We know that the perimeter is 72 meters, so we have
[tex] y+2x+(2x+6)=72 \iff y+4x+6=72 \iff y=-4x+66 [/tex]
Now we can use the pythagorean theorem: the sum of the squares of the legs is the square of the hypotenuse:
[tex] (2x)^2+(2x+6)^2 = y^2 \iff (2x)^2+(2x+6)^2 = (-4x+66)^2 [/tex]
Expand the squares to get
[tex] 8 x^2 + 24 x + 36 = 16 x^2 - 528 x + 4356 [/tex]
Bring everything to the right hand side to get
[tex] 8 x^2 - 552 x + 4320 = 0 [/tex]
Divide both sides by 8:
[tex] x^2 - 69x + 540 = 0 [/tex]
This equation has solutions [tex] x=9,\ x=60 [/tex]
This solutions would yield the following side lengths:
[tex] x=9\implies\begin{cases} 2x = 18\\2x+6=24\\-4x+66=30\end{cases}[/tex]
[tex] x=60\implies\begin{cases} 2x = 120\\2x+6=126\\-4x+66=-174\end{cases}[/tex]
Since we can't have negative side lengths, we can only accept the solution x=9.
In fact, you can check that 18,24,30 is a pythagorean triplet, i.e.
[tex] 18^2+24^2=30^2 [/tex]
