Help with Trigonometry 45-45-90 triangle

I have a right isosceles triangle with the hypotenuse of 11. How do I find the two congruent sides? Please explain and show work.
In simplest radical form

Pythagorean Theorem: a^2+b^2=c^2

x^2+x^2 = 11^2
2x^2 = 121
2x^2 / 2 = 121 / 2
x^2 = 121/2
√x^2 = √(121/2)
x = √121 / √2
x = 11 / √2
x = 11√2 / √2√2
x = 11√2 / 2

Answer: The other two sides are each 11√2 / 2 units long.

a = b in pythagorean thrm so

a^2 + a^2 = 11^2

2a^2 = 121

a^2 = 121/2

a = sqrt(121/2) = 11/sqrt(2) = (11sqrt(2))/2 = 11/2(sqrt(2)

With a 45-45-90, the other sides are equal to hypotenuse/sqrt(2)