, 2013

Let $s$ be the side length of each of the 60 identical squares.

Since the diagonal of each of the squares has length $2$, then by the Pythagorean Theorem, $s^2+s^2=2^2$ or $2s^2=4$, which gives $s^2=2$ or $s=\sqrt{2}$, since $s>0$.

Now $PQ=5s$ and $PS=12s$, so since $QS>0$, then by the Pythagorean Theorem, $$QS = \sqrt{PQ^2 + PS^2}=\sqrt{(5s)^2+(12s)^2}=\sqrt{25s^2+144s^2}=\sqrt{169s^2}=13s$$ Since $QS=13s$ and $s=\sqrt{2}$, then $QS=13\sqrt{2} \approx 18.38$.

Of the given choices, this is closest to 18.