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22 Dec 2020, 06:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
87% (01:23) correct
13%
(03:03)
wrong
based on 31
sessions
History
Date
Time
Result
Not Attempted Yet
In the figure below, the area of the shaded section of the circle is 33π. What is the diameter of the circle?
(A) 6
(B) 2π
(C) 9
(D) 12
(E) 4π
(A) 6
(B) 2π
(C) 9
(D) 12
(E) 4π
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2020-12-18_18-46-55.png [ 19.49 KiB | Viewed 1444 times ]
Updated on: 22 Dec 2020, 19:33
Originally posted by CrackverbalGMAT on 22 Dec 2020, 06:39.
Last edited by CrackverbalGMAT on 22 Dec 2020, 19:33, edited 1 time in total.
Last edited by CrackverbalGMAT on 22 Dec 2020, 19:33, edited 1 time in total.
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In terms of angles, the area of the shaded section represents 360 - 30 = 330 degrees
Area = \( 33 \pi = \frac{\theta}{360} * \pi r^2 = \frac{330}{360} * \pi r^2 = \frac{33}{36} \pi r^2\)
Therefore \(r^2 = 36\) or r = 6
Diameter = 2r = 12
Option D
Arun Kumar
Area = \( 33 \pi = \frac{\theta}{360} * \pi r^2 = \frac{330}{360} * \pi r^2 = \frac{33}{36} \pi r^2\)
Therefore \(r^2 = 36\) or r = 6
Diameter = 2r = 12
Option D
Arun Kumar
General Discussion
22 Dec 2020, 19:21
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Area of the circle=πr2
This area will be sum of shaded +non-shaded region.
Shaded area given=33π
Non-shaded area=θ/360*πr2 where θ=30degree
33π +(30/360)*πr2=πr2
33π=(11/12)πr2
Solve and get r=6
Therefore diameter=12
Option D
This area will be sum of shaded +non-shaded region.
Shaded area given=33π
Non-shaded area=θ/360*πr2 where θ=30degree
33π +(30/360)*πr2=πr2
33π=(11/12)πr2
Solve and get r=6
Therefore diameter=12
Option D
