Two circles each have radius $10 \mathrm{~cm}$. They overlap so that each contains exactly $25 \%$ of the other's circumference, as shown. The area of the shaded region is closest to
(A) $57.08 \mathrm{~cm}^{2}$
(B) $55.24 \mathrm{~cm}^{2}$
(C) $51.83 \mathrm{~cm}^{2}$
(D) $54.17 \mathrm{~cm}^{2}$
(E) $53.21 \mathrm{~cm}^{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-057.jpg?height=227&width=355&top_left_y=1949&top_left_x=1335)
Label points $A$ and $B$, the points of intersection of the two circles, and point $O$, the centre of the left circle.

Construct line segment $A B$, which by symmetry divides the shaded area in half.

Construct radii $O A$ and $O B$ with $O A=O B=10 \mathrm{~cm}$.

Since each circle contains $25 \%$ or $\frac{1}{4}$ of the other circle's circumference, $\angle A O B=\frac{1}{4} \times 360^{\circ}=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-269.jpg?height=310&width=453&top_left_y=1688&top_left_x=1511)

Thus, the area of sector $A O B$ is $\frac{1}{4}$ of the area of the entire circle, or $\frac{1}{4} \pi r^{2}=\frac{1}{4} \pi 10^{2}=25 \pi \mathrm{cm}^{2}$.

The area of $\triangle A O B$ is $\frac{O A \times O B}{2}=\frac{10 \times 10}{2}=50 \mathrm{~cm}^{2}$.

The area remaining after $\triangle A O B$ is subtracted from sector $A O B$ is equal to half of the shaded area. Thus, the shaded area is $2 \times(25 \pi-50) \approx 2 \times(28.5398)=57.0796 \mathrm{~cm}^{2}$.

The area of the shaded region is closest to $57.08 \mathrm{~cm}^{2}$.