Two identical squares, $A B C D$ and $P Q R S$, have side length 12. They overlap to form the 12 by 20 rectangle $A Q R D$ shown. What is the area of the shaded rectangle $P B C S$ ?
(A) 24
(B) 36
(C) 48
(D) 72
(E) 96

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-056.jpg?height=299&width=398&top_left_y=1672&top_left_x=1319)
Solution 1

Since $A Q=20$ and $A B=12$, then $B Q=A Q-A B=20-12=8$.

Thus, $P B=P Q-B Q=12-8=4$.

Since $P S=12$, the area of rectangle $P B C S$ is $12 \times 4=48$.

## Solution 2

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-267.jpg?height=290&width=417&top_left_y=1647&top_left_x=1540)

The sum of the areas of squares $A B C D$ and $P Q R S$ is $2 \times(12 \times 12)=2 \times 144=288$.

The area of rectangle $A Q R D$ is $12 \times 20=240$.

The sum of the areas of $A B C D$ and $P Q R S$ is equal to the sum of the areas of $A P S D, P B C S$, $P B C S$, and $B Q R C$.

The area of rectangle $A Q R D$ is equal to the sum of the areas of $A P S D, P B C S$, and $B Q R C$. Therefore, the sum of the areas of $A B C D$ and $P Q R S$, minus the area of $A Q R D$, is the area of $P B C S$.

Thus, the shaded rectangle $P B C S$ has area $288-240=48$.

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