Derive the formula for the volume of the frustum of a cone.

#### Solution

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base.

Let *r*_{1} and *r*_{2} be the radii of the ends of the frustum of the cone and *h* be the height of the frustum of the cone.

In ΔABG and ΔADF, DF||BG

∴ ΔABG ∼ ΔADF

DF/BG = AF/AG = AD/AB

`r_2/r_1 = (h_1-h)/h_1 = (l_1-l)/l_1`

`r_2/r_1 =1 -h/h_1 =1 -l/l_1`

`1-h/h_1= r_2/r_1`

`h/h_1 =1 -r_2/r_1 = (r_1-r_2)/r_1`

`h_1/h= r_1/(r_1-r_2)`

`h_1 = (r_1h)/(r_1-r_2)`

Volume of frustum of cone = Volume of cone ABC − Volume of cone ADE

`=1/3pir_1^2h_1 - 1/3pir_2^2(h_1-h)`

`=pi/3[r_1^2h_1-r_2^2(h_1-h)]`

`=pi/3[r_1^2((hr_1)/(r_1-r_2))-r_2^2((hr_1)/(r_1-r_2)-h)]`

`=pi/3[((hr_1^3)/(r_1-r_2))-r_2^2((hr_1-hr_1+hr_2)/(r_1-r_2))]`

`=pi/3[(hr_1^3)/(r_1-r_2)-(hr_2^3)/(r_1-r_2)]`

`=pi/3h[(r_1^3-r_2^3)/(r_1-r_2)]`

`=pi/3h[((r_1-r_2)(r_1^2+r_2^2+r_1r_2))/(r_1-r_2)]`

`= 1/3pih[r_1^2+r_2^2+r_1r_2]`