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O is the center of the semicircle. If angle BCO = 30 and BC

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O is the center of the semicircle. If angle BCO = 30 and BC  [#permalink]

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New post 26 Mar 2015, 18:56
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O is the center of the semicircle. If angle BCO = 30 and BC \(= 6\sqrt{3}\), what is the area of triangle CBO?

A. \(4\sqrt{3}\)
B.\(6\sqrt{3}\)
C. \(9\sqrt{3}\)
D. \(12\sqrt{3}\)
E. \(24\sqrt{3}\)

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O is the center of the semicircle. If angle BCO = 30 and BC  [#permalink]

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New post 26 Mar 2015, 20:32
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PareshGmat wrote:
Attachment:
triacirc_q1.png


O is the center of the semicircle. If angle BCO = 30 and BC \(= 6\sqrt{3}\), what is the area of triangle CBO?

A. \(4\sqrt{3}\)
B.\(6\sqrt{3}\)
C. \(9\sqrt{3}\)
D. \(12\sqrt{3}\)
E. \(24\sqrt{3}\)


Since BCO = 30 degrees and angle ABC = 90 degrees (angle in a semi circle), angle CAB = 60 degrees and we are dealing with 30-60-90 triangle (ABC). So sides are in the ratio \(1:\sqrt{3}:2\).
Side BC which is opposite 60 degrees in \(6\sqrt{3}\) so AB = 6 and AC = 12. Hence radius of the circle OC (= OA = OB) is 6.
Since OA = OB, angle OAB = angle ABO = 60 degrees. Triangle ABO is equilateral with area \(= (\sqrt{3}/4) * 6^2 = 9\sqrt{3}\)

Area of triangle ABC \(= (1/2)*AB*BC = (1/2)*6*6\sqrt{3} = 18\sqrt{3}\)

Area of triangle BCO = Area of ABC - Area of ABO \(= 18\sqrt{3} - 9\sqrt{3} = 9\sqrt{3}\)


Another Method:

Note that the two triangles which form the triangle ABC will have equal area because their altitude will be the same (from A to BC) and theirs bases have the same length (OB - OC). So half the area of ABC will be the required area.
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Re: O is the center of the semicircle. If angle BCO = 30 and BC  [#permalink]

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New post 26 Mar 2015, 22:45
Hi PareshGmat,

The "key" to this Multi-Shape Geometry question is the rule that when a triangle has it's vertices on the circumference of the triangle AND one of its sides is the diameter of the circle, then that triangle is a RIGHT TRIANGLE. From there, you can use a variety of right triangle rules (depending on whatever additional information you're given) to answer whatever question is asked.

From the additional information, you know that we're dealing with a 30/60/90 triangle and you know the length of one of the sides. You can now figure out the lengths of the other sides and the area of the triangle.

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Re: O is the center of the semicircle. If angle BCO = 30 and BC  [#permalink]

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New post 31 Dec 2016, 00:24
can find directly area of CBO

attitude^2=6^2-(3*sqrt3)^2=9, so attitude=3

A=(6*sqrt3*3)/2=9*sqrt3

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Re: O is the center of the semicircle. If angle BCO = 30 and BC  [#permalink]

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Re: O is the center of the semicircle. If angle BCO = 30 and BC &nbs [#permalink] 30 Oct 2018, 18:29
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