, 2014

Suppose that $R$ is the point at the bottom of the solid directly under $Q$ and $S$ is the back left bottom corner of the solid (unseen in the problem’s diagram).

Since $QR$ is perpendicular to the bottom surface of the solid, then $\triangle PRQ$ is right-angled at $R$ and so $PQ^2 = PR^2 + RQ^2$.

We note also that $\triangle PSR$ is right-angled at $S$, since the solid is made up of cubes.

Therefore, $PR^2 = PS^2 + SR^2$.

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This tells us that $PQ^2 = PS^2 + SR^2 + RQ^2$.

Since the edge length of each cube in the diagram is 1, then $PS=1$, $SR=4$, and $RQ=3$.

Therefore, $PQ^2 = 1^2 + 4^2 + 3^2 = 26$.

Since $PQ >0$, then $PQ = \sqrt{26}$.