Ratio

Let the side be $a$ . Area of square will be $a^{2}$ . So the area of rhombus must be $\frac{a^{2}}{2}$ . Let diagonals be denoted by $d_{1}$ and $d_{2}$ , $d_{1}>d_{2}$

$\frac{1}{2}d_{1}d_{2}=\frac{a^{2}}{2}$ $\Rightarrow d_{1}d_{2}=a^{2}$ $\color{#D61F06}{\text{.......(1)}}$

By pythagoras theorem,

$\big (\frac{d_{1}}{2} \big )^{2}+\big (\frac{d_{2}}{2} \big )^{2}=a^{2}$

$\Rightarrow d_{1}^{2}+d_{2}^{2}=4a^{2}$ $\color{#D61F06}{\text{.......(2)}}$

$\color{#D61F06}{\text{(2)}}+2×\color{#D61F06}{\text{(1)}}$

$(d_{1}+d_{2})^{2}=6a^{2}$

$d_{1}+d_{2}=\sqrt{6}a$ $\color{#D61F06}{\text{.......(3)}}$

$\color{#D61F06}{\text{(2)}}-2×\color{#D61F06}{\text{(1)}}$

$(d_{1}-d_{2})^{2}=2a^{2}$

$d_{1}-d_{2}=\sqrt{2}a$ $\color{#D61F06}{\text{.......(4)}}$

Divide $\color{#D61F06}{(3)}$ by $\color{#D61F06}{(4)}$ , and apply componendo and dividendo to get

$\frac{d_{1}}{d_{2}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}=\boxed{3.732}$