area of figure in sector of intersecting circles I need to find an area of blue part of figure APBC. I draw line segments between B and C, between C and A, and got equilateral triangle. I'm stuck here. Please help. Thanks.

|AB| = a, P is midpoint of segment AB

As achille hui says in the coments, $$(1-r)^2  = |AF|^2 =  \frac{1}{2^2} + |PF|^2 =  \frac{1}{2^2} + |DF|^2 - \frac{1}{4^2}  = \frac{1}{2^2} + \left(r + \frac14\right)^2 - \frac{1}{4^2}.$$ Solve for $r$ and you get $r=\dfrac{3}{10}.$
Since area of intersection between the two larger circles is $\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}$ (see here), shaded area is
\begin{align}
&\dfrac{1}{2}\left(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}\right)-\pi\left(\dfrac{1}{4}\right)^2-\pi\left(\dfrac{3}{10}\right)^2\\
\\
& = \dfrac{217 \pi }{1200}-\dfrac{\sqrt{3}}{4}\\
\end{align}
