A semicircle has a diameter of 86 meters. What is the area, in square meters, of the semicircle?43pi86pi924.5pi172pi

1. TRANSLATE the problem information

• Given information:
  • Diameter = 86 meters
  • Need to find area of semicircle in square meters
• What this tells us: We need the radius to use the circle area formula, and we need to account for this being half a circle

2. INFER the approach

• To find semicircle area, we need the full circle area first, then divide by 2
• Circle area requires radius, so we must convert diameter to radius first

3. TRANSLATE diameter to radius

• Since \(\mathrm{diameter = 2 × radius}\), we have: \(\mathrm{radius = diameter ÷ 2}\)
• \(\mathrm{r = 86 ÷ 2 = 43}\) meters

4. SIMPLIFY to find full circle area

• Area of circle = \(\mathrm{πr^2}\)
• Area = \(\mathrm{π(43)^2}\)
  = \(\mathrm{π(1849)}\)
  = \(\mathrm{1849π}\) square meters (use calculator for \(\mathrm{43^2}\))

5. INFER and SIMPLIFY for semicircle

• Semicircle area = (full circle area) ÷ 2
• Semicircle area = \(\mathrm{1849π ÷ 2}\)
  = \(\mathrm{924.5π}\) square meters (use calculator)

Answer: (C) \(\mathrm{924.5π}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students confuse diameter and radius, plugging the diameter directly into the area formula: \(\mathrm{A = π(86)^2 = 7396π}\), then dividing by 2 to get \(\mathrm{3698π}\). This doesn't match any answer choice, leading to confusion and guessing.

Second Most Common Error:

Missing conceptual knowledge about semicircles: Students find the full circle area correctly (\(\mathrm{1849π}\)) but forget that a semicircle is half a circle, selecting Choice (D) is not \(\mathrm{1849π}\), but they might make computational errors or select a wrong choice through other reasoning mistakes.

Third Error Path:

Inadequate SIMPLIFY execution: Students set up the problem correctly but make computational errors when calculating \(\mathrm{43^2}\) or when dividing 1849 by 2, potentially leading them to select Choice (A) \(\mathrm{43π}\) if they somehow use just the radius value, or other incorrect choices.

The Bottom Line:

This problem tests whether students can systematically work through a multi-step geometry problem involving unit conversion (diameter to radius), formula application, and the concept that semicircles represent exactly half of their corresponding complete circles.