2. Rectangle $R_{0}$ has sides of lengths 3 and 4 . Rectangles $R_{1}, R_{2}$, and $R_{3}$ are formed such that:
- all four rectangles share a common vertex $P$,
- for each $n=1,2,3$, one side of $R_{n}$ is a diagonal of $R_{n-1}$,
- for each $n=1,2,3$, the opposite side of $R_{n}$ passes through a vertex of $R_{n-1}$ such that the center of $R_{n}$ is located counterclockwise of the center of $R_{n-1}$ with respect to $P$.

Compute the total area covered by the union of the four rectangles.
Answer:
30
Solution: Let $A B C D$ be $R_{0}$ such that $\overline{A B}=3$ and $\overline{B C}=4$. Then, let $\overline{A C}$ be a side length of $R_{1}$ and let the other two vertices be $E$ and $F$ such that $B$ lies on segment $E F$. Notice that the area of $\triangle A B C$ is both half of the area of $R_{0}$ and half of the area of $R_{1}$. This means forming $R_{1}$ adds half of the area of $R_{0}$ to the union of rectangles. Similarly, forming $R_{2}$ adds half of the area of $R_{1}$ to the union of all rectangles, and the same for $R_{3}$. This means the total area of the union of rectangles is given by
$$
\left[R_{0}\right]+\frac{1}{2}\left[R_{1}\right]+\frac{1}{2}\left[R_{2}\right]+\frac{1}{2}\left[R_{3}\right]=\left[R_{0}\right]+\frac{1}{2}\left[R_{0}\right]+\frac{1}{2}\left[R_{0}\right]+\frac{1}{2}\left[R_{0}\right]=\frac{5}{2}\left[R_{0}\right]=\frac{5}{2}(3 \cdot 4)=30 .
$$

Note that in the above equation, $[X]$ denotes the area of shape $X$.