A solid right circular cone is cut into two parts at the middle of its height by a plane parallel to its base. Find the ratio of the volume of the smaller cone to the whole cone.

Given: A solid right circular cone is cut into two parts at the middle of its height by a plane parallel to its base.

To do: To find the ratio of the volume of the smaller cone to the whole cone.

Solution:

Let the radius of the given cone be $r$ and $h$.

$AD=h$ and $DC=r$

$\therefore AG=\frac{h}{2}$

In $\vartriangle AGF$ and $\vartriangle ADC$

$\angle AFG=\angle ACD$

$\because EF||BC$

$\therefore \angle AGE=\angle ADC=90^o$

$\vartriangle AGF\sim \vartriangle ADC$

$\Rightarrow \frac{AG}{AD}=\frac{GF}{DC}$

$\Rightarrow \frac{\frac{h}{2}}{h}=\frac{GF}{DC}$

$\Rightarrow \frac{GF}{DC}=\frac{1}{2}$

$\Rightarrow \frac{GF}{r}=\frac{1}{2}$

$\Rightarrow GF=\frac{r}{2}$

$\Rightarrow \frac{Volume\ of\ the\ smaller\ cone}{Volume\ of\ whole\ cone}=\frac{\frac{1}{3}\times \pi\ ( \frac{r}{2})^2\times( \frac{h}{2})}{\frac{1}{3}\times \pi r^2h}$

$=\frac{1}{8}$

$=1:8$

Thus, the ratio of the volume of the smaller cone to the whole cone is $1:8$.

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Updated on: 10-Oct-2022

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