Question

If $R$ and $r$ are the radii of the circumcircle and incircle of a regular polygon of $n$ sides, each side being of length $a$, then $a$ is equal to

• A. $2(R+r)\sin \left(\frac{{\displaystyle \pi }}{{\displaystyle 2n}}\right)$
• B. $2(R+r)\tan \left(\frac{{\displaystyle \pi }}{{\displaystyle 2n}}\right)$
• C. $2(R+r)$
• D. None of these

Answer 1

The correct option is B

$2(R+r)\tan \left(\frac{{\displaystyle \pi }}{{\displaystyle 2n}}\right)$

Step 1: Let us first draw a figure,

To show the radii of the circumcircle $R$ and incircle $r$ of a regular polygon of $n$ sides having length $a$,

In the above figure, generally assuming the number of sides of the regular polygon is $n$.

From the figure, the angle subtended by each side of length $a$ is $\frac{2\pi }{n}$, that is angle $BFC=\frac{2\pi }{n}$

From the figure above, in triangle $FCM,$

$\begin{array}{c}\frac{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{R}=\sin \left(\frac{\pi }{n}\right)\\ a=2R\sin \left(\frac{\pi }{n}\right).......\left(i\right)\end{array}$

Similarly,

$\begin{array}{c}\frac{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{r}=\tan \left(\frac{\pi }{n}\right)\\ a=2r\tan \left(\frac{\pi }{n}\right).......\left(ii\right)\end{array}$

Step 2: Convert the above relation in terms of radii of circumcircle $R$ and incircle $r$ , then calculate the length of polygon.

From equation (i) $R=\frac{a}{2\sin \left(\frac{\pi }{n}\right)}$ and from equation (ii) $r=\frac{a}{2\tan \left(\frac{\pi }{n}\right)}$

Adding $R$ and $r$, get

$\begin{array}{c}\Rightarrow R+r=\frac{a}{2\sin \left(\frac{\pi }{n}\right)}+\frac{a}{2\tan \left(\frac{\pi }{n}\right)}\\ \Rightarrow R+r=\frac{a}{2\sin \left(\frac{\pi }{n}\right)}+\frac{a\cos \left(\frac{\pi }{n}\right)}{2\sin \left(\frac{\pi }{n}\right)}\\ \Rightarrow R+r=\frac{a}{2\sin \left(\frac{\pi }{n}\right)}\left(1+\cos \left(\frac{\pi }{n}\right)\right)\\ \Rightarrow a=2\left(R+r\right)\frac{\sin \left(\frac{\pi }{n}\right)}{\left(1+\cos \left(\frac{\pi }{n}\right)\right)}...........\left(iii\right)\end{array}$

Step 3: Use the trigonometric formula $\sin 2\theta =2sin\theta \cos \theta$ and $\cos 2\theta =2{\cos }^2\theta -1$,in the equation (iii),

The length $a$ of the polynomial is obtained as

$\begin{array}{c}\therefore a=2(R+r)\frac{2\sin \left(\frac{\pi }{2n}\right)\cos \left(\frac{\pi }{2n}\right)}{\left(1+2{\cos }^2\left(\frac{\pi }{2n}\right)-1\right)}\\ =2(R+r)\frac{\sin \left(\frac{\pi }{2n}\right)\cos \left(\frac{\pi }{2n}\right)}{{\cos }^2\left(\frac{\pi }{2n}\right)}\\ =2(R+r)\frac{\sin \left(\frac{\pi }{2n}\right)}{\cos \left(\frac{\pi }{2n}\right)}\\ =2(R+r)\tan \left(\frac{\pi }{2n}\right)\end{array}$

Therefore, $a=2(R+r)\tan \left(\frac{\pi }{2n}\right)$,

Hence, the correct option is (B).