To find the area of the field in terms of $x$, we proceed as follows:

Step 1: Define the dimensions of the rectangle

Let:

• $w$ = width of the field
• $l$ = length of the field

The problem states that the width is half the length: $w = \frac{l}{2}$

Step 2: Use the perimeter formula

The perimeter $P$ of a rectangle is given by: $P = 2l + 2w$

The problem states that the field is surrounded by $x$ meters of fence: $x = 2l + 2w$

Substitute $w = \frac{l}{2}$: $x = 2l + 2\left(\frac{l}{2}\right)$

Simplify: $x = 2l + l$ $x = 3l$

Solve for $l$: $l = \frac{x}{3}$

Step 3: Calculate the width

Using $w = \frac{l}{2}$: $w = \frac{\frac{x}{3}}{2} = \frac{x}{6}$

Step 4: Calculate the area

The area $A$ of a rectangle is given by: $A = l \times w$

Substitute $l = \frac{x}{3}$ and $w = \frac{x}{6}$: $A = \frac{x}{3} \times \frac{x}{6}$

Simplify: $A = \frac{x^2}{18}$

Final Answer:

The area of the field in terms of $x$ is: $A = \frac{x^2}{18}$

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Related Questions:

1. What would the area be if $x$ were a specific value (e.g., 12 meters)?
2. How does the area change if the relationship between length and width is altered?
3. How would the formula change if the field were square?
4. Can we express the perimeter in terms of the area instead of $x$?
5. What is the physical interpretation of $x^2/18$ in this context?

Tip:

Always double-check unit consistency when working with geometric formulas.