7.2. A father had three sons, and he left them 9 ares of land as an inheritance - a rectangle measuring 25 m $\times 36$ m. The brothers decided to divide the land into three rectangular plots - three ares for each brother. How many options are there for the division (in terms of the length and width of the plots), and in which of the options will the total length of the fences between the plots be the smallest?
Answer: 4 options; the smallest length is 49 m in the option of dividing into a plot of $25 \times 12$ and two plots of $12.5 \times 24$. Solution. Let $ABCD$ be the original rectangle; $AB=25, BC=36$. Since it has 4 vertices and 3 plots, two vertices must belong to one plot. First, consider the case where the vertices of the shorter side (say, $AB$) belong to such a plot $ABNM$ (see figure). In this case, the remaining part - rectangle $MNCD$ - needs to be divided into two equal rectangles, which can be done in two ways (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_b28578248d01d32ec2f4g-1.jpg?height=478&width=512&top_left_y=1416&top_left_x=594)
C $\quad$ B $\quad \mathrm{N}$
C

![](https://cdn.mathpix.com/cropped/2024_05_06_b28578248d01d32ec2f4g-1.jpg?height=394&width=445&top_left_y=1496&top_left_x=1154)

Similarly, in the second case (when $B, C$ belong to one plot), there can be two division options (see figure). From the condition of equal areas, we get the plot sizes: in the first case - a) three plots $25 \times 12$ or b) one plot $25 \times 12$ and two plots $12.5 \times 24$; in the second case - a) three plots $\frac{25}{3} \times 36$ or b) one plot $\frac{25}{3} \times 36$ and two plots $\frac{50}{3} \times 18$.
B
a)
C
B
b)
C
![](https://cdn.mathpix.com/cropped/2024_05_06_b28578248d01d32ec2f4g-1.jpg?height=316&width=1058&top_left_y=2307&top_left_x=570)

The total length of the fences in these four options will be 50 or 49 (in the first case) and 72 or $\frac{158}{3}$ (in the second case). Therefore, 49 is the smallest length.