Notes
two half-squares and a circle solution

Solution to the Two Half-Squares and a Circle Puzzle

Solution by Properties of Chords and Pythagoras' Theorem

In the above diagram, point $O$ is the middle of the circle, $I$ is the midpoint of $E D$ and $H$ is the midpoint of $A B$. The point $G$ is where the extension of $E F$ meets $A B$, and likewise $K$ is where the extension of $E D$ meets $B C$.

As $E D$ is a chord of the circle, the centre of the circle lies on the perpendicular line through its midpoint (namely, $I$). The same argument applied to $A B$ shows that the centre also lies on the perpendicular line through $H$. As $E D$ and $A B$ are parallel, these perpendicular lines are also parallel. Since the centre lies on both, they must therefore be the same line. So the points $I$, $O$, and $H$ lie on a straight line that is perpendicular to both $E D$ and $A B$.

This establishes $E D$ as lying symmetrically over $A B$, and so the excess on one side, which is the length of $A G$, is equal to the excess on the other, which is the length of $D K$. Since $D K C$ is an isosceles right-angled triangle, $C K$ has the same length as $D K$.

The line $A J$ is a chord with the property that angle $J \hat{B} A$ is a right-angle, so as the angle in a semi-circle is a right-angle, $A J$ must be a diameter. By the same result, angle $A \hat{D} J$ is also a right-angle, so since angle $E \hat{D} F$ is $45^\circ$, angle $J \hat{D} K$ is $45^\circ$. This means that triangle $D K J$ is also an isosceles right-angled triangle and so $K J$ also has the same length as $D K$.

Then the length of $B J$ is that of $B C$ with two lengths of $D K$ removed. This is the same as the length of $E D$.

Let $a$ be the length of $E D$ (and thus of $B J$) and $b$ the length of $A B$. The shaded region has area $\frac{1}{2}(a^2 + b^2)$. Triangle $A B J$ is a right-angled triangle with hypotenuse $2$ and other side lengths $a$ and $b$, so by Pythagoras' theorem, $a^2 + b^2 = 2^2 = 4$. Hence the shaded area is $2$.

Solution by Invariance Principle

There are two special cases of this diagram. One in which both triangles are the same size, and one where one triangle has shrunk to zero size.

In this version, the total shape makes a square inside the circle. The diagonal of the square coincides with a diameter, since the area of a square is half the square of its diagonal, this establishes the area as $\frac{1}{2} \cdot 2^2 =2$.

In this version, the total shape is an isosceles right-angled triangle with one side as a diameter of the circle, so the area is $\frac{1}{2} \cdot 2 \cdot 2 = 2$.