Area Fiesta!!

Using the diagram above $\overline{AC} = \sqrt{2}a = \sqrt{2}x + x + \dfrac{x}{2} + \dfrac{x}{\sqrt{2}} \implies$

$4a = (6 + 3\sqrt{2})x \implies x = \dfrac{4a}{6 + 3\sqrt{2}} = \dfrac{4a}{3\sqrt{2}(\sqrt{2} + 1)}$ $= \dfrac{2\sqrt{2}(\sqrt{2} - 1)}{3}a$

$= \dfrac{4 - 2\sqrt{2}}{3}a$ and $\overline{CQ} = \dfrac{x}{2} + \dfrac{x}{\sqrt{2}} =$ $\dfrac{2 + \sqrt{2}}{2\sqrt{2}}x = \dfrac{\sqrt{2}(\sqrt{2} + 1)}{2\sqrt{2}}x =$

$\dfrac{\sqrt{2} + 1}{2}x = (\dfrac{\sqrt{2} + 1}{2})(\dfrac{4 - 2\sqrt{2}}{3})a = \dfrac{\sqrt{2}}{3}a$

$\triangle{FCG}$ is a right isosceles triangle $\implies \overline{FQ} \cong \overline{QG} \implies \triangle{FQC} \cong \triangle{CQG} \implies$

$A_{\triangle{FCG}} = 2 A_{\triangle{FQC}} = 2(\dfrac{1}{2}(\overline{CQ})^2) = \dfrac{2}{9}a^2$

and $A_{\triangle{AEH}} = 2A_{\triangle{AEP}} = 2(\dfrac{1}{2})(\sqrt{2}x)^2 = 2x^2 = \dfrac{2}{9}(4 - 2\sqrt{2})^2a^2 =$

$\dfrac{2}{9}(24 - 16\sqrt{2})a^2$

Let $A_{s} = A_{\triangle{FCG}} + A_{\triangle{AEH}} = \dfrac{2}{9}(25 - 16\sqrt{2})a^2$

The area of circle $A_{c} = \pi r^2 = \pi(\dfrac{x}{2})^2 = \dfrac{\pi}{9}(2 - \sqrt{2})^2a^2 =$ $\dfrac{2\pi}{9}(3 - 2\sqrt{2})a^2$

Let $A^{*} = A_{s} + A_{c} = \dfrac{2}{9}(25 - 16\sqrt{2} + (3 - 2\sqrt{2})\pi)a^2 \implies$

$A_{T} = a^2 - A^{*} = \dfrac{32\sqrt{2} - 41 - 2(3 - 2\sqrt{2})\pi}{9}a^2 \implies$

$\dfrac{A_{T}}{A_{ABCD}} = \dfrac{32\sqrt{2} - 41 - 2(3 - 2\sqrt{2})\pi}{9} \approx \boxed{0.35297886967}$ .