Notes
two semi-circles inside a semi-circle solution

Two Semi-Circles Inside a Semi-Circle

Solution by Angle Between a Radius and Tangent, Pythagoras' Theorem, and Properties of Chords

In the above diagram, point $O$ is the centre of the larger semi-circle while points $C$ and $B$ are the centres of the shaded ones. Let $r$ be the radius of the smaller semi-circles and $R$ the radius of the largest one. Then $r$ is the length of $B D$ and of $C A$, while $B C$ has length $2 r$. Both $O D$ and $O E$ have length $R$.

As $A B$ is tangent to the left-hand semi-circle, angle $B \hat{A} C$ is $90^\circ$ since it is the angle between a radius and tangent. Therefore, triangle $C A B$ is right-angled and applying Pythagoras' theorem to it shows that $A B$ has length $\sqrt{3} r$. Then $A D$ has length $(1 + \sqrt{3}) r$ and so $O A$ has length $(1 + \sqrt{3})r - R$.

Applying Pythagoras' theorem to triangle $C A O$ shows that the length of $O C$ is given by:

Since $E F$ is a chord of the outer semi-circle, and $C$ is its midpoint, the line $C O$ is perpendicular to $E F$ and so triangle $E C O$ is right-angled. Applying Pythagoras' theorem to this yields the equation:

The area of the larger semi-circle is therefore $\frac{1}{2} \pi R^2 = \frac{3}{2} \pi r^2$. The shaded area is $\pi r^2$. Therefore the fraction that is shaded is $\frac{2}{3}$.