# The question is below?

## If ${I}_{n}$ be the area of n-sided regular polygon inscribed in a circle of unit radius and ${O}_{n}$ be the area of the regular polygon circumscribing the given circle prove that ${I}_{n} = \left(\frac{{O}_{n}}{2}\right) \left[1 + \sqrt{1 - {\left(\frac{2 {I}_{n}}{n}\right)}^{2}}\right]$

Jun 6, 2018 Given that an n-sided regular polygon is inscribed in a circle of unit radius i.e. $R = 1$. Another regular polygon of n sides also circumscribes this until circle. Let circum radius of larger polygon be $R '$

Let $\theta$ be the angle subtended by each side of a an n-sided regular polygon at the center of it i.e at the circumcenter of the regular polygon.

Now area of the inner smaller polygon will be

${I}_{n} = n \cdot \frac{1}{2} {R}^{2} \sin \theta = \frac{n}{2} \sin \theta$

So $\sin \theta = \frac{2 {I}_{n}}{n}$

Now $R = R ' \cos \left(\frac{\theta}{2}\right)$

$\implies R ' = \frac{1}{\cos} \left(\frac{\theta}{2}\right)$

Again the area of the larger outer polygon will be

${O}_{n} = \frac{n}{2} {\left(R '\right)}^{2} \sin \theta$

So
${I}_{n} / {O}_{n} = \frac{1}{R '} ^ 2$

=>I_n/O_n=1/((1/cos(theta/2))^2

$\implies {I}_{n} = {O}_{n} / 2 \cdot 2 {\cos}^{2} \left(\frac{\theta}{2}\right)$

$\implies {I}_{n} = {O}_{n} / 2 \left(1 + \cos \theta\right)$

$\implies {I}_{n} = {O}_{n} / 2 \left(1 + \sqrt{1 - {\sin}^{2} \theta}\right)$

$\implies {I}_{n} = {O}_{n} / 2 \left[1 + \sqrt{1 - {\left(\frac{2 {I}_{n}}{n}\right)}^{2}}\right]$