, 2025

Draw line segments from $A$ to
$B$, $B$ to $C$, $C$
to $D$, $D$ to $E$, $E$
to $F$, $F$ to $G$, $G$
to $H$, and $H$ to $A$, as shown.

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The line segments $BC$, $DE$, $EF$, $FG$, $GH$, and $HA$ each have a length of $2$ units.

Hence, the radii of the semicircles with these diameters are all $1$, and the areas of the circles with
these diameters are all $\dfrac{1}{2}\pi(1)^2=\dfrac{\pi}{2}$.

The line segment $AB$ is the
hypotenuse of a triangle with legs of length $1$ and $2$.

By the Pythagorean Theorem, the length of $AB$ is $\sqrt{1^2+2^2}=\sqrt{5}$.

The radius of the semicircle with diameter $AB$ is $\dfrac{\sqrt{5}}{2}$, so its area is $\dfrac{1}{2}\pi\left(\dfrac{\sqrt{5}}{2}\right)^2=\dfrac{5\pi}{8}$.

By similar reasoning, the area of the semicircle with diameter $CD$ is also $\dfrac{5\pi}{8}$.

The area of the figure can be computed as the area of hexagon $ABCDEH$ plus the areas of the semicircles
with diameters $AB$, $CD$, $EF$, and $GH$, minus the areas of the semicircles
with diameters $BC$, $DE$, $FG$, and $AH$.

We have already computed the areas of the semicircles, so we now need to
compute the area of hexagon $ABCDEH$.

This hexagon can be viewed as a $\$3\times$
6$ rectangle with two “corners” removed. These “corners” are
right-angled triangles with hypotenuses $AB$ and $CD$.

The legs of these two triangles have length $1$ and $2$, so their areas are each $\dfrac{1}{2}\times1\times2=1$.

Thus, the area of hexagon $ABCDEH$ is
$3\times 6-2\times 1=16$.

Using the areas of the semicircles computed earlier, we can now compute
the area of the figure as $$16 +\frac{5\pi}{8} + \frac{5\pi}{8} + \frac{\pi}{2} + \frac{\pi}{2} - \frac{\pi}{2} - \frac{\pi}{2} - \frac{\pi}{2} - \frac{\pi}{2}=16+\dfrac{\pi}{4}\approx 16.78539$$ Thus, $x\approx 16.78539$, so $100x\approx 1678.539$. Rounding to the
nearest integer, we get $y=1679$, so
the answer is $1+6+7+9=23$.