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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Karishma

Veritas Prep GMAT Instructor

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