We have a triangle $\triangle ABC$ and a point $K$ on $BC$ such that $AK$ is an altitude to $\triangle ABC$. If $AC = 8,$ $BK = 2$, and $CK = 3,$ then what is $AB$?

ABJeIIy Aug 2, 2024

#1**+1 **

Well, this problem can be solved by the pythagorean thereom.

First off, note that AK is the altitude. This gives us multiple right triangles to work with.

We have

\(AK = \sqrt { AC^2 - CK^2 } = \sqrt { 8^2 - 3^2 } = \sqrt { 55 } \\ AB = \sqrt { AK^2 + BK^2 } = \sqrt { 55 + 4 } = \sqrt { 59 } \)

So the answer is sqrt59.

"Thanks! :)

NotThatSmart Aug 2, 2024