Right Triangle

catmg's answer:

(x)^2 + (x+y)^2 = (x + 2y)^2 (this is the pythagreon theorum)

x^2 + x^2 + y^2 + 2xy = x^2 + 4y^2 + 4xy

2x^2 + y^2 + 2xy = x^2 + 4y^2 + 4xy

x^2 = 3y^2 + 2xy

So we're looking for xy, and we could solve for it, but let's just say x = 1 for simplicity.

1^2 = 3*y^2 + 2(1)y

1 = 3y^2 + 2y. 

y = 1/3. 

So y/x = 1/3/1 = 1/3. 

I hope this helped. :)))

=^._.^=

here's link

https://web2.0calc.com/questions/right-triangle_21

Taking this from

x^2 =  3y^2 + 2xy

x^2  - 2xy  - 3y^2  = 0

(x - 3y) ( x + y) = 0

x + y = 0    not possible

x - 3y = 0

x = 3y

y / x =   1 /  3