# What is the area of a hexagon with 4 cm long sides?

Nov 16, 2015

$S = 24 \sqrt{3}$

#### Explanation:

Obviously, this question is about a regular 6-sided polygon. That means that all sides are equal (4 cm long each) and all inside angles equal to each other. That's what regular means, without this word the problem is not fully specified.

Every regular polygon has a center of rotational symmetry. If we rotate it around this center by ${360}^{o} / N$ (where $N$ is the number of its sides), the result of this rotation will coincide with the original regular polygon.

In case of a regular hexagon $N = 6$ and ${360}^{o} / N = {60}^{o}$. Therefore, each of the six triangles that are formed by connecting its center with all six vertices is an equilateral triangle with a side equaled to 4 cm. The area of this hexagon is six times greater than the area of such a triangle.

In an equilateral triangle with a side $d$ the altitude $h$ can be calculate from the Pythagorean Theorem as
${h}^{2} = {d}^{2} - {\left(\frac{d}{2}\right)}^{2} = \left(\frac{3}{4}\right) {d}^{2}$
Therefore, $h = \mathrm{ds} q r t \frac{3}{2}$

Area of such a triangle is
$A = \frac{d \cdot h}{2} = {d}^{2} \frac{\sqrt{3}}{4}$

From this the area of the regular hexagon with a side $d$ is
$S = 6 A = {d}^{2} \frac{3 \sqrt{3}}{2}$
For $d = 4$ the area is
$S = 16 \frac{3 \sqrt{3}}{2} = 24 \sqrt{3}$