Question:

Step 1: To find the maximum area of the rectangle $PQRS$ that inscribes the rectangle $ABCD$, we need to understand how the area of $PQRS$ changes with respect to the angle $\theta$ that the sides of $ABCD$ make with the sides of $PQRS$. The area will be maximum when the dimensions of $PQRS$ are such that they can just contain $ABCD$ while being as large as possible. Step 2: Given that the sides of $ABCD$ are 2 and 4, we can visualize or deduce geometrically that the area of $PQRS$ will be maximum when $ABCD$ is placed in such a way that its vertices touch the sides of $PQRS$ and the angle $\theta$ between the sides of $ABCD$ and $PQRS$ is $45^\circ$. This is because, at $45^\circ$, the projection of the sides of $ABCD$ onto the sides of $PQRS$ results in the largest possible dimensions for $PQRS$ that can still contain $ABCD$. Step 3: When $\theta = 45^\circ$, the sides of $PQRS$ can be determined using the projections of the sides of $ABCD$. Considering the geometry, the length $a$ and width $b$ of $PQRS$ will be such that $a$ and $b$ are equal due to the symmetry of the $45^\circ$ angle, and they can be calculated using the Pythagorean theorem applied to the right-angled triangles formed by the projections of $ABCD$ onto $PQRS$. This results in $a = b = \sqrt{2^2 + 4^2}/\sqrt{2} = \sqrt{20} = 2\sqrt{5}$, but considering the actual configuration and maximizing area, we should directly consider the relationship of $a$ and $b$ with the given sides and the angle, leading to $a + b = 6\sqrt{2}$ as the correct relationship for maximum area. Step 4: To find $(a+b)^2$, we simply square the sum of $a$ and $b$. Given $a + b = 6\sqrt{2}$, we have: $(a+b)^2 = (6\sqrt{2})^2 = 36 \times 2 = 72$. Step 5: Therefore, the final answer is that $(a+b)^2$ is equal to $72$, which corresponds to Option 1. The final answer is $\boxed{72}$.