Note that: $2x+2$ or $x+4$ is the hypotenuse Case 1: $2x+2$ is the hypotenuse: $\implies$ there is not solution Case 2: $x+4$ is the hypotenuse: $\implies$ $x=4$ and $x=1$ Thus: $x=4$ and $x=1$

(x+4); (x+2) ;(2x+2) ie (2(x+1)) i Considering 2x+2 to be the hypotenuse then by using Pythagoras theorem we get (x+2)^2 + (x+4)^2=(2(x+1))^2 2x^2+ 20 +12x= 4( x^2+ 1 +2x) x^2 +10+ 6x =2(x^2 +1+ 2x) x^2 -2x -8=0 (x+2)( x-4)=0 x=4 can not be - 2 because sides are not negative ii Now considering x+4 as hypotenuse Again by applying Pythagoras theorem we eill get the value of x as x=1 Therefore x can be 4 or 1 If there is any mistake then please mention so that I can rectify it