+130071 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
