Cutting a Quarter Circular Region by Half

Reflect the quarter circle and label it as follows:

Then the area of the sector enclosed by $AOB$ is $A_S = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{\theta}{2}$ , and the area of $\triangle AOB$ is $A_T = \frac{1}{2} r^2 \sin \theta = \frac{1}{2} \sin \theta$ . The difference between these two areas must equal half the area of the semicircle $A_O = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi$ , so $A_S - A_T = \frac{1}{2}A_O$ or $\frac{\theta}{2} - \frac{1}{2} \sin \theta = \frac{1}{2} \cdot \frac{1}{2} \pi$ , which solves numerically to $\theta \approx 2.3099$ .

From $\triangle AOC, x = \cos \frac{\theta}{2}$ , and substituting the value of $\theta$ from above gives $x \approx \cos \frac{2.3099}{2} \approx \boxed{0.404}$ .