Wedged in equilateral triangles.

Let $m$ be a side of square $ABCD$ .

Let $A_{e}$ be the area of equilateral $\triangle{DFE}$ $\implies 4A_{e} = 4(\dfrac{1}{2})(a)(\dfrac{\sqrt{3}}{2}a) = \sqrt{3}a^2$

Let $A_{I}$ be the area of $\triangle{GFD}$

The height $h_{\triangle{GFD}} = (m - a)\sin(60^{\circ}) = \dfrac{\sqrt{3}}{2}(m - a) \implies$

$4A_{I} = 4(\dfrac{1}{2})(m - a)^2(\dfrac{\sqrt{3}}{2}) = \sqrt{3}(m - a)^2$

Using the law of cosines on $\triangle{GFD}$ with included $\angle{FGD} \implies$

$a^2 = 2(m - a)^2 (\dfrac{3}{2}) = 3(m - a)^2 = 3a^2 -6ma + 3m^2 \implies 2a^2 - 6ma + 3m^2 = 0$

$\implies a = \dfrac{3 \pm \sqrt{3}}{2}m$

$a = \dfrac{3 + \sqrt{3}}{2}m \implies m - a = -(1 + \sqrt{3})m < 0$

$a = \dfrac{3 - \sqrt{3}}{2}m \implies m - a = \dfrac{\sqrt{3} - 1}{2}m$

$\therefore$ we drop $a = \dfrac{3 + \sqrt{3}}{2}m$ and choose $a = \dfrac{3 - \sqrt{3}}{2}m$

$\implies 4A_{e} = \dfrac{\sqrt{3}}{4}(3 - \sqrt{3})^2 m^2 = \dfrac{3\sqrt{3}}{2}(2 - \sqrt{3}) m^2$

and

$4A_{I} = \sqrt{3}(m - a)^2 = \dfrac{\sqrt{3}}{4}(\sqrt{3} - 1)^2 m^2 = \dfrac{\sqrt{3}}{2}(2 - \sqrt{3})m^2$

$\implies A_{s} = m^2 - (4A_{e} + 4A_{I}) = (1 - (2\sqrt{3}(2 - \sqrt{3})))m^2 =$ $(1 - (4\sqrt{3} - 6))m^2$

$= (7 - 4\sqrt{3})m^2$

$\implies \dfrac{A_{s}}{A_{\triangle{ABCD}}} = (7 - 4\sqrt{3}) = \alpha - \beta\sqrt{\lambda} \implies \alpha + \beta + \lambda = \boxed{14}$ .