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A circle is inscribed in an equilateral triangle $ A B C $ is side $ 12 \mathrm{~cm} $, touching its sides. Find the radius of the inscribed circle and the area of the shaded part."
Given:
A circle is inscribed in an equilateral triangle \( A B C \) of side \( 12 \mathrm{~cm} \), touching its sides.
To do:
We have to find the radius of the inscribed circle and the area of the shaded part.
Solution:
Length of each side of the equilateral triangle $ABC (a)= 12\ cm$
This implies,
Area of the equilateral triangle $=\frac{\sqrt{3}}{4} a^{2}$
$=\frac{\sqrt{3}}{4}(12)^{2}$
$=\frac{1.732 \times 12 \times 12}{4}$
$=62.352 \mathrm{~cm}^{2}$
Draw $AD \perp BC$.
This implies,
$OD=\frac{1}{3} (AD)$ [Since, $O$ is the centroid of the equilateral triangle]
$=\frac{1}{3} \times \frac{\sqrt{3}}{2} \times 12$
$=\frac{\sqrt{3}}{6} \times 12$
$=2 \sqrt{3}$
Radius of the incircle $r=OD$
$=2 \sqrt{3} \mathrm{~cm}$
Area of incircle $=\pi r^{2}$
$=\frac{22}{7} \times(2 \sqrt{3})^{2}$
$=\frac{22}{7} \times 12$
$=\frac{264}{7}$
$=37.714 \mathrm{~cm}^{2}$
Therefore,
Area of the shaded region $=62.352-37.714$
$=24.638 \mathrm{~cm}^{2}$
The radius of the inscribed circle is $2\sqrt3\ cm$ and the area of the shaded part is $24.638\ cm^2$.