$ 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 $54 \mathrm{~cm}^{2}$.
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