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 find the value of a.
A. $2(R + r)\sin \left( {\dfrac{\pi }{{2n}}} \right)$
B. $2(R + r)\tan \left( {\dfrac{\pi }{{2n}}} \right)$
C. $2(R + r)$
D. None of these

Answer 1

Hint: In order to solve the question, first find the required angle. Then using the figure write the values of R and r. Finally, find the sum of the radii of the circumcircle and incircle to find the value of a.

Formula Used:
$\tan \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
$\sin 2\theta = 2\sin \theta \cos \theta $
$\cos 2\theta = 2{\cos ^2}\theta - 1$

Complete step by step solution:
Given that
The radii of the circumcircle are R.
The radii of the incircle are r.
Consider the figure given below.

Image: Regular polygon
Here the total angle is $\dfrac{{2\pi }}{n}$.
Then the half-angle will be
$\dfrac{{2\pi }}{{2n}} = \dfrac{\pi }{n}$
From the figure, we can write that
$\dfrac{{\dfrac{a}{2}}}{R} = \sin \dfrac{\pi }{n}$
  $a = 2R\sin \dfrac{\pi }{n}$. . . . . . (1)
Also, we can write that
$\dfrac{{\dfrac{a}{2}}}{r} = \tan \dfrac{\pi }{n}$
  $a = 2r\tan \dfrac{\pi }{n}$. . . . . . (2)
From equation (1), we can find that
$R = \dfrac{a}{{2\sin \dfrac{\pi }{n}}}$
And from equation (2), we can write that
$r = \dfrac{a}{{2\tan \dfrac{\pi }{n}}}$
$r = \dfrac{{a\cos \dfrac{\pi }{n}}}{{2\sin \dfrac{\pi }{n}}}$ ( since $\tan \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$)
Next, we will find the sum $R + r$, that is
$R + r = \dfrac{a}{{2\sin \dfrac{\pi }{n}}} + \dfrac{{a\cos \dfrac{\pi }{n}}}{{2\sin \dfrac{\pi }{n}}}$
$R + r = \dfrac{a}{{2\sin \dfrac{\pi }{n}}}\left( {1 + \cos \dfrac{\pi }{n}} \right)$
$a = 2\left( {R + r} \right)\left( {\dfrac{{\sin \dfrac{\pi }{n}}}{{1 + \cos \dfrac{\pi }{n}}}} \right)$
$a = 2\left( {R + r} \right)\left( {\dfrac{{2\sin \dfrac{\pi }{{2n}} \times \cos \dfrac{\pi }{{2n}}}}{{1 + 2{{\cos }^2}\dfrac{\pi }{{2n}} - 1}}} \right)$. . . . . . (Since $\sin 2\theta = 2\sin \theta \cos \theta $ and $\cos 2\theta = 2{\cos ^2}\theta - 1$)
$a = 2\left( {R + r} \right)\left( {\dfrac{{\sin \dfrac{\pi }{{2n}}}}{{\cos \dfrac{\pi }{{2n}}}}} \right)$
$a = 2\left( {R + r} \right)\tan \dfrac{\pi }{{2n}}$

Option ‘B’ is correct

Note: Students can get confused while taking the angle. Always start solving by drawing a rough figure using the given information. Care should be taken while applying trigonometric identities.