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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 $AB$?

 Jun 2, 2024
 #2
avatar+806 
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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! :)

 Jun 2, 2024

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