+219 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 $AB$?
+1926 We knowt hat AK is the altitude, so we have
\(AK = \sqrt{AC^2 - CK^2} = \sqrt{8^2 - 3^2} = \sqrt{55}\) from the pythagorean theorem.
We can now find AB, with
\(AB = \sqrt{AK^2 + BK^2} = \sqrt{ 55 + 4 } = \sqrt { 59 } \)
so AB is \(\sqrt{59}\)
Thanks! :)
