Notes
four stacked squares in a circle in a square solution

Four Stacked Squares in a Circle in a Square

Solution by Angle in a Semi-Circle and Pythagoras' Theorem

The line segments $A D$ and $B C$ are parallel, so their perpendicular bisectors are also parallel. As both are chords, the centre lies on both of them, and so these perpendicular bisectors are actually the same line. As the line segments are of equal length, the line segment $D C$ is then parallel to this perpendicular bisector. Triangle $A D C$ is then a right-angled triangle and so $A C$ is a diameter since the angle in a semi-circle is a right-angle.

Let $x$ be the length of a side of a small square and let $y$ be the side length of the large square. Then $D C$ has length $3 x$ and the diameter of the circle is $y$. Applying Pythagoras' theorem shows that:

Therefore one small square is one tenth of the large square, so $\frac{2}{5}$ths of the large square is shaded.

Solution by Intersecting Chords Theorem

With $x$ and $y$ as above, $A G$ and $G D$ have length $\frac{x}{2}$, $E G$ has length $\frac{y - 3 x}{2}$ and $G F$ has length $\frac{y + 3 x}{2}$. Applying the intersecting chords theorem gives

Which gives the fraction shaded as $\frac{2}{5}$ as before. and the same length, and are both chords of the circle. The perpendicular bisectors of both pass through the centre of the circle, and since the chords are parallel the perpendicular bisectors are also parallel.