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

Answer :

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 ∆ ABG ∆ ADF,

Here, DF||BG

So, $$∆ \ ABG \ \sim \ ∆ \ ADF$$

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

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

$$\Rightarrow \frac{r_2}{r_1} \ = \ 1 \ - \ \frac{h}{h_1} \ = \ 1 \ - \ \frac{l}{l_1}$$

$$\Rightarrow 1 \ - \ \frac{h}{h_1} \ = \ \frac{r_2}{r_1}$$

$$\Rightarrow \ \frac{h}{h_1} \ = \ \frac{r_1 \ - \ r}{r_1}$$

$$\Rightarrow h_1 \ = \ \frac{r_1 h}{r_1 \ - \ r_2}$$

The total volume of frustum of the cone will be = Volume of cone ABC – Volume of cone ADE

$$= \ \frac{1}{3} \pi (r_1)^2 h_1 \ – \ \frac{1}{3} \pi (r_2)^2 (h_1 \ – \ h)$$

$$= \ ( \frac{ \pi}{3})[(r_1)^2 h_1 \ - \ (r_2)^2 (h_1 \ - \ h)]$$

$$= \ ( \frac{ \pi}{3})[ ( \frac{(r_1)^3 h}{r_1 \ - \ r_2} ) \ - \ (r_2)^2 ( \frac{r_1 h}{r_1 \ - \ r_2} \ - \ h)]$$

$$= \ ( \frac{ \pi}{3})[ ( \frac{(r_1)^3 h}{r_1 \ - \ r_2} ) \ - \ (r_2)^2 \ ( \frac{hr_1 \ - \ hr_1 \ + \ hr_2}{r_1 \ - \ r_2} ) ]$$

$$= \ \frac{ \pi }{3} h [ \frac{(r_1)^3 \ - \ (r_2)^3}{r_1 \ - \ r_2} ]$$

$$= \ \frac{ \pi }{3} h [ \frac{(r_1 \ - \ r_2)(r_1 ^2 \ + \ r_2^2 \ + \ r_1 r_2}{r_1 \ - \ r_2} ]$$

$$\therefore$$ Volume of frustum of the cone $$= \ \frac{1}{3} \pi h \ [r_1^2 \ + \ r_2^2 \ + \ r_1 r_2 \ ]$$