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In the below figure, an equilateral triangle $ A B C $ of side $ 6 \mathrm{~cm} $ has been inscribed in a circle. Find the area of the shaded region. (Take $ \pi=3.14) $"
Given:
An equilateral triangle \( A B C \) of side \( 6 \mathrm{~cm} \) has been inscribed in a circle.
To do:
We have to find the area of the shaded region.
Solution:
Length of the side of equilateral triangle $\mathrm{ABC}=6 \mathrm{~cm}$
Area of the equilateral triangle $=\frac{\sqrt{3}}{4} \times(6)^{2}$
$=\frac{\sqrt{3}}{4} \times 36$
$=9 \sqrt{3} \mathrm{~cm}^{2}$
$=9 \times 1.732$
$=15.588 \mathrm{~cm}^{2}$
Draw $AD \perp BC$ which passes through the centre of the circle at $O$.
This implies,
$\mathrm{OA}=\frac{2}{3} \mathrm{AD}$ ($O$ is the centroid)
$=\frac{2}{3} \times \frac{\sqrt{3}}{2} \times 6$
$=2 \sqrt{3}\ cm$
Radius of the circle $r=2 \sqrt{3} \mathrm{~cm}$
Therefore,
Area of the circle $=\pi r^{2}$
$=\frac{22}{7} \times(2 \sqrt{3})^{2}$
$=\frac{22}{7} \times 12$
$=\frac{264}{7}$
$=37.714 \mathrm{~cm}^{2}$
Area of the shaded portion $=37.714-15.588$
$=22.126 \mathrm{~cm}^{2}$
The area of the shaded region is $22.126\ cm^2$.