Question

A cone of radius 4 cm is divided into two parts by drawing a plane through the mid point of its axis and parallel to its base . Compare the volumes of two parts.

Answer 1

Radius of the cone, R = 4 cm
Let the height be H.
Since the plane divides the cone into two parts through the mid point so, a small cone and a frustum will be formed.

$OC=CA=\frac{H}{2}$

Let the radius of the smaller cone be r cm.

In ∆OCD and ∆OAB,

∠OCD = ∠OAB (90°)

∠COD = ∠AOB (Common)

∴∆OCD ∼ ∆OAB (AA Similarly criterion)
$$\begin{gathered} \Rightarrow \frac{OA}{OC}=\frac{AB}{CD}=\frac{OB}{OD}\left(correspondingsidesareproportional\right) \\ \Rightarrow \frac{H}{{\displaystyle \frac{H}{2}}}=\frac{4}{r} \\ \Rightarrow r=2cm \end{gathered}$$

Now volume of the smaller cone = $\frac{1}{3}\mathrm{\pi }{\left(\mathrm{CD}\right)}^2\times \mathrm{OC}=\frac{{\displaystyle 1}}{{\displaystyle 3}}\mathrm{\pi }{\left(2\right)}^2\times \frac{\mathrm{H}}{2}=\frac{2\mathrm{\pi H}}{3}\mathrm{cm}$
Height of the frustum of the cone = $\frac{H}{2}$

Volume of frustum of cone = $\frac{1}{3}\mathrm{\pi h}\left[{\mathrm{r}}_1^2+{\mathrm{r}}_1{\mathrm{r}}_2+{\mathrm{r}}_2^2\right]$
$$\begin{gathered} =\frac{1}{3}\mathrm{\pi }\left(\frac{\mathrm{H}}{2}\right)\left[{\left(4\right)}^2+{\left(2\right)}^2+4\times 2\right] \\ =\frac{14\mathrm{\pi H}}{3}{\mathrm{cm}}^3 \end{gathered}$$
$$\begin{gathered} \frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{cone}}{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{frustum}\mathrm{of}\mathrm{the}\mathrm{cone}}=\frac{\frac{2\mathrm{\pi H}}{3}}{{\displaystyle \frac{14\mathrm{\pi H}}{3}}} \\ \Rightarrow \frac{{\displaystyle \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{cone}}}{{\displaystyle \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{frustum}\mathrm{of}\mathrm{the}\mathrm{cone}}}=\frac{1}{7} \end{gathered}$$
Hence, the ratio of the volumes of the two parts will be 1 : 7.