problem: A triangle has vertices at coordinates $(1, 2), (7, 10)$ and $(1, 12)$.  What is the number of units in the length of the shortest side of the triangle?
solution: We must find the distance between each pair of points.
  
  The distance between $(1, 2)$ and $(1, 12)$ is simply 10, since these two points have the same $x$-coordinate.
  
  The distance between $(1, 2)$ and $(7, 10)$ is \[\sqrt{(1-7)^2 + (2-10)^2} = \sqrt{36 + 64} = 10.\]
  
  The distance between $(7, 10)$ and $(1, 12)$ is \[\sqrt{(7 - 1)^2 + (10 - 12)^2} = \sqrt{36 + 4} = 2\sqrt{10}.\]
  
  Out of 10, 10, and $2\sqrt{10}$, $2\sqrt{10}$ is the shortest value. We know this because $\sqrt{10} > \sqrt{9}$, so $\sqrt{10} > 3$, so $2\sqrt{10} < (\sqrt{10})^2 = 10$. Thus, the shortest side of the triangle has length $\boxed{2\sqrt{10}}$.