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03 Jun 2025, 11:26
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
33% (00:47) correct
66% (00:41) wrong based on 3 sessions
History
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Time
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Not Attempted Yet
The diameter of circle O is d, and the area is a.
Quantity A
\(\frac{πd^2}{2}\)
Quantity B
a
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Quantity A
\(\frac{πd^2}{2}\)
Quantity B
a
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
03 Jun 2025, 12:41
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Diameter of circle is d
Area of circle =pie d^2
Quantity A = 0.5 Quantity B
Hence Quantity B is greater
Area of circle =pie d^2
Quantity A = 0.5 Quantity B
Hence Quantity B is greater
14 Jun 2025, 02:13
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OE
[STEP 1]: Analyze the centered information and the quantities.
In this problem, you are given additional information: the sentence that tells you the diameter of circle O is d and the area is a. This is important information because it gives you a key to unlocking this question. Given that information, you can tell that you are comparing the area, a, of circle O and a quantity that includes the diameter of the same circle. If you’re thinking about the formula for calculating area given the diameter, you’re thinking right!
[STEP 2]: Approach strategically.
Make Quantity B look more like Quantity A by rewriting a, the area of the circle, in terms of the diameter, d. The area of any circle equals \(πr^2\), where r is the radius. Because the radius is half the diameter, you can substitute d/2 for r in the area formula to get \(a= πr^2 = π (\frac{d}{2})^2\) in Quantity B. Simplifying, you get \(\frac{πd^2}{4.}\)
Because both quantities contain π, we could compare \(\frac{d^2}{2}\) to \(\frac{d^2}{4}\) . But let’s take it one step further. You know that d is a distance and must be a positive number. That makes it possible to divide both quantities \(\frac{d^2}{2}\) and \(\frac{d^2}{4}\), and by \(d^2 \)and then just compare 1/2 to 1/4. This makes it easy to see that Quantity A is always greater because \(\frac{1}{2}>\frac{1}{4}.\)
Choice (A) is correct.
[STEP 1]: Analyze the centered information and the quantities.
In this problem, you are given additional information: the sentence that tells you the diameter of circle O is d and the area is a. This is important information because it gives you a key to unlocking this question. Given that information, you can tell that you are comparing the area, a, of circle O and a quantity that includes the diameter of the same circle. If you’re thinking about the formula for calculating area given the diameter, you’re thinking right!
[STEP 2]: Approach strategically.
Make Quantity B look more like Quantity A by rewriting a, the area of the circle, in terms of the diameter, d. The area of any circle equals \(πr^2\), where r is the radius. Because the radius is half the diameter, you can substitute d/2 for r in the area formula to get \(a= πr^2 = π (\frac{d}{2})^2\) in Quantity B. Simplifying, you get \(\frac{πd^2}{4.}\)
Because both quantities contain π, we could compare \(\frac{d^2}{2}\) to \(\frac{d^2}{4}\) . But let’s take it one step further. You know that d is a distance and must be a positive number. That makes it possible to divide both quantities \(\frac{d^2}{2}\) and \(\frac{d^2}{4}\), and by \(d^2 \)and then just compare 1/2 to 1/4. This makes it easy to see that Quantity A is always greater because \(\frac{1}{2}>\frac{1}{4}.\)
Choice (A) is correct.
