A pole is situated at the centre of a regular hexagonal park. The...

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,~$\pia^{2} = 27 ⇒ a = 3\sqrt{\dfrac{3}{\pi}}$ meters ... \left(1\right)$<br /><br /><img src="/api/proxy-image?url=https$AOT,$right angled at$O,$we have<br /><br />$\tan \left(\dfrac{\pi}{3}\right) = \dfrac{OT}{OA} ⇒ OT = OAtan\dfrac{\pi}{3} ⇒ OT = 3\sqrt{\dfrac{3}{\pi}} \times \sqrt{3}$<br /><br />$∴ OT = \dfrac{9}{\sqrt{\pi}}$meters<br /><br />$∴$~Height of the tower$= \dfrac{9}{\sqrt{\pi}}$ meters.