2019 AMC 12B Problem 15

Place $F=(3,1);$ the circle of radius $2$ then passes through $E$ and $G,$ and the middle semicircle lies entirely inside it

Subtract from the circle’s area the parts of the three semicircles that fall inside it

Put $A=(0,0),$ $\ B=(2,0),$ $\ C=(4,0),$ $\ D=(6,0),$ so the semicircles are centered at $(1,0),(3,0),(5,0)$ and their tops are $E=(1,1),$ $\ F=(3,1),$ $\ G=(5,1).$ The circle has center $F=(3,1)$ and radius $2,$ so it passes through $E$ and $G,$ and has area $4\pi.$

The middle semicircle $BFC$ lies entirely inside the circle, removing area $\dfrac{\pi}{2}.$ For the left semicircle, the overlap starts at $E$ and includes its right-hand quarter-circle, except for the region below the large circle. The large circle meets the $x$-axis at $x=3-\sqrt3.$ The excluded area is $$\begin{aligned} I &=\int_1^{3-\sqrt3}1\,dx\\ &\quad-\int_1^{3-\sqrt3} \sqrt{4-(x-3)^2}\,dx\\ &=2-\dfrac{\sqrt3}{2}-\dfrac{\pi}{3}. \end{aligned}$$ Therefore the overlap has area $$\begin{aligned} R&=\dfrac{\pi}{4}-I\\ &=\dfrac{7\pi}{12}-2+\dfrac{\sqrt3}{2}. \end{aligned}$$ By symmetry, the right semicircle contributes the same overlap.

The shaded area is $$4\pi-\dfrac{\pi}{2}-2R=\dfrac{7}{3}\pi-\sqrt3+4.$$ Hence $a=7,\ b=3,\ c=3,\ d=4,$ so $a+b+c+d=17.$

Thus, E is the correct answer.