Question

The radius and height of a cone are in the ratio of $4:3$. Find the curved surface area if the volume of cone is $269\frac{1}{2}c{m}^3$

Answer 1

Given let 'r' be radius and 'h' be height of cone

then $[\frac{r}{h}=\frac{4}{3}]$ ...(1)

and volume of cone $=269\frac{1}{2}c{m}^3$

$\Rightarrow$ now volume$=\frac{1}{3}\pi r^{2}h=269\frac{1}{2}$

from (1) $[h=\frac{3}{4}r]$

$\frac{1}{3}\pi r^2\times \frac{3r}{4}=\frac{539}{2}$

$r^3=\frac{4\times 539}{2\times \pi }=343.033343$

$[r=7cm]$ now $h=\frac{3}{4}\times 7=\frac{21}{4}cm$

now CSA of cone$\pi rl=\pi r\sqrt{r^2+h^2}$

$=\pi r\sqrt{r^2+h^2}=\mathrm{3.14.7}\sqrt{49+(\frac{21}{4}{)}^2}$

$[CSA=27.44c{m}^2]$

Hence curved surface area of cone is$27.44c{m}^2$