Question

If $p$ is the perimeter of an equilateral triangle inscribed in a circle, the area of the circle is:

• A. $\frac{\pi p^2}{3}$
• B. $\frac{\pi p^2}{9}$
• C. $\frac{\pi p^2}{27}$
• D. $\frac{\pi p^2}{81}$
• E. $\frac{\pi p^2\sqrt{3}}{27}$

Answer 1

The correct option is C $\frac{\pi p^2}{27}$

Let $△ABC$ be the equilateral $△$ inscribed in a circle of radius 'r'.

$△ABC$ is an equilateral $△$, therefore each angle be $60°$. AD, BE and CF are the perpendicular on side BC, AC and AB respectively

$\therefore AE=EC=\frac{1}{2}AC$

Perimeter of $△ABC=p$ then each side is $\frac{p}{3}$

In $△OAE,\cos 30°=\frac{AE}{OA}=\frac{p/6}{r}\Rightarrow r=\frac{p}{6\cos 30°}=\frac{2p}{6\sqrt{3}}=\frac{p}{3\sqrt{3}}$

Area of circle$=\pi r^2=\pi \times {\left(\frac{p}{3\sqrt{3}}\right)}^2=\frac{\pi p^2}{27}$