Sum up their area and you have the area of your regular $n$-gon. Take such an $n$-gon, and divide it into ideal triangles. Well, the area is maximal if all corner angles are zero, i.e. So you are asking about a bound for the area of a regular polygon, right? Find corresponding results for regular $n$-gons.Since these are contained within a completely ideal triangle (of area $\pi$), you know their area to be finite as well. You can also have a single ideal corner point, or two of them. (I know that in the limit as the vertices go to infinity, there are even ideal hyperbolic triangles in which all three angles are $0°$, and so the area is finite by $(\pi−A−B−C)R^2$) Also deduce that triangles with angle zero exist, in a certain sense, and that their area is finite.The interior angles of a regular polygon always measure less than 180 degrees so two of them cannot. Since there are 360 degrees in a circle, if m regular n sided polygons fit together at a single vertex this means that. (this part is easy.)Īngle sum formula tells you as much. We are given that the interior angles of Pn each measure 180×(n2) n degrees. You can then consider this triangle as the $2m$ th part of a regular $m$-gon with corner angles $\fracn$ for each $n\geq7$. $$ \alpha=\frac\pi m \qquad \beta=\frac\pi n \qquad \gamma=\frac\pi 2 $$Įxactly if $\alpha+\beta+\gamma=\pi$, i.e. In the Euclidean plane, you get a right triangle with corner angles Why there are infinitely many regular tessellations of the hyperbolic plane? polygon is either an equilateral triangle, a square or a regular hexagon.
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