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01 Mar 2017, 15:22
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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:
95%
(hard)
Question Stats:
51% (02:59) correct
49%
(03:28)
wrong
based on 232
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In the coordinate plane, a circle with its center on the negative x-axis has a radius of 12 units, and passes through (0, 6) and (0, – 6). What is the area of the part of this circle in the first quadrant?
(A) \(6\pi - 12\sqrt{3}\)
(B) \(12\pi - 12\sqrt{3}\)
(C) \(12\pi - 18\sqrt{3}\)
(D) \(24\pi - 18\sqrt{3}\)
(E) \(24\pi - 36\sqrt{3}\)
This is a question from a set of 15 challenging GMAT Quant practice questions. For the whole collection, as well as the OE with diagram for this question, see:
Challenging GMAT Math Practice Questions
Mike
(A) \(6\pi - 12\sqrt{3}\)
(B) \(12\pi - 12\sqrt{3}\)
(C) \(12\pi - 18\sqrt{3}\)
(D) \(24\pi - 18\sqrt{3}\)
(E) \(24\pi - 36\sqrt{3}\)
This is a question from a set of 15 challenging GMAT Quant practice questions. For the whole collection, as well as the OE with diagram for this question, see:
Challenging GMAT Math Practice Questions
Mike
01 Mar 2017, 21:01
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The circle passes through (0, 6) and (0, – 6). Therefore, if we draw a chord between these two points, the length of the chord will be 12.
Again, the radius is 12. So if we connect the center and (0, 6) and (0, – 6), we get an equilateral triangle which has sides of length 12. Hence the angle created at the center by the chord is 60.
So, the area OAB = π(12)^2 * 60/360 = π*12*12*(1/6) = π*12*2 = 24π.
Now, the area of the triangle = (\sqrt{3} /4)*12*12 = \sqrt{3}*3*12 = 36\sqrt{3}.
So, the area of the of the part of this circle in the first quadrant = 1/2(24π - 36\sqrt{3}) = (12π - 18\sqrt{3})
Cheers.
Again, the radius is 12. So if we connect the center and (0, 6) and (0, – 6), we get an equilateral triangle which has sides of length 12. Hence the angle created at the center by the chord is 60.
So, the area OAB = π(12)^2 * 60/360 = π*12*12*(1/6) = π*12*2 = 24π.
Now, the area of the triangle = (\sqrt{3} /4)*12*12 = \sqrt{3}*3*12 = 36\sqrt{3}.
So, the area of the of the part of this circle in the first quadrant = 1/2(24π - 36\sqrt{3}) = (12π - 18\sqrt{3})
Cheers.
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General Discussion
02 Mar 2017, 11:34
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saswata4s
So, the area of the of the part of this circle in the first quadrant = 1/2(24π - 36\sqrt{3}) = (12π - 18\sqrt{3})
Cheers.
Dear saswata4s,Cheers.
Thank you very much for finding this typo.
You are 100% correct. I corrected the answer choices above. Thanks again,
Mike McGarry
