In the diagram, square $P Q R S$ has side length 42 and is divided into four non-overlapping rectangles. If each of these four rectangles has the same perimeter, what is the area of the shaded rectangle?
(A) 252
(B) 432
(C) 441
(D) 490
(E) 540

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-041.jpg?height=425&width=463&top_left_y=232&top_left_x=1292)
Suppose that the height of the top left rectangle is $2 x$.

Since square $P Q R S$ has side length 42 , then the bottom rectangle has height $42-2 x$.

Since the width of the bottom rectangle is 42 (the side length of the square), then the perimeter of the bottom rectangle is $2(42)+2(42-2 x)=168-4 x$.

Suppose that the width of the top left rectangle is $y$.

Since each of the small rectangles has the same perimeter, then $2(2 x)+2 y=168-4 x$. (That is, the perimeter of the top left rectangle equals the perimeter of the bottom rectangle.)

Thus, $2 y=168-8 x$ or $y=84-4 x$.

Since the side length of the square is 42 , then the width of the top right rectangle is $42-(84-4 x)=4 x-42$.

Since the two rectangles on the top right have the same perimeter and the same width, then they must have the same height, which is equal to half of the height of the top left rectangle.

Thus, the height of each of the two rectangles on the right is $x$.

Finally, the perimeter of the top right rectangle must equal also equal $168-4 x$.

Thus, $2(4 x-42)+2 x=168-4 x$ or $8 x-84+2 x=168-4 x$ and so $14 x=252$ or $x=18$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-082.jpg?height=464&width=545&top_left_y=1844&top_left_x=1495)

This tells us that the shaded rectangle has dimensions $x=18$ by

$4 x-42=4(18)-42=30$ and so its area is $18 \times 30=540$.

(We can check by substitution that the resulting rectangles in the original diagram are $6 \times 42$, $36 \times 12$ and $30 \times 18$, which all have the same perimeter.)