A rectangular field has a perimeter of 140 meters. The length of the field is 2 meters longer than its...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
A rectangular field has a perimeter of \(140\) meters. The length of the field is \(2\) meters longer than its width. What is the area of the field, in square meters?
1,156
1,200
1,224
1,260
1. TRANSLATE the problem information
- Given information:
- Perimeter = 140 meters
- Length = width + 2 meters
- Need to find: area
- What this tells us: We have one constraint (perimeter) and one relationship between dimensions that should let us find both length and width.
2. INFER the approach
- Since we have two unknowns (length and width) but know their relationship, we can use one variable
- Let width = \(\mathrm{w}\), then length = \(\mathrm{w + 2}\)
- Use the perimeter constraint to create an equation we can solve
3. TRANSLATE into mathematical expressions
- Perimeter formula: \(\mathrm{P = 2(length + width)}\)
- Substitute our expressions: \(\mathrm{140 = 2[(w + 2) + w]}\)
4. SIMPLIFY to solve for width
- \(\mathrm{140 = 2(w + 2 + w)}\)
- \(\mathrm{140 = 2(2w + 2)}\)
- \(\mathrm{140 = 4w + 4}\)
- \(\mathrm{136 = 4w}\)
- \(\mathrm{w = 34}\) meters
5. Find length and calculate area
- Width = 34 meters
- Length = \(\mathrm{34 + 2 = 36}\) meters
- Area = \(\mathrm{length \times width = 36 \times 34 = 1,224}\) square meters
Answer: (C) 1,224
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skills: Students attempt to guess-and-check with the answer choices instead of setting up the algebraic relationship systematically.
They might try working backwards from each area value, but without the proper setup, they get confused about which dimensions correspond to which areas and end up guessing. This leads to random answer selection rather than systematic solution.
Second Most Common Error:
Poor SIMPLIFY execution: Students set up the equation correctly but make arithmetic errors when solving \(\mathrm{4w + 4 = 140}\).
Common mistakes include getting \(\mathrm{w = 32}\) (forgetting to subtract 4 first) or \(\mathrm{w = 36}\) (confusing which value is width vs length). If they get \(\mathrm{w = 32}\), then length = 34, giving area = \(\mathrm{32 \times 34 = 1,088}\), which doesn't match any answer choice and causes confusion.
The Bottom Line:
This problem tests whether students can translate a word problem into algebraic relationships and then execute the algebra correctly. The key insight is recognizing that one variable can represent both dimensions through their given relationship.
1,156
1,200
1,224
1,260