Notes
two semi-circles in a semi-circle solution

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

Solution by angle in a semi-circle, similar triangles, and Pythagoras' theorem

In the above diagram, $O$ is the centre of the left-hand green semi-circle and $D$ the centre of the right-hand one. The point of contact, $E$, lies on the line $O D$.

Angle $A \hat{C} B$ is a right-angle as it is the angle in a semi-circle. Angle $O \hat{F} C$ is a right-angle as it is the angle between a radius and tangent. The line segments $O F$ and $C D$ are radii of the two semi-circles and so are the same length. The quadrilateral $O D C F$ is therefore a rectangle.

The line segment $O D$ is therefore parallel to $A B$ and so triangles $A C B$ and $O D B$ are similar. Since $D B$ is half of $C B$, $O B$ is half of $A B$ and is therefore a radius of the large semi-circle.

In the triangle $O D B$, the length of $O D$ is twice that of $D B$, which is a radius of the smaller semi-circle. Applying Pythagoras' theorem to this triangle gives the following relationship between the radii of the semi-circles:

Multiplying this through by $\frac{1}{2} \pi$ shows that the area of the larger semi-circle is $5$ times that of the green ones, and so has area $20$.