Area of a Regular Polygon

Calculates the area of a regular n-gon as n × side² ÷ (4 × tan(180° ÷ n)). Each interior angle is 180° × (n − 2) ÷ n. The polygon needs at least 3 sides.

Explanation

A regular polygon has equal sides and equal angles; equilateral triangles, squares and regular hexagons are all examples. The number of sides $n$ and the side length $a$ are enough to pin down the area.

S = \dfrac{n a^2}{4 \tan\left(\dfrac{180^\circ}{n}\right)}

$S$ — the area
$n$ — the number of sides, an integer of 3 or more
$a$ — the side length

Each interior angle is $\dfrac{180^\circ (n - 2)}{n}$ , and the perimeter is $n a$ .

Example

With the defaults, $n = 6$ and $a = 4$ , a regular hexagon. Since $\tan 30^\circ = \dfrac{1}{\sqrt{3}} \approx 0.5774$ ,

S = \dfrac{6 \times 4^2}{4 \tan 30^\circ} = \dfrac{96}{2.3094} \approx 41.5692

The interior angle is $\dfrac{180^\circ \times 4}{6} = 120^\circ$ and the perimeter is $6 \times 4 = 24$ .

Where the formula comes from

Join the centre to every vertex and the polygon splits into $n$ congruent isosceles triangles. Each has base $a$ and height equal to the apothem, the distance from the centre to a side, which is $\dfrac{a}{2 \tan(180^\circ / n)}$ . Multiplying the area of one triangle by $n$ gives the formula above.

Watch out

The number of sides has to be a whole number of at least 3; decimals and values below 3 are rejected. Do not confuse the interior angle, the angle between two adjacent sides and $120^\circ$ in the example, with the central angle $\dfrac{360^\circ}{n}$ , which is $60^\circ$ here.