A pentagon inscribed in a square

$AC$ is an axis of symmetry of the compound shape, hence, $\triangle AEF$ is an isosceles right triangle. Thus,
$AE=\dfrac{a}{\sqrt{2}}$ and $\angle AEF=45{}^\circ$ . Since for the angle of the regular pentagon we have $\angle FEG=108{}^\circ$ , we get

$\begin{aligned} & \angle BEG=180{}^\circ -\left( \angle AEF+\angle FEG \right)=180{}^\circ -\left( 45{}^\circ +108{}^\circ \right)=27{}^\circ \\ & \Rightarrow EB=a\cdot \cos 27{}^\circ \\ \end{aligned}$ Consequently,

$\begin{aligned} AE+EB=AB & \Rightarrow \frac{a}{\sqrt{2}}+a\cdot \cos 27{}^\circ =1 \\ & \Rightarrow a=\frac{1}{\frac{1}{\sqrt{2}}+\cos 27{}^\circ } \\ & \Rightarrow a\approx 0.62573786 \\ \end{aligned}$ For the answer,

$\left\lfloor {{10}^{5}}a \right\rfloor =\boxed{62573}.$

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Here is a proof of the fact that $\triangle AEF$ is an isosceles right triangle, without taking symmetry for granted:
Let $\angle AEF=\theta{}^\circ$ , $AE={{x}_{1}}$ , $EB={{x}_{2}}$ , $AF={{y}_{1}}$ , $FD={{y}_{2}}$ . Then, $\angle BEG=180{}^\circ -\left( 108{}^\circ +\theta {}^\circ \right)=72{}^\circ -\theta {}^\circ$ $\angle AFE=90{}^\circ -\theta{}^\circ$ $\angle DFI=180{}^\circ -\left( 108{}^\circ +90{}^\circ -\theta {}^\circ \right)=\theta {}^\circ -18{}^\circ$ Hence, ${{x}_{1}}=a\cdot \cos \left( \theta {}^\circ \right)$ ${{x}_{2}}=a\cdot \cos \left( 72{}^\circ -\theta {}^\circ \right)$ ${{y}_{1}}=a\cdot \sin \left( \theta {}^\circ \right)$ ${{y}_{2}}=a\cdot \cos \left( \theta {}^\circ -18{}^\circ \right)$ Now, we can construct an equation for $\theta$ :

$\begin{aligned} {{x}_{1}}+{{x}_{2}}=1={{y}_{1}}+{{y}_{2}}& \Leftrightarrow \bcancel{a}\cdot \cos \left( \theta {}^\circ \right)+\bcancel{a}\cdot \cos \left( 72{}^\circ -\theta {}^\circ \right)=\bcancel{a}\cdot \sin \left( \theta {}^\circ \right)+\bcancel{a}\cdot \cos \left( \theta {}^\circ -18{}^\circ \right) \\ & \Leftrightarrow \cos \left( \theta {}^\circ \right)+\cos \left( 72{}^\circ \right)\cos \left( \theta {}^\circ \right)+\sin \left( 72{}^\circ \right)\sin \left( \theta {}^\circ \right)=\sin \left( \theta {}^\circ \right)+\cos \left( \theta {}^\circ \right)\cos \left( 18{}^\circ \right)+\sin \left( \theta {}^\circ \right)\sin \left( 18{}^\circ \right) \\ & \Leftrightarrow \cos \left( \theta {}^\circ \right)\left[ 1+\cos \left( 72{}^\circ \right)-\cos \left( 18{}^\circ \right) \right]=\sin \left( \theta {}^\circ \right)\left[ 1+\sin \left( 18{}^\circ \right)-\sin \left( 72{}^\circ \right) \right] \\ & \Leftrightarrow \cos \left( \theta {}^\circ \right)=\sin \left( \theta {}^\circ \right),\ \ \ \ \ \green {\text{ since }\cos \left( 72{}^\circ \right)=\sin \left( 18{}^\circ \right)\text{ and }\cos \left( 18{}^\circ \right)=\sin \left( 72{}^\circ \right)} \\ & \Leftrightarrow \tan \left( \theta {}^\circ \right)=1 \\ & \Leftrightarrow \theta =45 \\ \end{aligned}$ Hence, $\triangle AEF$ is an isosceles right triangle.

Note: @Atomsky Jahid This last statement implies that $AC$ is, indeed, an axis of symmetry of the compound shape.