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22 Dec 2020, 06:30
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Difficulty:
15%
(low)
Question Stats:
87% (01:23) correct
13%
(03:03)
wrong
based on 31
sessions
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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 1509 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
