A rectangle has an area of 63 square meters and a length of 9 meters. What is the width, in...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{Area = 63\text{ square meters}}\)
  • \(\mathrm{Length = 9\text{ meters}}\)
  • Need to find: width (in meters)

2. INFER the approach needed

• We need the rectangle area formula since we have area and length, but need width
• The area formula will let us set up an equation to solve for the unknown width

3. TRANSLATE the relationship into an equation

• Rectangle area formula: \(\mathrm{A = length \times width}\)
• Substituting our values: \(\mathrm{63 = 9 \times w}\)

4. SIMPLIFY to solve for width

• To isolate w, divide both sides by 9:
  \(\mathrm{63 \div 9 = w}\)
• \(\mathrm{w = 7}\)

Answer: A. 7

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER reasoning: Students recognize they need to do something with 63 and 9, but don't think about which operation makes sense for the area relationship. Instead, they perform subtraction because they see two numbers and want to "find the difference."

\(\mathrm{63 - 9 = 54}\)

This leads them to select Choice B (54).

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up \(\mathrm{63 = 9w}\) but then multiply instead of divide, thinking "to get rid of the 9, I multiply both sides by 9."

\(\mathrm{63 \times 9 = 567}\)

This leads them to select Choice D (567).

The Bottom Line:

This problem tests whether students can connect the area formula to algebraic problem-solving. The key insight is that when you know area and one dimension, division (not addition, subtraction, or further multiplication) gives you the missing dimension.