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?
Let the radius of the cone be rx and height be hx
Then the volume of cone
$\mathrm{v}_{\mathrm{x}}=\frac{1}{3} \pi \mathrm{r}_{\mathrm{X}}^{2} \mathrm{~h}_{\mathrm{x}}$
Now,
Radius of the reduced cone = rx/2
Therefore volume of reduced cone vy
$=\frac{1}{3} \pi\left(\frac{r}{2}\right)^{2} * h_{\mathrm{x}}$
$=\frac{1}{12} * \pi * \mathrm{r}_{\mathrm{x}} \mathrm{x}^{2} * \mathrm{~h}_{\mathrm{x}}$
$\therefore \frac{\mathrm{V}_{\mathrm{y}}}{\mathrm{V}_{\mathrm{x}}}=\frac{\frac{1}{12} \pi \mathrm{r}^{2} \mathrm{~h}}{\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}}$
$=\frac{3}{12}=\frac{1}{4}$
Therefore the ratio between the volumes of the reduced and the original cone is 1: 4.