cylinder. It touches the top, bottom and the curved surface 
of the cylinder. If radius of the base of the cylinder is ‘r’, 
(1) What is the ratio of the radii of the sphere and the  
       cylinder ?
(2) What is the ratio of the curved surface area of the  
       cylinder and the surface area of the sphere ?
(3) What is the ratio of the volumes of the cylinder and the 
      sphere ?
Fig. 7.3
The length, breadth and height of 
an oil can are 20 cm, 20 cm and 30 cm 
respectively as shown in the adjacent 
figure. 
How much oil will it contain ?
   (1 litre = 1000 cm
3
)  
The adjoining figure shows the 
measures of a Joker’s cap. How much 
cloth is needed to make such a cap ?
Let’s think.
r
r
Ex. (1)
Ex. (2)
142
As shown in the above figures, take a ball and a beaker of the same radius as that 
of the ball. Cut a strip of paper of length equal to the diameter of the beaker. Draw 
two lines on the strip dividing it into three equal parts. Stick it on the beaker straight 
up from the bottom. Fill water in the beaker upto the first mark of the strip from the 
bottom. Push the ball in the beaker slowly so that it touches its bottom. Observe how 
much the water level rises. 
Y ou will notice that the water level has risen exactly upto the total height of the strip. 
Try to understand how we get the formula for the volume of a the sphere. The shape 
of the beaker is cylindrical. 
Therefore, the volume of the part of the beaker upto height 2r can be obtained by 
the formula of volume of a cylinder. Let us assume that the volume is V. 
 \ V = pr
2
h = p ´ r
2
 ´ 2r = 2pr
3 
 (
?? h = 2r)
But V = volume of the ball + volume of the water which was already in the beaker.
 = volume of the ball + 
1
3
 ´ 2pr
3
 
\ volume of the ball = V - 
1
3
 ´ 2pr
3
          
= 2pr
3 
- 
2
3
pr
3
     = 
6pr
3 
- 2pr
3
3
 = 
4pr
3 
3
 
Hence we get the formula of the volume of a sphere as V = 
4
3
pr
3
(Now you can find the answer of the question number 3 relating to figure 7.3)
Activity :
2r
2r
2r
r
Fig. 7.5 Fig. 7.6 Fig. 7.4
Page 4


140
7
Mensuration
•  Mixed examples on 
 surface area and 
 volume of different 
 solid figures
Let’s recall.
   Last year we have studied surface area and volume of some three dimentional 
figures. Let us recall the formulae to find the surface areas and volumes.
No. Three dimensional 
figure
Formulae
1 . Cuboid
     
Lateral surface area   = 2h ( l + b )
Total surface area  = 2  (lb + bh + hl )
Volume  = lbh
2 .  Cube
           
Lateral surface area =  4l
2
Total surface area = 6l
2
Volume = l
3
3 . Cylinder
                 
Curved surface area =  2prh
Total surface area = 2pr ( r + h )
Volume = pr
2
h
4 . Cone
        
Slant height (l)  = 
hr
22
+
Curved surface area = prl 
Total surface area = pr (r + l)
Volume = 
1
3
 ´ pr
2
h
l
l
h
b
• Arc of circle - length of arc 
• Area of a sector 
• Area of segment of a circle
l
h
h
r
r
Let’s study.
141
Fig 7.1
30 cm
20 cm
20 cm
Fig 7.2
21 cm
10 cm
No.
Three dimensional 
figure
Formulae
5. Sphere
          
Surface area = 4 pr
2
Volume = 
4
3
 pr
3
6.
Hemisphere 
         
Curved surface area = 2pr
2
Total surface area of a solid hemisphere = 3pr
2
Volume = 
2
3
 pr
3
Solve the following examples
As shown in the adjacent figure, a sphere is placed in a 
cylinder. It touches the top, bottom and the curved surface 
of the cylinder. If radius of the base of the cylinder is ‘r’, 
(1) What is the ratio of the radii of the sphere and the  
       cylinder ?
(2) What is the ratio of the curved surface area of the  
       cylinder and the surface area of the sphere ?
(3) What is the ratio of the volumes of the cylinder and the 
      sphere ?
Fig. 7.3
The length, breadth and height of 
an oil can are 20 cm, 20 cm and 30 cm 
respectively as shown in the adjacent 
figure. 
How much oil will it contain ?
   (1 litre = 1000 cm
3
)  
The adjoining figure shows the 
measures of a Joker’s cap. How much 
cloth is needed to make such a cap ?
Let’s think.
r
r
Ex. (1)
Ex. (2)
142
As shown in the above figures, take a ball and a beaker of the same radius as that 
of the ball. Cut a strip of paper of length equal to the diameter of the beaker. Draw 
two lines on the strip dividing it into three equal parts. Stick it on the beaker straight 
up from the bottom. Fill water in the beaker upto the first mark of the strip from the 
bottom. Push the ball in the beaker slowly so that it touches its bottom. Observe how 
much the water level rises. 
Y ou will notice that the water level has risen exactly upto the total height of the strip. 
Try to understand how we get the formula for the volume of a the sphere. The shape 
of the beaker is cylindrical. 
Therefore, the volume of the part of the beaker upto height 2r can be obtained by 
the formula of volume of a cylinder. Let us assume that the volume is V. 
 \ V = pr
2
h = p ´ r
2
 ´ 2r = 2pr
3 
 (
?? h = 2r)
But V = volume of the ball + volume of the water which was already in the beaker.
 = volume of the ball + 
1
3
 ´ 2pr
3
 
