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 = 3 × (shortest leg)
  • Area = 1,156√2 square units
  • Need to find diameter of circle

2. INFER the key geometric relationship

• For a right triangle inscribed in a circle, the hypotenuse is always the diameter of the circle
• This comes from the theorem that an angle inscribed in a semicircle is a right angle

3. TRANSLATE the side relationships

• Let shortest leg = a
• Then hypotenuse h = 3a (given)
• Need to find the other leg using Pythagorean theorem

4. INFER and SIMPLIFY to find the other leg

• Pythagorean theorem: \(\mathrm{a^2 + b^2 = h^2}\)
• Substitute \(\mathrm{h = 3a}\):
  \(\mathrm{a^2 + b^2 = (3a)^2 = 9a^2}\)
• Solve for \(\mathrm{b^2}\):
  \(\mathrm{b^2 = 9a^2 - a^2 = 8a^2}\)
• Therefore:
  \(\mathrm{b = \sqrt{8a^2} = \sqrt{4 \times 2 \times a^2} = 2\sqrt{2}a}\)

5. TRANSLATE and SIMPLIFY using the area formula

• Area = \(\mathrm{\frac{1}{2} \times base \times height = \frac{1}{2} \times a \times 2\sqrt{2}a = \sqrt{2}a^2}\)
• Set equal to given area:
  \(\mathrm{\sqrt{2}a^2 = 1{,}156\sqrt{2}}\)
• Divide both sides by \(\mathrm{\sqrt{2}}\):
  \(\mathrm{a^2 = 1{,}156}\)
• Take square root:
  \(\mathrm{a = 34}\)

6. INFER the final answer

• Since hypotenuse = diameter and \(\mathrm{h = 3a}\)
• Diameter = \(\mathrm{3 \times 34 = 102}\)

Answer: 102

Why Students Usually Falter on This Problem

Most Common Error Path:

Missing conceptual knowledge: Students don't know that a right triangle inscribed in a circle has its hypotenuse as the diameter. Without this key insight, they can't connect the triangle problem to finding the circle's diameter. This leads to confusion about how to even begin the problem, causing them to get stuck and guess.

Second Most Common Error:

Weak INFER reasoning: Students correctly set up \(\mathrm{h = 3a}\) but fail to use the Pythagorean theorem to find the third side in terms of 'a'. Instead, they might try to use the area formula with two unknowns (a and b), creating an unsolvable equation. This leads to algebraic confusion and incorrect calculations.

The Bottom Line:

This problem requires connecting circle geometry with right triangle properties, then managing a multi-step algebraic solution. The key insight about inscribed right triangles is often the make-or-break moment for student success.