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# A joker’s cap is in the form of a right circular cone of base radius $7cm$ and height $24cm$. Find the area of the sheet required to make $10$ such caps. (Assume $\mathrm{\pi }=\frac{22}{7}$)

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Solution

## Step 1: Find the slant height of the conical capCurved surface area of cone is $\mathrm{\pi rl}$.Since, Radius of conical cap ($r$) $=7cm$ Height of conical cap ($h$) $=24cm$Area of the sheet required to make $1$cap$=\mathrm{\pi rl}$ (where $l$ is the slant height and $l=\sqrt{{r}^{2}+{h}^{2}}$)Now, $⇒l=\sqrt{{7}^{2}+{24}^{2}}\phantom{\rule{0ex}{0ex}}⇒l=\sqrt{49+576}\phantom{\rule{0ex}{0ex}}⇒l=\sqrt{625}\phantom{\rule{0ex}{0ex}}⇒l=25$Step 2: Find the area of sheet require to make $10$ such capsArea of the sheet required to make $1$cap $=\mathrm{\pi rl}$ $=\frac{22}{7}×7×25\phantom{\rule{0ex}{0ex}}=22×25\phantom{\rule{0ex}{0ex}}=550c{m}^{2}$So, area of the sheet required to make $10$ such caps$=10×550=5500c{m}^{2}$Hence, area of the required sheet is $5500c{m}^{2}$  Suggest Corrections  1      Similar questions