The perimeter of a rectangular field is 380 yd. The length is 30 yd longer than the width. Find the dimensions.

Let's define the dimensions of the rectangular field.

- Let \( w \) be the width of the field.
- The length \( l \) is given as \( l = w + 30 \) (since the length is 30 yards longer than the width).

The formula for the perimeter \( P \) of a rectangle is:

\[
P = 2l + 2w
\]

We're told the perimeter is 380 yards, so:

\[
380 = 2(w + 30) + 2w
\]

Now, let's solve for \( w \):

1. Distribute the terms:
\[
380 = 2w + 60 + 2w
\]

2. Combine like terms:
\[
380 = 4w + 60
\]

3. Subtract 60 from both sides:
\[
320 = 4w
\]

4. Divide by 4:
\[
w = 80
\]

So, the width \( w \) is 80 yards.

Now, substitute \( w = 80 \) into the expression for the length:
\[
l = 80 + 30 = 110
\]

Thus, the dimensions of the field are:
- Width: 80 yards
- Length: 110 yards

### Final Answer:
- Width: 80 yards
- Length: 110 yards

Do you want more details or have any questions?

Here are 5 relative questions for further practice:
1. What is the area of the field given its dimensions?
2. If the perimeter was 500 yards instead, what would the dimensions be?
3. How would the dimensions change if the length were twice the width instead of 30 yards longer?
4. What is the diagonal length of the field?
5. How would the formula for the perimeter change if the field were a square?

**Tip:** When solving for dimensions in perimeter problems, remember to express both the length and width in terms of one variable before solving.

Learn how to find the dimensions of a rectangular field with a perimeter of 380 yards, where the length is 30 yards longer than the width. This problem involves key algebraic concepts, including solving linear equations and understanding geometric properties of rectangles. Suitable for students in Grades 6-8.

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