## Task 2 - 240722

A garden of rectangular shape is exactly $13 \mathrm{~m}$ longer than it is wide. To completely fence it, exactly $92 \mathrm{~m}$ of fencing is needed.

a) Calculate the area of the garden!

b) The garden is to be completely divided into beds and paths, with the following conditions to be met:

Each bed has the shape of a rectangle with side lengths of $3 \mathrm{~m}$ and $1 \mathrm{~m}$. Between any two adjacent beds and between the fence and the beds, there is a $25 \mathrm{~cm}$ wide path everywhere.

Investigate whether there is a division of the garden that meets these conditions! If this is the case, determine the number of beds for such a division!
a) If $a$ and $b$ are the length and width of the garden in meters, respectively, then $a=13+b$ (1) and, since the half-perimeter is $92 \mathrm{~m} / 2 = 46 \mathrm{~m}$, $a+b=46$. (2)

Substituting $a$ from (1) into (2) yields $13+2b=46$, i.e., $b=16.5$ and thus from (1) $a=29.5$.

The area of the garden is therefore $16.5 \mathrm{~m} \cdot 29.5 \mathrm{~m} = 486.75 \mathrm{~m}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0478.jpg?height=431&width=625&top_left_y=233&top_left_x=721)

Figure a (not to scale)

b) If a path 25 cm wide is initially laid along two adjacent sides of the garden (shaded in Figure a), a rectangle remains with a length of $29.25 \mathrm{~m}$ and a width of $16.25 \mathrm{~m}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0478.jpg?height=280&width=528&top_left_y=865&top_left_x=293)

If this rectangle can be divided into smaller rectangles with side lengths of $3.25 \mathrm{~m}$ and $1.25 \mathrm{~m}$ (Figure b), then an arrangement of beds that meets the required conditions can be achieved.

One such arrangement is shown in Figure a, which can be confirmed because $29.25 / 3.25 = 9$ and $16.25 / 1.25 > 13$. In this division, the number of beds is $9 \cdot 13 = 117$.