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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+106539 
+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

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