If `I_n` is the area of `n-s i d e d` 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=(O_n)/2(sqrt(1+((2I_n)/n)^2))`

Explanation

Let OAB be one triangle out of n of a n-sided polygon inscribed in a circle of radius 1. Then,
`angleAOB = (2pi)/(n)`
`OA = OB = 1`

`:. " Area " (DeltaAOB) = (1)/(2) 1 xx 1 xx sin ((2pi)/(n))`
`= (1)/(2) sin ((2pi)/(n))`
Area of n -sided polygon `= I_(n) = (n)/(2) sin.(2pi)/(n)`....(i)
Similarly, let O'A'B' be one of the triangle out of n of n-sided polygon escribing the circle of unit radius.
Then, in `Detla O'B'A', cos.(pi)/(n) = (1)/(O'B')`
or `O'B' = sec.(pi)/(n)`
Area `(Delta O'B'A') = (1)/(2) (O'B')^(2) sin ((2pi)/(n))`
`= (1)/(2) sec^(2) ((pi)/(n)) sin ((2pi)/(n))`
Therefore, the area of n-sided polygon is given by
`(O_(n)) = (n)/(2) sec^(2).(pi)/(n) sin.(2pi)/(n)` (ii)
From Eqs. (i) and (ii), we get
`(I_(n))/(O_(n)) = ((n)/(2) sin.(2pi)/(n))/((n)/(2) sec^(2).(pi)/(n) sin.(2pi)/(n)) = (1)/(sec^(2).(pi)/(n))`
or `I_(n) = (cos^(2).(pi)/(n)) O_(n)`
`= (O_(n))/(2) [1 + cos ((2pi)/(n))]`
`= (O_(n))/(2) [1 + sqrt(1 - sin^(2).(2pi)/(n))]`
`= (O_(n))/(2) [1 - sqrt(1 -((2I_(n))/(n))^(2))]`