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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. $2(R + r)$
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  • $\begingroup$ Please use a title that describes your problem. What have you done on the problem? Have you tried any examples? $\endgroup$ – Michael Burr Apr 12 '17 at 12:45
  • $\begingroup$ 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. $\endgroup$ – d k Apr 12 '17 at 12:53
  • $\begingroup$ Can sm1 plz help $\endgroup$ – d k Apr 12 '17 at 13:31
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    $\begingroup$ txtspeak should be avoided on this site. $\endgroup$ – Michael Burr Apr 12 '17 at 13:33
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HINT: a regular hexagon is composed of $6$ equilateral triangles of side $R$ and $r=h$ the altitude.

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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

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