$ A B C D E F $ is a regular hexagon with centre $ O $. If the area of triangle $ O A B $ is 9 $ \mathrm{cm}^{2} $, find the area of the hexagon."

Given:

\( A B C D E F \) is a regular hexagon with centre \( O \).

The area of triangle \( O A B \) is 9 \( \mathrm{cm}^{2} \).

To do: 

We have to find the area of the hexagon.

Solution:

We get six equal equilateral triangles by joining the vertices of the hexagon with $O$.

Area of the equilateral triangle $OAB = 9\ cm^2$.

This implies,

Area of hexagon $= 9 \times 6\ cm^2$

$= 54\ cm^2$

The area of the hexagon is $54 \mathrm{~cm}^{2}$.