Question.
A joker's cap is in the form of right circular cone of base radius $7 \mathrm{~cm}$ and height $24 \mathrm{~cm}$. Find the area of the sheet required to make 10 such caps. [ Assume $\left.\pi=\frac{22}{7}\right]$
Solution:
Radius $(r)$ of conical cap $=7 \mathrm{~cm}$
Height $(h)$ of conical cap $=24 \mathrm{~cm}$
Slant height ( $l$ ) of conical cap $=\sqrt{r^{2}+h^{2}}$
$=\left[\sqrt{(7)^{2}+(24)^{2}}\right] \mathrm{cm}=(\sqrt{625}) \mathrm{cm}=25 \mathrm{~cm}$
CSA of 1 conical cap $=\pi r l$
$=\left(\frac{22}{7} \times 7 \times 25\right) \mathrm{cm}^{2}=550 \mathrm{~cm}^{2}$
CSA of 10 such conical caps $=(10 \times 550) \mathrm{cm}^{2}=5500 \mathrm{~cm}^{2}$
Therefore, $5500 \mathrm{~cm}^{2}$ sheet will be required.
Radius $(r)$ of conical cap $=7 \mathrm{~cm}$
Height $(h)$ of conical cap $=24 \mathrm{~cm}$
Slant height ( $l$ ) of conical cap $=\sqrt{r^{2}+h^{2}}$
$=\left[\sqrt{(7)^{2}+(24)^{2}}\right] \mathrm{cm}=(\sqrt{625}) \mathrm{cm}=25 \mathrm{~cm}$
CSA of 1 conical cap $=\pi r l$
$=\left(\frac{22}{7} \times 7 \times 25\right) \mathrm{cm}^{2}=550 \mathrm{~cm}^{2}$
CSA of 10 such conical caps $=(10 \times 550) \mathrm{cm}^{2}=5500 \mathrm{~cm}^{2}$
Therefore, $5500 \mathrm{~cm}^{2}$ sheet will be required.