In the diagram, $A B C$ is a quarter of a circle with radius 8 . A semi-circle with diameter $A B$ is drawn, as shown. A second semi-circle with diameter $B C$ is also drawn. The area of the shaded region is closest to
(A) 22.3
(B) 33.5
(C) 25.1
(D) 18.3
(E) 20.3

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-029.jpg?height=346&width=336&top_left_y=1643&top_left_x=1317)
Let the area of the shaded region that lies outside of both semicircles be $X$.

Let the area of the shaded region that lies inside of both semi-circles be $Y$.

The sum of the areas of both semi-circles counts the shaded area $Y$ twice (since the area of overlap of the semi-circles is $Y$ ).

Therefore, if we subtract $Y$ from the sum of the areas of both

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-145.jpg?height=339&width=330&top_left_y=1560&top_left_x=1559)
semi-circles, and add $X$, we get the area of the quarter-circle $A B C$.

That is, the area of quarter-circle $A B C$ is equal to

(the area of the semi-circle drawn on $A B)+($ the area of the semi-circle drawn on $B C)-Y+X$. The area of quarter-circle $A B C$ is $\frac{1}{4} \pi(8)^{2}=16 \pi$.

The area of the semi-circle drawn on $A B$ is $\frac{1}{2} \pi(4)^{2}=8 \pi$.

The area of the semi-circle drawn on $B C$ is also $8 \pi$.

Thus, $16 \pi=8 \pi+8 \pi-Y+X$ or $16 \pi=16 \pi-Y+X$, and so $Y=X$.

We build square $D B E F$ so that $D$ is 4 vertical units from $B$, and $E$ is 4 horizontal units to the right of $B$. Next, we will show that $F$ lies on both semi-circles.

Since $A B C$ is a quarter of a circle, then $\angle A B C=90^{\circ}$.

Beginning at point $B$, we move up vertically 4 units to point $D$, and then move right 4 units in a direction perpendicular to $A B$.

After these two moves, we arrive at the point labelled $F$.

Since the diameter of the semi-circle drawn on $A B$ has length 8 , then the radius of this semi-circle is 4 .

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-146.jpg?height=341&width=350&top_left_y=228&top_left_x=1557)

Therefore, $D$ is the centre of this semi-circle (since $D B=4$ ), and $F$

lies on this semi-circle (since $D F=4$ ).

Beginning again at point $B$, we move right 4 units to point $E$, and then move up vertically 4 units in a direction perpendicular to $B C$.

Since these are the same two moves we made previously (up 4 and right 4), then we must again arrive at $F$.

Since the diameter of the semi-circle drawn on $B C$ has length 8 , then the radius of this semicircle is 4 .

Therefore, $E$ is the centre of this semi-circle (since $E B=4$ ), and $F$ lies on this semi-circle (since $E F=4$ ).

The two semi-circles intersect at exactly one point (other than point $B$ ).

Since we have shown that $F$ lies on both semi-circles, then $F$ must be this point of intersection of the two semi-circles.

Therefore, $D B E F$ is a square with side length 4 , and $F$ is the point of intersection of the two semi-circles.

Finally, we find the value of $Y$.

First we construct $B F$, the diagonal of square $D B E F$.

By symmetry, $B F$ divides the shaded area $Y$ into two equal areas.

Each of these equal areas, $\frac{Y}{2}$, is equal to the area of $\triangle B E F$ subtracted from the area of the quarter-circle $B E F$.

That is, $\frac{Y}{2}=\frac{1}{4} \pi(4)^{2}-\frac{1}{2}(4)(4)$, and so $\frac{Y}{2}=4 \pi-8$, or $Y=8 \pi-16$.

The area of the shaded region is $X+Y=2 Y=16 \pi-32$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-146.jpg?height=347&width=334&top_left_y=1453&top_left_x=1557)

Of the answers given, $16 \pi-32$ is closest to 18.3.

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