Circles and Squares

You could draw an equilateral triangle by connecting A and B to the center point. That makes an equilateral triangle with a side of 2, so the area of the overlapping part is $\frac{2\pi}{3}-\sqrt3$ . The area of the circle is $4\pi$ , and the area of the square is $4$ . The area of the whole figure is area of the circle - overlapping part + square which is equal to $4\pi-(\frac{2\pi}{3}-\sqrt3)+4$ which is $\frac{10\pi}{3}+\sqrt{3}+4$ . Therefore, $a+b+c+d=10+3+3+4=\boxed{20}$

$\large A_{sector}=\dfrac{60}{360}\pi r^2=\dfrac{60}{360}\pi (2^2)=\dfrac{2}{3}\pi$

$\large A_{triangle}=\dfrac{s^2\sqrt{3}}{4}=\dfrac{2^2\sqrt{3}}{4}=\sqrt{3}$

$\large A_{segment}=A_{sector}-A_{triangle}=\dfrac{2}{3}\pi-\sqrt{3}$

let $\large A$ be the total area of the figure

$\large\text{A = Area of the square + Area of the circle - Area of the segment}$

$\large A=2^2+\pi (2^2) - \left(\dfrac{2}{3}\pi-\sqrt{3}\right)=\dfrac{10\pi}{3}+\sqrt{3}+4$

Finally,

$\large a+b+c+d=10+3+3+4=$ $\color{#3D99F6}\large\boxed{20}$