## Perimeter of a Regular Polygon Inscribed in a Circle Radius r

Suppose a regular polygon with
$n$
sides is inscribed in a circle of radius
$r$
, with the vertices of the polygon touching the circle.
From the centre of the circle lines are drawn to each vertex, and
$n$
triangles are formed. The angle of each triangle at the centre of the circle is
$\frac{360}{n} \equiv \frac{2 \pi}{n}$
, and the length of the side opposite this angle is is
\begin{equation} \begin{aligned} a &= \sqrt{r^2+r^2-2 \times r \times r \times cos( \frac{2 \pi}{n}} \\ &= r \sqrt{2-2 cos( \frac{2 \pi}{n}} \\ &= r \sqrt{2} \sqrt{1-cos( \frac{2 \pi}{n}} \\ &= r \sqrt{2} \sqrt{2 sin^2 \frac{\pi}{n}} \\ &= 2 r sin \frac{\pi}{n} \end{aligned} \end{equation}
. There are
$n$
such triangles, so the perimeter of the polygon is
$n \times 2 r sin \frac{\pi}{n}$
.
As the number of sides gets larger
$sin (\frac{\pi}{n})$
tends to
$\frac{\pi}{n}$
, so the perimeter of the polygon tends to
$n \times 2r \frac{\pi}{n}=2 \pi r$
.