B1. A square is divided into six rectangles as shown in the picture. These rectangles do not all have the same shape, but they do all have the same area. Given that side $A B$ has a length of 5.

What is the length of $B C$?

Note: the picture is not drawn to scale.

![](https://cdn.mathpix.com/cropped/2024_04_17_9e808ca4ac6657f5611fg-3.jpg?height=322&width=331&top_left_y=273&top_left_x=1522)
B1.

The six rectangles have equal area. Rectangles $c$ and $d$ are twice as high as rectangle $a$ and therefore twice as narrow. They thus have a width of $\frac{5}{2}$. Rectangle $e$ then has a width of $\frac{5}{2} + \frac{5}{2} + 5 = 10$ and is therefore half as high as rectangle $a$. But then rectangle $f$ is exactly $\frac{5}{2}$ times as high as rectangle $a$ and must therefore have a width of $\frac{5}{5 / 2} = 2$. The entire square therefore has sides of length $5 + \frac{5}{2} + \frac{5}{2} + 2 = 12$. Since the square is $\frac{5}{2}$ times as high as rectangle $a$, it follows that the height of rectangle $a$ is equal to $|B C| = \frac{12}{5 / 2} = \frac{24}{5}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_8a1f196dae4ba99bbc23g-3.jpg?height=371&width=380&top_left_y=180&top_left_x=1475)