Partially Shaded Semicircle

Let the radius of the semicircle be $1$ . Then $AC=CD=DB=\dfrac 23$ and the area of the semicircle is $A_{\text{semicircle}} = \dfrac \pi 2$ . And the area of the shaded region is:

$\begin{aligned} A_{\text{shaded}} & = A_{\text{pink sector}} + 2A_{\text{red }\triangle} \\ & = \frac {2 \sin^{-1} \frac 13}{2\pi} \times \pi + 2 \times \frac 12 \times \frac 13 \times \sqrt{1-\left(\frac 13\right)^2} \\ & = \sin^{-1} \frac 13 + \frac {\sqrt 8}9 \end{aligned}$

Therefore $\dfrac {A_{\text{shaded}}}{A_{\text{semicircle}}} = \dfrac {\sin^{-1} \frac 13 + \frac {\sqrt 8}9}{\frac \pi 2} \approx \boxed{41.6} \%$ .