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The centroid of \triangle ABC is G.  Find AG.

 

 Dec 30, 2023
 #1
avatar+126665 
+1

We can find the  x  coordinate of A thusly

 

12^2   - x^2  = 17^2 - (18 - x)^2

 

144 - x^2  = 289 - (x^2 - 36x + 324)

 

144 = -35 + 36x

 

179 = 36x

 

x = 179 / 36

 

And we can find the y coordinate of A as follows

 

sqrt [ 12^2 - (179/36)^2 ] =  sqrt  [154583 ] / 36

 

So  A =  ( 179/36, sqrt [ 154583] / 36 )

 

Let the midpoint of  BC  = (9,0)  = D

 

AG will be 2/3 of the distance AD

 

So

 

AG  =  (2/3)sqrt  [ ( 179/36 - 9)^2  +  (154583/36^2) ]  = sqrt [ 542] / 3   ≈ 7.76

 

 

cool cool cool

 Dec 30, 2023
 #2
avatar+2 
+1

OMG Cphill! There are other methods besides coordinate geometry that are more efficient and easier to follow.

Holtran  Dec 31, 2023
edited by Holtran  Dec 31, 2023

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