Question

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

Answer 1

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 $\angle ABG$ and $\angle ADF$, $DF\parallel BG$
Therefore, $\angle ABG\sim \angle ADF$
$\frac{DF}{BG}=\frac{AF}{AG}=\frac{AD}{AB}$

$\Rightarrow \frac{r_2}{r_1}=\frac{h_1-h}{h_1}=\frac{l_1-l}{1_1}$

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

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

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

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

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

Volume of frustum of cone = Volume of cone ABC- Volume of cone ADE

$=\frac{1}{3}\pi r_1^2{h}_1-\frac{1}{3}\pi r_2^2(h_1-h)$

$=\frac{\pi }{3}[r_1^2{h}_1-r_2^2(h_1-h\left)\right]$

$=\frac{\pi }{3}[r_1^2\left(\frac{r_{1}h}{r_1-r_2}\right)-r_2^2(\frac{r_{1}h}{r_1-r_2}-h)]$

$=\frac{pi}{3}[\left(\frac{r_1^{3}h}{r_1-r_2}\right)-r_2^2(h{r}_1-h{r}_1+h{r}_2)]$

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

$=\frac{\pi }{3}h\left[\frac{r_1^3-r_2^3}{r_1-r_2}\right]$

$=\frac{\pi }{3}h\left[\frac{(r_1-r_2)\left(r_1^2\right)+r_2^2+r_1{r}_2}{r_1-r_2}\right]$

$=\frac{1}{3}\pi h[r_1^2+r_2^2+r_1{r}_2]$