What proportion of the quarter circle is shaded?

Question

Interesting yet challenging quiz I found on a website. My answer is a $\frac{1}{ \sqrt{2}}$.

After I assumed the semicircle has radius $r\sin{45}$, where $r$ is the radius of the quarter circular part.

Any objections or comment?

Answer 1

Your result is not correct. Let $R$ be the radius of the quarter circle and $r$ be the radius of the half-circle. Then, by Pythagoras theorem applied to $\triangle OO'C$, $$R^2=r^2+d^2$$ where $d=\sqrt{2}r$ is the distance of the two centers $O$ and $O'$ (note that $\angle O'OH=45^{\circ}$).

Can you take it from here?

Answer 2

Let $R$ be the radius of the outer circle and $M=(r,r)$ be the center of the brown semidisc. Then $|OM|=\sqrt{2} r$ and therefore $R^2=3r^2$. The ratio of the areas then comes to ${2\over3}$.