# Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained

**Solution:**

A frustum of a cone with h as height, l as the slant height, r_{1} and r_{2} radii of the ends where r_{1} > r_{2}

To Prove:

CSA of the frustum of the cone = 1/3πh (r_{1}² + r_{2}² + r_{1}r_{2})

where r_{1}, r_{2}, and h are the radii and height of the frustum of the cone respectively.

Construction:

Extend side BC and AD of the frustum of cone to meet at O.

Proof:

The frustum of a cone can be viewed as a difference of two right circular cones OAB and OCD.

Let h_{1} and l_{1} be the height and slant height of cone OAB and h_{2} and l_{2} be the height and slant height of cone OCD respectively.

In ΔAPO and ΔDQO

ΔAPO = ΔDQO = 90° (Since both cones are right circular cones)

∠AOP = ∠DOQ (Common)

Therefore, ΔAPO ∼ ΔDQO ( A.A criterion of similarity)

AP/DQ = AO/DO = OP/OQ (Corresponding sides of similar triangles are proportional)

⇒ r_{1}/r_{2} = l_{1}/l_{2} = h_{1}/h_{2}

⇒ r_{1}/r_{2} = h_{1}/h_{2} or ⇒ r_{2}/r_{1} = h_{2}/h_{1}

Subtracting 1 from both sides

r_{1}/r_{2}- 1 = h_{1}/h_{2} - 1

(r_{1} - r_{2})r_{2} = (h_{1} - h_{2})h_{2}

(r_{1} - r_{2})r_{2} = h/h_{2} ....(i)

h_{2} = hr_{2}/(r_{1} - r_{2})

or ⇒ r_{2}/r_{1} = h_{2}/h_{1}

Subtracting 1 from both sides we get

r_{2}/r_{1} - 1 = h_{2}/h_{1} - 1

(r_{2} - r_{1})/r_{1} = (h_{2} - h_{1})/h_{1}

(r_{1} - r_{2})/r_{1} = (h_{1} - h_{2})/h1

(r_{1} - r_{2})/r_{1} = h/h_{1}

h_{1} = hr_{1}/(r_{1} - r_{2}) ....(ii)

Volume of frustum of cone = Volume of cone OAB - Volume of cone OCD

= 1/3 πr_{1}²h_{1} - 1/3 πr_{1}²h_{2}

= 1/3π [r_{1}²h_{1} - r_{2}²h_{2}]

= 1/3π [r_{1}² x hr_{1}/(r_{1} - r_{2}) - r_{1}² × hr_{2}(r_{1} - r_{2})] [Using (i) and (ii)]

= 1/3π [hr_{1}³/(r_{1} - r_{1}) - hr_{2}³/(r_{1} - r_{2})]

= 1/3πh [(r_{1}³ - r_{2}³)/(r_{1} - r_{2})]

= 1/3πh [(r_{1} - r_{2})(r_{1}² + r_{1}²r_{2}² + r_{2}²)/(r_{1} - r_{2})] [(a³ - b³) = (a - b)(a² + ab + b²)]

= 1/3πh (r_{1}² + r_{1}² + r_{1}²r_{1}²)

Volume of the cone = 1/3πh (r_{1}² + r_{2}² + r_{1}r_{2})

Hence proved

**Video Solution:**

## Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained

### NCERT Solutions for Class 10 Maths - Chapter 13 Exercise 13.5 Question 7 :

Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained

The formula for the volume of the frustum of a cone is (1/3)πh (r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})