Circle, square and equilateral triangle inscribed in a right triangle.

$m\angle{KDI} = 60^{\circ} \implies m\angle{KDO} = 30^{\circ} \implies m\angle{DOK} = 60^{\circ}$

For $\triangle{KDO}:$

$l = \tan(60^{\circ}) = \sqrt{3} \implies \overline{DE} = \sqrt{3} + 1 = \overline{EF} \implies$

The area of the square $\boxed{A_{s} = (\sqrt{3} + 1)^2 = 4 + 2\sqrt{3}}$

For $\triangle{JOG}:$

$m\angle{HGJ} = 120^{\circ}$ and $m\angle{OJG} = m\angle{OHG} = 90^{\circ} \implies$ $m\angle{JOH} = 60^{\circ}$ in quadrilateral $JOHG$

$\implies m\angle{JOG} = 30^{\circ} \implies m\angle{JGO} = 60^{\circ} \implies \dfrac{1}{n} = \tan(60^{\circ}) = \sqrt{3} \implies$

$n = \dfrac{1}{\sqrt{3}} = \overline{JG} = \overline{HG} \implies \overline{GF} = \overline{EF} + \overline{EH} + \overline{HG} = \sqrt{3} + 2 + \dfrac{1}{\sqrt{3}} =$

$\dfrac{4 +2\sqrt{3}}{\sqrt{3}} \implies \boxed{A_{\triangle{GFM}} = \dfrac{\sqrt{3}}{4}(\dfrac{4 + 2\sqrt{3}}{\sqrt{3}})^2 = \dfrac{7\sqrt{3} + 12}{3}} \implies$

$A_{T} = \dfrac{13\sqrt{3} + 24}{3} = \dfrac{\alpha\sqrt{\beta} + \lambda}{\beta} \implies \alpha + \beta + \lambda = \boxed{40}$