Geometry finding area problem

Question

A regular 2N -sided polygon of perimeter L has its vertices lying on a circle. Find the radius of the circle and the area of the polygon.

Answer 1

We use radian notation. The same argument can be carried out in degree notation. It is slightly less messy-looking to deal with the general regular polygon with $n$ vertices. So from now on $n$ is used as an abbreviation for $2N$.

Join the vertices to the centre of the circle. This divides the polygon into $n$ triangles, each with central angle $\frac{2\pi}{n}$. Take a typical such triangle, and call it $\triangle AOB$, where $O$ is the centre of the circle.

Drop a perpendicular from $O$ to $AB$, meeting $AB$ at $P$. This divides $\triangle AOB$ into two right triangles.

If $r$ is the radius of the circle, then by trigonometry using $\triangle AOP$, we find that $AB$ has length $2r\sin\left(\frac{\pi}{n}\right)$.

It follows that $$L=(n)2r\sin\left(\frac{\pi}{n}\right).$$ We can solve this equation for $r$.

For the area, note that $AB$ has length $\frac{L}{n}$. Let $h$ be the height of $\triangle AOB$ with respect to the base $AB$.

By trigonometry, $$\frac{h}{\frac{L}{2n}}=\cot\left(\frac{\pi}{n}\right).$$ So we now know the base $AB$ and height $h$ of $\triangle AOB$, and hence its area. Multiply by $n$ to get the area of the polygon.