Since it is a right triangle then you can use Pythagoras' theorem, that is,
[tex]\begin{gathered} a^2+b^2=c^2 \\ \text{ Where c is the hypotenuse and} \\ a\text{ and b are the legs of the right triangle} \end{gathered}[/tex]
So, in this case, you have
[tex]\begin{gathered} a=7 \\ b=9 \\ c=x \end{gathered}[/tex][tex]\begin{gathered} a^2+b^2=c^2 \\ \text{ Replacing values} \\ (7)^2+(9)^2=c^2 \\ 49+81=c^2 \\ 130=c^2 \\ \text{ Apply square root to both sides of the equation} \\ \sqrt[]{130}=\sqrt[]{c^2} \\ \sqrt[]{130}=c=x \\ \text{ or} \\ 11.4=c=x \end{gathered}[/tex]
Therefore, the value of x in simplest radical form is
[tex]\sqrt[]{130}[/tex]
