It's All Squares !!

In the diagram above the four red triangles are all similar.

$MC = \dfrac{\sqrt{5}}{2} \implies A_{\triangle{MBC}} = \dfrac{1}{2}(\dfrac{1}{2})(1) = \dfrac{1}{4}$ and $\triangle{AMS} \sim \triangle{MBC} \implies$

$\dfrac{1}{2x} = 2 \implies x = \dfrac{1}{4} \implies$ $A_{\triangle{AMS}} = \dfrac{1}{2}(\dfrac{1}{2})(\dfrac{1}{4}) = \dfrac{1}{16}$

$\triangle{CTD} \sim \triangle{MBC} \implies \dfrac{\sqrt{5}}{2} = \dfrac{1}{y} \implies y = \dfrac{2}{\sqrt{5}}$

$\implies DT = \sqrt{1 - \dfrac{4}{5}} = \dfrac{1}{5} \implies A_{\triangle{CDT}} = \dfrac{1}{2}(\dfrac{2}{\sqrt{5}})(\dfrac{1}{\sqrt{5}}) = \dfrac{1}{5}$

and $RD = \dfrac{\sqrt{5}}{2} - \dfrac{1}{\sqrt{5}} = \dfrac{3}{2\sqrt{5}}$ and $\triangle{SRD} \sim \triangle{MBC} \implies$

$\dfrac{2\sqrt{5}}{3} = \dfrac{1}{2z} \implies z = \dfrac{3}{4\sqrt{5}} \implies$ $A_{\triangle{SRD}} = \dfrac{1}{2}(\dfrac{3}{4\sqrt{5}})(\dfrac{3}{2\sqrt{5}}) = \dfrac{9}{80}$

$\implies A_{red} = \dfrac{1}{4} + \dfrac{1}{16} + \dfrac{1}{5} + \dfrac{9}{80} = \dfrac{5}{8}$

The total area $A_{T} = A_{QMCP} + A_{\triangle{MBC}} + A_{\triangle{AMS}} = \dfrac{5}{4} + \dfrac{1}{4} + \dfrac{1}{16} = \dfrac{25}{16}$

$\implies$ the desired area $A = \dfrac{A_{red}}{A_{T}} = \dfrac{2}{5} = \boxed{0.4}$ .