Proof. 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
$\begin{array}{rl} A & \displaystyle =n\left( \frac{1}{2}as \right) \\ & \\ & \displaystyle =\frac{1}{2}a(ns) \\ & \\ & \displaystyle =\frac{1}{2}aP. \end{array}$
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.
\begin{pspicture}(-2,-2)(2,2) \psline[linecolor=blue](0,0)(0,-1.732) \psline[linecolor=red](-2,0)(2,0) \psline[linecolor=red](-1,-1.732)(1,1.732) \psline[linecolor=red](1,-1.732)(-1,1.732) \pspolygon(2,0)(1,1.732)(-1,1.732)(-2,0)(-1,-1.732)(1,-1.732) \end{pspicture}
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