Since $\angle W A X=90^{\circ}$ regardless of the position of square $A B C D$, then $A$ always lies on the semi-circle with diameter $W X$.

The centre of this semi-circle is the midpoint, $M$, of $W X$.

To get from $P$ to $M$, we must go up 4 units and to the left 3 units (since $W X=2$ ), so $P M^{2}=3^{2}+4^{2}=25$ or $P M=5$.

Since the semi-circle with diameter $W X$ has diameter 2 , it has radius 1 , so $A M=1$.

So we have $A M=1$ and $M P=5$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-191.jpg?height=431&width=331&top_left_y=153&top_left_x=954)

Therefore, the maximum possible length of $A P$ is $5+1=6$, when $A, M, P$ lie on a straight line.
In the diagram, four squares of side length 2 are placed in the corners of a square of side length 6 . Each of the points $W, X, Y$, and $Z$ is a vertex of one of the small squares. Square $A B C D$ can be constructed with sides passing through $W, X, Y$, and $Z$. The maximum possible distance from $A$ to $P$ is closest to
(A) 5.2
(B) 5.4
(C) 5.6
(D) 5.8
(E) 6.0