smc

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 $\boxed{\frac{1}{2}s(2-\sqrt{3})}$ when $s=1$.