Area and Perimeter Word Problems

Using the rectangle formulas in reverse: solving for a missing side from an area constraint and from a perimeter constraint, and why the two give different answers.

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Formulas earn their keep when a measurement is missing. If you know the area or the perimeter but not one of the sides, the formula becomes an equation — and solving it recovers the side you need.

Concept The two formulas
Area:  A = l \times w .
Perimeter:  P = 2l + 2w .

Read left to right they compute a result. Read the other way, with the result known and a side unknown, they become equations to solve.

Concept The situation
2 m 2 m 2 m 6 m of frontage ?
Three shops stand side by side on a street, each 2 m wide. Together they form a single rectangle with  3 \times 2 = 6 m of frontage. The depth is unknown.
Example Given the area

The permitted area is 18 m². How deep may the shops be?

Let  x be the depth. The frontage is 6 m.
 6x = 18 .
Divide both sides by 6:  x = 3 .
Check:  6 \times 3 = 18 .
⟹ a depth of 3 m
Example Given the perimeter

Now suppose the permitted perimeter is 20 m instead. How deep may the shops be?

Let  y be the depth, with the frontage still 6 m.
 2 \times 6 + 2y = 20 .
Simplify:  12 + 2y = 20 .
Subtract 12 from both sides:  2y = 8 .
Divide both sides by 2:  y = 4 .
Check:  2 \times 6 + 2 \times 4 = 12 + 8 = 20 .
⟹ a depth of 4 m

This one needed two steps rather than one, because the perimeter formula contains both a multiplier and a constant.

Concept Why the answers differ

Same shops, same frontage — yet one constraint permits 3 m of depth and the other permits 4 m. The two limits measure different things.

An area limit caps the floor space inside.
A perimeter limit caps the boundary around.

With a depth of 4 m the area would be  6 \times 4 = 24 m², well over the 18 m² allowed in the first problem. So always check which quantity the constraint actually restricts before choosing a formula.

Note Mistakes to avoid
Using the area formula when the limit is on the perimeter, or the reverse.
Forgetting to combine the three shop widths into one frontage of 6 m.
Dividing by 2 before subtracting the constant in the perimeter equation.
Doubling only one side in P = 2l + 2w.
Reporting the answer without its unit, or with the wrong one.
Summary
  1. A formula with a missing side becomes an equation to solve.
  2. An area constraint gives a one-step equation: 6x = 18, so x = 3.
  3. A perimeter constraint gives a two-step equation: 12 + 2y = 20, so y = 4.
  4. Clear the constant before dividing by the coefficient.
  5. Area limits floor space; perimeter limits the boundary — they give different answers.