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
A square is inscribed in a circle whose diameter is \(58\) inches. What is the perimeter, in inches, of the square?
- \(58\)
- \(58\sqrt{2}\)
- \(116\)
- \(116\sqrt{2}\)
\(58\)
\(58\sqrt{2}\)
\(116\)
\(116\sqrt{2}\)
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.
\(58\)
\(58\sqrt{2}\)
\(116\)
\(116\sqrt{2}\)