In an isosceles right triangle, the altitude from the right angle to the hypotenuse has length 9 units. What is...

1. TRANSLATE the problem information

• Given information:
  • Isosceles right triangle (45-45-90 triangle)
  • Altitude from right angle to hypotenuse = 9 units
  • Need to find hypotenuse length

2. INFER the strategic approach

• Key insight: We can calculate the triangle's area in two different ways
• Method 1: Using the two equal legs
• Method 2: Using the hypotenuse and the given altitude
• Setting these equal will give us a relationship we can solve

3. TRANSLATE geometric properties to algebra

• Let \(\mathrm{x}\) = length of each leg
• Let \(\mathrm{h}\) = length of hypotenuse
• In a 45-45-90 triangle: \(\mathrm{h = x\sqrt{2}}\)

4. INFER the area relationship

• Area using legs: \(\mathrm{A = \frac{1}{2}x^2}\)
• Area using hypotenuse: \(\mathrm{A = \frac{1}{2} \times h \times 9}\)
• These must be equal: \(\mathrm{\frac{1}{2}x^2 = \frac{1}{2} \times h \times 9}\)

5. SIMPLIFY the area equation

• Cancel \(\mathrm{\frac{1}{2}}\): \(\mathrm{x^2 = 9h}\)
• Substitute \(\mathrm{h = x\sqrt{2}}\): \(\mathrm{x^2 = 9(x\sqrt{2})}\)
• Divide both sides by \(\mathrm{x}\): \(\mathrm{x = 9\sqrt{2}}\)

6. SIMPLIFY to find the hypotenuse

• \(\mathrm{h = x\sqrt{2}}\)
• \(\mathrm{h = 9\sqrt{2} \times \sqrt{2}}\)
• \(\mathrm{h = 9 \times 2}\)
• \(\mathrm{h = 18}\)

Answer: C. 18

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize they can use the area relationship to create a solvable equation. Instead, they try to work directly with the geometric relationships without establishing clear mathematical connections. This leads to confusion about how the altitude measurement relates to finding the hypotenuse, causing them to guess randomly among the answer choices.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students set up the correct equation \(\mathrm{x^2 = 9x\sqrt{2}}\) but make algebraic errors when solving. They might forget to divide by \(\mathrm{x}\), or incorrectly handle the \(\mathrm{\sqrt{2}}\) terms, potentially getting \(\mathrm{x = 9}\) instead of \(\mathrm{x = 9\sqrt{2}}\). This leads them to calculate \(\mathrm{h = 9\sqrt{2}}\), causing them to select Choice B (\(\mathrm{9\sqrt{2}}\)).

The Bottom Line:

This problem requires recognizing that altitude problems in special right triangles often involve area relationships. The key breakthrough is realizing you can calculate area two ways and set them equal—once you have that insight, it's straightforward algebra to find the answer.