+1839 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 $AB$?
+1897 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.
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