2007/08 AHSMC Part 1 Problem 3

Since $AB \le 1$ and $BC \le 2$, the area of triangle $ABC$ is at most $2$. This maximum value can be attained when $AB = 1$ and $BC = 2$ and are perpendicular to each other. Now $CA = \sqrt{5}$, and we indeed have $2 \le CA \le 3$. The answer is (C).