+0  
 
0
17
1
avatar+1072 

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 $AB$?

 Jan 13, 2024
 #1
avatar+128732 
+1

                        A

                                     8

 

B        2           K         3               C

 

Because AK is an altitude,  we  can manipulate the Pythagorean Theorem twice

 

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

 

AB = sqrt [ AK^2  + BK^2 ] = sqrt  [ 55  + 4 ] = sqrt [ 59 ]

 

 

cool cool cool

 Jan 13, 2024

1 Online Users