Notes
rectangle, circle, and triangle solution

Solution to the Rectangle, Circle, and Triangle Puzzle

Solution by Properties of Chords, the Intersecting Chords Theorem, and Similar Triangles

With the points labelled as above, triangle $A B E$ has a right-angle at $B$ so since the angle in a semi-circle is $90^\circ$, $A E$ is a diameter of the circle. Then as $B D$ is a chord of the circle and $A E$ is perpendicular to this chord and passes through the centre of the circle, it must pass through the midpoint of $B D$. Therefore $A D$ has the same length as $A B$, which is $4$.

Applying the intersecting chords theorem to the chords $B C$ and $A C$ shows that:

So the length of $E C$ is $\frac{5}{3}$ meaning that the length of $B E$ is $\frac{4}{3}$. Therefore $B A$ is $3$ times the length of $B E$.

Triangles $B F E$ and $B A F$ are both similar to triangle $B A E$, so the length of $B F$ is $4$ times that of $F E$, and of $A F$ is $4$ times of $B F$. Putting these together, $B F$ is $\frac{4}{17}$ of the length of $A E$.

As the rectangle and triangle have the same “height”, the fraction of the rectangle that is shaded is given by comparing half the length of $B D$ to the length of $G H$. Half the length of $B D$ is the length of $B F$, and the length of $B D$ is a diameter of the circle, which has the same length as $A E$. Therefore $\frac{4}{17}$ths of the rectangle is shaded.