Dynamic Square Geometry

Let the side length of the purple and cyan square be $x$ and $y$ , respectively. Since, $CM=BE=x,\; \angle BEO=\angle CMO=90^{\circ}$ and $OC=OB=1$ , by RHS congruence criterion , $\triangle CMO\cong \triangle BEO \\ \implies OE=OM=\frac{x}{2}\;\;\;\;\;\;\;\;\;\;\;\;$ Using the Pythagorean Theorem , in $\triangle CMO$ , $\begin{aligned} OM^2+CM^2&=OC^2 \\ \frac{x^2}{4}+x^2&=1 \\ \therefore \; x&=\frac{2}{\sqrt{5}} \end{aligned}$ Note, $\angle ACB=\angle MCB=90^{\circ}$ since angle in a semicircle is right angle and $MCBE$ is a square. Hence, the points $A,M,C$ are collinear and $M$ is the foot of the perpendicular from the center $O$ to the chord $AC$ . Therefore, $\begin{aligned} MA&=MC \\ y\sqrt{2}&=x\\ \therefore \; y&=\sqrt{\frac{2}{5}} \end{aligned}$ The sum of the area of the two squares is, $x^2+y^2=\dfrac{4}{5}+\dfrac{2}{5}=\boxed{\dfrac{6}{5}}$

I took a screenshot when the purple square's side intersected the centre And Then just by looking at the pic I assumed the ans to be $\boxed{\frac{6}{5}}$ which was correct:D