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If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone?
Given:
The radius of the base of a cone is halved, keeping the height same.
To do:
We have to find the ratio of the volume of the reduced cone to that of the original cone.
Solution:
Let $r$ be the radius and $h$ be the height of the original cone.
This implies,
Volume of the original cone $=\frac{1}{3} \pi r^2h$
By halving the radius and keeping the height same,
Volume of the new cone $=\frac{1}{3} \pi (\frac{r}{2})^{2} h$
$=\frac{1}{3} \pi \frac{r^{2}}{2} h$
$=\frac{1}{4}(\frac{1}{3} \pi r^{2} h)$
Therefore,
Ratio between the two cones $=\frac{1}{3} \pi r^{2} h: \frac{1}{4}(\frac{1}{3} \pi r^{2} h)$
$=1: \frac{1}{4}$
$=4: 1$
The ratio of the volume of the reduced cone to that of the original cone is $1: 4$.
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