Area of Regular Hexagon/Proof 1

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

From Regular Hexagon is composed of Equilateral Triangles, it follows that a regular hexagon can be dissected into six congruent equilateral triangles:

Let $\AA_T$ be the area of the bottom triangle.

Then by Area of Equilateral Triangle:

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

As $H$ consists of six congruent triangles, it follows that:

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

$\blacksquare$