Question

$I_n$ is the area of n sided regular polygon inscribed in
a circle of unit radius and $O_n$ be the area of the
polygon circumscribing the given circle, prove that
$ I_n = \frac{ O_n}{2} \Bigg ( 1 + \sqrt { 1 - \bigg( \frac{ 2 I_n}{ n} \bigg)^2 } |bigg) $

Answer 1

We know that, $ I_n = \frac{ n}{2} r^2 \, sin \, \frac{ 2 \pi}{ n} $
$$ [since, $I_n$ is area of regular polygon]
$\Rightarrow $ $ $ $ \frac{ 2 I_n }{ n} = \, sin \, \frac{ 2 \pi}{ n} $ $ $ $ [ \because r = 1 ] $ ...(i)
and $$ $ O_n = nr^2 \, tan \, \frac{ \pi}{ 2} $
$$ [since, $O_n$ is area of circumscribing polygon]
$ \frac{ O_n}{n} = tan \, \frac{ \pi}{ n} $ $$ ... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$ \frac{ 2 I_n }{ O_n} = \frac{ sin \, \frac{ 2 \pi }{ n} }{ tan \, \frac { \pi}{ n}} $
$\Rightarrow \frac{ I_n}{ O_n} = \, cos^2 \, \frac{ \pi}{ n} = \frac{ 1 + cos \, \frac{ 2 \pi}{ n}} { 2} $
$\therefore \frac{ I_n}{ O_n} = \frac{ 1 + \sqrt { 1 - (2 I_n / n)^2 }}{ 2} $ $$ [ from Eq. (i)]
$\Rightarrow I_n = \frac{ O_n}{ 2} \, (1 + \sqrt{ (1 - 2I_n / n)^2) } $