By joining four identical trapezoids, which have non-parallel sides of equal length and bases measuring 50 and $30 \mathrm{~cm}$, as shown in the given figure, we can form a square of $2500 \mathrm{~cm}^{2}$ in area, which has a square "hole" in the middle. What is the

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-027.jpg?height=189&width=594&top_left_y=585&top_left_x=1279)
area, in $\mathrm{cm}^{2}$, of each of the four trapezoids?
(a) 200
(b) 250
(c) 300
(d) 350
(e) 400
The correct answer is (e).

By joining the four trapezoids, we form a square with a side length of $50 \mathrm{~cm}$ and, therefore, an area of $2500 \mathrm{~cm}^{2}$. Since the "hole" square has a side length of $30 \mathrm{~cm}$, its area is $30 \times 30=900 \mathrm{~cm}^{2}$. Thus, the area of each of the four trapezoids, in $\mathrm{cm}^{2}$, is given by $(2500-900) \div 4=1600 \div 4=400$.

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-081.jpg?height=328&width=300&top_left_y=1452&top_left_x=1503)