Math Problem Statement

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.

Solution

Let's define the variables:

  • Let the length of the rectangle be ll.
  • Since the length is an even integer, let l=2nl = 2n for some integer nn, where nn 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+10w = 5(2n + 2) = 10n + 10.

The formula for the perimeter PP of a rectangle is:

P=2(l+w)P = 2(l + w)

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

2(l+w)=2602(l + w) = 260

Substitute l=2nl = 2n and w=10n+10w = 10n + 10 into the perimeter formula:

2(2n+10n+10)=2602(2n + 10n + 10) = 260

Simplify the equation:

2(12n+10)=2602(12n + 10) = 260

Divide both sides by 2:

12n+10=13012n + 10 = 130

Subtract 10 from both sides:

12n=12012n = 120

Solve for nn:

n=10n = 10

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

l=2n=2(10)=20 feetl = 2n = 2(10) = 20 \text{ feet} w=10n+10=10(10)+10=100+10=110 feetw = 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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Geometry

Formulas

Perimeter of a rectangle: P = 2(l + w)
Width in terms of length: w = 5(2n + 2)

Theorems

Basic Perimeter Theorem

Suitable Grade Level

Grades 6-8