Four congruent rectangles and a square are assembled without overlapping to form a large square, as shown. Each of the rectangles has a perimeter of $40 \mathrm{~cm}$. The total area of the large square is
(A) $160 \mathrm{~cm}^{2}$
(B) $200 \mathrm{~cm}^{2}$
(C) $400 \mathrm{~cm}^{2}$
(D) $800 \mathrm{~cm}^{2}$
(E) $1600 \mathrm{~cm}^{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-076.jpg?height=319&width=315&top_left_y=716&top_left_x=1298)
Suppose that each of the smaller rectangles has a longer side of length $x \mathrm{~cm}$ and a shorter side of length $y \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-145.jpg?height=388&width=382&top_left_y=1465&top_left_x=926)

Since the perimeter of each of the rectangles is $40 \mathrm{~cm}$, then $2 x+2 y=40$ or $x+y=20$.

But the side length of the large square is $x+y \mathrm{~cm}$.

Therefore, the area of the large square is $(x+y)^{2}=20^{2}=400 \mathrm{~cm}^{2}$.

ANswer: (C)