Question

In the given figure O is the centre of the circle chord AB = OA and the area of triangle AOB is $4\sqrt{3}c{m}^2$ Then area of circle is

• A. $16\pi c{m}^2$
• B. $4\pi c{m}^2$
• C. $24\sqrt{3}c{m}^2$
• D. $24\pi c{m}^2$

Answer 1

The correct option is A $16\pi c{m}^2$

Let radius of circle $=r$

then $OA=OB=r$

Given $AB=OA$

So $AB=OA=OB=r$

It means Triangle OAB is an equilateral triangle.

Area of an Equilateral triangle OAB $=\frac{\sqrt{3}}{4}{r}^2$

Given Area of triangle OAB $=4\sqrt{3}c{m}^2$

So, $\frac{\sqrt{3}}{4}{r}^2=4\sqrt{3}$

$r^2=4\sqrt{3}\frac{4}{\sqrt{3}}=16c{m}^2$

Now, Area of Circle $=\pi r^2=\pi \left(16\right)=16\pi c{m}^2$