Example

Finding Rectangle Dimensions When Width Is Defined in Terms of Length

Apply the geometry problem-solving strategy when one dimension of a rectangle is described relative to the other, requiring both unknowns to be expressed through a single variable before substituting into the perimeter formula.

Problem: The width of a rectangle is two feet less than the length. The perimeter is 5252 feet. Find the length and width.

  1. Read: A rectangle has a perimeter of 5252 ft, and its width is 22 feet less than its length.
  2. Identify: The length and width of the rectangle.
  3. Name: Since the width is described in terms of the length, let LL = the length. The phrase "two feet less than the length" translates to L2L - 2 = the width. Draw and label the figure with length LL and width L2L - 2.
  4. Translate: Write the perimeter formula and substitute:

P=2L+2WP = 2L + 2W

52=2L+2(L2)52 = 2L + 2(L - 2)

  1. Solve: Distribute: 52=2L+2L452 = 2L + 2L - 4. Combine like terms: 52=4L452 = 4L - 4. Add 44 to both sides: 56=4L56 = 4L. Divide by 44: 564=4L4\frac{56}{4} = \frac{4L}{4}, so 14=L14 = L. The length is 1414 feet. Find the width: L2=142=12L - 2 = 14 - 2 = 12. The width is 1212 feet.
  2. Check: 14+12+14+12=5214 + 12 + 14 + 12 = 52, and 52=5252 = 52 \checkmark
  3. Answer: The length is 1414 feet and the width is 1212 feet.

When neither dimension is given directly but one is described relative to the other, both unknowns must be expressed through a single variable before substituting into the formula. This produces an equation with like terms (2L+2L2L + 2L) that must be combined before isolating the variable — a key step that distinguishes this type of problem from ones where one dimension is already a known number. Note that the figure is not drawn until Step 3, after the variable expression for the width is established, so that one side can be labeled with the expression L2L - 2.

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Updated 2026-04-21

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