Notes
six equal area rectangles solution

Six Equal Area Rectangles

Solution by Chord Properties, Equilateral Triangles, and Area of a Sector

With the points labelled as in the above diagram, $O$ is the centre of the circle and $E$ the midpoint of $B E$. So also $F$ is the midpoint of $A D$. Let $r$ be the length of its radius.

The horizontal rectangles are all congruent, since they have the same width and area, so the height of two of them is a radius, or $\frac{r}{2}$. Their width is twice the radius, so their area is $r^2$. Hence $r^2 = 3$.

In the right-angled triangle $O F C$, the length of $O F$ is half that of $O C$, and so this is half an equilateral triangle. In particular, angle $B \hat{O} C$ is $60^\circ$. So then angle $C \hat{O} D$ is $30^\circ$.

Since $F$ is the midpoint of the chord $A D$, angle $O \hat{F} D$ is a right-angle. Angle $E \hat{O} D$ is $30^\circ + 30^\circ = 60^\circ$, and so triangle $O F D$ is also half an equilateral triangle. Since $O D$ and $O C$ have the same length, triangles $O F D$ and $C E O$ are congruent. As they overlap in triangle $O F G$, this means that triangle $O G D$ and quadrilateral $F E C G$ have the same area.

This then means that the region $F E C D$ (with $C D$ the arc) and the sector $O C D$ have the same area. The total yellow region comprises four congruent copies of $F E C D$, and so their area is the same as four sectors with central angle $30^\circ$, equivalently one sector with central angle $120^\circ$. Hence the yellow regions have total area: