The perimeter of a rectangle is 260 feet. Find the length and width if the length is an even integer and the width is 5 times the next consecutive even integer.


Let's define the variables:

- Let the length of the rectangle be \( l \).
- Since the length is an even integer, let \( l = 2n \) for some integer \( n \), where \( n \) is the half of the length.
- The width is 5 times the next consecutive even integer, so the width is \( w = 5(2n + 2) = 10n + 10 \).

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

\[
P = 2(l + w)
\]

We are given that the perimeter is 260 feet. Therefore:

\[
2(l + w) = 260
\]

Substitute \( l = 2n \) and \( w = 10n + 10 \) into the perimeter formula:

\[
2(2n + 10n + 10) = 260
\]

Simplify the equation:

\[
2(12n + 10) = 260
\]

Divide both sides by 2:

\[
12n + 10 = 130
\]

Subtract 10 from both sides:

\[
12n = 120
\]

Solve for \( n \):

\[
n = 10
\]

Now, substitute \( n = 10 \) back into the expressions for length and width:

\[
l = 2n = 2(10) = 20 \text{ feet}
\]
\[
w = 10n + 10 = 10(10) + 10 = 100 + 10 = 110 \text{ feet}
\]

Thus, the length is 20 feet, and the width is 110 feet.

Would you like more details or explanations?

Here are 5 related questions:
1. How would the solution change if the perimeter were 300 feet?
2. What happens if the width is expressed as a function of the length in a different way?
3. How do you calculate the area of the rectangle with these dimensions?
4. What is the ratio of the width to the length for this rectangle?
5. How would this change if the length were the one dependent on the width?

**Tip:** For word problems, clearly defining variables helps in translating the problem into equations.

The perimeter of a rectangle is 260 feet. Find the length and width if the length is an even integer and the width is 5 times the next consecutive even integer.

Learn how to find the length and width of a rectangle with a perimeter of 260 feet, where the length is an even integer and the width is 5 times the next consecutive even integer. This problem involves key algebraic and geometric concepts, suitable for Grades 6-8.

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