In a right triangle, the lengths of the two legs are x units and \(\mathrm{(x - 10)}\) units. The area...

1. TRANSLATE the problem information

• Given information:
  • Right triangle with legs of length \(\mathrm{x}\) and \(\mathrm{(x - 10)}\) units
  • Area = 72 square units
• What this tells us: We need to use the triangle area formula with these specific leg lengths

2. INFER the approach

• Since we have a right triangle, the two legs are perpendicular and can serve as base and height
• We'll use the area formula \(\mathrm{A = \frac{1}{2} \times base \times height}\) to create an equation
• This will give us a quadratic equation to solve for x

3. TRANSLATE into mathematical equation

Set up: \(\mathrm{\frac{1}{2} \times x \times (x - 10) = 72}\)

4. SIMPLIFY the equation

• Multiply both sides by 2: \(\mathrm{x(x - 10) = 144}\)
• Expand the left side: \(\mathrm{x^2 - 10x = 144}\)
• Move everything to one side: \(\mathrm{x^2 - 10x - 144 = 0}\)

5. SIMPLIFY by factoring

• We need two numbers that multiply to -144 and add to -10
• Those numbers are -18 and 8: \(\mathrm{(-18) \times 8 = -144}\) and \(\mathrm{(-18) + 8 = -10}\)
• Factor: \(\mathrm{(x - 18)(x + 8) = 0}\)
• Solutions: \(\mathrm{x = 18}\) or \(\mathrm{x = -8}\)

6. APPLY CONSTRAINTS to select valid solution

• Since x represents a length, it must be positive, so \(\mathrm{x \neq -8}\)
• Also, since one leg is \(\mathrm{(x - 10)}\), we need \(\mathrm{x \gt 10}\) for both legs to be positive
• Therefore: \(\mathrm{x = 18}\)

7. INFER verification step

• Check: If \(\mathrm{x = 18}\), legs are 18 and 8 units
• Area = \(\mathrm{\frac{1}{2} \times 18 \times 8 = 72}\) ✓

Answer: C (18)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Students struggle with factoring the quadratic \(\mathrm{x^2 - 10x - 144 = 0}\), either making sign errors when finding factor pairs or incorrectly expanding \(\mathrm{(x - 18)(x + 8)}\).

They might factor incorrectly as \(\mathrm{(x - 12)(x + 12)}\) or \(\mathrm{(x - 16)(x + 9)}\), leading to wrong solutions like \(\mathrm{x = 12}\) or \(\mathrm{x = 16}\). This confusion could lead them to select Choice B (10) if they incorrectly think \(\mathrm{x = 12}\) and round down, or get stuck and guess.

Second Most Common Error:

Poor APPLY CONSTRAINTS reasoning: Students find both solutions \(\mathrm{x = 18}\) and \(\mathrm{x = -8}\) but fail to recognize that \(\mathrm{x = -8}\) creates a negative leg length (\(\mathrm{x - 10 = -18}\)), making it geometrically impossible.

They might select \(\mathrm{x = -8}\) and then realize something's wrong, leading to confusion and random guessing among the choices.

The Bottom Line:

This problem combines algebraic manipulation skills with geometric reasoning. Success requires not just solving the quadratic, but also understanding that geometric measurements must make physical sense in the context of the problem.