**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?

**Solution:**

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.