SOLUTION: A right triangle has sides measuring 2x-1, 3x-13, and 3x-4 units. What is the value of x? I'm thinking I should try to do Pythag's theorem on it like (3x-13)^2 + (2x-1)^2 = (3x-4)

Algebra ->  Triangles -> SOLUTION: A right triangle has sides measuring 2x-1, 3x-13, and 3x-4 units. What is the value of x? I'm thinking I should try to do Pythag's theorem on it like (3x-13)^2 + (2x-1)^2 = (3x-4)      Log On


   

Question 230460: A right triangle has sides measuring 2x-1, 3x-13, and 3x-4 units. What is the value of x?
I'm thinking I should try to do Pythag's theorem on it like (3x-13)^2 + (2x-1)^2 = (3x-4)^2, but i'm not sure. everytime I do that I always get a different answer. Also, I'm not really sure which equation to use for the legs and which to use for the hypotenuse. Could you help me? Thanks!
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
Test these terms by replacing x with any number greater than 13/3=4.33.
Try 5
3*5-13=15-13=2 the short side
2*5-1=10-1=9 the second side.
3*5-4=15-4=11 the hypotenuse.
(3x-13)^2 + (2x-1)^2 = (3x-4)^2
9x^2-78x+169+4x^2-4x+1=9x^2-24x+16
9x^2+4x^2-9x^2-78x-4x+24x+169+1-16=0
4x^2-58x+154=0
(x-11)(4x-14)=0
x-11=0
x=11 ans.
4x-14=0
4x=14
x=14/4
x=3.5 ans.