Notes
two quarter circles in a rectangle in a semi-circle solution

Solution to the Two Quarter Circles in a Rectangle in a Semi-Circle Puzzle

Solution by Pythagoras' theorem and Area Scale Factor

In the above diagram, point $O$ is the centre of the larger semi-circle.

The line segment $D B$ passes through the point where the two quarter circles touch so its length is twice the radius of the quarter circles. Line segment $A D$ is a radius, so triangle $D A B$ is a right-angled triangle with hypotenuse twice the length of one of its sides. Pythagoras' theorem then shows that $A B$ has length $\sqrt{3}$ times the length of $A D$. Then $O A$ is half this, so applying Pythagoras' theorem to triangle $O A D$ shows that $O D$ is $\frac{\sqrt{7}}{2}$ times the length of $A D$. Since area scales as the square of the length scale factor, the outer semi-circle is then $\frac{7}{4}$ the area of the two quarter circles, so the shaded region is $\frac{4}{7}$ths of the total region.