\ volume of the ball = V - 
1
3
 ´ 2pr
3
          
= 2pr
3 
- 
2
3
pr
3
     = 
6pr
3 
- 2pr
3
3
 = 
4pr
3 
3
 
Hence we get the formula of the volume of a sphere as V = 
4
3
pr
3
(Now you can find the answer of the question number 3 relating to figure 7.3)
Activity :
2r
2r
2r
r
Fig. 7.5 Fig. 7.6 Fig. 7.4
143
?????????????? Solved Examples ?????????????
Ex. (1)  The radius and height of a cylindrical water reservoir is 2.8 m and 3.5 m  
  respectively. How much maximum water can the tank hold ? A person needs 
  70 litre of water per day. For how many persons is the water sufficient for  
  a day ? (p =  
22
7
)
Solution : (r) = 2.8 m,     ( h) = 3.5 m,  p =  
22
7
                  Capacity of the water reservoir = Volume of the cylindrical reservoir
   = pr
2
h
   = 
22
7
 ´ 2.8 ´ 2.8 ´ 3.5
   = 86.24 m
3
   = 86.24 ´ 1000    (
?? 1m
3
 = 1000 litre)
   = 86240.00 litre.
  \ the reservoir can hold 86240 litre of water. 
   The daily requirement of water of a person is 70 litre.
  \ water in the tank is sufficient for 
86240
70
 = 1232 persons.
Ex. (2)  How many solid cylinders of radius 10 cm and height 6 cm can be made by  
  melting a solid sphere of radius 30 cm ? 
Solution :       Radius of a sphere, r = 30 cm
  Radius of the cylinder, R = 10 cm
  Height of the cylinder, H = 6 cm
  Let the number of cylinders be n.
  Volume of the sphere = n ´ volume of a cylinder
               \  n = 
Volume of the sphere
Volume of a cylinder
                         = =
()
()
4
3
3
2
p
p
r
RH
         = 
=
×()
×
4
3
30
10 6
3
2
  = 
=
×××
××
4
3
30 30 30
10 10 6
  = 60 
  \ 60 cylinders can be made .
Page 5


140
7
Mensuration
•  Mixed examples on 
 surface area and 
 volume of different 
 solid figures
Let’s recall.
   Last year we have studied surface area and volume of some three dimentional 
figures. Let us recall the formulae to find the surface areas and volumes.
No. Three dimensional 
figure
Formulae
1 . Cuboid
     
Lateral surface area   = 2h ( l + b )
Total surface area  = 2  (lb + bh + hl )
Volume  = lbh
2 .  Cube
           
Lateral surface area =  4l
2
Total surface area = 6l
2
Volume = l
3
3 . Cylinder
                 
Curved surface area =  2prh
Total surface area = 2pr ( r + h )
Volume = pr
2
h
4 . Cone
        
Slant height (l)  = 
hr
22
+
Curved surface area = prl 
Total surface area = pr (r + l)
Volume = 
1
3
 ´ pr
2
h
l
l
h
b
• Arc of circle - length of arc 
• Area of a sector 
• Area of segment of a circle
l
h
h
r
r
Let’s study.
141
Fig 7.1
30 cm
20 cm
20 cm
Fig 7.2
21 cm
10 cm
No.
Three dimensional 
figure
Formulae
5. Sphere
          
