Given that$, \text{Area (EFGH)} = 62.5\% \;\text{Area (ABCD)}$
$\Rightarrow \text{Area (EFGH)} = \frac{5}{8}\;\text{Area(ABCD)}$
Let $\text{EB} = 1$ and $\text{CG} = r.$
Similarly, the rest other dimensions are also of lengths $1$ or $r$ units
Applying pythagoras theorem on $\triangle \text{DHG}$ we get $\text{GH} = \sqrt{(1+r^{2})}$
$\text{Area(ABCD)} = (1+r)^2$ and $\text{Area(EFGH)} = \left(\sqrt{(1+r^{2})}\right)^2 = (1 + r^{2})$
$\Rightarrow (1+r^{2}) = \frac{5}{8}(1+r)^{2}$
$\Rightarrow (1 + r^{2}) = \frac{5}{8}(1+r^{2}+2r)$
$\Rightarrow 8 + 8r^{2} = 5 + 5r^{2} + 10r$
$\Rightarrow 3r^{2} – 10r + 3 = 0$
$\Rightarrow 3r^{2} – 9r – r + 3 = 0$
$\Rightarrow 3r(r – 3) – 1(r – 3) = 0$
$\Rightarrow (r-3)(3r-1) = 0$
$\Rightarrow r = 3,\frac{1}{3}$
As $\text{CG} > \text{EB}$
So, $r = 3$
$\therefore$ The ratio $\text{EB} : \text{CG} = 1 : r = 1 : 3$
Option (D)