A right triangle has a hypotenuse with a length of 25 inches. The length of one of the triangle's legs...

1. TRANSLATE the problem information

• Given information:
  • Right triangle with hypotenuse = 25 inches
  • One leg is 17 inches longer than the other leg
  • Need to find the area
• What this tells us: We need to find both leg lengths first, then calculate area

2. TRANSLATE the relationships into algebra

• Let the shorter leg = \(\mathrm{x}\) inches
• Then the longer leg = \(\mathrm{x} + 17\) inches
• The hypotenuse = 25 inches

3. INFER the approach using Pythagorean theorem

• Since we have a right triangle, we can use: \(\mathrm{leg_1}^2 + \mathrm{leg_2}^2 = \mathrm{hypotenuse}^2\)
• This gives us: \(\mathrm{x}^2 + (\mathrm{x} + 17)^2 = 25^2\)

4. SIMPLIFY the equation through algebraic steps

• Expand the left side:
  \(\mathrm{x}^2 + (\mathrm{x}^2 + 34\mathrm{x} + 289) = 625\)
• Combine like terms:
  \(2\mathrm{x}^2 + 34\mathrm{x} + 289 = 625\)
• Move all terms to one side:
  \(2\mathrm{x}^2 + 34\mathrm{x} - 336 = 0\)
• Divide by 2:
  \(\mathrm{x}^2 + 17\mathrm{x} - 168 = 0\)

5. SIMPLIFY by factoring the quadratic

• We need two numbers that multiply to -168 and add to 17
• Those numbers are 24 and -7: \((\mathrm{x} + 24)(\mathrm{x} - 7) = 0\)
• This gives us \(\mathrm{x} = -24\) or \(\mathrm{x} = 7\)

6. APPLY CONSTRAINTS to select the valid solution

• Since length cannot be negative: \(\mathrm{x} = 7\) inches
• The legs are 7 inches and \(7 + 17 = 24\) inches

7. Calculate the area

• Area = (1/2) × base × height = (1/2) × 7 × 24 = 84 square inches
  \(\mathrm{Area} = \frac{1}{2} \times \mathrm{base} \times \mathrm{height}\)
  \(= \frac{1}{2} \times 7 \times 24\)
  \(= 84\) square inches

Answer: B. 84

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students struggle to set up the relationship between the legs algebraically. They might try to guess and check with answer choices rather than creating variables, or they set up the wrong relationship (like making both legs equal to \(\mathrm{x} + 17\)).

Without proper variable setup, they can't apply the Pythagorean theorem systematically. This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students make algebraic errors when expanding \((\mathrm{x} + 17)^2\) or when factoring the quadratic equation \(\mathrm{x}^2 + 17\mathrm{x} - 168 = 0\). Common mistakes include:

• Expanding incorrectly as \(\mathrm{x}^2 + 17^2\) instead of \(\mathrm{x}^2 + 34\mathrm{x} + 289\)
• Making sign errors when moving terms or factoring
• Getting the wrong factors and solving for incorrect \(\mathrm{x}\) values

These calculation errors lead to wrong leg lengths and ultimately wrong areas, causing them to select Choice A (72) or Choice C (168).

The Bottom Line:

This problem tests whether students can bridge word problems with quadratic equations. The key challenge is translating the leg relationship into algebra, then executing multi-step algebraic manipulation without errors.