We have a triangle $\triangle ABC$ and a point $K$ on $BC$ such that $AK$ is an altitude to $\triangle ABC$. If $AC = 9,$ $BK = \sqrt{5}$, and $CK = 4,$ then what is $AB$?

If AK is an altitude, then angle AKC is a right angle

By the Pythagorean Theorem...

AK = √ [ AC^2 - CK^2 ] = √ [ 9^2 - 4^2 ] = √ [ 81 - 16] = √65

Also...angle AKB is right....which implies that

AB = √ [ AK^2 + BK^2 ] = √ [ (√ 65)^2 + (√5)^2 ] = √ [65 + 5 ] = √70 units^2