Rectangle Dimensions: Using Perimeter and Length-Width Relationships

Geometry 7th-8th Grade
PROBLEM
A rectangular room is 6 meters longer than it is wide, and its perimeter is 28 meters. Find the dimension of the room.

What You Will Learn

  • How to translate word relationships into algebraic expressions
  • Setting up and solving linear equations from geometric constraints
  • Using the perimeter formula strategically with one variable
  • Verifying solutions by checking both the constraint and the relationship
  • Recognizing when dimension problems require exactly two pieces of information

Let's Draw It

Before diving into algebra, let's visualize what we know about this rectangular room:

A rectangular room is 6 meters longer than it is wide, and its perimeter is 28 meters. Find the dimension of the room.

The diagram shows our unknown width w and the length expressed as w + 6. The perimeter constraint gives us the equation we need to solve for w.

Solution: Method 1 — The Perimeter Formula Approach

When we know a relationship between dimensions and the perimeter, we can express everything in terms of one variable.

Step 1 — Define the variable

Let w = the width of the rectangular room in meters.

Step 2 — Express the length

Since the room is 6 meters longer than it is wide:

Length = w + 6

Step 3 — Apply the perimeter formula

For any rectangle, the perimeter is P = 2L + 2W. Substituting our known values:

28 = 2(w + 6) + 2w

Step 4 — Solve for the width

Distribute and simplify:

28 = 2w + 12 + 2w
28 = 4w + 12
16 = 4w
w = 4

Step 5 — Find the length

Using our relationship from Step 2:

Length = w + 6 = 4 + 6 = 10

Solution: Method 2 — The Half-Perimeter Method

Sometimes it's cleaner to work with the sum of length and width rather than the full perimeter formula.

Step 1 — Find the half-perimeter

Since the perimeter is 28 meters, the sum of length and width is:

Length + Width = 28 ÷ 2 = 14

Step 2 — Set up the equation

Let w = width. Then length = w + 6, so:

w + (w + 6) = 14

Step 3 — Solve for width

Combine like terms:

2w + 6 = 14
2w = 8
w = 4

Step 4 — Find the length

Length = w + 6 = 4 + 6 = 10 meters.

Both methods give us the same answer, but this approach often involves simpler arithmetic when working with perimeter problems.

The dimensions of the room are: Width = 4 meters, Length = 10 meters

Verification

Let's check our answer against both conditions:

Check the perimeter

P = 2L + 2W = 2(10) + 2(4) = 20 + 8 = 28 ✓

Check the length-width relationship

Length - Width = 10 - 4 = 6 ✓

Both conditions are satisfied, confirming our solution is correct.

Watch Out For These

✗ Confusing length and width in the setup

Some students write "width = length + 6" instead of "length = width + 6". Always read carefully: "6 meters longer than wide" means we add 6 to the width to get the length.

✗ Using the wrong perimeter formula

Writing P = L + W instead of P = 2L + 2W. Remember that perimeter means going all the way around the rectangle, so we need both lengths and both widths.

✗ Forgetting to find both dimensions

Solving for w = 4 and stopping there. The problem asks for "the dimension" (plural implied) - you need both length and width.

The Pattern Behind This

Rectangle dimension problems follow a predictable structure that you can apply to similar situations:

If length = width + d, and perimeter = P, then:
width = (P - 4d) ÷ 4
length = width + d

In our problem, d = 6 and P = 28, so width = (28 - 24) ÷ 4 = 1. Wait, that doesn't match!

Let me recalculate the general formula. If P = 2(w + d) + 2w = 4w + 2d, then w = (P - 2d) ÷ 4.

With our values: width = (28 - 12) ÷ 4 = 4 ✓

This formula is a handy shortcut, but understanding the setup process is more important than memorizing it.

How to Spot This Problem Type

Rectangle dimension problems typically include these key phrases:

  • "[X] meters longer than it is wide" or "[X] feet more than the width"
  • "The length is [X] times the width" or "twice as long as it is wide"
  • "Perimeter is [number]" or "distance around is [number]"
  • "Find the dimensions" or "Find the length and width"

The key insight is that you need exactly two pieces of information: one constraint (like perimeter) and one relationship between the dimensions. With just the perimeter, there are infinitely many possible rectangles. With just the relationship, you can't determine the actual size. Together, they give you exactly one solution.

