A hemisphere has a volume of 144pi. What is the radius of the hemisphere?45612

1. TRANSLATE the problem information

• Given information:
  • Hemisphere volume = \(144\pi\)
  • Need to find radius

2. INFER the approach

• To find radius from volume, we need the hemisphere volume formula
• A hemisphere is half a sphere, so its volume formula is \(\mathrm{V} = \frac{2}{3}\pi\mathrm{r}^3\)
• Strategy: Set up equation and solve for r

3. SIMPLIFY through algebraic steps

• Set up the equation: \(144\pi = \frac{2}{3}\pi\mathrm{r}^3\)
• Divide both sides by \(\pi\): \(144 = \frac{2}{3}\mathrm{r}^3\)
• Multiply both sides by 3/2: \(216 = \mathrm{r}^3\)
• Take the cube root: \(\mathrm{r} = \sqrt[3]{216} = 6\)

Answer: C (6)

Why Students Usually Falter on This Problem

Most Common Error Path:

Missing conceptual knowledge: Using the wrong volume formula

Students often confuse hemisphere and sphere formulas, using \(\mathrm{V} = \frac{4}{3}\pi\mathrm{r}^3\) instead of \(\mathrm{V} = \frac{2}{3}\pi\mathrm{r}^3\). Following this path:

• \(144\pi = \frac{4}{3}\pi\mathrm{r}^3\)
• \(144 = \frac{4}{3}\mathrm{r}^3\)
• \(108 = \mathrm{r}^3\)
• \(\mathrm{r} = \sqrt[3]{108} \approx 4.76\)

Since 4.76 is closest to 5, this may lead them to select Choice B (5).

Second Most Common Error:

Weak SIMPLIFY execution: Arithmetic errors in algebraic manipulation

Students might correctly set up \(216 = \mathrm{r}^3\) but struggle with the cube root calculation, either miscalculating \(\sqrt[3]{216}\) or getting confused about which perfect cubes they should know. This leads to confusion and guessing among the remaining choices.

The Bottom Line:

This problem tests whether students remember the distinction between sphere and hemisphere formulas, then requires confident execution of algebraic steps including cube roots.