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\(O\) is the center of the semicircle. If angle \(BCO = 30\) and \(BC =6\sqrt3,\) what is the area of triangle \(ABO?\)

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\(O\) is the center of the semicircle. If angle \(BCO = 30\) and \(BC =6\sqrt3,\) what is the area of triangle \(ABO?\)

by Gmat_mission » Sun Apr 26, 2020 12:01 pm

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\(O\) is the center of the semicircle. If angle \(BCO = 30\) and \(BC =6\sqrt3,\) what is the area of triangle \(ABO?\)

A. \(4\sqrt3\)
B. \(6\sqrt3\)
C. \(9\sqrt3\)
D. \(12\sqrt3\)
E. \(24\sqrt3\)

[spoiler]OA=C[/spoiler]

Source: Magoosh
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Re: \(O\) is the center of the semicircle. If angle \(BCO = 30\) and \(BC =6\sqrt3,\) what is the area of triangle \(ABO

by Scott@TargetTestPrep » Sun May 03, 2020 5:27 am
Gmat_mission wrote: ↑
Sun Apr 26, 2020 12:01 pm
 triacirc_q1.png

\(O\) is the center of the semicircle. If angle \(BCO = 30\) and \(BC =6\sqrt3,\) what is the area of triangle \(ABO?\)

A. \(4\sqrt3\)
B. \(6\sqrt3\)
C. \(9\sqrt3\)
D. \(12\sqrt3\)
E. \(24\sqrt3\)

[spoiler]OA=C[/spoiler]

Source: Magoosh
Since triangle ABC is inscribed in a semicircle, it’s a right triangle. Since angle BCO is 30 degrees, triangle ABC is a 30-60-90 right triangle. Since BC = 6√3, AB = 6. Thus, the area of triangle ABC = ½ x 6√3 x 6 = 18√3.

Notice that radius BO splits triangle ABC into two smaller triangles: ABO and CBO. Both of these triangles have equal bases (notice AO = CO since they are both radii of the semicircle), and they also have the same height (which can be dropped from point B to AC). Therefore, triangles ABO and CBO have the same area. Since their combined area is the area of triangle ABC, the area of triangle ABO is half that of triangle ABC. Thus, the area of triangle ABO is ½ x 18√3 = 9√3.

Answer: C.

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