In a right triangle, the hypotenuse is 13 centimeters, and one leg is 7 centimeters longer than the other leg....

1. TRANSLATE the problem information

• Given information:
  • Right triangle with hypotenuse = 13 cm
  • One leg is 7 cm longer than the other leg
  • Need to find the shorter leg

• Variable setup: Let \(\mathrm{x}\) = shorter leg, then \(\mathrm{x + 7}\) = longer leg

2. INFER the mathematical approach

• Since we have a right triangle, the Pythagorean theorem applies
• We'll set up the equation: \(\mathrm{(shorter\,leg)^2 + (longer\,leg)^2 = (hypotenuse)^2}\)
• This will create a quadratic equation in terms of \(\mathrm{x}\)

3. Apply the Pythagorean theorem

• Set up: \(\mathrm{x^2 + (x + 7)^2 = 13^2}\)
• SIMPLIFY by expanding: \(\mathrm{x^2 + x^2 + 14x + 49 = 169}\)
• Combine like terms: \(\mathrm{2x^2 + 14x + 49 = 169}\)
• Rearrange: \(\mathrm{2x^2 + 14x - 120 = 0}\)
• Divide by 2: \(\mathrm{x^2 + 7x - 60 = 0}\)

4. INFER the solution method and solve

• This quadratic doesn't factor easily, so use the quadratic formula
• With \(\mathrm{a = 1, b = 7, c = -60}\):
  \(\mathrm{x = \frac{-7 \pm \sqrt{49 + 240}}{2}}\)
  \(\mathrm{= \frac{-7 \pm \sqrt{289}}{2}}\)
  \(\mathrm{= \frac{-7 \pm 17}{2}}\)
• This gives \(\mathrm{x = 5}\) or \(\mathrm{x = -12}\)

5. APPLY CONSTRAINTS to select the final answer

• Since length must be positive in a geometric context, \(\mathrm{x = 5}\)
• Check: If shorter leg = 5, then longer leg = 12
• Verification: \(\mathrm{5^2 + 12^2}\)
  \(\mathrm{= 25 + 144}\)
  \(\mathrm{= 169}\)
  \(\mathrm{= 13^2}\) ✓

Answer: B (5)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students struggle to set up the algebraic relationship between the legs. They might set up the legs as \(\mathrm{x}\) and \(\mathrm{7}\) (instead of \(\mathrm{x}\) and \(\mathrm{x+7}\)), thinking "one leg is 7" rather than "one leg is 7 MORE than the other."

This leads to the incorrect equation \(\mathrm{x^2 + 7^2 = 13^2}\), giving \(\mathrm{x^2 = 169 - 49 = 120}\), so \(\mathrm{x = \sqrt{120} \approx 10.95}\). This doesn't match any answer choice, causing confusion and guessing.

Second Most Common Error:

Inadequate APPLY CONSTRAINTS reasoning: Students correctly solve the quadratic but fail to reject the negative solution \(\mathrm{x = -12}\). They might think both solutions are valid or get confused about which represents the "shorter" leg.

This may lead them to select Choice D (12) by mistakenly thinking the negative solution somehow converts to the longer leg.

The Bottom Line:

This problem requires careful translation of the relationship between the legs and systematic algebraic manipulation. The key insight is recognizing that "7 longer than the other" means adding 7 to a variable, not treating 7 as a standalone measurement.