We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
70
1
avatar+58 

Hi forum!

Man I haven't been active in a while.

 

Okay, here's the problem. 
 

The median of a triangle connects a vertex of the triangle to the midpoint of the side opposite the vertex. In triangle ABC, we have AB=3 and AC=4. Side BC and the median from A to BC have the same length. What is the length of BC?

 

I used the Law of Cosines to try to find the answer.

I got close: basically, BC squared is the same as the cosine of angle A.

But that's where I get stuck; I can't find angle A from the given information(or maybe I'm just not looking hard enough!).

 

Thanks guys! - BasicMaths

 Oct 11, 2019
edited by BasicMaths  Oct 11, 2019
 #1
avatar+104969 
+2

We can use the Law of Cosines twice to find BC

 

Let the intersection of the median drawn from A to BC =  D

 

The  angles formed by the intersection of  the median with BC  will be supplementary...so, their cosines will have opposite signs.....so cos BDA  =  - cos CDA  .....so  -cos BDA  = cos CDA

 

So we have that

3^2 = [ (1/2) BC]^2 + (BC)^2  - 2[ (1/2)BC] [BC] cos BDA    (1)

 

4^2  = [ (1/2)BC]^2 + (BC)^2  - 2[(1/2)BC] [BC] cos CDA

4^2 = [ (1/2)BC]^2  + (BC)^2  - 2[ (1/2)BC][BC] (-cos BDA)       

4^2  = [ (1/2)BC]^2  + (BC)^2  + 2[ (1/2)BC][BC] (cos BDA)    (2)

 

Add (1) and (2)   and we get that

 

25  =  2[ (1/2)BC]^2 + 2[BC]^2      simplify

 

25  = 2 (1/4)BC^2 + 2BC^2

 

25  = (1/2)BC^2 + 2BC^2

 

25  = (2.5)BC^2        divide both sides by 2.5

 

10  = BC^2

 

BC  = √10

 

Here's a pic :

 

 

 

 

cool cool cool

 Oct 11, 2019
edited by CPhill  Oct 11, 2019
edited by CPhill  Oct 11, 2019

15 Online Users

avatar