Rhombus Square

Let the square and the rhombus each have side length $s$ . Since the diagonals of a rhombus are always perpendicular to each other, its diagonals will have lengths $2x, 2\sqrt{s^2-x^2}.$ The rhombus' area is simply the sum of four congruent right triangles, and the required area equation computes to:

$\frac{s^2}{2} = 4[\frac{1}{2} x\sqrt{s^2-x^2}]$ (i)

which we can obtain a quadratic equation according to:

$s^4 = 16x^2(s^2-x^2) \Rightarrow s^4 - 16x^2s^2 + 16x^4 = 0 \Rightarrow s^2 = \frac{16x^2 \pm \sqrt{256x^4 - 4(1)(16x^4)}}{2} \Rightarrow s^2 = (8\pm 4\sqrt{3})x^2$ (ii).

The ratio of the rhombus' longer diagonal to its shorter one equals:

$R = \frac{2\sqrt{s^2-x^2}}{2x} = \frac{\sqrt{ (8 \pm 4\sqrt{3})x^2 - x^2}}{x} = \sqrt{7 \pm 4\sqrt{3}}$

and since we require $R > 1$ , we admit only the positive root $\Rightarrow R = \sqrt{7+4\sqrt{3}} \approx \boxed{3.732}.$