In a right triangle, the lengths of the two legs differ by 6 units.The area of the triangle is 27.5...

1. TRANSLATE the problem information

• Given information:
  • Right triangle with legs differing by 6 units
  • Area = 27.5 square units
  • Need to find shorter leg length

• What this tells us: If shorter leg = \(\mathrm{x}\), then longer leg = \(\mathrm{x + 6}\)

2. INFER the approach

• We have area and a relationship between the legs
• Use the right triangle area formula to create an equation
• This will give us a quadratic equation to solve

3. Set up the area equation

• Area formula: \(\mathrm{A = \frac{1}{2} \times leg_1 \times leg_2}\)
• Substitute: \(\mathrm{27.5 = \frac{1}{2} \times x \times (x + 6)}\)

4. SIMPLIFY to solve for x

• Multiply both sides by 2: \(\mathrm{55 = x(x + 6)}\)
• Expand: \(\mathrm{55 = x^2 + 6x}\)
• Rearrange: \(\mathrm{x^2 + 6x - 55 = 0}\)

5. SIMPLIFY by factoring

• Need two numbers that multiply to -55 and add to 6
• Those numbers are 11 and -5: \(\mathrm{(x + 11)(x - 5) = 0}\)
• Solutions: \(\mathrm{x = -11}\) or \(\mathrm{x = 5}\)

6. APPLY CONSTRAINTS to select final answer

• Since length must be positive in real-world context: \(\mathrm{x = 5}\)

Answer: 5

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students often struggle with setting up the leg relationship correctly. They might use variables like \(\mathrm{x}\) and \(\mathrm{y}\) without establishing that \(\mathrm{y = x + 6}\), or worse, they might set up the legs as \(\mathrm{x}\) and 6 (thinking the longer leg is just 6 units).

This leads to incorrect equations and completely wrong solutions, causing them to get stuck and abandon systematic solving in favor of guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up the initial equation but make algebraic errors during the solving process. Common mistakes include sign errors when expanding \(\mathrm{x(x + 6)}\), incorrectly factoring the quadratic, or computational errors when multiplying by 2.

This leads to wrong intermediate steps and ultimately incorrect final answers, though they may arrive at plausible-looking integer values that seem reasonable.

The Bottom Line:

This problem combines algebraic translation skills with quadratic solving, requiring students to maintain accuracy through multiple computational steps while remembering to apply real-world constraints at the end.