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We have that  the sides are

2 , 2r , 2r^2          so.....by the Pythagorean Theorem

2^2  + (2r)^2  = (2r^2)^2

4  + 4r^2  = 4r^4         divide through by 4

1  + r^2  = r^4        rearrange

r^4 - r^2  - 1  =   0             let  r^2   = m    and we have

m^2 - m - 1  = 0            complete the square on m

m^2 - m + 1/4  = 1 + 1/4

(m - 1/2)^2 = 5/4           take the positive root

m - 1/2  = √5 / 2

m = [ 1 + √5 ] / 2 =  r^2      ⇒  this is  known as  the irrational number  "Phi"

So  

m = r^2 = Phi

√m  = r  =  √Phi

So.....the side lengths are

2 ,  2 √Phi , 2Phi

So....the hypotenuse is    2Phi  units in length

Proof :

2^2  + (2√Phi)^2 = (2Phi)^2

4 + 4Phi  =  4Phi^2        divide through by 4

1 + Phi   = Phi^2     which is an identity