# What is the area of this hexagon?

Jan 9, 2016

Most likely, this question is about a regular hexagon with a given size $a$ of a side.

$S = \frac{3 \sqrt{3}}{2} {a}^{2}$

#### Explanation:

If we connect a center of a regular hexagon with all $6$ vertices, we will get $6$ equilateral triangles with each side equal to $a$.

The proof is trivial. Since this hexagon is regular, the distances from its center to all vertices are equal to each other. Hence, all these triangles are isosceles with a top being in the center of a hexagon.
In addition, an angle at the top of each triangle is $\frac{1}{6}$ of ${360}^{o}$, that is ${60}^{o}$.
Two other angles of each triangle, therefore, are also equal to ${60}^{o}$.
That makes each isosceles triangle an equilateral one.

An equilateral triangle with a side $a$ has an altitude equal to
$\sqrt{{a}^{2} - {\left(\frac{a}{2}\right)}^{2}} = a \frac{\sqrt{3}}{2}$

Therefore, its area is
${S}_{\Delta} = \frac{1}{2} {a}^{2} \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} {a}^{2}$

The area of a hexagon in $6$ times greater, that is
${S}_{h e x a g o n} = \frac{6 \sqrt{3}}{4} {a}^{2} = \frac{3 \sqrt{3}}{2} {a}^{2}$