An isosceles right triangle has a perimeter of 94 + 94sqrt(2) inches. What is the length, in inches, of one...

1. TRANSLATE the problem information

• Given information:
  • Isosceles right triangle
  • Perimeter = \(94 + 94\sqrt{2}\) inches
  • Need to find: length of one leg

2. INFER the triangle properties and set up the equation

• In an isosceles right triangle:
  • Two legs have equal length (call this t)
  • Hypotenuse = \(\mathrm{t}\sqrt{2}\) (this is a key property!)
• Perimeter = leg + leg + hypotenuse = \(\mathrm{t} + \mathrm{t} + \mathrm{t}\sqrt{2} = \mathrm{t}(2 + \sqrt{2})\)
• Set up equation: \(\mathrm{t}(2 + \sqrt{2}) = 94 + 94\sqrt{2}\)

3. SIMPLIFY to solve for t

• Divide both sides by \((2 + \sqrt{2})\): \(\mathrm{t} = \frac{94 + 94\sqrt{2}}{2 + \sqrt{2}}\)
• To rationalize, multiply by \(\frac{2 - \sqrt{2}}{2 - \sqrt{2}}\):
  \(\mathrm{t} = \frac{(94 + 94\sqrt{2})(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})}\)

4. SIMPLIFY the multiplication

• Numerator: \((94 + 94\sqrt{2})(2 - \sqrt{2}) = 188 - 94\sqrt{2} + 188\sqrt{2} - 188 = 94\sqrt{2}\)
• Denominator: \((2 + \sqrt{2})(2 - \sqrt{2}) = 4 - 2 = 2\)
• Therefore: \(\mathrm{t} = \frac{94\sqrt{2}}{2} = 47\sqrt{2}\)

Answer: B. \(47\sqrt{2}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students often assume that since it's an isosceles right triangle, all three sides are equal, forgetting that the hypotenuse in a right triangle is always longer than the legs.

If they set up the equation as \(3\mathrm{t} = 94 + 94\sqrt{2}\), they get \(\mathrm{t} = \frac{94 + 94\sqrt{2}}{3}\), which doesn't match any answer choice. This leads to confusion and guessing.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly set up \(\mathrm{t}(2 + \sqrt{2}) = 94 + 94\sqrt{2}\) but struggle with rationalizing the denominator. They might try to divide directly and get confused by the radical expressions.

This computational difficulty may lead them to select Choice C (94) by incorrectly thinking the leg length should be simpler, or they abandon the systematic solution and guess.

The Bottom Line:

This problem tests both conceptual understanding of special right triangles and algebraic manipulation with radicals. Students must remember the specific property that in an isosceles right triangle, the hypotenuse equals leg × \(\sqrt{2}\), then successfully work through the rationalization process.