An isosceles right triangle has an area of 450 square inches. What is the perimeter, in inches, of this triangle?

1. TRANSLATE the problem information

• Given information:
  • Isosceles right triangle with area = 450 square inches
  • Need to find perimeter

• What this tells us: Two legs are equal, one angle is \(\mathrm{90°}\)

2. INFER the approach

• In an isosceles right triangle, both legs have the same length
• The legs form the base and height for area calculation
• We'll find leg length first, then hypotenuse using Pythagorean theorem

3. SIMPLIFY to find leg length

• Area formula: \(\mathrm{Area = \frac{1}{2} \times base \times height}\)
• For our triangle: \(\mathrm{Area = \frac{1}{2} \times leg \times leg = \frac{leg^2}{2}}\)
• Set up equation: \(\mathrm{\frac{leg^2}{2} = 450}\)
• Solve: \(\mathrm{leg^2 = 900}\), so \(\mathrm{leg = 30\text{ inches}}\)

4. SIMPLIFY to find hypotenuse

• Use Pythagorean theorem: \(\mathrm{hypotenuse^2 = leg_1^2 + leg_2^2}\)
• Substitute: \(\mathrm{hypotenuse^2 = 30^2 + 30^2 = 900 + 900 = 1800}\)
• Take square root: \(\mathrm{hypotenuse = \sqrt{1800} = \sqrt{900 \times 2} = 30\sqrt{2}}\)

5. Calculate perimeter

• Perimeter = \(\mathrm{leg_1 + leg_2 + hypotenuse}\)
• Perimeter = \(\mathrm{30 + 30 + 30\sqrt{2} = 60 + 30\sqrt{2}}\)

Answer: \(\mathrm{C\text{ }(60 + 30\sqrt{2})}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that in an isosceles right triangle, both legs serve as base and height for the area formula, or they don't realize both legs are equal.

They might try to use \(\mathrm{Area = \frac{1}{2} \times base \times height}\) with unknown values for both base and height, leading to confusion about how to set up the equation. This leads to getting stuck and guessing among the answer choices.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students make calculation errors when simplifying \(\mathrm{\sqrt{1800}}\), either not recognizing it equals \(\mathrm{30\sqrt{2}}\) or making arithmetic mistakes.

They might calculate \(\mathrm{\sqrt{1800}}\) incorrectly as \(\mathrm{60\sqrt{2}}\) or approximate it as a decimal, leading them to select Choice \(\mathrm{D\text{ }(60 + 60\sqrt{2})}\) or give up on the radical form entirely.

The Bottom Line:

The key insight is recognizing that "isosceles right triangle" gives you crucial structural information—the legs are equal AND they form the perpendicular sides needed for area calculation. Without this connection, students can't even begin the solution process.