Rectangle Dimensions: Using Perimeter and Length-Width Relationships

Skills This Problem Builds

• Translating a comparison into algebra. "6 meters longer than" becomes + 6. This phrase pattern shows up in nearly every word problem you will meet.
• Choosing an anchor variable. Naming the smaller or "base" quantity w and describing everything else in terms of it keeps you to one unknown.
• Reading the perimeter formula as a story. Perimeter counts every side, so 2L + 2W means "two lengths and two widths." That structure is also what makes the half-perimeter shortcut work.
• Solving the same problem three ways. One-variable algebra, a bar-model argument, and a sum-and-difference system all land on the same answer. Seeing why they agree is a real sign of understanding.
• Checking against the story. A correct-looking solution should satisfy both clues in the problem, not just one.

Let's Draw It

Before writing any algebra, sketch what the words describe. We don't know the numbers yet, so the sides get labels in terms of one unknown, w.

Opposite sides of a rectangle are equal, so the sketch has only two different lengths: w and w + 6.

Solution: Method 1 — Width as the Anchor Variable

The sentence "6 meters longer than it is wide" describes the length in terms of the width, so the width is the natural thing to name. Do that and you need only one unknown.

Step 1 — Name the unknown

Let w be the width of the room in meters.

Step 2 — Express the length using the width

The length is 6 meters more than the width, so

Step 3 — Write the perimeter equation

The perimeter is the distance all the way around: two lengths plus two widths. Substitute our expressions and set the total equal to 28.

Step 4 — Simplify and solve

Distribute the 2, combine the w terms, then undo the operations one at a time.

Step 5 — Go back and find the length

We solved for the width, but the question asks for both dimensions. Use the relationship from Step 2:

Solution: Method 2 — Half the Perimeter and the Bar Model

This approach needs almost no algebra. It relies on one observation: if you walk halfway around a rectangle, you cover exactly one length and one width.

Step 1 — Cut the perimeter in half

The full trip is 28 meters, so half of it is 14 meters:

Step 2 — Draw those 14 meters as a bar

The length is the width plus an extra 6. So the 14-meter bar is made of one width, another width, and the extra 6 meters:

Total bar: 14 meters.

Step 3 — Remove the extra piece

Take away the 6-meter extra from the total and what remains is two equal widths:

The same answer, found with arithmetic you could do on a napkin. Notice that this is Method 1 in disguise: dividing by 2 first and then subtracting 6 are the same inverse operations, just in a friendlier order.

Solution: Method 3 — A Sum-and-Difference System

If you prefer two variables, translate each clue into its own equation. Let L be the length and W the width.

Add the two equations. The W terms cancel, which is exactly why this elimination step works so well:

Subtracting the equations instead would eliminate L and give 2W = 8, so W = 4 directly. Either route works.

Verification

The problem gave two clues, so the answer must pass two tests.

We can also substitute into the Method 1 equation, 2(w + 6) + 2w = 28, with w = 4: 2(10) + 8 = 28 ✓. Both clues hold, so the dimensions are right.

Sanity Check: Testing Widths by Hand

Algebra is not the only way in. Because the length always equals the width plus 6, you can simply try widths and watch the perimeter climb until it reaches 28:

Each time the width grows by 1 meter, the length grows by 1 too, so the perimeter grows by 4 meters. That is why the equation simplified to 4w + 12: the 4 is the growth per meter, and the 12 is the "6 + 6" built into the two longer sides. The table also shows the answer is unique, since no other width can hit 28 exactly.

What Trips People Up

The Pattern Behind This

Nothing here depends on the numbers 6 and 28. Suppose a rectangle has perimeter P and its length exceeds its width by d. Half the perimeter is P/2, and the bar-model argument from Method 2 gives a general recipe:

For our room: P/4 = 7 and d/2 = 3, so the width is 7 − 3 = 4 and the length is 7 + 3 = 10. Read it as a story: the dimensions sit symmetrically around the "average side" P/4, one d/2 below it and one d/2 above.

If You See These Words...

• "___ longer than it is wide", "twice as long as wide", "3 more than the width": one dimension is described using the other. Name the described-from one as your variable.
• "perimeter", "fence around", "border", "trim", "baseboard": the total is 2L + 2W.
• "Find the dimensions": you owe the reader two numbers, one for each side.

Same structure, different costume: "Two numbers add to 14 and one is 6 larger than the other. Find them." That is exactly the system in Method 3, with the rectangle removed.

Beyond the Textbook

• Flooring and trim. Installers often know the total length of baseboard they have bought (a perimeter) and need to figure out what room sizes it can cover.
• Fencing a garden or pen. A fixed amount of fence and a desired shape (for example, "longer than wide") pins down the rectangle, and the same equations apply.
• Screen and picture frames. Frame makers work from a perimeter of molding and an aspect ratio or length-to-width gap to determine what size frame the material will build.

When the Answer Isn't Neat

Real rooms are rarely whole numbers of meters. Keep the 6-meter gap but change the perimeter to 30 meters. The method is identical:

Dividing gives w = 4.5, so the room is 4.5 m by 10.5 m. A decimal is not a sign of a mistake. It only means the numbers did not happen to divide evenly. The setup is the same, and only the arithmetic at the end changes.

Where This Leads

Replace the perimeter clue with an area clue and the equation becomes w(w + 6) = area. That is a quadratic equation, the gateway to factoring and the quadratic formula in Algebra 1. Perimeter is linear (the unknown appears to the first power); area is quadratic, and that difference is the next big step. The fourth What-If below previews it using factor pairs.

Try These Variations

These four problems build on each other: a new set of numbers, a different relationship, a change to the room, and finally a switch from perimeter to area. Attempt each on paper before revealing the solution.