7. Two circles with the same radius intersect at $X$ and $Y$. Let $P$ be a point on an arc $X Y$ of one circle that is inside the other. Given that the segment $X Y$ is 3 units long and that the angle $X \widehat{P} Y$ measures $120^{\circ}$, what is the area of the intersection between the two circles?
(A) $2\left(\pi-\frac{1}{4} \sqrt{3}\right)$
(B) $3(\pi-\sqrt{3})$
(C) $\frac{1}{2}(3 \pi-\sqrt{3})$
(D) $2\left(\pi-\frac{2}{3} \sqrt{3}\right)$
(E) $2\left(\pi-\frac{3}{4} \sqrt{3}\right)$.
7. The answer is (E). The intersection of the two circles is divided by $X Y$ into two congruent figures, bounded by the arcs $X Y$, so to get the result, it is enough to calculate the area of one of these two figures and multiply by two. Let $O$ be the center of the circle on which $P$ lies; by the properties of central and inscribed angles, $X \widehat{O} Y = X \widehat{P} Y = 120^{\circ}$. Therefore, triangle $X O Y$ is isosceles with base angles of $30^{\circ}$, and thus $O X = O Y = \frac{2}{\sqrt{3}} \cdot \frac{X Y}{2} = \sqrt{3}$. The area bounded by the segment $X Y$ and the arc on which $P$ lies can be calculated by difference

![](https://cdn.mathpix.com/cropped/2024_04_17_5df6f2fb8919533f8a6eg-09.jpg?height=445&width=597&top_left_y=1405&top_left_x=1272)
as the area of the circular sector $X O Y$ minus the area of triangle $X O Y$.

Since the angle at $O$ is $120^{\circ}$, the area of the circular sector is $\frac{120^{\circ}}{360^{\circ}} \pi O X^{2} = \pi$; in triangle $X O Y$, the height relative to $X Y$ is $\frac{X O}{2}$, and thus its area is $\frac{3 \sqrt{3}}{4}$. Therefore, the required area is $2\left(\pi - \frac{3 \sqrt{3}}{4}\right)$.