Notes
three overlapping rectangles ii solution

Three Overlapping Rectangles II

Solution by Similar Triangles

With the points labelled as in the above diagram, let $a$ and $b$ be the lengths of the sides of the orange rectangle, with $a$ the length of $A G$ and $b$ the length of $A B$.

The triangles $G A C$ and $C D E$ are similar, with $G$ corresponding to $C$, because they are both right-angled triangles and the angles $G \hat{C} A$ and $D \hat{C} E$ add up to $90^\circ$ using angles on a straight line.

Let $c$ be the length of $A C$ and $d$ of $C D$. Then the ratios $a : c$ and $d : b$ are equal, meaning that $a b = c d$. Then also the length of $A D$ is $a + b$ and $c + d$, so these are also equal. This is enough to show that $c$ and $d$ are equal to $a$ and $b$ in some order, and from the diagram it must be that $c = a$ and $d = b$.

(This can also be seen using circle geometry. Draw a circle centred on the midpoint of the line segment joining $E G$ with $E G$ as diameter. Since angle $E \hat{B} G$ is a right-angle - this is because the rectangles are similar so their diagonals are at right-angles - the point $B$ lies on this circle. Since also angle $E \hat{C} G$ is a right-angle, the point $C$ also lies on this circle. Then by symmetry, the lengths of $A B$ and $C D$ are equal.)

This then shows that the triangles $G A C$ and $E D C$ are isosceles right-angled triangles, and so the length of $G C$ is $\sqrt{2} a$ while the length of $E C$ is $\sqrt{2} b$. So the area of the pink rectangle is $2 a b = 24$.

Solution by Invariance Principle

The orange rectangles can be varied, providing their area remains $12$. By drawing them as squares, as in the following diagram, the area of the pink rectangle (now also a square) is clearly the same as the total area of the orange squares, thus $24$.