2014 AMC 10B Problem 22

The distance from the center of the square to the center of the semicircles can be found as a hypotenuse of a right triangle. One of the legs is from the center of the square to the center of one of the sides which is of distance $1.$ The other leg is from the center of the side to the center of one of the semicircles which is of distance $\dfrac 12.$ This also shows that the radius of the semicircles is $\dfrac 12.$ Therefore, the distance from the center of the square to the center of the semicircle is $$\sqrt{1^2 + \left(\dfrac 12\right)^2} = \dfrac {\sqrt 5}2.$$ Then we subtract $\dfrac 12$ for the radius of the semicircle. This makes the radius of the circle $$\dfrac{\sqrt 5 -1}2 .$$ Thus, the correct answer is B .