The length of a rectangle is 3yd longer than its width.
If the perimeter of the rectangle is 50yd, find its length and width.
 


Let's denote the width of the rectangle as \( w \) (in yards). Since the length is 3 yards longer than the width, the length can be written as \( l = w + 3 \).

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

\[
P = 2l + 2w
\]

Given that the perimeter is 50 yards, we can substitute the values into the formula:

\[
50 = 2(w + 3) + 2w
\]

Now, let's solve this equation step by step:

1. Expand the equation:
   \[
   50 = 2w + 6 + 2w
   \]

2. Combine like terms:
   \[
   50 = 4w + 6
   \]

3. Subtract 6 from both sides:
   \[
   44 = 4w
   \]

4. Divide by 4:
   \[
   w = 11
   \]

Now that we know the width is 11 yards, we can find the length:

\[
l = w + 3 = 11 + 3 = 14
\]

Thus, the width is 11 yards, and the length is 14 yards.

Would you like further details or have any questions?

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Here are 5 related questions to expand your understanding:

1. What would the dimensions be if the perimeter was 60 yards instead of 50?
2. How would the dimensions change if the length was 5 yards longer than the width instead of 3?
3. What is the area of the rectangle given the length and width?
4. Can you express the width in terms of the length if the perimeter is given?
5. How would the solution change if the perimeter was given in meters instead of yards?

**Tip**: Always double-check your perimeter formula to avoid errors, especially when setting up your equation!

The length of a rectangle is 3 yards longer than its width. If the perimeter of the rectangle is 50 yards, find its length and width.

This problem explains how to find the length and width of a rectangle where the perimeter is 50 yards, and the length is 3 yards longer than the width. The solution involves basic algebra and geometry, making it suitable for students in Grades 6-8.

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