Question

Derive the formula for the volume of the frustum of a cone, given to you in Section $13.5$ using the symbols as explained.

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 radil of the ends of the frustum of the cone

and $h$ be the height of the frustum of the cone

In $△ABG$ and $△ADR,DR\parallel BG$

$\therefore △ABG\sim △ADF$

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

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

$\frac{l-l}{l_1}=\frac{r_2}{r_1}$

$\frac{l}{l_1}=\frac{r_1-r_2}{r_1}$

$\frac{l_1}{l}=\frac{r_1}{r_1-r_2}$

$l_1=\frac{r_{1}l}{r_1-r_2}$

$CSA$ of frustum $DECB=CSA$ of cone $ABC-CSA$ of cone $ADE$

$=\pi r_1{l}_1-\pi r_2(l_1-l)$

$=\pi r_1\left(\frac{l{r}_1}{r_1-r_2}\right)-\pi r_2[\frac{r_{1}l}{r_1-r_2-l}-l]$

$=\frac{\pi {r_1}^{2}l}{r_1-r_2}-\pi r_2\left[\frac{r_{1}l-r_{1}l+r_{2}l}{r_1-r_2}\right]$

$\frac{\pi {r_1}^{2}l}{r_1-r_2}-\frac{\pi {r_2}^{2}l}{r_1-r_2}$

$=\frac{\pi r\left(\right(r_1{)}^2-{r_2}^2)}{r_1-r_2}$

CSA frustum $=\pi l(r_1+r_2)$

Total surface area $=CSA$ of frustum $+$ Area of upper conedar end $+$ Area of lower conedar end

$=\pi (r_1+r_2)l+\pi {r_2}^2+\pi {r_1}^2$

$=\pi \left(\right(r_1+r_2)l+{r_2}^2+{r_1}^2)$