### Instruction: 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 $AB$?### Response: I know that an altitude of a triangle forms a right angle with the base, so $\angle AKC$ and $\angle AKB$ are both $90^\circ$.
That means $\triangle AKC$ and $\triangle AKB$ are both right triangles, and I can use the Pythagorean theorem to find $AK$ and $AB$.
Let me start with $\triangle AKC$.
By the Pythagorean theorem, I have $AK^2 + CK^2 = AC^2$.
Plugging in the given values, I get $AK^2 + 4^2 = 9^2$.
Simplifying, I get $AK^2 = 9^2 - 4^2 = 65$.
Taking the square root of both sides, I get $AK = \sqrt{65}$.
Now I can use $\triangle AKB$ to find $AB$.
By the Pythagorean theorem again, I have $AK^2 + BK^2 = AB^2$.
Plugging in the values I know, I get $\sqrt{65}^2 + \sqrt{5}^2 = AB^2$.
Simplifying, I get $65 + 5 = AB^2$.
Taking the square root of both sides, I get $AB = \sqrt{70}$.