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

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

To do:

We have to find the ratio of their volumes.

Solution:

Ratio of the heights of the two cones $=1:3$

Ratio of their radii $= 3: 1$

Let the radius of the first cone $(r_1)$ be $x$ and that of the second cone $(r_2)$ be $3x$/

Similarly,

Let the height of the first cone $(h_1)$ be $3y$ and that of the second cone be $(h_2)$

Therefore,

Volume of the first cone $=\frac{1}{3} \pi (r_1)^{2} h_1$

$=\frac{1}{3} \pi(x)^{2} \times 3 y$

$=\frac{1}{3} \pi x^{2} \times 3 y$

$=\pi x^{2} y$

Volume of the second cone $=\frac{1}{3} \pi(3 x)^{2} \times y$

$=\frac{1}{3} \pi \times 9 x^{2} y$

$=3 \pi x^{2} y$

Ratio of the volumes of the two cones $=\pi x^{2} y: 3 \pi x^{2} y$

$=1: 3$