The volume of a right circular cone is $ 9856 \mathrm{~cm}^{3} $. If the diameter of the base is $ 28 \mathrm{~cm} $, find
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone.

 Given:

The volume of a right circular cone is $9856\ cm^3$.

The diameter of the base is $28\ cm$.

To do:

We have to find

(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone.

Solution:

 Volume of the right circular cone $= 9856\ cm^3$

Diameter of the base of the cone $= 28\ cm$

This implies,

Radius of the cone $(r)=\frac{28}{2}$

$=14 \mathrm{~cm}$

We know that,

Volume of a cone of height $h$ and base radius $r$ is $\frac{1}{3}\pi r^2h$

Therefore,

Height of the cone $(h)=\frac{\text { Volume } \times 3}{\pi r^{2}}$

$=\frac{9856 \times 3 \times 7}{22 \times 14 \times 14}$

$=48\ cm$

(ii) Slant height of the cone $(l)=\sqrt{r^{2}+h^{2}}$

$=\sqrt{(14)^{2}+(48)^{2}}$

$=\sqrt{196+2304}$

$=\sqrt{2500}$

$=50 \mathrm{~cm}$

(iii) Curved surface area of the cone $=\pi r l$

$=\frac{22}{7} \times 14 \times 50$

$=2200 \mathrm{~cm}^{2}$