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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 $=\pi r l$
$=\frac{22}{7} \times 14 \times 50$
$=2200 \mathrm{~cm}^{2}$