17. The answer is $(\mathbf{C})$. Referring to the figure, we observe that triangle $A P Q$ is equilateral and has a side length of $10 \mathrm{~m}$. We can then calculate its height $A H: \overline{A H}=\frac{\sqrt{3}}{2} \overline{P Q}=$ $\frac{\sqrt{3}}{2} 10 \mathrm{~m}=5 \sqrt{3} \mathrm{~m}$, and $\overline{A K}=\overline{C H}=\overline{K H}-\overline{A H}=(10-5 \sqrt{3}) \mathrm{m}$. Therefore, we also have: $\overline{A C}=\overline{A H}-\overline{C H}=(5 \sqrt{3}-(10-5 \sqrt{3})) \mathrm{m}=10(\sqrt{3}-1) \mathrm{m}$. The quadrilateral $A B C D$ is a square, as it has all sides of equal length and diagonals of equal length, due to the symmetry properties of the figure. The length of its side $A B$ is equal to the length of its diagonal $A C$ divided by $\sqrt{2}$, that is: $\overline{A B}=\frac{10(\sqrt{3}-1)}{\sqrt{2}} \mathrm{~m}=5(\sqrt{6}-\sqrt{2}) \mathrm{~m}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_d9e56cae5925cb0c0feag-5.jpg?height=574&width=617&top_left_y=1672&top_left_x=774)
17) In the square in the figure, four circular arcs have been drawn, each with its center at one of the vertices of the square and a radius equal to the side of the square, which measures $10 \mathrm{~m}$. What is the distance between points $A$ and $B$?
(A) $3(\sqrt{6}-1) \mathrm{m}$,
(B) $5 \mathrm{~m}$
(C) $5(\sqrt{6}-\sqrt{2}) \mathrm{m}$
(D) $8(\sqrt{3}-1) \mathrm{m}$, (E) none of the above.

![](https://cdn.mathpix.com/cropped/2024_04_17_260a987a255c499fc983g-2.jpg?height=418&width=426&top_left_y=1510&top_left_x=973)