Question

Two cones have their heights in the ratio $1:3$ and the radii of their bases in the ratio $3:1$. Show that their volumes are in the ratio $3:1$.

Answer 1

Step 1: Find the volume of both the cones.

Let the height of first cone be $h$ and the height of second cone be $3h$.

Similarly, let the base radius of first cone be $3r$ and the base radius of second cone be $r$.

Formula used: Volume of the cone $=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

Volume $\left(V_1\right)$ of the first cone having height $h$ and base radius $3r$ $=\frac{1}{3}\mathrm{\pi }\times {\left(3\mathrm{r}\right)}^2\times \mathrm{h}$

$=\frac{1}{3}\mathrm{\pi }\times 9r^2\times \mathrm{h}$

$=3{\mathrm{\pi r}}^2\mathrm{h}$

Volume $\left(V_2\right)$ of the second cone having height $3h$ and base radius $r$ $=\frac{1}{3}\mathrm{\pi }\times r^2\times 3\mathrm{h}$

$=\mathrm{\pi }{r}^2\mathrm{h}$

Step 2: Find the ratio of volumes of first and second cone.

$\Rightarrow \frac{V_1}{V_2}=\frac{3{\mathrm{\pi r}}^2\mathrm{h}}{{\mathrm{\pi r}}^2\mathrm{h}}$

$\Rightarrow V_1:V_2=3:1$

Hence, proved that the ratio of volumes of first and second cone is $3:1$.