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Answer :

Let ABC be a cone. From the cone the frustum DECB is cut by a plane parallel to its base. Here, r_{1} and r_{2} are the radii of the frustum ends of the cone and h be the frustum height.

Now, consider the \(\triangle \) ABG \(\triangle \) ADF,

Here, DF||BG

So, \( \triangle \ ABG \ \sim \ \triangle \ ADF \)

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

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

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

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

\( = \pi (r_1 \ + \ r_2)l \ + \ \pi (r_2)^2 \ + \ \pi (r_1)^2 \)

\(\therefore \) Surface area of frustum \(= \ \pi [r_1 \ + \ r_2)l \ + \ (r_1)^2 \ + \ (r_2)^2] \)

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