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Join the centre of the unit circle to adjacent vertices of the regular polygon. Each of the $n$ resulting triangles has two sides of length $1$ and included angle $2\pi /n$. Bisecting one gives a right triangle, so each polygon side, a chord of a circle, has length $2\sin (\pi /n)$. Thus the perimeter is $2n\sin (\pi /n)$.

Each of these triangles has area $\frac{1}{2}\sin (2\pi /n)$, so the polygon's area is $\frac{n}{2}\sin (2\pi /n)$. Comparing the two expressions gives