Line $EG$ passes through the center $F$ of the radius-$2$ circle, so the shaded upper semicircle has area $2\pi$.

The chord $XZ$ lies on $AD$, one unit from $F$. Thus $$\angle XFZ=2\arccos\left(\frac12\right)=\frac{2\pi}{3}.$$ The shaded circular segment below $AD$ has area $$\begin{aligned} &\frac12(2^2)\left(\frac{2\pi}{3}\right)\\ &\quad-\frac12(2)(2)\sin\left(\frac{2\pi}{3}\right)\\ &=\frac{4\pi}{3}-\sqrt3. \end{aligned}$$

The portion between $EG$ and $AD$ consists of four congruent pieces of the following form.

Each piece is a unit square with a quarter of a unit circle removed, so the four pieces have total area $$4\left(1-\frac\pi4\right)=4-\pi.$$

The total shaded area is therefore $$\begin{aligned} &2\pi+\left(\frac{4\pi}{3}-\sqrt3\right)\\ &\quad+(4-\pi)\\ &=\frac{7\pi}{3}-\sqrt3+4. \end{aligned}$$ Hence $a=7$, $b=3$, $c=3$, and $d=4$, giving $a+b+c+d=17$.

Thus, the answer is E .