Let's find this length first
6√2 is the hypotenuse, then
[tex]\begin{gathered} (6\sqrt{2})^2=6^2+y^2 \\ \\ y^2=(6\sqrt{2})^2-6^2 \\ \\ y^2=36\cdot2-36 \\ \\ y^2=36 \\ \\ y=\sqrt{36}=6 \end{gathered}[/tex]
Then we can find x because
[tex]\begin{gathered} x^2=y^2+12^2 \\ \\ x^2=6^2+12^2 \\ \\ x^2=36+144 \\ \\ x^2=180 \\ \\ x=\sqrt{180} \\ \\ x=6\sqrt{5} \end{gathered}[/tex]
The length of x is
[tex]x=6\sqrt{5}[/tex]
