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

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 $

$\blacksquare$

Proof 2

A regular hexagon is a regular 6-sided polygon.

Therefore:

\(\ds \AA\)

\(=\)

\(\ds \dfrac 1 4 \times 6 \times s^2 \times \cot \dfrac \pi 6\)

Area of Regular Polygon

\(\ds \)

\(=\)

\(\ds \dfrac 3 2 \times s^2 \times \sqrt 3\)

Cotangent of $30 \degrees$

\(\ds \)

\(=\)

\(\ds \dfrac {3 \sqrt 3} 2 s^2\)

$\blacksquare$