Answer:
[tex]\large\boxed{x=8\sqrt2}[/tex]
Step-by-step explanation:
[tex]\bold{METHOD\ 1}\\\\\text{Use the Pythagorean theorem:}\\\\leg^2+leg^2=hypotenuse^2\\\\\text{We have}\ leg=8,\ leg=8,\ hypotenuse=x.\\\text{Substitute:}\\\\x^2=8^2+8^2\\\\x^2=2(8^2)\to x=\sqrt{2(8^2)}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\x=\sqrt2\cdot\sqrt{8^2}\qquad\text{use}\ \sqrt{a^2}=a\\\\x=8\sqrt2[/tex]
[tex]\bold{METHOD\ 2}\\(look\ at\ the\ picture)\\\\\text{We have}\ a=8.\ \text{Therefore}\ a\sqrt2=8\sqrt2[/tex]
