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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.

Bhoot
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2 Answers2

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Hint:
Which triangles are equilateral?

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Eclipse Sun
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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 $$\frac{\sqrt 3}4\left((a_1+a_2+a_3)^2-a_1^2-a_3^2-a_5^2\right) =\frac{\sqrt 3}4\left((a_1+a_3)a_2-a_5^2\right)$$

Hagen von Eitzen
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  • Your RHS of the equation is incorrect. For example, take every variable with value 1. LHS = const6, RHS = const1 – Daksh Shah Apr 08 '20 at 13:45