areageometrypolygons
There are many formulas for finding the area of a regular polygon- but this is the one I am interested in:
$$A=\frac{S^2n}{4\tan(\frac{\pi}{n})}$$
where $n$ is the number of sides, and $S$ is the length of each side. Like I said, I would like to know how it is derived.
Hint: If you cut the equilateral triangle into two by bisecting an angle, you make two right triangles. Can you identify the base and height?
Added: Look at the right triangles and try to find x
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.
Best Answer
Hint: "Triangulate" your polygon, i.e. decompose it in $n$ triangles and measure the area of one of them (they are all same). Note that they are all isosceles (because the polygon is regular) and you can easily compute their angles. So you may take the altitude of it passing through the center of the polygon and then apply basic trigonometric formulas.