Notes
four quarter circles in a circle solution

Solution to the Four Quarter Circles in a Circle Puzzle

Solution by Pythagoras' Theorem

In the above diagram then $O$ is the centre of the circle, $M$ is the midpoint of $A B$, and $N$ is the midpoint of $C D$. The distances $a$, $b$, $c$, and $d$ are measured from the intersection point $P$, and $r$ is the radius of the outer circle.

As $N$ is the midpoint of $C D$, $O N$ is perpendicular to $C D$ and similarly $O M$ is perpendicular to $A B$. Since angle $B \hat{P} D$ is a right-angle, the quadrilateral $O M P N$ is a rectangle. In particular, $O M$ and $N P$ have the same length as each other, and $O N$ and $M P$ have the same length as each other.

Since $N$ is the midpoint of $C D$, $N D$ has length $\frac{c + d}{2}$ and $N P$ has length $\frac{d - c}{2}$. Similarly, $B M$ has length $\frac{a + b}{2}$ and $M P$ has length $\frac{b - a}{2}$.

Triangle $O N D$ is a right-angled triangle so applying Pythagoras' theorem results in:

This simplifies to

A similar argument applied to triangle $O M B$ leads to

Taken together, these imply that $c d = a b$ (this could also be seen through use of the intersecting chords theorem) and so the original equation becomes

The length $a$ is the hypotenuse of an isosceles right-angled triangle with radii of the yellow quarter circle forming the other two sides. This radius is therefore of length $\frac{a}{\sqrt{2}}$ so the area of that quarter circle is

The total area of the quarter circles is therefore

Since the area of the outer circle is $\pi r^2$, the shaded area is thus half of the area of the full circle.