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Let's break down and solve the mathematical problem based on the text and diagram provided in the image.

### Problem Breakdown:
- **The terrain** is a square \( ABCD \) with a side length of 12 meters.
- A **peg \( P \)** is placed on the side \( [AB] \), and the triangle \( APD \) represents the surface that is to be cultivated.
- **\( x \)** is defined as the length \( AP \).

### Questions:
1. **What is the total area of the terrain?**
   - Since the terrain is a square, the area is:
   \[
   \text{Area of the square} = \text{side length}^2 = 12^2 = 144 \text{ m}^2
   \]

2. **(a) To which set does \( x \) belong?**
   - \( x \) represents the distance \( AP \) and is a length along the side \( AB \) of the square, which has a maximum of 12 meters. Thus:
   \[
   x \in [0, 12]
   \]
   
   **(b) Express the area \( S(x) \) of the surface to be cultivated as a function of \( x \).**
   - The triangle \( APD \) has a base \( x \) and a height equal to the side of the square \( (12 \, \text{m}) \). The area of a triangle is given by:
   \[
   S(x) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times 12 = 6x
   \]
   
3. **What is the nature of the function \( S(x) \)? Why can we describe it as a continuous linear growth phenomenon?**
   - The function \( S(x) = 6x \) is a linear function because it is in the form \( f(x) = ax \), where \( a = 6 \), which describes a straight line. Since the area increases continuously as \( x \) increases, this is a linear growth phenomenon.

4. **Construct the table of variations of \( S(x) \). What is the maximum area of the surface to be cultivated?**
   - \( S(x) = 6x \) increases as \( x \) increases. The maximum value occurs when \( x = 12 \), giving:
   \[
   S(12) = 6 \times 12 = 72 \, \text{m}^2
   \]
   Thus, the maximum area of the cultivated surface is 72 square meters.

5. **(a) What will be the area of the cultivated surface if the peg is placed 2.50 m from point \( A \)?**
   - Substitute \( x = 2.5 \) into the function \( S(x) = 6x \):
   \[
   S(2.5) = 6 \times 2.5 = 15 \, \text{m}^2
   \]
   
   **(b) Where should the peg be placed so that the cultivated surface area is at least one-fifth of the total area of the terrain?**
   - One-fifth of the total area of the terrain is:
   \[
   \frac{1}{5} \times 144 = 28.8 \, \text{m}^2
   \]
   To find where \( S(x) = 28.8 \), solve the equation:
   \[
   6x = 28.8 \implies x = \frac{28.8}{6} = 4.8 \, \text{m}
   \]
   Therefore, the peg should be placed 4.8 meters from point \( A \) for the cultivated area to be at least 28.8 square meters.

### Summary:
- The total area of the terrain is \( 144 \, \text{m}^2 \).
- The function for the cultivated area is \( S(x) = 6x \), a linear function.
- The maximum cultivated area is \( 72 \, \text{m}^2 \).
- If the peg is placed 2.5 m from \( A \), the cultivated area is \( 15 \, \text{m}^2 \).
- To have at least one-fifth of the total area, the peg should be placed 4.8 m from \( A \).

Would you like more details on any specific part? Here are some questions that build on this problem:

1. How would the function change if the terrain was a rectangle instead of a square?
2. What if the triangle had a curved boundary, such as a parabola, instead of a straight line?
3. How would the function change if the peg could be placed on any side of the square?
4. Can you describe a situation where this problem could model something in the real world?
5. How would the results change if the side length of the square was different?

**Tip**: When dealing with area-related problems involving triangles or other shapes, always check the formula for the specific shape's area.

Un terrain a la forme d’un carré ABCD de 12 mètres de côté. Un piquet P est placé sur le côté [AB]. Le triangle APD représente la surface à cultiver dans le terrain. On note x la longueur AP.

This problem involves finding the area of a triangular surface within a square of side 12m. The task includes calculating areas, expressing functions in terms of x, and identifying linear growth phenomena. Suitable for Grades 8-10.

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