First find the hypotenuse of the triangle with side lengths sqrt 2 and sqrt 2:
(sqrt2)^2 + (sqrt2)^2 = x^2
x^2 = 4
x = 2
Since x is one of the legs of the triangle with side lengths y and 1, we can use the Pythagorean Theorem again:
1^2 + 2^2 = y^2
y^2 = 5
y = sqrt5, or choice (B).