# Two Overlapping Semi-Circles +-- {.image} [[TwoOverlappingSemiCircles.png:pic]] > The larger semicircle has twice the area of the smaller one. What’s the angle? =-- ## Solution by [[chord|Chord Properties]] and [[Pythagoras' Theorem]] +-- {.image} [[TwoOverlappingSemiCirclesLabelled.png:pic]] =-- 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. +-- {.image} [[TwoOverlappingSemiCirclesSimplified.png:pic]] =-- 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 $45^\circ$.