Question

Two similiar cones have volumes 12$\pi$ cu. units and 96 $\pi$ cu. units. If the curved surface area of the smaller cone is 15 $\pi$ sq. units, what is the curved surface area of the larger one

Answer 1

Let the radius of cone one is $r$ and cone two is $R$ and height of cone one is $h$ and cone two is $H$

We know that volume of cone =$\frac{1}{3}\pi r^{2}h$

Taking ratios of volume of two cones

$\therefore \frac{12\pi }{96\pi }=\frac{\pi r^3}{\pi R^3}$

$\Rightarrow \frac{1}{8}=\frac{r^3}{R^3}$

$\Rightarrow \frac{1}{4}=\frac{r^2}{R^2}$

$\frac{CA{S}_1}{CA{S}_2}=\frac{\pi rl}{\pi RL}=\frac{r^2}{R^2}$ (the cone are similar )

$\therefore \frac{15\pi }{CA{S}_2}=\frac{1}{4}$

$\Rightarrow CA{S}_2=4\times 15\pi =60\pi$ sq unit