Notes
multiple semi-circles ii solution

Multiple Semi-Circles II

Solution by Pythagoras' Theorem and the Area of a Circle

Consider first this diagram where all the arcs are semi-circles with parallel diameters and centres at $O$, $B$, and $C$.

Since the diameters are parallel, $B$ is the midpoint of the chord and so triangle $O B A$ is right-angled. Therefore, Pythagoras' theorem applies to triangle $A B O$. Writing the radii of the circles as $a$, $b$, $c$ with $a$ the length of $A B$, $b$ of $B O$, and $c$ of $O A$ then this means that:

The area of a semi-circle is $\frac{1}{2}\pi$ times the square of its radius, so from the above then:

which uses the fact that $C D$ and $B O$ have the same length. This means that the area of the largest semi-circle is the sum of the areas of the two smaller ones.

In terms of the coloured regions, this means that:

And hence the purple region (which is known as a semi-circular lune) has the same area as the orange.

This applies to the problem as follows.

Using the above calculation of the area of a semi-circular lune, the area of region $A$ is equal to the area of the union of regions $B$ and $C$. Similarly, the area of region $D$ is equal to the area of the union of regions $E$, $B$, and $C$, $F$. Therefore, the yellow and red regions have the same area.