Class 10 – Mensuration

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  • Volume and surface area of combined solids
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  • Conversion of solids
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Combination of solids: The solids formed by two or more basic solids are called combination of solids.  

i)               Container        

ii)              Circus tent            iii) Playing toy (lattu)


Surface area of composite solids:

i) Consider the container seen in figure. To find the surface area of such a solid. The solid is made up of a cylinder with two hemispheres stuck at either end. It would look like the container after putting the pieces all together.

Consider the surface of the newly formed object, only the curved surfaces of the two hemispheres are seen and the curved surface of the cylinder. So, the total surface area of the new solid is the sum of the curved surface areas of each of the individual parts.
   TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere.


ii) Now consider the playing toy from examples. First, take a cone and a hemisphere and being their flat faces together. Here, the base radius of the cone equal to the radius of the hemisphere, for the top is to have a smooth surface. So, the steps would be as shown in figure.

The surface area of the top, which consists of the CSA of the hemisphere and the CSA of the cone.
Total surface area of the toy = CSA of hemisphere + CSA of cone
Now, consider some examples.

Example: A circus tent is in the form of a cone over a cylinder. The diameter of the base is 9 m and the height of cylindrical part is 4.8 m and the total height of the tent is 10.8 m. How much canvas cloth is required for the tent?

Solution:See the adjacent figure

Height of the tent                       = 10.8 m

Height of the cylindrical portion = 4.8 m

Height of the cone                      = 6 m

Radius of the base                    

  Slant height of the cone (l)                                                                                                                                           


       Slant height of the cone (l

  Canvas required for conical portion of tent = curved surface area of the cone


Canvas required for the cylindrical part of the tent


Total canvas required = curved surface area of the cone + curved surface area of cylinder


Total canvas required = 241.84 sq.m

Example: Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take )

Solution:This top is exactly like the object we have discussed in examples. So, TSA of the toy = CSA of hemisphere + CSA of cone 
Now, the curved surface area of the hemisphere = 2πr2 sq.units 


Also, the height of the cone = height of the top – height (radius) of the hemispherical part 

So, the slant height of the cone (l)  
Therefore, CSA of cone =
πrl sq.units 

This gives the surface area of the top as 

Description:‘Total surface area of the top’ is not the sum of the total surface areas of the cone and hemisphere.

Volume of a combination of solids: For calculating volume of combined solids.

Split the solid shapes into fundamental solid shapes. Find the volumes separately and the result to get the volume of combined solid.

Example: A solid toy is in the form of right circular cylinder with hemispherical shape at one end and a cone at the other end. Their common diameter is 4.2 cm and the height of the cylindrical and conical portions are 12 cm and 7 cm respectively. Find the volume of the solid toy. .

Solution: Height of the conical portion = 7 cm

Diameter of the conical portion = 4.2 cm

Volume of the conical portion =


                 = 32.34 cm3 ………(1)

Height of the cylinder = 12 cm

Volume of the cylinder

                = 166.32 cm3 …..(2)

Volume of the hemisphere     =

On adding (1), (2) and (3), 

Volume of the solid toy = volume of cone + volume of cylinder + volume of hemisphere

     = 32.34 + 166.32 + 19.40 = 218.06 cm3.


1.              Volume of water released by a pipe/canal = Area of cross-section × Rate of flow × Time.

2.              In case of a pipe, the cross-section is usually a circle.

3.              In case of canal, the cross-section is usually a rectangle or a trapezium.

4.              If rate of flow is given in km/h, it can be converted into m/sec by multiplying with

5.              If time is given in minutes or hours, it must be converted into seconds because rate of flow is in m/sec.

6.              Volume of water standing in a field = Area of field × Height of standing water.

7.              If area of filed is given in hectares, it must be converted into m2 by multiplying with 10,000.

8.              Height of standing water must be taken in metres. If given in cm it must be converted to metre by dividing by 100.