Circle P has a circumference of 12pi inches. Circle Q has an area of 49pi square inches. What is the...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
Circle P has a circumference of \(12\pi\) inches. Circle Q has an area of \(49\pi\) square inches. What is the total area, in square inches, of circles P and Q?
\(61\pi\)
\(85\pi\)
\(121\pi\)
\(193\pi\)
1. TRANSLATE the problem information
- Given information:
- Circle P: \(\mathrm{circumference = 12π}\) inches
- Circle Q: \(\mathrm{area = 49π}\) square inches
- Find: total area of both circles
2. INFER the approach
- Circle Q's area is already given, but Circle P's area is not
- To find Circle P's area, I need its radius first
- I can get the radius from the circumference using \(\mathrm{C = 2πr}\)
3. SIMPLIFY to find Circle P's radius
- Set up equation: \(\mathrm{12π = 2πr}\)
- Divide both sides by \(\mathrm{2π}\): \(\mathrm{r = 6}\) inches
4. Calculate Circle P's area
- Area = \(\mathrm{πr^2 = π(6)^2 = 36π}\) square inches
5. Add the areas together
- Total area = Circle P area + Circle Q area
- Total area = \(\mathrm{36π + 49π = 85π}\) square inches
Answer: \(\mathrm{85π}\) (Choice B)
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students don't recognize they need to find Circle P's radius first. They might try to work directly with the circumference to find area, or get confused about what to do with the given information.
This leads to confusion and guessing among the answer choices.
Second Most Common Error:
Poor SIMPLIFY execution: Students set up the circumference equation correctly (\(\mathrm{12π = 2πr}\)) but make algebraic errors when solving for r, such as getting \(\mathrm{r = 3}\) or \(\mathrm{r = 12}\) instead of \(\mathrm{r = 6}\).
If they get \(\mathrm{r = 3}\), Circle P's area becomes \(\mathrm{π(3)^2 = 9π}\), leading to total area = \(\mathrm{9π + 49π = 58π}\). If they get \(\mathrm{r = 12}\), Circle P's area becomes \(\mathrm{π(12)^2 = 144π}\), leading to total area = \(\mathrm{144π + 49π = 193π}\). This may lead them to select Choice (D) (\(\mathrm{193π}\)).
The Bottom Line:
This problem tests whether students can connect circumference and area through the radius. The key insight is recognizing that radius is the bridge between these two circle measurements.
\(61\pi\)
\(85\pi\)
\(121\pi\)
\(193\pi\)