Area of Regular Polygons

Where the formula comes from

Draw segments from the center of a regular polygon to each vertex. A regular $n$-gon splits into $n$ congruent triangles. Each triangle has a side of the polygon as its base ($s$) and the apothem as its height ($a$), so each triangle's area is $\displaystyle \frac{1}{2}sa$.

Add up all $n$ triangles: $\displaystyle A = n \cdot \frac{1}{2}sa = \frac{1}{2}a(ns)$. But $ns$ — the number of sides times the side length — is just the perimeter $P$. That gives $\displaystyle A = \frac{1}{2}aP$. The hexagon below shows the cut: six radii split it into six congruent triangles, each with base $s$ and height the apothem $a$.

Using the formula

Most problems give you the apothem plus either the perimeter or the side length. If you get the side length, find the perimeter first: $P = ns$. Then substitute into $\displaystyle A = \frac{1}{2}aP$.

Keep the two lengths straight: the apothem goes from the center to the midpoint of a side, while the radius goes from the center to a vertex. The apothem is always the shorter of the two, and it's the one the area formula wants.

Worked examples

Try one yourself

Common questions