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Let's solve the problem step by step.

### Problem:
- The perimeter of a rectangular field is 380 yards.
- The length is 90 yards longer than the width.
- We need to find the dimensions of the field.

### Variables:
Let the width of the rectangle be \( w \) yards.
Then, the length of the rectangle is \( w + 90 \) yards.

### Perimeter Formula:
The perimeter \( P \) of a rectangle is given by:
\[
P = 2(\text{Length} + \text{Width})
\]
Substituting the given values:
\[
380 = 2((w + 90) + w)
\]

### Simplifying the Equation:
\[
380 = 2(2w + 90)
\]
\[
380 = 4w + 180
\]
Now, subtract 180 from both sides:
\[
200 = 4w
\]
Divide both sides by 4:
\[
w = 50
\]

### Finding the Length:
The length is:
\[
w + 90 = 50 + 90 = 140
\]

### Final Dimensions:
- Width = 50 yards
- Length = 140 yards

Would you like more details, or do you have any other questions?

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Here are five related questions to extend this concept:
1. How would the problem change if the perimeter were 500 yards?
2. Can you calculate the area of this rectangular field using the dimensions?
3. What if the length were instead double the width? How would that affect the dimensions?
4. How would the formula for the perimeter change if this were a square instead of a rectangle?
5. What would the dimensions be if the length were 80 yards longer than the width?

**Tip:** Always verify the solution by plugging the dimensions back into the perimeter formula to check correctness.

The perimeter of a rectangular field is 380 yd. The length is 90 yd longer than the width. Find the dimensions.

Learn how to solve for the dimensions of a rectangular field where the perimeter is 380 yards, and the length is 90 yards longer than the width. This problem uses basic algebra and the perimeter formula for rectangles. Ideal for students in Grades 6-8.

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