+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 Since AK is the altitude, so we can use the pythaogrean thereom.
We have'
\(AK = \sqrt { AC^2 - CK^2 } = \sqrt { 8^2 - 3^2 } = \sqrt { 55 } \\ AB = \sqrt { AK^2 + BK^2 } = \sqrt { 55 + 4 } = \sqrt { 59 }\)
Thanks! :)
