13) The answer is (E). Applying to the figure shown in the text two rotations, the first of $120^{\circ}$ and the second of $240^{\circ}$ with respect to the center of the circle to which the arc belongs, we obtain the figure shown alongside. From it, it is evident that the required area $S$ is $\frac{1}{3}$ of the area obtained by subtracting from the area of an equilateral triangle with side 6 the area of the circle inscribed in it, which has a radius equal to $\frac{1}{3}$ of the height of the triangle. Therefore, we have:

![](https://cdn.mathpix.com/cropped/2024_04_17_88883ec2564951b28c79g-2.jpg?height=361&width=415&top_left_y=1015&top_left_x=1511)

$$
S=\frac{1}{3}\left(\frac{1}{2} \cdot 6 \cdot \frac{6 \sqrt{3}}{2}-\pi\left(\frac{1}{3} \cdot \frac{6 \sqrt{3}}{2}\right)^{2}\right)=\frac{1}{3}(9 \sqrt{3}-3 \pi)=3 \sqrt{3}-\pi .
$$
13) Knowing that the equilateral triangle in the figure has a side length of 3 and that the arc of the circle is tangent to two sides, what is the area of the gray figure?
(A) $\sqrt{3}-\frac{\pi}{6}$
(B) $\pi-\sqrt{3}$
(C) $2 \sqrt{3}-\pi$
(D) $\frac{\sqrt{3}+\pi}{6}$
(E) $3 \sqrt{3}-\pi$

![](https://cdn.mathpix.com/cropped/2024_04_17_f693bde306ad8eee0381g-2.jpg?height=239&width=250&top_left_y=537&top_left_x=1180)