# NCERT Chapter Summary: Surface Areas and Volumes

So far, you have been dealing with figures that can be easily drawn on our notebooks or blackboards. These are called plane figures.

It would be interesting to see what happens if you cut out many of these plane figures of the same shape and size from cardboard sheet and stack them up in a vertical pile. By this process, you obtain some solid figures such as a cuboid, a cylinder, cube, and sphere.

Important Formulas

1. Surface area of a cuboid = 2(lb + bh + hl)

2. Surface area of cube = 6a2

3. Curved surface area of cylinder = 2πrh

4. Total surface area of cylinder = 2πr(r + h)

5. Curved surface area of cone = πrl

6. Total surface area of right circular cone = πrl + πr2 = πr(l + r)

7. Surface area of sphere = 4πr2

8. Curved surface area of hemisphere = 2πr2

9. Total surface area of hemisphere = 3πr2

10. Volume of cuboid = l × b × h

11. Volume of cube = a3

12. Volume of cylinder = πr2h

13. Volume of cone = 1/3 × πr2h

14. Volume of sphere = 4/3 × πr3

15. Volume of hemisphere = 2/3 × πr3

### Class 10

In this chapter, you will study how to determine the surface area of an object formed by combining any two of the basic solids - cuboid, cone, cylinder, sphere and hemisphere.

Also, how to find the volume of objects formed by combining any two of a cuboid, cone, cylinder, sphere and hemisphere.

Given a right circular cone, which is sliced through by a plane parallel to its base, when the smaller conical portion is removed, the resulting solid is called a Frustum of a Right Circular Cone. The formulae involving the frustum of a cone are:

Volume of a frustum of a cone = 1/3 × πh × (r12 + r22 + r1r2)

Curved surface area of a frustum of a cone = πl(r1 + r2)

Total surface area of frustum of a cone = πl(r1 + r2) + π(r12 + r22)