Notes
concentric semi-circles solution

Concentric Semi-Circles

Solution by Lengths in a Square

Let the radii of the three circles be $b$, $c$, and $d$ in increasing order, so $O B$ has length $b$, $O C$ has length $c$, and $O D$ has length $d$. Since $O A$ is also a radius of the middle circle, $O A$ also has length $c$ but it also has length $b + 2$. So $c = b + 2$.

Then $O C$ is the diagonal of a square with side length $b$, so using the relationship between lengths in a square, $c = \sqrt{2}b$. Putting this together with the above, $b + 2 = \sqrt{2} b$ so $b = \frac{2}{\sqrt{2} - 1}$. Then $c = \frac{2 \sqrt{2}}{\sqrt{2} - 1}$.

Since $O D$ is the diagonal of a square with side length $c$, $d = \sqrt{2} c = \frac{4}{\sqrt{2} - 1}$. Then $E F$ has length $d - c$ which is $\frac{4}{\sqrt{2} - 1} - \frac{2 \sqrt{2}}{\sqrt{2} - 1} = 2\sqrt{2}$. The square therefore has area $8$.

Solution by Similarity

In the above diagram, the half of the diagram on the right is similar to the half of the diagram on the left. The scale factor is $\sqrt{2}$ since $O E$ is the same length as $O C$ and this is $\sqrt{2}$ times the length of $O B$. Therefore, the area scale factor is $2$ and so the square with side $E F$ has double the area of that with side $A B$, hence has area $8$>