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Point O is the center of the semicircle. If ∠ BCO = 30 ° and

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Point O is the center of the semicircle. If ∠ BCO = 30 ° and [#permalink] New post 30 Nov 2018, 17:25
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Point O is the center of the semicircle. If ∠ BCO = 30 ° and \(BC = 6 \sqrt{3}\) what is the area of triangle ABO?

Attachment:
GRE exam - Point O is the center of the semicircle..jpg
GRE exam - Point O is the center of the semicircle..jpg [ 9.5 KiB | Viewed 317 times ]


A. \(4 \sqrt{3}\)

B. \(6 \sqrt{3}\)

C. \(9 \sqrt{3}\)

D. \(12 \sqrt{3}\)

E. \(24 \sqrt{3}\)
[Reveal] Spoiler: OA

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Last edited by Carcass on 03 Dec 2018, 11:13, edited 1 time in total.
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Re: Point O is the center of the semicircle. If ∠ BCO = 30 ° and [#permalink] New post 03 Dec 2018, 10:54
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Missing information from the prompt.

Should state: BC = 6√3
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Re: Point O is the center of the semicircle. If ∠ BCO = 30 ° and [#permalink] New post 03 Dec 2018, 11:14
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Thank you so much. Did not show up.

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Re: Point O is the center of the semicircle. If ∠ BCO = 30 ° and [#permalink] New post 03 Dec 2018, 19:26
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Carcass wrote:
Point O is the center of the semicircle. If ∠ BCO = 30 ° and \(BC = 6 \sqrt{3}\) what is the area of triangle ABO?

Attachment:
GRE exam - Point O is the center of the semicircle..jpg


A. \(4 \sqrt{3}\)

B. \(6 \sqrt{3}\)

C. \(9 \sqrt{3}\)

D. \(12 \sqrt{3}\)

E. \(24 \sqrt{3}\)


triangle ABC is a right angled triangle at A, and ∠ BCO = 30 °, so ABC is 30-60-90 triangle and sides are in ratio 1:\(\sqrt{3}\):2
now opposite 60 is side \(BC = 6 \sqrt{3}\), so AB=6 and AC=12=2*radius.... radius = 6

Let us see OAB..
As we can see AB=6 and OA=OB=radius=6
so it is an equilateral triangle with side 6..
area = \((\sqrt{3}/4)*6^2=9\sqrt{3}\)

C
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Re: Point O is the center of the semicircle. If ∠ BCO = 30 ° and   [#permalink] 03 Dec 2018, 19:26
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