Question

What is the area of the circle inscribed in an equilateral triangle of side $24$ cm?

• A. $24\pi c{m}^2$
• B. $36\pi c{m}^2$
• C. $48\pi c{m}^2$
• D. $18\pi c{m}^2$

Answer 1

The correct option is C $48\pi c{m}^2$

$△ABC$ is an equilateral triangle and a circle is inscribe in it with centre O. The angle bisectors of $\angle B$ and $\angle C$ meet at point O. $\therefore$ $\angle OBD=\angle OCD={30}^0$ $OD$ is the radius and $BC$ is a tangent to the incircle.

$\Rightarrow OD\perp BC$ hence $△OBD$ is a right angled triangle.

$\angle BOD={60}^0$, therefore $OD=\frac{1}{\sqrt{3}}BD=\frac{1}{2\sqrt{3}}BC=\frac{24}{2\sqrt{3}}=\frac{12}{\sqrt{3}}cm$

$\therefore$ Area of incircle $=\pi r^2=\pi \times {\left(\frac{12}{\sqrt{3}}\right)}^2=\frac{144\pi }{3}=48\pi {cm}^2$