A joker's cap is in the form of a 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.

AcademicMathematicsNCERTClass 9

Given:

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

To do:

We have to find the area of the sheet required to make 10 such caps.

Solution:

The radius of the base of the conical cap $(r) = 7\ cm$

Height of the cap $(h) = 24\ cm$

Therefore,

Slant height of the cap $=\sqrt{r^{2}+h^{2}}$

$=\sqrt{(7)^{2}+(24)^{2}}$

$=\sqrt{49+576}$

$=\sqrt{625}$

$=25 \mathrm{~cm}$

Area of the slant surface $=\pi r l$

$=\frac{22}{7} \times 7 \times 25$

$=550 \mathrm{~cm}^{2}$

Area of 10 such caps $=550 \times 10$

$=5500 \mathrm{~cm}^{2}$

Therefore, the area of the sheet required to make 10 such caps is $5500 \mathrm{~cm}^{2}$.

raja
Updated on 10-Oct-2022 13:46:36

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