Question

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?

Answer 1

Let r be the radius and h be the height of the cone,then volume=$\frac{1}{3}\pi r^{2}h$

By halving the radius and height remaining the same, new volume = $\frac{1}{3}\pi {\left(\frac{r}{2}\right)}^{2}h=\frac{1}{3}\pi \frac{r^2}{4}h=\frac{1}{4}\left(\frac{1}{3}\pi r^{2}h\right)$

$\therefore Ratiobetweenthevolumeoftwocones=\frac{1}{3}\pi r^{2}h:\frac{1}{4}\left(\frac{1}{3}\pi r^{2}h\right)=1:\frac{1}{4}=4:1\therefore Ratiobetweenthenewconeandtheoriginalcone=1:4$