7) Two coplanar circles of radius 1 are arranged such that the circumference of each passes through the center of the other. What is the area of the intersection of the two circles?
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{\sqrt{3}+\pi}{2}$
(C) $\frac{\pi}{2}$
(D) $\sqrt{3}$
(E) none of the above.
7) The answer is $(\mathrm{E})$.

The area of the intersection is given by the area of two equilateral triangles with side length 1 and the 4 circular sectors like the dashed one in the figure. The area of a sector is obtained as the difference between the area of the slice $O O^{\prime} A$, which is $1 / 6$ of the area of the circle with radius 1, and that of the equilateral triangle with side length 1. Therefore, the area is equal to

$$
4\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)+2 \frac{\sqrt{3}}{4}=\frac{2}{3} \pi-\frac{\sqrt{3}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_0a7649c17dc24dfa1af2g-1.jpg?height=385&width=503&top_left_y=2183&top_left_x=1436)