jmc

We must draw first! We first apply the Pythagorean Theorem to $△AKC$ to find that $A{K}^2=A{C}^2-K{C}^2=9^2-4^2=65.$ Now, we apply the Pythagorean Theorem to $△AKB$ to find that $A{B}^2=A{K}^2+B{K}^2=65+(\sqrt{5}{)}^2=70.$ Therefore, $AB=\boxed{\sqrt{70}}.$