There are no. of ways to solve it:

From the diagram we know it is a regular hexagon

and formula of the area of a regular hexagon = \((3\sqrt{3} * side^2)\) * \(\frac{1}{2}\)

= \((3\sqrt{3} * 6^2)\) * \(\frac{1}{2}\) = \(54\sqrt{3}\).


Now if we donot know the formula then, please refer to the diagram below:

Now we can write : y° + y° + y° + y° + y° + y° = 360°
or y° = 60°.

Now we can see the regular hexagon is divided into 6 triangle.
So each triangle will be equilateral triangle since it is inscribed in a regular hexagon.

Now area of Equilateral triangle = \(\sqrt{3} * (side^2)\) * \(\frac{1}{4}\) = \(\sqrt{3} * (6^2)\) * \(\frac{1}{4}\)

=\(9\sqrt{3}\)

Therefore for 6 triangles we have = 6* \(9\sqrt{3}\) = \(54\sqrt{3}\).

Now if we donot know the area of equilateral triangle then we can always find out:

Take any one triangle inscribed in the regular hexagon:-
SO the area of th triangle = \(\frac{1}{2} * base * altitude\)

Now we need to find out the altitude, which can be found out by Pythagoras theorem, as we draw a line from the midpoint of the hexagon to the base of the triangle as shown in diagram,It divides the base at the midpoint as it is a equilateral triangle. (Also a 30 - 60 - 90 triangle will also have the same result)

So \(altitude^2 = 6^2 - 3^2\) = 27

or altitude = \(3\sqrt{3}\)

Now area = \(\frac{1}{2} * 6 * 3\sqrt{3}\) = \(9\sqrt{3}\)

Now area of all the six triangle = 6 * \(9\sqrt{3}\) = \(54\sqrt{3}\) = Area of the hexagon.

** Please note this applies only for a regular hexagon

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