Notes
square in a semi-circle in a square solution

Square in a Semi-Circle in a Square

Solution by Properties of Chords and Pythagoras' Theorem

Since the top edge of the smaller square is a chord of the semi-circle, the centre of the circle lies on its perpendicular bisector and so is the midpoint of the bottom edge. This means that, in the above diagram, the length of $O A$ is half that of $A C$. Writing $x$ for the length of $O A$, this means that $O B$ is $2 x$. The area of the smaller square is then $4 x^2$.

Let $y$ be the length of $O C$. Since $O A C$ is a right-angled triangle, applying Pythagoras' theorem shows that $y^2 = x^2 + (2 x)^2 = 5 x^2$. As $O B$ has the same length as $O C$, the area of the outer square is then $4 y^2$. The outer square is therefore $5$ times bigger than the smaller, and so the fraction that is shaded is $\frac{1}{5}$th.