Area Frenzy.

$A_{EDF} = a + 2$

$A_{ACF} = \dfrac{1}{2}(x - 2)y = a + 1 \implies$

$\boxed{(1): xy - 2y = 2(a + 1)}$

$A_{\triangle{ABE}} = \dfrac{1}{2}(y - (a + 2))x = a \implies$

$\boxed{(2): xy - (a + 2)x = 2a}$

Subtraction equation $(2)$ from equation $(1)$ we obtain:

$(a + 2)x - 2y = 2 \implies x = \dfrac{2(1 + y)}{a + 2} \implies$

Replacing $x = \dfrac{2(1 + y)}{a + 2}$ in equation $(1)$ we obtain:

$y^2 - (a + 1)y - (a + 1)(a + 2) = 0 \implies$ $y = \dfrac{a + 1 \pm \sqrt{(a + 1)(5a + 9)}}{2}$

$a + 1 < \sqrt{5a^2 + 14a + 9}$ , so drop negative root and choose

$y = \dfrac{a + 1 + \sqrt{5a^2 + 14a + 9}}{2} \implies$

$x = \dfrac{a + 3 + \sqrt{5a^2 + 14a + 9}}{a + 2}$

$\implies$ The area of the square $A_{s} = 3a + 3 + \sqrt{5a^2 + 14a + 9} \implies$

The desired area $A = 3a + 3 + \sqrt{5a^2 + 14a + 9} - (3a + 3) = \sqrt{5a^2 + 14a + 9} = 4\sqrt{6}$

$\implies 5a^2 + 14a + 9 = 96 \implies 5a^2 + 14a - 87 = 0 \implies (a - 3)(5a + 29) = 0$

$a > 0 \implies a = \boxed{3}$ .