= Solution
<Gauss-Bonnet theorem> gives the area of a curvature-$-1$ triangle as $\pi-(\alpha+\beta+\gamma)$. Split the regular polygon into $2n$ congruent right triangles. If $\theta$ is a polygon half-angle, the right-triangle identity and part (a) give
$$
\tan\theta
=\frac1{\tan(\pi/n)\cosh R}
=\frac{1-r^2}{1+r^2}\cot\frac\pi n.
$$
The polygon area is $(n-2)\pi-2n\theta$. Rewriting $\theta=\pi/2-\cot^{-1}(\tan\theta)$ yields
$$
\boxed{
A_n(r)=2n\left[
\cot^{-1}\!\left(\frac{1-r^2}{1+r^2}\cot\frac\pi n\right)
-\frac\pi n\right]}.
$$
Back to article page
