+0  
 
-1
53
1
avatar

In triangle ABC, AB = 13, BC = 14, and AC = 15. Let M be the midpoint of \(\overline{BC}\). Find AM.

 Apr 9, 2020
 #1
avatar+111329 
+1

          A

 

 13                 15    

 

B       14                C

 

 

Let  B  = (0,0)   and C  = ( 14,0)

 

So......M  =  (7,0)  ....so   BM  =  7

 

Using  the Law  of Cosines

 

AC^2 = AB^2  + BC^2  - 2(AB * BC) cos (ABC)

 

15^2  =  13^2  + 14^2  - 2(13 * 14)  cos (ABC)

 

[ 15^2  - 13^2  - 14^2 ] / [ -2 ( 13 * 14) ]  = cos ABC

 

5/13  =  cos ABC

 

 

And applying this again we have  that

 

AM  = √ [ BM^2 + BA^2  - 2 (BM * BA) cos ABC ]

 

AM = √ [ 7^2  + 13^2 - 2 ( 7 * 13) (5/13) ]

 

AM  = √[ 49 + 169 - 2*7*5 ]

 

AM = √ [ 49 + 169 - 70]

 

AM = √148  = 2√37  units

 

 

cool cool cool

 Apr 9, 2020

17 Online Users

avatar