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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.
Given:
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 $=\pi x^{2} y: 3 \pi x^{2} y$
$=1: 3$