A triangle has vertices at coordinates (2,2), (5,6) and (6,2).  What is the number of units in the length of the longest side of the triangle?
We must find the distance between each pair of points.

The distance between $(2, 2)$ and $(6, 2)$ is 4, since these two points have the same $y$-coordinate.

The distance between $(2, 2)$ and $(5, 6)$ is $\sqrt{(5 - 2)^2 + (6 - 2)^2} = \sqrt{9 + 16} = 5$.

The distance between $(5, 6)$ and $(6, 2)$ is $\sqrt{(6 - 5)^2 + (2 - 6)^2} = \sqrt{1 + 16} = \sqrt{17}$.

Out of 4, 5, and $\sqrt{17}$, 5 is the largest value. Thus, the longest side of the triangle has length $\boxed{5}$.