+0  
 
0
143
1
avatar+15 

We have a triangle $\triangle ABC$ and a point $K$ on $BC$ such that $AK$ is an altitude to $\triangle ABC$. If $AC = 9,$ $BK = \sqrt{5}$, and $CK = 4,$ then what is $AB$?

 Aug 16, 2018
 #1
avatar+98128 
+1

If AK is an  altitude, then angle AKC  is   a right angle

 

By the Pythagorean Theorem...

 

AK  = √  [ AC^2  - CK^2 ] = √ [ 9^2  - 4^2 ] = √ [ 81 - 16] = √65

 

Also...angle  AKB is right....which implies that

 

AB  = √ [ AK^2 + BK^2 ] = √ [  (√ 65)^2 + (√5)^2 ] =  √ [65 + 5 ] =    √70  units^2

 

 

 

cool cool cool

 Aug 16, 2018

19 Online Users

avatar
avatar