Inside square $ABCD$ with side $s$, quarter-circle arcs with radii $s$ and centers at $A$ and $B$ are drawn. These arcs intersect at a point $X$ inside the square. How far is $X$ from the side of $CD$?
$\text{(A) } \tfrac{1}{2} s(\sqrt{3}+4)\quad \text{(B) } \tfrac{1}{2} s\sqrt{3}\quad \text{(C) } \tfrac{1}{2} s(1+\sqrt{3})\quad \text{(D) } \tfrac{1}{2} s(\sqrt{3}-1)\quad \text{(E) } \tfrac{1}{2} s(2-\sqrt{3})$
All answers are proportional to $s$, so for ease, let $s=1$.
Let $ABCD$ be oriented so that $A(0, 0), B(1, 0), C(1, 1), D(0, 1)$.
The circle centered at $A$ with radius $1$ is $x^2 + y^2 = 1$.  The circle centered at $B$ with radius $1$ is $(x - 1)^2 + y^2 = 1$.  Solving each equation for $1 - y^2$ to find the intersection leads to $(x - 1)^2 = x^2$, which leads to $x = \frac{1}{2}$.
Plugging that back in to $x^2 + y^2 = 1$ leads to $y^2 = 1 - \frac{1}{4}$, or $y = \pm \frac{\sqrt{3}}{2}$.  Since we want the intersection within the square where $0 < x, y < 1$, we take the positive solution, and the intersection is at $X(\frac{1}{2}, \frac{\sqrt{3}}{2})$.
The distance from $X$ to side $CD$, which lies along the line $y=1$, is $1 - \frac{\sqrt{3}}{2}$, or $\frac{2  -\sqrt{3}}{2}$.  This is answer $\fbox{E}$ when $s=1$.