Half an Annulus

The radii of the three circles are shown on the diagram, where we are trying to find $r.$

Using the whole annulus and the darker grey annulus

The area of the whole annulus is the area of the full circle (raduis $14$ cm) without the area of the empty circle (radius $2$ cm), so it is given by $\pi14^2-\pi2^2=196\pi-4\pi=192\pi$ cm$^2.$

The area of the darker grey annulus is the area within the dotted circle without the area in the empty circle - so $\pi r^2-\pi2^2=\pi r^2-4\pi$ cm$^2.$

This area is equal to half of the whole annulus, so $\pi r^2-4\pi=\frac{1}{2}\times192\pi\Rightarrow \pi r^2-4\pi=96\pi.$

So $\pi r^2=100\pi$, so $r^2=100$, so $r=10.$

Using the whole annulus and the lighter grey annulus

The area of the whole annulus is the area of the full circle (raduis $14$ cm) without the area of the empty circle (radius $2$ cm), so it is given by $\pi14^2-\pi2^2=196\pi-4\pi=192\pi$ cm$^2.$

The area of the lighter grey annulus is the area within the full circle without the area in the dotted circle - so $\pi14^2-\pi r^2=196\pi-\pi r^2$ cm$^2.$

This area is equal to half of the whole annulus, so $196\pi-\pi r^2=\frac{1}{2}\times192\pi\Rightarrow 196\pi-\pi r^2=96\pi.$

So $196\pi-96\pi=\pi r^2\Rightarrow\pi r^2=100\pi$, so $r^2=100$, so $r=10.$

Using the lighter grey annulus and the darker grey annulus

The area of the lighter grey annulus is the area within the full circle without the area in the dotted circle - so $\pi14^2-\pi r^2=196\pi-\pi r^2$ cm$^2.$

The area of the darker grey annulus is the area within the dotted circle without the area in the empty circle - so $\pi r^2-\pi2^2=\pi r^2-4\pi$ cm$^2.$

These two areas are equal, so $196\pi-\pi r^2=\pi r^2-4\pi.$ Solving this equation for $r$, $$\begin{align}196\pi-\pi r^2&=\pi r^2-4\pi\\

\Rightarrow 196\pi +4\pi&=\pi r^2+\pi r^2\\

\Rightarrow 200\pi &=2\pi r^2\\

\Rightarrow 100&=r^2\\

\Rightarrow 10&=r\end{align}$$