Look at the picture.
ΔABD and ΔBCD are similar, therefore:
[tex]\dfrac{y}{5}=\dfrac{11}{y}\ \ \ |\cross\ multiply\\\\y^2=55\to y=\sqrt{55}[/tex]
Use Pythagorean theorem:
[tex]x^2=5^2+(\sqrt{55})^2\\\\x^2=25+55\\\\x^2=80\to x=\sqrt{80}\\\\x=\sqrt{16\cdot5}\\\\x=\sqrt{16}\cdot\sqrt5\\\\\boxed{x=4\sqrt5}[/tex]
