9) In the rhombus in the figure, triangles $A B C$ and $A C D$ are equilateral and have a side length of $1 \mathrm{~m}$. If we rotate the rhombus by $60^{\circ}$ about vertex $A$, what is the area of the surface covered by the rhombus during the rotation?
(A) $\frac{\pi}{2} \mathrm{~m}^{2}$
(E) $2 \mathrm{~m}^{2}$.
(B) $1 \mathrm{~m}^{2}$, (C) $\pi \mathrm{~m}^{2}$,
(D) $\frac{\pi}{3} \mathrm{~m}^{2}$,
9. The answer is (A).

Suppose the rhombus rotates counterclockwise. Draw the semicircle $\mathcal{C}$ with center $A$, passing through $B, C$, and $D$, as indicated in the figure, and let $E$ be the point diametrically opposite to $D$ on $\mathcal{C}$. During the rotation, each of the vertices $B, C$, and $D$ moves on $\mathcal{C}$. Moreover, since the rotation is $60^{\circ}$, vertex $D$ goes to $C$, vertex $C$ goes to $B$, and vertex $B$ goes to $E$.

![](https://cdn.mathpix.com/cropped/2024_04_17_e415f128e14398de33bcg-3.jpg?height=243&width=440&top_left_y=1592&top_left_x=865)

Overall, the rhombus covers, during the rotation, the area enclosed by the semicircle and the diameter $D E$, which is half of a circle with a radius of $1 \mathrm{~m}$, whose area is $\frac{\pi}{2} \mathrm{~m}^{2}$. [Problem proposed by A. Sambusetti.]