Question:A right triangle is inscribed in a circle so that each vertex lies on the circle. The hypotenuse of the...

1. TRANSLATE the problem information

• Given information:
  • Right triangle inscribed in a circle (all vertices on circle)
  • Hypotenuse = 2 × (shortest leg)
  • Area = 512√3 square units
  • Need to find: diameter of circle

2. INFER the approach and set up variables

• Let the shorter leg = a, longer leg = b, hypotenuse = c
• From the given relationship: \(\mathrm{c = 2a}\)
• We need to find the relationship between all sides, then use the area to solve for specific values

3. INFER the relationship between the legs using Pythagorean theorem

• Since it's a right triangle: \(\mathrm{a^2 + b^2 = c^2}\)
• Substitute \(\mathrm{c = 2a}\): \(\mathrm{a^2 + b^2 = (2a)^2}\)
• SIMPLIFY: \(\mathrm{a^2 + b^2 = 4a^2}\), so \(\mathrm{b^2 = 3a^2}\)
• Therefore: \(\mathrm{b = a\sqrt{3}}\)

4. INFER how to use the area information

• Area of triangle = \(\mathrm{\frac{1}{2} \times base \times height = \frac{1}{2} \times a \times b}\)
• Substitute \(\mathrm{b = a\sqrt{3}}\): \(\mathrm{Area = \frac{1}{2} \times a \times a\sqrt{3} = \frac{a^2\sqrt{3}}{2}}\)

5. SIMPLIFY to solve for the shorter leg

• Set up the equation: \(\mathrm{\frac{a^2\sqrt{3}}{2} = 512\sqrt{3}}\)
• Divide both sides by √3: \(\mathrm{\frac{a^2}{2} = 512}\)
• Multiply both sides by 2: \(\mathrm{a^2 = 1024}\)
• Take the square root: \(\mathrm{a = 32}\)

6. INFER the final answer using Thales' theorem

• Calculate hypotenuse: \(\mathrm{c = 2a = 2(32) = 64}\)
• Key insight: For a right triangle inscribed in a circle, the hypotenuse is the diameter of the circle
• Therefore, diameter = 64

Answer: 64

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may misinterpret "hypotenuse is twice the length of its shortest leg" and set up incorrect relationships like \(\mathrm{b = 2a}\) or confuse which leg is shorter.

This leads them to derive wrong side relationships, get incorrect values when solving the area equation, and arrive at wrong diameter values. This causes confusion and guessing among the available answer choices.

Second Most Common Error:

Missing conceptual knowledge about Thales' theorem: Students correctly find the hypotenuse length but don't realize that for a right triangle inscribed in a circle, the hypotenuse equals the diameter.

They may try to use other circle formulas or get stuck trying to find the radius first, leading to incomplete solutions and random answer selection.

The Bottom Line:

This problem requires connecting multiple geometric concepts (Pythagorean theorem, triangle area, and circle properties) in sequence. Students who struggle with translating the given relationship or don't know Thales' theorem will have difficulty completing the solution chain.