A
B
C
D
E
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.Gmat_mission wrote: ↑Sun Apr 26, 2020 12:01 pmtriacirc_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
