How do I find the number of sides of a regular polygon given only its side length and area? How do I find the number of sides of a regular polygon given only  its side length and area?
Absolutely no IDEA where to start. Anyone?
 A: Let $n$ be the number of sides; let $A$, $B$ be the two endpoints of one side of length $s$; let $O$ be the center.  Then $\angle AOB = \dfrac{360^\circ}{n}$.  Take $AB$ to be the base of the triangle $AOB$.  Then the height $h$ satisfies
$$
\frac{s/2}{h} = \frac{\text{opposite}}{\text{adjacent}} = \tan \left(\frac 1 2 \cdot \frac{360^\circ}{n}\right).
$$
Given that you can find $h$.  Then
$$
\text{polygon area} = n\cdot(\text{area of triangle $AOB$}).
$$
And so on . . .
I think the bottom line can be expressed without transcendental functions like the tangent.  I'll think about how to do that later.
A: The approach I was describing in my comment (which was all I had time to write then) is similar to the one given by Michael Hardy, but does not require finding the altitude of a triangle.  The regular polygon with $ \ n \ $ sides ("regular $ \ n-$ gon") is divided up into $ \ n \ $ isosceles triangles arranged around the centroid of the figure, giving them an "apex angle" of $ \ \frac{2 \pi}{n} \ . $ The base of each triangle is a side $ \ s \ $ of the polygon, and the other (congruent) legs of the triangle will be said to have length $ \ L \ . $

The Law of Cosines gives us 
$$  s^2 \ = L^2 \ + \ L^2 \ - \ 2 \cdot L \cdot L \cdot \cos \frac{2 \pi}{n} \  = \ 2  L^2 \ (1 \ - \ \cos \frac{2 \pi}{n} ) $$
and the "included angle" formula for the area of a triangle yields
$$ A_{tri} \ = \ \frac{1}{2} \cdot L \cdot L \cdot \sin \frac{2 \pi}{n} \ . $$
From these results, we can write the area of the triangle in terms of $ \ n \ $ and $ \ s \ $ as
$$ A_{tri} \ = \ \frac{1}{2} \  \left[ \frac{s^2}{2  \ (1 \ - \ \cos \frac{2 \pi}{n} )}  \right] \cdot \sin \frac{2 \pi}{n} \ =  \   \left[ \frac{\sin \frac{2 \pi}{n}}{4  \ (1 \ - \ \cos \frac{2 \pi}{n} )}  \right] \cdot s^2 \ . $$ 
The polygon comprises $ \ n \ $ of these triangles, so its area is 
$$ A(n) \ = \ \left[  \frac{n \ \sin \frac{2 \pi}{n}}{4  \ (1 \ - \ \cos \frac{2 \pi}{n} )}  \right] \cdot s^2 \ .  \ \  \ \mathbf{ [1] }$$
Using the "small-angle approximations" for the trigonometric functions of the apex angle as $ \ n \ \rightarrow \ \infty \ , $ we find that
$$ A(n) \ \rightarrow \ \left[  \frac{n \ \cdot  \frac{2 \pi}{n}}{4  \ (1 \ - \ [ \ 1 \ - \ \frac{1}{2}\left( \frac{2 \pi}{n} \right)^2 \ ] \ )}   \right] \cdot s^2 \ = \ \frac{n^2 \cdot s^2}{4 \pi} \ , $$
producing the relation described by Henry (with the appropriate dimension -- I believe he is using unit side lengths).
Equation 1 above will give us the usual area formulas for equilateral triangles, squares, etc., but the enclosed area also tends to infinity as $ \ n \ $ does, since we are using a fixed side length.  If we instead consider the perimeter $ \ p \ $ for the polygon, and write the share of that perimeter represented by each side as $ \ s = \frac{p}{n} \ , $ then we may also express our result as
$$ A(n) \ = \ \left[  \frac{n \ \sin \frac{2 \pi}{n}}{4  \ (1 \ - \ \cos \frac{2 \pi}{n} )}  \right] \cdot \left( \frac{p}{n} \right)^2 \ = \ \left[  \frac{ \sin \frac{2 \pi}{n}}{4n  \ (1 \ - \ \cos \frac{2 \pi}{n} )}  \right] \cdot p^2 \ \rightarrow \ \frac{p^2}{4 \pi} \ . $$
The limit does indeed give the relation between the area and circumference of a circle.
A: If the area is $A$ and there are $n$ sides then $A=\dfrac{n}{4\tan\frac{\pi}{n}} \approx \dfrac{n^2}{4\pi}$ for large $n$ so $n \approx \sqrt{4 \pi A}$.  Note that for a circle, the circumference is the square root of the product of $4\pi$ and the area, so this makes sense.
In fact, looking at small values, we need to take the ceiling so  $n=\bigg\lceil \sqrt{4 \pi A} \bigg\rceil$.
