Question

A circle is inscribed in an equilateral triangle ABC with side $12$ cm, touching its sides. Find the radius of the inscribed circle and the area of the shaded part.

Answer 1

$\Rightarrow$ Here, $AB=BC=AC=12cm$

$\Rightarrow$ Let $OP=OR=OQ=r$

$\Rightarrow$ We have $O$ as the incenter and $OP,OQ$ and $OR$ are equal.

$\Rightarrow$ $ar\left(△ABC\right)=ar\left(△OAB\right)+ar\left(△OBC\right)+ar\left(△OCA\right)$

$\frac{\sqrt{3}}{4}\times (side{)}^2=(\frac{1}{2}\times OP\times AB)+(\frac{1}{2}\times OQ\times BC)+(\frac{1}{2}\times OR\times AC)$

$\Rightarrow$ $\frac{\sqrt{3}}{4}\times (12{)}^2=(\frac{1}{2}\times r\times 12)+(\frac{1}{2}\times r\times 12)+(\frac{1}{2}\times r\times 12)$

$\Rightarrow$ $\frac{\sqrt{3}}{4}\times (12{)}^2=3(\frac{1}{2}\times 12\times r)$

$\therefore$ $r=\frac{36\sqrt{3}}{18}$

$\therefore$ $r=2\sqrt{3}cm$

$\Rightarrow$ Area of the shaded region = Area of $△ABC$ - Area of circle.

$\Rightarrow$ Area of the shaded region $=\frac{\sqrt{3}}{4}\times (12{)}^2-\frac{22}{7}\times (2\sqrt{3}{)}^2$

$\Rightarrow$ Area of the shaded region $=(62.35-37.71)c{m}^2=24.64c{m}^2$