By joining four identical trapezoids with bases $30 \mathrm{~cm}$ and $50 \mathrm{~cm}$ and equal non-parallel sides, as shown in the figure, we can form a square with an area of $2500 \mathrm{~cm}^{2}$, with a square "hole" in the middle. What is the area of each trapezoid, in $\mathrm{cm}^{2}$?
(A) 200
(B) 250
(C) 300
(D) 350
(E) 400
The correct answer is (E).

![](https://cdn.mathpix.com/cropped/2024_05_01_dbce7bc2a200e234c3e2g-11.jpg?height=328&width=304&top_left_y=1267&top_left_x=347)

By combining the four trapezoids, we form a square with a side length of $50 \mathrm{~cm}$, and thus an area of $2500 \mathrm{~cm}^{2}$. Since the "hole" square has a side length of $30 \mathrm{~cm}$, its area is $30 \times 30=900 \mathrm{~cm}^{2}$. Therefore, the area of each of the 4 trapezoids, in $\mathrm{cm}^{2}$, is

$$
(2500-900) \div 4=1600 \div 4=400
$$