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)
  • Longer leg = \(\sqrt{3}\) × shorter leg
  • Area = \(450\sqrt{3}\) square units
  • Need to find diameter

• What this tells us: We can use the leg relationship to set up variables

2. TRANSLATE into algebraic expressions

• Let shorter leg = x units
• Then longer leg = \(\mathrm{x}\sqrt{3}\) units
• Area formula: \(\mathrm{A} = \frac{1}{2} \times \mathrm{leg_1} \times \mathrm{leg_2}\)

3. SIMPLIFY to solve for the shorter leg

• Set up equation: \(\frac{1}{2} \times \mathrm{x} \times \mathrm{x}\sqrt{3} = 450\sqrt{3}\)
• This gives us: \(\frac{\mathrm{x}^2\sqrt{3}}{2} = 450\sqrt{3}\)
• Divide both sides by \(\sqrt{3}\): \(\frac{\mathrm{x}^2}{2} = 450\)
• Multiply by 2: \(\mathrm{x}^2 = 900\)
• Take square root: \(\mathrm{x} = 30\) units

4. SIMPLIFY using Pythagorean theorem

• We now know: shorter leg = 30, longer leg = \(30\sqrt{3}\)
• Hypotenuse = \(\sqrt{30^2 + (30\sqrt{3})^2}\)
• Calculate: \(\sqrt{900 + 2700} = \sqrt{3600} = 60\) units

5. INFER the connection to the circle

• Key insight: For any right triangle inscribed in a circle, Thales' theorem tells us the hypotenuse equals the diameter
• Therefore: diameter = 60 units

Answer: D (60)

Why Students Usually Falter on This Problem

Most Common Error Path:

Missing conceptual knowledge: Thales' theorem
Many students successfully find the hypotenuse length (60) but don't know how to connect this to the circle's diameter. They might think they need to find the radius first or use a different circle formula entirely.

This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Weak SIMPLIFY execution: Radical manipulation errors
When solving \(\frac{\mathrm{x}^2\sqrt{3}}{2} = 450\sqrt{3}\), students might incorrectly handle the \(\sqrt{3}\) terms. Common mistakes include forgetting to divide both sides by \(\sqrt{3}\) or making arithmetic errors with the resulting equation.

This may lead them to select Choice A (30) if they mistake the shorter leg for the diameter.

The Bottom Line:

This problem combines algebra with a specific geometric theorem (Thales' theorem) that many students haven't fully internalized. Success requires both computational accuracy and recognizing the key relationship between inscribed right triangles and their circumscribed circles.