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

(N/A) 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 $\triangle ABG$ and $\triangle ADF,$ $DF \parallel BG.$
$\therefore \triangle ABG \sim \triangle ADF.$
$\frac{DF}{BG} = \frac{AF}{AG} = \frac{AD}{AB}.$
$\frac{r_{2}}{r_{1}} = \frac{h_{1}-h}{h_{1}} = 1 - \frac{h}{h_{1}}.$
$\frac{h}{h_{1}} = 1 - \frac{r_{2}}{r_{1}} = \frac{r_{1}-r_{2}}{r_{1}}.$
$h_{1} = \frac{r_{1}h}{r_{1}-r_{2}}.$
Volume of frustum of cone $=$ Volume of cone $ABC$ $-$ Volume of cone $ADE.$
$= \frac{1}{3} \pi r_{1}^{2} h_{1} - \frac{1}{3} \pi r_{2}^{2}(h_{1}-h).$
$= \frac{\pi}{3} [r_{1}^{2} h_{1} - r_{2}^{2}(h_{1}-h)].$
$= \frac{\pi}{3} [r_{1}^{2} (\frac{r_{1}h}{r_{1}-r_{2}}) - r_{2}^{2} (\frac{r_{1}h}{r_{1}-r_{2}} - h)].$
$= \frac{\pi}{3} [\frac{r_{1}^{3}h}{r_{1}-r_{2}} - r_{2}^{2} (\frac{r_{1}h - r_{1}h + r_{2}h}{r_{1}-r_{2}})].$
$= \frac{\pi}{3} [\frac{r_{1}^{3}h - r_{2}^{3}h}{r_{1}-r_{2}}].$
$= \frac{\pi h}{3} [\frac{r_{1}^{3} - r_{2}^{3}}{r_{1}-r_{2}}].$
$= \frac{\pi h}{3} [\frac{(r_{1}-r_{2})(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})}{r_{1}-r_{2}}].$
$= \frac{1}{3} \pi h (r_{1}^{2} + r_{2}^{2} + r_{1}r_{2}).$