Question

The angle of elevation of the top of a vertical pole when observed from each vertex of a regular hexagon is$\frac{\pi }{3}$. If the area of the circle circumscribing the hexagon be $A{m}^2,$ then the area of the hexagon is

• A. $\frac{3\sqrt{3}}{8}A{m}^2$
• B. $\frac{\sqrt{3}}{\pi }A{m}^2$
• C. $\frac{3\sqrt{3}}{4\pi }A{m}^2$
• D. $\frac{3\sqrt{3}}{2\pi }A{m}^2$

Answer 1

The correct option is D $\frac{3\sqrt{3}}{2\pi }A{m}^2$

Let a be the side of the regular hexagon
Now, for the equilateral triangle FOA, we have
$OF=OA=AF=a$
Hence area of the circle $A=\pi a^2\Rightarrow a=\sqrt{\frac{A}{\pi }}$
Now, area of the hexagon $=\frac{3\sqrt{3}}{2}{a}^2$
$=\frac{3\sqrt{3}}{2\pi }A{m}^2$