Question

A pole is situated at the centre of a regular hexagonal park. The angle of elevation of the top of the vertical pole when observed from each vertex of the hexagon is $\frac{\pi }{3}.$ If the area of the circle circumscribing the hexagon is $27 \, m^{2}$ , then the height of the tower is

• A. $3\sqrt{\frac{3}{\pi }} \, m$
• B. $\frac{3}{\sqrt{\pi }} \, m$
• C. $\sqrt{\frac{3}{\pi }} \, m$
• D. $\frac{9}{\sqrt{\pi }} \, m$

Answer 1

Answer: (D)

Let $a$ be the side of the regular hexagon
Now, for the equilateral triangle $FOA,$ we have
$OF=OA=AF=a$
Hence, area of circle, $\pi a^{2}=27\Rightarrow a=3\sqrt{\frac{3}{\pi }}$ meters ... $\left(1\right)$

In right-angled triangle $AOT,$ right angled at $O,$ we have
$tan\left(\frac{\pi }{3}\right)=\frac{O T}{O A}\Rightarrow OT=OA \, tan\frac{\pi }{3}\Rightarrow OT=3\sqrt{\frac{3}{\pi }}\times \sqrt{3}$
$\therefore OT=\frac{9}{\sqrt{\pi }}$ meters
$\therefore $ Height of the tower $=\frac{9}{\sqrt{\pi }}$ meters.