Problem 5. Rectangle $ABCD$ is composed of three congruent rectangles (see drawing). If the perimeter of each of these rectangles is $60 \mathrm{~cm}$, what is the area of the square that has the same perimeter as rectangle $ABCD$?

![](https://cdn.mathpix.com/cropped/2024_06_05_1b044dba0a21c223c4c1g-2.jpg?height=228&width=271&top_left_y=251&top_left_x=1191)
Solution. From the drawing, it is clear that one side of each of the small rectangles must be twice as long as the other. If the length of the shorter side is $x$, then the longer side is $2 x$, so the perimeter of one of them is $2(x+2 x)=60$. From this, $x=10 \mathrm{~cm}$. The perimeter of rectangle $A B C D$ is: $2(3 x+2 x)=10 x=100 \mathrm{~cm}$, and so is the perimeter of the square. The length of the side of the square is: $100: 4=25 \mathrm{~cm}$, so its area is: $P=25^{2}=625 \mathrm{~cm}^{2}$.

## V section