Calculating perimeter of n-sided regular polygons using only height? Say you were drawing n-sided regular polygons on a square grid (first side you drew being flat always). You wanted any polygon you drew to be 100 units high, aka the uppermost point being y = 100. With a square, this would mean you would want the perimeter to be 400, because if each side was 100 then from top to bottom it would be 100.
I know a rough approximation of some answers, so for # of sides, 4 = 100, 5 = ~65, 6 = ~58, 7 = ~45, 8 = ~41.5, 9 = ~35, etc.
I'm hoping there is some sort of formula for this, where given the desired total height and # of sides, you could come up with a measure for the perimeter and thus each side length.
I tried playing with Pythagorean theorem, triangle areas, apothems, etc. but I don't have a firm enough grasp on that stuff to put something together. 
EDIT: I think I got it, perimeter = n*height*tan(180/n)
 A: Let $r$ be the radius of the circumcircle of the polygon. From this you get the perimeter as
$$s=2nr\sin\tfrac\pi n$$
This is becuse you can divide your polygon into $n$ isosceles triangles, and dividing each gives $2n$ congruent right triangles, each of which has a central angle of $\alpha=\frac{360°}{2n}$ and a hypothenuse of length $r$. This is the green triangle in the following image:

But what is the height of the polygon, in terms of $r$? For even $n$ you'd get
$$h=2r\cos\tfrac\pi n$$
since from bottom edge to midpoint as well as from midpoint to top edge, you have one of the legs of such a right triangle. So for even $n$ you get
$$s = 2n\frac{h}{2\cos\tfrac\pi n}\sin\tfrac\pi n=nh\,\tan\tfrac\pi n$$
This is the result you had yourself already. But for odd $n$, the length from center to top vertex is $r$, so your total height is
$$h = r\left(1+\cos\tfrac\pi n\right)$$
and the perimeter is correspondingly
$$s=2n\frac{h}{1+\cos\tfrac\pi n}\sin\tfrac\pi n$$
Wolfram Alpha will help you find a shorter formula for that as well:
$$s=2n\frac{h}{1+\cos\tfrac\pi n}\sin\tfrac\pi n=2nh\cot\tfrac\pi{2n}$$
I guess you can find a geometric explanation for that formula if you consider the regular $2n$-gon, but I'll leave it at that for now.
