How do I determine the length of the sides of a regular pentagon using the circumradius? I'm in the process of designing a fort for a story, and I want the fort to be a regular pentagon.  I need the corners of the fort to be a certain distance from the center, but I haven't been able to find a formula that lets me determine the length of the sides of a regular pentagon using the circumradius.  Plenty of sites explain how to determine the circumradius of a regular pentagon, the apothem of a regular pentagon, and the area of a regular pentagon, but none have explained how to determine the length of the sides of a regular pentagon using either the circumradius or the apothem.  I'm not exactly proficient with this kind of stuff, so I could really use some advice.
 A: If $R$ is the radius, then the side length will be $2R \sin 36^{\circ}$ because the angle subtended by any side on to the center will be $72^{\circ}$. Now consider the triangle with one vertex at the center and one side as the side of the pentagon and the other two sides as the radius. Hopefully you can take it from here.

added stuff:
$\angle CFD=\frac{360^{\circ}}{5}=72^{\circ}$. $FC=FD=R$ (the radius). So $\triangle CFD$ is isosceles. This means $\angle CFG=\angle GFD=36^{\circ}$.
Now consider the right triangle $\triangle FGC$. Then
$$\sin (\angle CFG)=\frac{GC}{FC}=\frac{x}{R}.$$
Now the side length $CD=2GC=2R \sin 36^{\circ}$.
A: In a regular pentagon $ABCDE$ with center $O$ and circumradius $R$ we have
$$ AB = 2R\sin 36^\circ $$
and $\frac{AC}{AB}=\varphi=\frac{1+\sqrt{5}}{2}$ by well-known similarities.  On the other hand $\frac{AC}{AB}=2\cos 36^\circ$, hence
$$ AB = 2 R \sqrt{1-\cos^2 36^\circ} = 2R\sqrt{1-\left(\frac{\varphi}{2}\right)^2}=R\sqrt{4-\varphi^2}=R\sqrt{3-\varphi}=R\sqrt{\frac{5-\sqrt{5}}{2}} $$
where $\sqrt{\frac{5-\sqrt{5}}{2}}\approx 1.17557$. The apothem equals $R\cos 36^\circ$, hence in terms of the apothem
$$ AB= 2a\tan 36^\circ = a \frac{\sqrt{3-\varphi}}{\varphi/2}=2a(\varphi-1)\sqrt{3-\varphi}=2a\sqrt{7-4\varphi}=2a\sqrt{5-2\sqrt{5}}$$
where $2\sqrt{5-2\sqrt{5}}\approx 1.453$.
