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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 $AB$?

 Apr 8, 2024
 #1
avatar+128794 
+1

                             A

                            

                                     8

 

             B     2       K      3       C

 

Since AK is an  altitude

AK = sqrt [ AC^2 - KC^2 ]   =  sqrt [ 8^2  - 3^2 ] = sqrt [ 55]

 

And

 

AB = sqrt [ AK^2  + BK^2 ]  =  sqrt [ (sqrt 55)^2 + 2^2 ] = sqrt [ 59]

 

 

cool cool cool

 Apr 8, 2024

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