Answer: [tex]2\sqrt{113}[/tex]
Work Shown
[tex]a^2 + b^2 = c^2 \text{ .... Pythagorean Theorem}\\\\c = \sqrt{a^2 + b^2}\\\\\text{x} = \sqrt{14^2 + 16^2}\\\\\text{x} = \sqrt{452}\\\\\text{x} = \sqrt{4*113}\\\\\text{x} = \sqrt{4}*\sqrt{113}\\\\\text{x} = 2\sqrt{113}\\\\[/tex]
I used GeoGebra to confirm the answer is correct.
