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 $\left.\pi=\frac{22}{7}\right]$
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 $2200 \mathrm{~cm}^{2}$.
(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 $2200 \mathrm{~cm}^{2}$.