Perimeter Error: Forgetting the Factor of 2
A student is solving an ISEE Mathematics Achievement problem: 'The length of a rectangle is 3 cm more than twice its width. The perimeter is 36 cm. What is the area of the rectangle?' The student writes: 'Let width = w. Length = 2w + 3. Perimeter = length + width = (2w + 3) + w = 36. So 3w + 3 = 36, 3w = 33, w = 11. Then length = 25. Area = 11 × 25 = 275 cm².' The student checks the answer choices and finds 275, but the correct answer is 65. The student is confused because the algebra seemed straightforward. The student has 2 minutes left and wants to find the mistake without redoing the entire problem. The student remembers that perimeter involves two lengths and two widths but is not sure how to apply that check quickly.
1. Identify the student’s error, solve the problem correctly, and explain why the student’s answer of 275 cm² is impossible for a rectangle with perimeter 36 cm.
Read the worked answer
The error is using Perimeter = length + width instead of Perimeter = 2(length + width). Correct: Let width = w. Length = 2w + 3. Perimeter = 2[(2w + 3) + w] = 2(3w + 3) = 6w + 6 = 36. Then 6w = 30, w = 5. Length = 2(5) + 3 = 13. Area = 5 × 13 = 65 cm². The student’s answer is impossible because a rectangle with perimeter 36 cm has semiperimeter 18 cm (length + width = 18). The student’s dimensions 11 and 25 sum to 36, which is the perimeter, not the semiperimeter; the actual perimeter of an 11-by-25 rectangle would be 72 cm. Also, area 275 cm² would require a much larger perimeter for a rectangle; the maximum area for a given perimeter is a square, and for P=36 the square side is 9, area 81, so 275 is far too large.
2. The student asks for a 30-second check that would catch this error on test day. Describe a quick verification method using the relationship between perimeter and area.
Read the worked answer
Quick check: after finding dimensions, compute length + width and compare to half the given perimeter. For P = 36, length + width must be 18. The student’s 11 + 25 = 36, not 18, so the dimensions are wrong. A second check: plug the found dimensions back into the perimeter formula: 2(11 + 25) = 72, not 36. A third check: area cannot exceed the square of (P/4): (36/4)² = 81; 275 > 81, so reject. On test day, always write the formula first-P = 2l + 2w-and after solving, substitute both dimensions into the original formula. This catches the common error of forgetting the factor of 2.