The surface area of a sphere is 169pi/4 square inches. What is the volume, in cubic inches, of the sphere?169pi/122197pi/962197pi/482197pi/24

1. TRANSLATE the problem information

• Given information:
  • Surface area = \(\frac{169\pi}{4}\) square inches
  • Need to find volume

• What this tells us: We have surface area and need volume, so radius will be the connecting link.

2. INFER the approach

• Strategy: Since both surface area and volume formulas involve radius, find radius first using surface area, then calculate volume.

• Start with surface area formula: \(\mathrm{S} = 4\pi\mathrm{r}^2\)

3. SIMPLIFY to find the radius

Set up equation: \(4\pi\mathrm{r}^2 = \frac{169\pi}{4}\)

Divide both sides by \(4\pi\):
\(\mathrm{r}^2 = \frac{169\pi}{4} \div 4\pi = \frac{169}{16}\)

Take square root: \(\mathrm{r} = \sqrt{\frac{169}{16}} = \frac{13}{4}\)

4. SIMPLIFY to find the volume

Apply volume formula: \(\mathrm{V} = \frac{4}{3}\pi\mathrm{r}^3\)

Substitute \(\mathrm{r} = \frac{13}{4}\):
\(\mathrm{V} = \frac{4}{3}\pi\left(\frac{13}{4}\right)^3\)

Calculate \(\left(\frac{13}{4}\right)^3 = \frac{2197}{64}\) (use calculator if needed)

\(\mathrm{V} = \frac{4}{3}\pi\left(\frac{2197}{64}\right) = \frac{8788\pi}{192}\)

Reduce fraction: \(\frac{8788\pi}{192} = \frac{2197\pi}{48}\)

Answer: C. \(\frac{2197\pi}{48}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Poor INFER skill: Students don't recognize that they need to find radius as an intermediate step. They might try to directly relate surface area to volume without understanding the radius connection.

This leads to confusion and random guessing among the answer choices.

Second Most Common Error:

Weak SIMPLIFY execution: Students correctly find \(\mathrm{r} = \frac{13}{4}\) but make arithmetic errors when calculating \(\left(\frac{13}{4}\right)^3\) or when simplifying the final fraction \(\frac{8788\pi}{192}\).

Common calculation errors include getting \(13^3\) wrong or not properly reducing the complex fraction, which may lead them to select Choice B \(\left(\frac{2197\pi}{96}\right)\) if they don't fully reduce their fraction.

The Bottom Line:

This problem tests whether students can work systematically through multi-step sphere problems, connecting surface area and volume through radius. The key insight is recognizing that radius is the bridge between these two formulas.