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The centroid of ABC is G.  Find BG.

 

 Apr 23, 2022
 #1
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I don't think this triangle is  possible

 

Let CM  be one median  ....its length =  (4)(3/2)   = 6

Let AN   be another median...its length =  (8) (3/2) = 12

 

Let BC  = x    and  BA  = y

 

And we have two right triangles formed

One is  CBM   with  legs  of x , y/2  and a  hypotenuse of 6

Another is  BNA  with legs of x/2, y and a hypotenuse of 12

 

By the Pythagorean Theorem

 

x^2 + (y/2)^2  =  6^2     ⇒  x^2  + y^2/4  =  36       ⇒          4x^2  + y^2  =  144   

(x/2)^2 + y^2 =  12^2      ⇒  x^2/4  +y^2  =  144    ⇒           x^2  + 4y^2  = 576   

 

The  solutions to these equations  are  (x, y)  = (0, 12)  or  (0 , -12)

 

This means that BC  =  0    (impossible)   and  BA =  12

 

But if BA  = 12, then  BM =  6

But this  means that in right triangle CBM,   CM =  6  and BM  = 6  which means that  one leg   is equal to a hypotenuse which is also impossible

 

cool cool cool

 Apr 23, 2022

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