# convex polygons in hyperbolic geometry

Does $\exists$ on the hyperbolic plane, a convex quadrilateral $Q$ and a convex pentagon $P$ with the same angle sum? I found this question to be rather interesting.

-
Yes. Both $0.{}$ – Will Jagy Apr 27 '13 at 5:50
Could you provide an example? – Richard Carpenter Apr 27 '13 at 5:59

Theorem:Gauss-Bonnet: An n-gon with angles $\alpha_1,\alpha_2,...,\alpha_n$ has area $(n-2)\pi-(\alpha_1,\alpha_2,...,\alpha_n).$ Now if a quadrilateral has all interior angle $\pi/2$ then its area is zero which in not possible. But for a pentagon this case is possible because an ideal pentagon ha all angles 0 and if we start to draw smaller pentagon the interior angles will became small up-to angle $3\pi/5$ which is greater than $\pi/2.$ Hence there will be a polygon yu desired. This logic applies for all polygon with more than 4 sides. – tessellation Apr 27 '13 at 6:40
Yes I am. Actually if you take a regular pentagon of area $\pi/2$ which will exists by the above argument its all angles will be $\pi/2$. – tessellation Apr 27 '13 at 6:52