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26 Mar 2015, 19:56
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
76% (02:59) correct
24%
(03:03)
wrong
based on 146
sessions
History
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Not Attempted Yet
Attachment:
triacirc_q1.png [ 14.51 KiB | Viewed 18469 times ]
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. \(24\sqrt{3}\)
26 Mar 2015, 21:32
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PareshGmat
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.
General Discussion
26 Mar 2015, 23:45
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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.
GMAT assassins aren't born, they're made,
Rich
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.
GMAT assassins aren't born, they're made,
Rich
