# The volume of a right circular cone is $9856 \mathrm{~cm}^{3}$.

Question.

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

$\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$

Solution:

(i) Radius of cone $=\left(\frac{28}{2}\right) \mathrm{cm}=14 \mathrm{~cm}$

Let the height of the cone be h.

Volume of cone $=9856 \mathrm{~cm}^{3}$

$\Rightarrow \frac{1}{3} \pi r^{2} h=9856 \mathrm{~cm}^{3}$

$\Rightarrow\left[\frac{1}{3} \times \frac{22}{7} \times(14)^{2} \times h\right] \mathrm{cm}^{2}=9856 \mathrm{~cm}^{3}$

$h=48 \mathrm{~cm}$

Therefore, the height of the cone is $48 \mathrm{~cm}$.

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

$=\left[\sqrt{(14)^{2}+(48)^{2}}\right] \mathrm{cm}$

$=[\sqrt{196+2304}] \mathrm{cm}$

$=50 \mathrm{~cm}$

Therefore, the slant height of the cone is $50 \mathrm{~cm}$.

(iii) CSA of cone $=\pi r$

$=\left(\frac{22}{7} \times 14 \times 50\right) \mathrm{cm}^{2}$

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

Therefore, the curved surface area of the cone is $2200 \mathrm{~cm}^{2}$.