jmc

The line $y=\frac{x}{2}$ will intersect the two vertical sides of the square, as shown below: The equation of the right side of the square is $x=a,$ so we have $y=\frac{x}{2}=\frac{a}{2},$ which means that the intersection point with the right side of the square is $\left(a,\frac{a}{2}\right).$ Similarly, the equation of the left side of the square is $x=-a,$ so we have $y=\frac{x}{2}=-\frac{a}{2},$ which means that the intersection point with the left side of the square is $\left(-a,-\frac{a}{2}\right).$ It follows that the sides of each quadrilateral have lengths $\frac{a}{2},$ $2a,$ $\frac{3a}{2},$ and $\sqrt{a^2+(2a{)}^2}=a\sqrt{5},$ by the Pythagorean theorem. Hence, the perimeter of the quadrilateral is $$\frac{a}{2}+2a+\frac{3a}{2}+a\sqrt{5}=\left(4+\sqrt{5}\right)a,$$and when this is divided by $a,$ we get $\boxed{4+\sqrt{5}}.$