Surface area = 4 pr
2
Volume = 
4
3
 pr
3
6.
Hemisphere 
         
Curved surface area = 2pr
2
Total surface area of a solid hemisphere = 3pr
2
Volume = 
2
3
 pr
3
Solve the following examples
As shown in the adjacent figure, a sphere is placed in a 
cylinder. It touches the top, bottom and the curved surface 
of the cylinder. If radius of the base of the cylinder is ‘r’, 
(1) What is the ratio of the radii of the sphere and the  
       cylinder ?
(2) What is the ratio of the curved surface area of the  
       cylinder and the surface area of the sphere ?
(3) What is the ratio of the volumes of the cylinder and the 
      sphere ?
Fig. 7.3
The length, breadth and height of 
an oil can are 20 cm, 20 cm and 30 cm 
respectively as shown in the adjacent 
figure. 
How much oil will it contain ?
   (1 litre = 1000 cm
3
)  
The adjoining figure shows the 
measures of a Joker’s cap. How much 
cloth is needed to make such a cap ?
Let’s think.
r
r
Ex. (1)
Ex. (2)
142
As shown in the above figures, take a ball and a beaker of the same radius as that 
of the ball. Cut a strip of paper of length equal to the diameter of the beaker. Draw 
two lines on the strip dividing it into three equal parts. Stick it on the beaker straight 
up from the bottom. Fill water in the beaker upto the first mark of the strip from the 
bottom. Push the ball in the beaker slowly so that it touches its bottom. Observe how 
much the water level rises. 
Y ou will notice that the water level has risen exactly upto the total height of the strip. 
Try to understand how we get the formula for the volume of a the sphere. The shape 
of the beaker is cylindrical. 
Therefore, the volume of the part of the beaker upto height 2r can be obtained by 
the formula of volume of a cylinder. Let us assume that the volume is V. 
 \ V = pr
2
h = p ´ r
2
 ´ 2r = 2pr
3 
 (
?? h = 2r)
But V = volume of the ball + volume of the water which was already in the beaker.
 = volume of the ball + 
1
3
 ´ 2pr
3
 
\ volume of the ball = V - 
1
3
 ´ 2pr
3
          
= 2pr
3 
- 
2
3
pr
3
     = 
6pr
3 
- 2pr
3
3
 = 
4pr
3 
3
 
Hence we get the formula of the volume of a sphere as V = 
4
3
pr
3
(Now you can find the answer of the question number 3 relating to figure 7.3)
Activity :
2r
2r
2r
r
Fig. 7.5 Fig. 7.6 Fig. 7.4
143
?????????????? Solved Examples ?????????????
Ex. (1)  The radius and height of a cylindrical water reservoir is 2.8 m and 3.5 m  
  respectively. How much maximum water can the tank hold ? A person needs 
  70 litre of water per day. For how many persons is the water sufficient for  
  a day ? (p =  
22
7
)
Solution : (r) = 2.8 m,     ( h) = 3.5 m,  p =  
22
7
                  Capacity of the water reservoir = Volume of the cylindrical reservoir
   = pr
2
h
   = 
22
7
 ´ 2.8 ´ 2.8 ´ 3.5
   = 86.24 m
3
   = 86.24 ´ 1000    (
?? 1m
3
 = 1000 litre)
   = 86240.00 litre.
  \ the reservoir can hold 86240 litre of water. 
   The daily requirement of water of a person is 70 litre.
  \ water in the tank is sufficient for 
86240
70
 = 1232 persons.
Ex. (2)  How many solid cylinders of radius 10 cm and height 6 cm can be made by  
  melting a solid sphere of radius 30 cm ? 
Solution :       Radius of a sphere, r = 30 cm
  Radius of the cylinder, R = 10 cm
  Height of the cylinder, H = 6 cm
  Let the number of cylinders be n.
  Volume of the sphere = n ´ volume of a cylinder
               \  n = 
Volume of the sphere
Volume of a cylinder
                         = =
()
()
4
3
3
2
p
p
r
RH
         = 
=
×()
×
4
3
30
10 6
3
2
  = 
=
×××
××
4
3
30 30 30
10 10 6
  = 60 
  \ 60 cylinders can be made .
144
Ex. (3)   A tent of a circus is such that its lower part is cylindrical and upper part  
  is conical. The diameter of the base of the tent is 48 m and the height of the  
  cylindrical part is 15 m. Total height of the tent is 33 m. Find area  
  of canvas required to make the tent. Also find volume of air in the tent. 
Solution :  Total height of the tent = 33 m.   
  Let height of the cylindrical part be H
 \ H = 15 m. 
  Let the height of the conical part be h 
 \h = (33-15) = 18 m. 
  Slant height of cone, l = rh
2
2
+
()
   = 
24 18
22
+
   = 
576 324 +
   = 900
                              = 30 m.
  Canvas required for tent  =  Curved surface area of the cylindrical part +  
      Curved surface area of conical part 
    = 2prH + prl
    = pr (2H + l)
    = 
22
7
 ´ 24 (2 ´ 15 + 30)
    = 
22
7
 ´ 24 ´ 60
    = 4525.71 m
2
 Volume of air in the tent = volume of cylinder + volume of cone
    = pr
2
H + 
1
3
 pr
2
h
    = pr
2
 
H+
?
?
?
?
?
?
1
3
h
    = 
22
7
 ´ 24
2
 (15 + 
1
3
 ´ 18)
    = 
22
7
 ´ 576 ´ 21
    = 38,016 m
3
  \ canvas required for the tent = 4525.71 m
2
         \ volume of air in the tent = 38,016 m
3
.  
Fig 7.7
18 m
24 m
15 m