A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24 cm.
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.
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