Trying to find the length $x$

Question

I'm asked to find the length $x$. To do that, I tried to draw perpendicular drawn from $DC$ to $AB$. However, it did not actually give anything useful. Any help is greatly appreciated.

Answer 1

Triangle $ACD$ is isosceles and right so by Pythagoras we have $AC = 8$ and $\angle ACD = 45$ so $\angle ACB = 90$ and then again by Pythagoras we have $$x^2 = 8^2+ 6^2\implies x=10$$

Answer 2

As triangle $ADC$ is isosceles we have angle $ACB=90^\circ$ and length $AC=\sqrt{(4\sqrt2)^2+(4\sqrt2)^2}=8$ by the pythagorean theorem. So then applying the pythagorean theorem again we have $x=\sqrt{8^2+6^2}=10$.