Quarantined With Circles

Here $\theta_1 = \frac{2 \pi }{2n} = \frac{\pi }{n}.$

As $OB$ is tangent to the circle with center at $A$, $\angle OBA = \frac{\pi }{2}.$ So $\theta_2 = \frac{\pi }{2} - \theta_1 $

Now, $\sin\theta_1 = \frac{r }{R + r} \\ \Rightarrow r = R\sin\theta_1 + r\sin\theta_1 \\ \Rightarrow r = \frac{R\sin\theta_1}{1 - \sin\theta_1}$

Now, area of $AOB$ triangle$\ =\frac{1}{2}*(r+R)*r\sin\theta_2 \ [using\ \Delta = \frac{1}{2}ab\sin\theta]$

Area of Ash region$\ =\pi R^2\ *\ \frac{\theta_1}{2\pi}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = R^2\ *\ \frac{\theta_1}{2}$

Area of Red region$\ =\pi r^2\ *\ \frac{\theta_2}{2\pi}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = r^2\ *\ \frac{\theta_2}{2}$

$\therefore\ $ $Area\ of\ blue\ region$ $\ = Area\ of\ AOB\ triangle\ - Ash\ region\ -\ Red\ region\\$
Finally, multiply this with $2n,$ total number of blue regions.