Notes
four quarter circles ii solution

Solution to the Four Quarter Circles II Puzzle

Solution by Pythagoras' Theorem and Area of a Circle

With the points labelled as above, $O$ is the centre of the semi-circle and left-hand quarter circle, while $A$ is the centre of the right-hand quarter circle. Let $a$, $b$, and $c$ be the lengths of their respective radii, so $O B$ has length $a$, $O A$ has length $b$, and $A B$ has length $c$.

Using the formula for the area of a circle, the area of the shaded region is:

Triangle $O A B$ is right-angled so Pythagoras' theorem applies, meaning that $a^2 = b^2 + c^2$. Substituting into Equation (1) shows that the shaded region has area:

and this is half of the area of the full semi-circle.

Solution by Invariance Principle

The inner quarter circles can be varied, and the configuration above shows that the shaded region comprises half of the overall diagram.