A right triangle is inscribed in a circle such that each vertex of the triangle lies on the circumference of...

1. TRANSLATE the problem information

• Given information:
  • Right triangle inscribed in circle (all vertices on circumference)
  • Hypotenuse = \(\mathrm{3 \times (shortest\, leg)}\)
  • Area = \(\mathrm{1{,}156\sqrt{2}}\) square units
  • Need to find: diameter of circle

2. INFER the key geometric relationship

• Since we have a right triangle inscribed in a circle, Thales' theorem tells us the hypotenuse must be the diameter of the circle
• This means: \(\mathrm{diameter = hypotenuse}\)

3. TRANSLATE into algebraic variables

• Let shortest leg = \(\mathrm{a}\)
• Then hypotenuse = \(\mathrm{3a}\)
• Let other leg = \(\mathrm{b}\)

4. INFER and SIMPLIFY the relationship between the legs

• Using Pythagorean theorem: \(\mathrm{a^2 + b^2 = (3a)^2}\)
• Expanding: \(\mathrm{a^2 + b^2 = 9a^2}\)
• Solving for \(\mathrm{b^2}\): \(\mathrm{b^2 = 9a^2 - a^2 = 8a^2}\)
• Therefore: \(\mathrm{b = 2a\sqrt{2}}\)

5. SIMPLIFY using the area formula

• Area = \(\mathrm{\frac{1}{2} \times base \times height = \frac{1}{2} \times a \times b}\)
• Substituting \(\mathrm{b = 2a\sqrt{2}}\):
  \(\mathrm{Area = \frac{1}{2} \times a \times (2a\sqrt{2}) = a^2\sqrt{2}}\)
• Setting equal to given area: \(\mathrm{a^2\sqrt{2} = 1{,}156\sqrt{2}}\)
• Dividing both sides by \(\mathrm{\sqrt{2}}\): \(\mathrm{a^2 = 1{,}156}\)
• Taking square root: \(\mathrm{a = \sqrt{1{,}156} = 34}\) (use calculator)

6. SIMPLIFY to find the diameter

• Hypotenuse = \(\mathrm{3a = 3 \times 34 = 102}\)
• Since hypotenuse = diameter: \(\mathrm{diameter = 102}\)

Answer: C) 102

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that Thales' theorem applies to this inscribed right triangle, so they don't realize the hypotenuse equals the diameter.

Without this connection, they may try to use other circle formulas or get confused about how the triangle relates to the circle. They might attempt to find the radius using circumference or area formulas that don't apply here, leading to incorrect calculations and potentially selecting Choice A (68) or getting stuck and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up the area equation \(\mathrm{a^2\sqrt{2} = 1{,}156\sqrt{2}}\) but make algebraic errors when solving.

They might forget to divide both sides by \(\mathrm{\sqrt{2}}\), leading to \(\mathrm{a^2 = 1{,}156\sqrt{2}}\) instead of \(\mathrm{a^2 = 1{,}156}\). This gives them \(\mathrm{a = \sqrt{1{,}156\sqrt{2}} \approx 40.4}\), making the hypotenuse \(\mathrm{\approx 121}\), which doesn't match any answer choice. This leads to confusion and guessing.

The Bottom Line:

This problem requires recognizing a key geometric theorem (Thales') that directly connects the triangle to its circumscribed circle. Without this insight, students get lost trying to force other geometric relationships that don't apply to this specific configuration.