Is it possible to determine the side length of a regular polygon if you know only the number of points it has, and the "width" (or possibly more geometrically accurately, the circumscribed circle diameter) of the shape?
The application is that I want to draw different shapes (i.e., with differing numbers of vertices) but I want each of them to have the same width, so I set constant the width of the shape, but new need to calculate the side length.
On
Alternative approach:
Given a regular polygon with $~n~$ sides, draw the line segment from each vertex to the center of the circle. You will then have $~n~$ triangles formed. Each triangle will be an isosceles triangle, with two of the sides equal to $~r = \dfrac{d}{2},~$ and the third side having the unknown length.
This unknown length may be computed as follows:
The angle opposite this unknown side has measure $\theta = ~\dfrac{360^\circ}{n}.~$
The reason is that the sum of the interior angles is $~360^\circ,~$ and there are $~n~$ triangles formed.
The Law of Cosines now kicks in.
Denoting the unknown length as $~C,~$ you have that
$C^2 = r^2 + r^2 - 2(r \times r)\cos(\theta) \implies $
$\displaystyle C^2 = 2r^2 \times \left[ ~1 - \cos\left(~\frac{360^\circ}{n} \right) ~\right].$
Yes. Imagine a spoke connecting each vertex to the center, dividing the polygon into triangles. Each triangle can be split into two right triangles, like in this picture I ripped from wikihow. Image
Now the radius of the circle is your hypotenuse, the central angle is 360÷2n so you can use sin of the angle to calculate the opposite leg length.