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area of regular polygon



Given a \PMlinkname{regular $n$-gon}{RegularPolygon} with apothem of length $a$ and \PMlinkname{perimeter}{Perimeter2} $P$, its area is


Given a regular $n$-gon $R$, line segments can be drawn from its center to each of its vertices.  This divides $R$ into $n$ congruent triangles.  The area of each of these triangles is $\displaystyle \frac{1}{2}as$, where $s$ is the length of one of the sides of the triangle.  Also note that the perimeter of $R$ is $P=ns$.  Thus, the area $A$ of $R$ is

A & \displaystyle =n\left( \frac{1}{2}as \right) \\
& \\
& \displaystyle =\frac{1}{2}a(ns) \\
& \\
& \displaystyle =\frac{1}{2}aP. \end{array}$

To illustrate what is going on in the proof, a regular hexagon appears below with each line segment from its center to one of its vertices drawn in red and one of its apothems drawn in blue.