16. The answer is $(\mathbf{A})$.

Draw the segments $D G, E H$ and $F G$ parallel to the sides of the square as shown in the figure; clearly, $F G=1$ and, since triangle $A B C$ is similar to triangle $F G H$, it follows that $H G=\frac{1}{2}$. Therefore, the area of triangle $F G H$ is $\mathcal{A}(F G H)=\left(\frac{1}{2}\right)^{2} \mathrm{~m}^{2}$, and the area of the shaded part is $4 \cdot \frac{1}{4}=1 \mathrm{~m}^{2}$. [Problem proposed by F. Mugelli]

![](https://cdn.mathpix.com/cropped/2024_04_17_652336b520aa23eab041g-5.jpg?height=385&width=365&top_left_y=127&top_left_x=1563)
16) Calculate the area of the shaded part in the figure, knowing that the side of the square is $2 \mathrm{~m}$ long and that the points of the star fall at the midpoints of the sides of the square.
(A) $1 \mathrm{~m}^{2}$
(B) $2 \mathrm{~m}^{2}$
(C) $\frac{1}{2} \mathrm{~m}^{2}$
(D) $\pi \mathrm{m}^{2}$
(E) $2 \sqrt{2} \mathrm{~m}^{2}$

![](https://cdn.mathpix.com/cropped/2024_04_17_9ee229095b4334063cc2g-2.jpg?height=273&width=306&top_left_y=1609&top_left_x=1070)