# Area of irregular hexagon with all angles = 120 degrees

I want to derive a formula to calculate the area of a irregular hexagon which is guaranteed to have all internal angles = 120 degrees. Please guide me how to proceed to form a general formula.

Let the side lengths be $a_1, \ldots, a_6$. If we add an isolateral triangle of side length $a_1$ to the first edge, one of side length $a_3$ to the third and one of side length $a_5$ to the fifth edge, we obtain an isolaeral triangle of side length $a_1+a_2+a_3=a_3+a_4+a_5=a_5+a_6+a_1$. The area of the hexagon is therefore $$\frac{\sqrt 3}4\left((a_1+a_2+a_3)^2-a_1^2-a_3^2-a_5^2\right) =\frac{\sqrt 3}4\left((a_1+a_3)a_2-a_5^2\right)$$

Hint:
Which triangles are equilateral?