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A square is inscribed in a circle whose diameter is 58 inches. What is the perimeter, in inches, of the...

GMAT Geometry & Trigonometry : (Geo_Trig) Questions

Source: Prism
Geometry & Trigonometry
Right triangles and trigonometry
HARD
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A square is inscribed in a circle whose diameter is \(58\) inches. What is the perimeter, in inches, of the square?

  1. \(58\)
  2. \(58\sqrt{2}\)
  3. \(116\)
  4. \(116\sqrt{2}\)
A

\(58\)

B

\(58\sqrt{2}\)

C

\(116\)

D

\(116\sqrt{2}\)

Solution

1. TRANSLATE the problem information

  • Given information:
    • Square is inscribed in a circle
    • Circle's diameter = 58 inches
    • Need to find: perimeter of the square

2. INFER the geometric relationship

  • Key insight: When a square is inscribed in a circle, the diagonal of the square equals the diameter of the circle
  • Therefore: diagonal of square = 58 inches
  • Strategy: Find the side length first, then calculate perimeter

3. INFER the formula connection

  • For any square with side length s: \(\mathrm{diagonal = s\sqrt{2}}\)
  • We have: \(\mathrm{58 = s\sqrt{2}}\)
  • Need to solve for s

4. SIMPLIFY to find the side length

  • From \(\mathrm{58 = s\sqrt{2}}\), divide both sides by \(\mathrm{\sqrt{2}}\):
  • \(\mathrm{s = \frac{58}{\sqrt{2}}}\)
  • Rationalize: \(\mathrm{s = \frac{58}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{58\sqrt{2}}{2} = 29\sqrt{2}}\)

5. INFER the final calculation

  • Perimeter of square = 4 × side length
  • \(\mathrm{P = 4s = 4(29\sqrt{2}) = 116\sqrt{2}}\)

Answer: \(\mathrm{(D) \ 116\sqrt{2}}\)


Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students may incorrectly assume that the side length of the square equals the diameter of the circle, rather than recognizing that the diagonal equals the diameter.

If they think \(\mathrm{side = diameter = 58}\), then they calculate \(\mathrm{perimeter = 4(58) = 232}\). Since this isn't an answer choice, this leads to confusion and guessing, possibly selecting \(\mathrm{(C) \ 116}\) as it seems 'close' to their reasoning.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly set up \(\mathrm{58 = s\sqrt{2}}\) but make algebraic errors when solving for s, such as getting \(\mathrm{s = 58\sqrt{2}}\) instead of \(\mathrm{s = 29\sqrt{2}}\).

This would lead them to calculate perimeter = \(\mathrm{4(58\sqrt{2}) = 232\sqrt{2}}\), which isn't an answer choice, causing them to get stuck and randomly select an answer.

The Bottom Line:

This problem tests spatial reasoning about inscribed figures and requires students to distinguish between radius, diameter, side length, and diagonal. The key breakthrough is recognizing the diagonal-diameter relationship for inscribed squares.

Answer Choices Explained
A

\(58\)

B

\(58\sqrt{2}\)

C

\(116\)

D

\(116\sqrt{2}\)

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