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A regular hexagon is inscribed in a circle. If the area of hexagon is $ 24 \sqrt{3} \mathrm{~cm}^{2} $, find the area of the circle. (Use $ \pi=3.14 $ )
Given:
A regular hexagon is inscribed in a circle.
The area of the hexagon is \( 24 \sqrt{3} \mathrm{~cm}^{2} \).
To do:
We have to find the area of the circle.
Solution:
Let $ABCDEF$ be the regular hexagon and $O$ is the centre.
We get six equal equilateral triangles by joining the vertices of the hexagon with $O$.
Let the radius of the circle $=$ Length of the side of the equilateral triangle $AOB=r$
Area of the hexagon $=24\sqrt{3}\ cm^2$
This implies,
$6\times\frac{\sqrt{3}}{4} r^{2}=24\sqrt{3}$
$\Rightarrow r^{2}=\frac{24\sqrt{3}\times 4}{6\sqrt{3}}$
$\Rightarrow r^{2}=16$
Area of the circle $=\pi r^{2}$
$=3.14 \times 16$
$=50.24 \mathrm{~cm}^{2}$
The area of the circle is $50.24 \mathrm{~cm}^{2}$.
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