Centroid of a triangle

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