Let ABC be a cone. From the cone the frustum DECB is cut by a plane parallel to its base. Here, r₁ and r₂ are the radii of the frustum ends of the cone and h be the frustum height.

Now, consider the ΔABG and ΔADF,

DF||BG

ΔABG ~ ΔADF

\(\frac{DF}{BG}\) = \(\frac{AF}{AG}\) = \(\frac{AD}{AB}\)

\(\frac{r_2}{r_1}\) = \(\frac{h_1 - h}{h_1}\) = \(\frac{l_1 - l}{l_1}\)

\(\frac{ l}{l_1}\) = \(1 - \frac{r_2}{r_1}\) = \(\frac{r_1 - r_2}{r_1}\)

Now, by rearranging we get,

\(l_1 = \frac{r_1 l}{r_1 - r_2}\)

The total surface area of frustum will be equal to the total CSA of frustum + the area of upper circular end + area of the lower circular end

= π(r₁ + r₂)l + πr₂² + πr₁²

∴ Surface area of frustum = π[(r₁ + r₂)l + r₁² + r₂²]