# Determining the side length of regular polygon given its circumradius and inradius

If $R$ and $r$ are, respectively, the circumradius and inradius of a regular polygon of $n$ sides, each side of length $a$, then $a$ is equal to: Options are:

1. $2(R+r)\sin\left(\dfrac{\pi}{2n}\right)$
2. $2(R +r)\tan\left(\dfrac{\pi}{2n}\right)$
3. $2(R + r)$
• Please use a title that describes your problem. What have you done on the problem? Have you tried any examples? – Michael Burr Apr 12 '17 at 12:45
• I tried using d relations between R n r.But it didn't work out. Dis ques was actually in a solution of triangles book so i thot using dat way might help. – d k Apr 12 '17 at 12:53
• Can sm1 plz help – d k Apr 12 '17 at 13:31
• txtspeak should be avoided on this site. – Michael Burr Apr 12 '17 at 13:33

HINT: a regular hexagon is composed of $6$ equilateral triangles of side $R$ and $r=h$ the altitude.

The answer to your question is none of the above. For any regular polygon of $k$, side of length $a$, cirumradius $R$, and inradius $r$, the interior angle is

$$\theta=\left(\frac{k-2}{k}\right)\pi$$

and

$$a=2r\tan \frac{\pi}{k}=2R\sin\frac{\pi}{k}$$

To finish things off,

$$\text{area}=\frac{1}{4}ks^2\cot\frac{\pi}{k}=kr^2\tan\frac{\pi}{k}=\frac{1}{2}kR^2\sin\frac{2\pi}{k}$$

Reference: CRC Standard Mathematical Tables and Formulae