Notes
two semi-circles diagonally across a square solution

Two Semi-Circles Diagonally Across a Square

Solution by Pythagoras' Theorem

With the points labelled as above, let $a$ be the diameter of the lower semi-circle and $b$ the diameter of the upper. The total shaded area is then $\frac{1}{8} \pi (a^2 + b^2)$.

Triangle $A O E$ is a right-angled triangle. The full diagonal of the square is then $a + b$, so the length of $A O$ is $\frac{a + b}{2}$. The length of $O E$ is $\frac{a + b}{2} - b = \frac{a - b}{2}$. Applying Pythagoras' Theorem to triangle $A O E$ yields the equation:

The left-hand side expands to

So the shaded area is:

Solution by Invariance Principle

The relative sizes of the semi-circles can be varied. In the first special case, the two are the same size, so the half diagonal of the square is of length $4$ and this is also the diameters of the semi-circles. This gives an area of $2 \times \frac{1}{2} \times \pi \times 2^2 = 4 \pi$.

In the second special case, the lower semi-circle is as large as it can be and the upper as small. This means that the diameter of the lower semi-circle is the diagonal of the square and the side of the square is of length $4$. The diagonal is therefore $4 \sqrt{2}$ and the area of the semi-circle is $\frac{1}{2} \pi (2 \sqrt{2})^2 = 4 \pi$.