Area of Regular Hexagon

Theorem
Let $H$ be a regular hexagon.

Let the length of one side of $H$ be $s$.

Let $\AA$ be the area of $H$.

Then:
 * $\AA = \dfrac {3 \sqrt 3} 2 s^2$

Proof 1
The regular hexagon can be subdivided into triangles:


 * Regular Hexagon.svg

Consider the bottom triangle. The angle $\alpha$ divides the full circle into 6 equal parts. Therefore:


 * $ \alpha = \dfrac\pi 3 $

Since the hexagon is regular, the remaining two angles within the triangle are equal. Call the angle $\beta$.

Since Sum of Angles of Triangle equals Two Right Angles,


 * $ \alpha + 2\beta = \pi $
 * $ 2\beta = \pi - \dfrac\pi 3 = \dfrac{2\pi} 3 $
 * $ \beta = \dfrac\pi 3 $

Therefore, the triangle is equilateral.

Let $\AA_T$ be the area of that triangle. Then by Area of Equilateral Triangle,


 * $ \AA_T = \dfrac{\sqrt 3} 4 s^2 $

Therefore,


 * $ \AA = 6\AA_T = \dfrac{6\sqrt 3} 4 s^2 = \dfrac{3\sqrt3} 2 s^2 $

Proof 2
A regular hexagon is a regular 6-sided polygon.

Therefore: