Relation between areas of regular polygons wrt constant height. If we have a regular $n$-gon with height 1 (midpoint to furthest vertex for odd-gons/midpoints to midpoints for even-gons), how does the area of different regular $n$-gons compare to each other from triangle (3-gon) to circle ($\infty$-gon)?
Is it true that for even-gons, the area decreases with increasing $n$, and that for odd gons, area increases with increasing $n$?
Can you put the areas in an ascending order?
 A: The height is $h=r+a$ if $n$ is odd and $h=2a$ if $n$ is even, where $r$ is the  circumradius and $a$ is the apothem, which is given by $a=r\cos\left(\frac{\pi}{n}\right)$.
The area is
$\displaystyle
A=na^2\tan{\tfrac{\pi}{n}}
$.
Use $h=1$ to express the area solely as function of $n$.
A: Lines from the center of a regular polygon, with n sides, divide it into n isosceles triangles with central angle $\frac{2\pi}{n}$.  If we take the length of a side to be s and the length of the two equal sides of the triangle to be l, then, by the cosine law, the $s^2= l^2+ l^2- 2l^3 cos\left(\frac{2\pi}{n}\right)= 2l^2(1- cos\left(\frac{2\pi}{n}\right)$.  Further, taking the length of a line from the center of the polygon to the midpoint of a side to be h, by the Pythagorean theorem, $l^2= h^2+ \frac{s^2}{4}$ so that $l^2= h^2+ \frac{s^2}{4}= h^2+ \frac{1}{2}l^2\left(1- cos\left(\frac{2\pi}{n}\right)\right)$ so that $h^2= \frac{1}{2}l^2\left(1+ cos\left(\frac{2\pi}{n}\right)\right)$ and, finally, $h= l\sqrt{\frac{1}{2}\left(1+ cos\left(\frac{2\pi}{n}\right)\right)}$.
Now in an odd sided polygon, with your "height" equal to 1, $h= \frac{1}{2}$ and $l= \frac{4}{1+ cos(\left(\frac{2\pi}{n}\right)}$ so that $s= 2\sqrt{l^2- h^2}= 2l\sqrt{\frac{1}{2}+ \frac{1}{2}cos\left(\frac{2\pi}{n}\right)}$.  The area of each of those n triangles is $\frac{1}{2}sl= \sqrt{\frac{1}{2}+ \frac{1}{2}cos\left(\frac{2\pi}{n}\right)}$ and the area of the entire polynomial is $n\sqrt{\frac{1}{2}+ \frac{1}{2}cos\left(\frac{2\pi}{n}\right)}$.
In an even sided polynomial, $l+ h= l\left(1+ \sqrt{1+ cos\left(\frac{2\pi}{n}\right)}\right)= 1$ so $l= \frac{1}{\left(1+ \sqrt{1+ cos\left(\frac{2\pi}{n}\right)}\right)}$.
