Notes
semi-circle in a square in a quarter circle solution

Solution to the Semi-Circle in a Square in a Quarter Circle Puzzle

Solution by Pythagoras' Theorem

The quadrilateral $A B C D$ is not given as a square, but can be shown to be one. Firstly, $A D$ and $B C$ are tangent to the circle at the ends of the diameter $A B$, so as the angle between a radius and tangent is a right-angle, the edges $A D$ and $C B$ are perpendicular to $A B$. The quadrilateral $A E C F$ has edges $A E$ and $A F$ of equal length as each other, as are $C E$ and $C F$, so is a kite. The line through $A$ and $C$ is therefore a line of symmetry, and since $F D$ is a continuation of $C F$ and $E B$ of $C E$ by the same lengths, reflection in the line $A C$ is also a line of symmetry of $A B C D$. Therefore $A D$ has the same length as $A B$, and angle $A \hat{D} C$ is also a right-angle. This is enough to establish $A B C D$ as a square.

This also shows that the total shaded region is a quarter circle.

Let $a$ be the radius of the orange semi-circle and $b$ the radius of the outer quarter circle. Then the quarter circle has area $\frac{1}{4} \pi b^2$ and the semi-circle has area $\frac{1}{2} \pi a^2$, this latter being equal to $10$. The length of an edge of the square is then $2 a$.

As $F$ is the midpoint of $D C$, the length of $D F$ is half of the length of the side of the square, so is $a$. Applying Pythagoras' theorem to the triangle $A D F$ shows that:

The area of the quarter circle is therefore $\frac{5}{4} \pi a^2 = \frac{5}{2} \cdot 10 = 25$. The blue area is then $25 - 10 = 15$.