An isosceles right triangle (where both legs are equal) has a perimeter of \(8(2 + \sqrt{2})\). What is the area...

1. TRANSLATE the problem information

• Given information:
  • Isosceles right triangle (both legs equal)
  • Perimeter = \(8(2 + \sqrt{2})\)
  • Need to find area

2. INFER the triangle relationships

• An isosceles right triangle is a 45-45-90 special right triangle
• If each leg has length s, then the hypotenuse has length \(s\sqrt{2}\)
• This gives us the side ratio \(s : s : s\sqrt{2}\)

3. TRANSLATE the perimeter into an equation

• Perimeter = leg + leg + hypotenuse
• Perimeter = \(s + s + s\sqrt{2} = 2s + s\sqrt{2}\)

4. SIMPLIFY by factoring

• Factor out the common term s:
• Perimeter = \(s(2 + \sqrt{2})\)

5. INFER the solution strategy

• We know the perimeter equals \(8(2 + \sqrt{2})\)
• Set up the equation: \(s(2 + \sqrt{2}) = 8(2 + \sqrt{2})\)

6. SIMPLIFY to find the leg length

• Divide both sides by \((2 + \sqrt{2})\):
• \(s = 8\)

7. APPLY the area formula

• Area of right triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)
• Since both legs equal s: Area = \(\frac{1}{2} \times s \times s = \frac{1}{2}s^2\)
• Area = \(\frac{1}{2} \times 8^2 = \frac{1}{2} \times 64 = 32\)

Answer: 32

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Not recognizing that an isosceles right triangle follows the 45-45-90 ratio, so students try to use the Pythagorean theorem instead.

Students might set up: \(s^2 + s^2 = h^2\) to find the hypotenuse, then get \(h = s\sqrt{2}\), but this takes much longer and creates opportunities for errors. Some students get confused during this process and make calculation mistakes or lose track of their variables.

This leads to confusion and potential guessing rather than the clean algebraic approach.

Second Most Common Error:

Poor SIMPLIFY execution: Setting up the equation correctly but making errors when solving \(s(2 + \sqrt{2}) = 8(2 + \sqrt{2})\).

Students might try to expand both sides instead of simply dividing by \((2 + \sqrt{2})\), leading to unnecessary complications like \(2s + s\sqrt{2} = 16 + 8\sqrt{2}\). This creates a more complex system that's prone to algebraic mistakes.

This may lead them to incorrect values for s and consequently wrong area calculations.

The Bottom Line:

Success depends on recognizing the special right triangle properties immediately and using the clean algebraic approach rather than getting bogged down in lengthy calculations.