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If O is the center of the circle and x = 160, what is the pe
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16 Jul 2019, 11:21
16 Jul 2019, 11:21
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81% (01:34) correct
18% (01:47) wrong based on 43 sessions
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#GREpracticequestion If O is the center of the circle and x = 160.jpg [ 30.38 KiB | Viewed 3508 times ]
If O is the center of the circle and x = 160, what is the perimeter of sector ABCO?
(A) \(18 + 8\pi\)
(B) \(18 + 16\pi\)
(C) \(36 + 8pi\)
(D) \(36 + 16\pi\)
(E) \(36 + 24\pi\)
Re: If O is the center of the circle and x = 160, what is the pe
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19 Jul 2019, 01:52
19 Jul 2019, 01:52
1
(theta/360)*2πr + 2r
(160/360)*2π18 + 36
16π + 36
Posted from my mobile device
(160/360)*2π18 + 36
16π + 36
Posted from my mobile device
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If O is the center of the circle and x = 160, what is the pe
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08 Feb 2021, 07:06
08 Feb 2021, 07:06
Carcass wrote:
Attachment:
#GREpracticequestion If O is the center of the circle and x = 160.jpg
If O is the center of the circle and x = 160, what is the perimeter of sector ABCO?
(A) \(18 + 8\pi\)
(B) \(18 + 16\pi\)
(C) \(36 + 8pi\)
(D) \(36 + 16\pi\)
(E) \(36 + 24\pi\)
Arc length \(=(\frac{n}{360})(2 \pi)(radius)\), where \(n =\) the central angle
So, the length of arc ABC \(=(\frac{160}{360})(2 \pi)(18)=(\frac{4}{9})(36 \pi)=16 \pi\)
Perimeter of sector ABCO = (arc ABC) + (length of side OC) + (length of side OA)
\(= (16 \pi) + (18) + (18)\)
\(= 16 \pi + 36\)
Answer: D
