Answer:
[tex]x = \dfrac{\sqrt{22}}{2}[/tex]
Step by step explanation:
[tex]\text{This is an isosceles right triangle}~ \left(45^{\circ} - 45^{\circ} - 90^{\circ}\right).\\\\\text{Apply Pythagorean theorem,}\\\\~~~~~~\left(\sqrt{11}\right)^2 =x^2 +x^2 \\\\\implies 11 = 2x^2 \\\\\implies x^2 = \dfrac{11}2\\\\\implies x = \sqrt{\dfrac{11}{2}}\\\\\implies x = \dfrac{\sqrt{11}}{\sqrt 2}\\\\\implies x = \dfrac{\sqrt{11}\sqrt 2}{2}\\\\\implies x = \dfrac{\sqrt{22}}{2}[/tex]
