A right triangle has sides of length 2sqrt(2), 6sqrt(2), and sqrt(80) units. What is the area of the triangle, in...

1. INFER which side is the hypotenuse

• Given sides: \(2\sqrt{2}\), \(6\sqrt{2}\), and \(\sqrt{80}\) units
• In a right triangle, the hypotenuse is always the longest side
• Compare the lengths:
  • \(2\sqrt{2} \approx 2.83\)
  • \(6\sqrt{2} \approx 8.49\)
  • \(\sqrt{80} = 4\sqrt{5} \approx 8.94\)
• Therefore \(\sqrt{80}\) is the hypotenuse, and \(2\sqrt{2}\) and \(6\sqrt{2}\) are the legs

2. INFER the area strategy

• For right triangles, we can use \(\mathrm{A} = \frac{1}{2}\mathrm{bh}\) where the base and height are the two legs
• This works because the legs are perpendicular to each other
• Use the legs: \(\mathrm{base} = 2\sqrt{2}\), \(\mathrm{height} = 6\sqrt{2}\)

3. SIMPLIFY the area calculation

• \(\mathrm{A} = \frac{1}{2}\mathrm{bh} = \frac{1}{2}(2\sqrt{2})(6\sqrt{2})\)
• Multiply the coefficients: \(\frac{1}{2}(2)(6) = \frac{1}{2}(12) = 6\)
• Multiply the radicals: \(\sqrt{2} \times \sqrt{2} = \sqrt{4} = 2\)
• Combine: \(\mathrm{A} = 6 \times 2 = 12\) square units

Answer: B. 12

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Not identifying which side is the hypotenuse correctly, or not recognizing that legs should be used for base and height.

Students might assume the first two sides given (\(2\sqrt{2}\) and \(6\sqrt{2}\)) are base and height without checking if this makes sense geometrically. Some might try to use the hypotenuse \(\sqrt{80}\) as either the base or height, leading to incorrect area calculations. This leads to confusion and potentially selecting any of the wrong answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Making arithmetic errors when multiplying the radicals.

Students might incorrectly calculate \((2\sqrt{2})(6\sqrt{2})\), perhaps computing \(\sqrt{2} \times \sqrt{2} = \sqrt{4}\) as just \(\sqrt{4}\) instead of 2, or making errors with the coefficients. They might get something like \(6\sqrt{4}\) and leave it unsimplified, or incorrectly simplify it to get a wrong numerical answer. This may lead them to select Choice A (\(8\sqrt{2} + \sqrt{80}\)) if they're trying to combine terms incorrectly.

The Bottom Line:

This problem tests whether students can identify the structure of a right triangle (which side is which) and then execute the area calculation correctly. The key insight is recognizing that in a right triangle, you use the perpendicular legs for the area formula, not the hypotenuse.