, 2025

Suppose the midpoints of $AD$
and $BC$ are $P$ and $Q$, respectively.

Join $P$ to $Q$ and label the intersection of $PQ$ with each circle $S$ and $T$, as shown.

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Since $AD=BC$, the semi-circles
have equal diameters, and thus equal radii, $r$, and so $AP=PD=PS=BQ=QC=TQ=r$.

The shortest distance between the two semi-circles is $ST=2$, and so $ABCD$ has dimensions $AD=2r$ and $AB=PQ=2r+2$.

The area of $ABCD$ is $AD\times AB=2r(2r+2)=224$.

Solving this equation, we get $$\begin{align*} 2r(2r+2)&=224\\ 4r^2+4r&=224\\ r^2+r-56&=0\\ (r+8)(r-7)&=0\end{align*}$$ and so $r=7$ (since $r>0$). The area of the shaded region is
the difference between the area of $ABCD$ and the combined areas of the two
semi-circles, or $\$224-\left(2\cdot\dfrac12\pi7^2\right)\approx$
70$.