Question

A solid cone of base radius 10 cm is cut into two parts through the midpoint of its height, by a plane parallel to its base. Find the ratio of the volumes of the two parts of the cone. [CBSE 2013]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{Radius}\mathrm{of}\mathrm{solid}\mathrm{cone},R=\mathrm{CP}=10\mathrm{cm}, \\ \mathrm{Let}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{solid}\mathrm{cone}\mathrm{be},\mathrm{AP}=H, \\ \mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{cone},\mathrm{QD}=r\mathrm{and} \\ \mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{cone}\mathrm{be},\mathrm{AQ}=h. \\ \mathrm{Also},\mathrm{AQ}=\frac{\mathrm{AP}}{2}\mathrm{i}.\mathrm{e}.h=\frac{H}{2}\mathrm{or}H=2h.....\left(i\right) \\ \mathrm{Now},\mathrm{in}∆\mathrm{AQD}\mathrm{and}∆\mathrm{APC}, \\ \angle \mathrm{QAD}=\angle \mathrm{PAC}\left(\mathrm{Common}\mathrm{angle}\right) \\ \angle \mathrm{AQD}=\angle \mathrm{APC}=90° \\ \mathrm{So},\mathrm{by}\mathrm{AA}\mathrm{criteria} \\ ∆\mathrm{AQD}~\mathrm{APC} \\ \Rightarrow \frac{\mathrm{AQ}}{\mathrm{AP}}=\frac{\mathrm{QD}}{\mathrm{PC}} \\ \Rightarrow \frac{h}{H}=\frac{r}{R} \\ \Rightarrow \frac{h}{2h}=\frac{r}{R}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ \Rightarrow \frac{1}{2}=\frac{r}{R} \\ \Rightarrow R=2r.....\left(\mathrm{ii}\right) \\ \mathrm{As}, \\ \mathrm{Volume}\mathrm{of}\mathrm{smaller}\mathrm{cone}=\frac{1}{3}\mathrm{\pi }{r}^{2}h \\ \mathrm{And}, \\ \mathrm{Volume}\mathrm{of}\mathrm{solid}\mathrm{cone}=\frac{1}{3}\mathrm{\pi }{R}^{2}H \\ =\frac{1}{3}\mathrm{\pi }{\left(2r\right)}^2\times \left(2h\right)\left[\mathrm{Using}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\right] \\ =\frac{8}{3}\mathrm{\pi }{r}^{2}h \\ \mathrm{So}, \\ \mathrm{Volume}\mathrm{of}\mathrm{frustum}=\mathrm{Volume}\mathrm{of}\mathrm{solid}\mathrm{cone}-\mathrm{Volume}\mathrm{of}\mathrm{smaller}\mathrm{cone} \\ =\frac{8}{3}\mathrm{\pi }{r}^{2}h-\frac{1}{3}\mathrm{\pi }{r}^{2}h \\ =\frac{7}{3}\mathrm{\pi }{r}^{2}h \\ \mathrm{Now},\mathrm{the}\mathrm{ratio}\mathrm{of}\mathrm{the}\mathrm{volumes}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{parts}=\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{cone}}{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{frustum}} \\ =\frac{\left({\displaystyle \frac{1}{3}}\mathrm{\pi }{r}^{2}h\right)}{\left({\displaystyle \frac{7}{3}}\mathrm{\pi }{r}^{2}h\right)} \\ =\frac{1}{7} \\ =1:7 \end{gathered}$$

So, the ratio of the volume of the two parts of the cone is 1 : 7.