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

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To do:

We have to derive the formula for the volume of the frustum of a cone.

Solution:

Let $ABC$ be a cone.

From the cone the frustum $DECB$ is cut by a plane parallel to its base.

$r_1$ and $r_2$ be the radii of the frustum ends of the cone and $h$ be the height of the frustum.

In $\triangle ABG$ and $\triangle ADF$

$DF \| BG$

Therefore,

$\triangle ABG \sim \triangle ADF$

This implies,

$\frac{D F}{B G}=\frac{A F}{A G}=\frac{A D}{A B}$

$\frac{r_{2}}{r_{1}}=\frac{h_{1}-h}{h_{1}}=\frac{l_{1}-l}{l_{1}}$

$\frac{r_{2}}{r_{1}}=1-\frac{h}{h_{1}}=1-\frac{l}{l_{1}}$

$1-\frac{h}{\mathrm{h}_{1}}=\frac{r_{2}}{r_{1}}$

$\frac{h}{\mathrm{h}_{1}}=1-\frac{r_{2}}{r_{1}}$

$\frac{h}{\mathrm{h}_{1}}=\frac{r_1-r_{2}}{r_{1}}$

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

$h_1=\frac{r_1h}{r_1-r_2}$

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

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

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

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

$=\frac{\pi}{3}[(\frac{h r_{1}^{3}}{r_{1}-r_{2}})-r_{2}^{2}(\frac{h r_{1}-h r_{1}+h r_{2}}{r_{1}-r_{2}})]$

$=\frac{\pi}{3}[\frac{h r_{1}^{3}}{r_{1}-r_{2}}-\frac{h r_{2}^{3}}{r_{1}-r_{2}}]$

$=\frac{\pi}{3} h[\frac{r_{1}^{3}-r_{2}^{3}}{r_{1}-r_{2}}]$

$=\frac{\pi}{3} h[\frac{(r_{1}-r_{2})(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2})}{r_{1}-r_{2}}]$

$=\frac{\pi}{3} h(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2})$

Updated on 10-Oct-2022 13:47:39