#### Question

Derive the formula for the curved surface area and total surface area of the frustum of 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 - 1/l_1`

`l - l/l_1= r_2/r_1`

`l/l_1 =1-r_2/r_1 =(r_1-r_2)/r_1`

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

`l_1 = (r_1l)/(r_1-r_2)`

CSA of frustum DECB = CSA of cone ABC − CSA cone ADE

`= pir_1l_1 - pir_2(l_1-l)`

`=pir_1((lr_1)/(r_1-r_2))-pir_2[(r_1l)/(r_1-r_2)-l]`

`= (pir_1^2l)/(r_1-r_2) - pir_2((r_1l-r_1l+r_2l)/(r_1-r_2))`

`=(pir_1^2l)/(r_1-r_2)-(pir_2^2l)/(r_1-r_2)`

`= pil[(r_1^2-r_2^2)/(r_1-r_2)]`

CSA of frustum = Π(r_{1} + r_{2})l

Total surface area of frustum = CSA of frustum + Area of upper circular end + Area of lower circular end

`= pi(r_1+r_2)l+pir_2^2+pir_1^2`

`=pi[(r_1+r_2)l+r_1^2+r_2^2]`