Semicircle in a quarter circle

Relevant wiki: Pythagorean Theorem

Let the radius of the quarter circle be 1 and the radius of the semicircle be $r$ . By symmetry, the line joining the center of the quarter circle and the center of the semicircle makes an angle of $45^\circ$ with the base of the quarter circle and the diameter of the semicircle is perpendicular to the line joining the centers as shown in the figure.

By Pythagorean theorem, we have:

$\begin{aligned} {\color{#3D99F6}\overline{OB}^2} + \color{#D61F06}\overline{AB}^2 & = 1^2 \\ {\color{#3D99F6}(r+r\cos 45^\circ)^2} + \color{#D61F06}(r-r\sin 45^\circ)^2 & = 1 \\ r^2\left(1+\frac 1{\sqrt 2}\right)^2 + r^2\left(1-\frac 1{\sqrt 2}\right)^2 & = 1 \\ 2 r^2\left(1+ \frac 12 \right) & = 1 \\ r^2 & = \frac 13 \\ \implies r & = \frac 1{\sqrt 3} \end{aligned}$

Therefore, the proportion of the shaded area of the quarter circle

$\begin{aligned} Q & = \dfrac {\text{Area of semicircle}}{\text{Area of the quarter circle}} \\ & = \frac {\frac 12 \times \pi \times \left(\frac 1{\sqrt 3} \right)^2}{\frac 14 \times \pi \times 1^2} = \frac {\frac \pi 6}{\frac \pi 4} = \frac 23 \approx \boxed{0.667} \end{aligned}$