An isosceles right triangle has a hypotenuse of length 58 inches. What is the perimeter, in inches, of this triangle?

1. TRANSLATE the problem information

• Given information:
  • Isosceles right triangle
  • Hypotenuse length = 58 inches
  • Need to find perimeter

2. INFER the solving strategy

• Since the triangle is isosceles right triangle, the two legs must be equal length
• To find perimeter, we need all three side lengths
• We have hypotenuse, so we need to find the leg lengths first
• Use Pythagorean theorem to find leg lengths

3. Set up the Pythagorean relationship

Let \(\mathrm{x}\) = length of each leg (since they're equal in isosceles right triangle)

Using \(\mathrm{a^2 + b^2 = c^2}\): \(\mathrm{x^2 + x^2 = 58^2}\)

4. SIMPLIFY to solve for x

• Combine like terms: \(\mathrm{2x^2 = 58^2}\)
• Divide both sides by 2: \(\mathrm{x^2 = \frac{58^2}{2}}\)
• Take square root:
  \(\mathrm{x = \sqrt{\frac{58^2}{2}}}\)
  \(\mathrm{x = \frac{58\sqrt{2}}{2}}\)
  \(\mathrm{x = 29\sqrt{2}}\)

5. INFER the final calculation needed

• Perimeter = sum of all three sides
• Perimeter = hypotenuse + leg₁ + leg₂ = \(\mathrm{58 + x + x = 58 + 2x}\)

6. SIMPLIFY the final expression

• Substitute \(\mathrm{x = 29\sqrt{2}}\):
  \(\mathrm{Perimeter = 58 + 2(29\sqrt{2})}\)
  \(\mathrm{Perimeter = 58 + 58\sqrt{2}}\)

Answer: C. \(\mathrm{58 + 58\sqrt{2}}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER reasoning about triangle properties: Students may forget that "isosceles right triangle" means the two legs are equal, and instead try to use other approaches or get confused about which sides are equal.

This leads to incorrect setup of equations and typically causes confusion, leading them to guess randomly among the choices.

Second Most Common Error:

Poor SIMPLIFY execution with radicals: Students correctly set up \(\mathrm{2x^2 = 58^2}\) but make algebraic errors when simplifying \(\mathrm{\sqrt{\frac{58^2}{2}}}\), perhaps getting confused about radical arithmetic and arriving at incorrect expressions.

This may lead them to select Choice A (\(\mathrm{29\sqrt{2}}\)) if they stop after finding just one leg length, or Choice B (\(\mathrm{58\sqrt{2}}\)) if they double the wrong expression.

The Bottom Line:

This problem tests whether students can connect the geometric property of isosceles right triangles to algebraic problem-solving, requiring both conceptual understanding and careful radical manipulation.