Two circled with radius $2R$ and $\sqrt{2}R$ intersect each other at points $A$ and $B$. The centres of both the circles are on the same side of $AB$. $O$ is the centre of the bigger circle and $\angle{AOB}$ is $60^\circ$. Find the area of the common region between two cirlces.
I solved this problem two times and every time I came to the same solution of $R^2(2\pi-\frac{\sqrt{3}}{2})$. However the answer provided is $R^2(\frac{13\pi}{6}+1-\sqrt{3})$ and unable to get to the same. Can someone please help with a solution for this? Please help !
PS : I am sorry as I am not able to put my solution approach here as I can not put up a figure to show my calculations.
Simplify the process with $R=1$
$AO'=\sqrt 2,AE=1,EO'=1", $ $\triangle AEO'$ is a $45^\circ-90^\circ-45^\circ$ triangle.
\begin{align}S_{ADE}=S_{ADO}-S_{AEO}=\frac{\pi}{3}-\frac{\sqrt 3}{2}\end{align}
\begin{align}S_{ACD}=S_{AO'C}-S_{ADE}-S_{AEO'}=\frac{\pi}{4}-(\frac{\pi}{3}-\frac{\sqrt 3}{2})-\frac{1}{2}=\sqrt3-\frac{\pi}{12}-\frac{1}{2}\end{align}
\begin{align}S=2\pi-\sqrt 3+\frac{\pi}{6}+1=\frac{13\pi}{6}-\sqrt 3+1\end{align}
Scale back R: \begin{align}S=R^2(\frac{13\pi}{6}-\sqrt 3+1)\end{align}