Where This Shows Up in Real Life

Rectangle dimension problems aren't just academic exercises:

  • Construction and architecture: Determining room sizes when you know the total perimeter of a building footprint and specific room proportions.
  • Landscaping: Designing rectangular gardens or lawns when you have a fixed amount of border material and aesthetic proportion requirements.
  • Manufacturing: Optimizing rectangular material cuts when you know the total edge length available and need specific aspect ratios.

What If?

1
Area Instead of Perimeter
A rectangular room is 6 meters longer than it is wide, and its area is 40 square meters. Find the dimensions.
Step 1 — Set up variables

Let w = width, so length = w + 6

Step 2 — Apply area formula

Area = length × width, so 40 = w(w + 6) = w² + 6w

Step 3 — Rearrange to standard form

w² + 6w - 40 = 0

Step 4 — Factor the quadratic

(w + 10)(w - 4) = 0, so w = -10 or w = 4

Step 5 — Choose the positive solution

Since width must be positive, w = 4 meters and length = 10 meters

Verification

Area = 4 × 10 = 40 ✓, Length - Width = 10 - 4 = 6 ✓

2
Different Relationship
A rectangular garden's length is twice its width. Its perimeter is 36 meters. What is the garden's area?
Step 1 — Express dimensions

Let w = width, so length = 2w

Step 2 — Use perimeter formula

36 = 2(2w) + 2w = 4w + 2w = 6w

Step 3 — Solve for width

w = 36 ÷ 6 = 6 meters

Step 4 — Find length

Length = 2w = 2(6) = 12 meters

Step 5 — Calculate area

Area = 12 × 6 = 72 square meters

Verification

Perimeter = 2(12) + 2(6) = 36 ✓, Length = 2 × width ✓

3
Reverse the Unknown
A rectangular room has a perimeter of 32 meters and an area of 60 square meters. What is the difference between its length and width?
Step 1 — Set up the system

Let length = L and width = W. We have 2L + 2W = 32 and LW = 60

Step 2 — Simplify perimeter equation

L + W = 16, so W = 16 - L

Step 3 — Substitute into area equation

L(16 - L) = 60, which gives 16L - L² = 60

Step 4 — Solve the quadratic

L² - 16L + 60 = 0 factors to (L - 10)(L - 6) = 0

Step 5 — Find both dimensions

If L = 10, then W = 6. If L = 6, then W = 10. Either way, the dimensions are 10m and 6m.

Step 6 — Calculate the difference

Length - Width = 10 - 6 = 4 meters

4
Adding a Border
The original room (4m × 10m) will have a 1-meter wide decorative border added around the outside. What will be the perimeter of the outer edge?
Step 1 — Understand the border

A 1-meter border goes around the outside, adding 1 meter on each side

Step 2 — Find the new width

New width = original width + 1m + 1m = 4 + 2 = 6 meters

Step 3 — Find the new length

New length = original length + 1m + 1m = 10 + 2 = 12 meters

Step 4 — Calculate the new perimeter

Perimeter = 2(12) + 2(6) = 24 + 12 = 36 meters

Alternative approach

The border adds 2 × 1 = 2 meters to each dimension, so it adds 2(2) + 2(2) = 8 meters to the perimeter. Original perimeter + 8 = 28 + 8 = 36 meters ✓

Frequently Asked Questions

How do you find dimensions of a rectangle when you know the perimeter and relationship between length and width? +
Set up a variable for one dimension, express the other in terms of the relationship, then use the perimeter formula P = 2L + 2W. In this problem, if width = w, then length = w + 6, so 28 = 2(w + 6) + 2w, which simplifies to 28 = 4w + 12, giving w = 4 meters.
What's the difference between using the full perimeter formula versus the half-perimeter approach? +
Both methods work equally well. The full perimeter uses P = 2L + 2W, while the half-perimeter recognizes that L + W = P/2. Here, L + W = 14, so with L = W + 6, we get W + (W + 6) = 14. The half-perimeter method often involves simpler arithmetic.
Why do rectangle dimension problems always have two equations? +
You need two pieces of information to find two unknowns (length and width). Usually one equation comes from the perimeter or area constraint, and the other from the relationship between dimensions. In this case, we have P = 28 (perimeter constraint) and L = W + 6 (relationship constraint).
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-05-20

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