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MENSURATION  

  

Class 10 Notes 

 

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

 

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. 

 

hemisphere + CSA of cylinder + CSA of other hemisphere. 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  

                                                                                                                sq.m 

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 = π rl sq.units  

This gives the surface area of the top as  

 

‘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 cm 3 ………(1) 

Height of the cylinder = 12 cm 

Volume of the cylinder  

= 166.32 cm 3 …..(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 cm 3 .  

 

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 m 2  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.   

 

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