$P$ is a point interior to rectangle $ABCD$ and such that $PA=3$ inches, $PD=4$ inches, and $PC=5$ inches. Then $PB$, in inches, equals:
$\textbf{(A) }2\sqrt{3}\qquad\textbf{(B) }3\sqrt{2}\qquad\textbf{(C) }3\sqrt{3}\qquad\textbf{(D) }4\sqrt{2}\qquad \textbf{(E) }2$
From point $P$, create perpendiculars to all four sides, labeling them $a, b, c, d$ starting from going north and continuing clockwise.  Label the length $PB$ as $x$:

We have $a^2 + b^2 = 5^2$ and $c^2 + d^2 = 3^2$, leading to $a^2 + b^2 + c^2 + d^2 = 34$.
We also have $a^2 + d^2 = 4^2$ and $b^2 + c^2 = x^2$, leading to $a^2 + b^2 + c^2 + d^2 = 16 + x^2$.
Thus, $34 = 16 + x^2$, or $x = \sqrt{18} = 3\sqrt{2}$, which is option $\boxed{\textbf{(B)}}$