Notes
two overlapping semi-circles solution

Two Overlapping Semi-Circles

Solution by Chord Properties and Pythagoras' Theorem

In the above diagram, $O$ is the centre of the smaller semi-circle and so is the midpoint of the chord $A B$. The point $Q$ is one of the points where the smaller semi-circle intersects with the diameter of the larger one.

Consider the following diagram, in which $P$ is the centre of the larger semi-circle.

As $A B$ is a chord of the larger circle and $O$ is its midpoint, the line $O P$ is perpendicular to $A B$. Let $a$, $b$, $c$ be the lengths of $O A$, $O P$, and $P A$ respectively. Then $a$ is the radius of the smaller semi-circle and $c$ of the larger. Since the larger has twice the area of the smaller, $\frac{1}{2} \pi c^2 = 2 \times \frac{1}{2} \pi a^2$ so $c^2 = 2 a^2$. Triangle $O P A$ is a right-angled triangle so from Pythagoras' theorem, $c^2 = a^2 + b^2$ which leads to $b^2 = a^2$ and hence (as both are lengths), $b = a$.

This means that point $P$ is a distance $a$ from $O$ and so also lies on the smaller semi-circle. This shows that $P$ of this diagram and $Q$ in the earlier diagram are actually the same point, so triangle $O Q A$ is an isosceles right-angled triangle and hence angle $O \hat{A} Q$ is $45^\circ$.