Bunuel wrote:
The circle with center O has a circumference of \(6\pi{\sqrt{3}}\). If AC is a diameter of the circle, what is the length of line segment BC?
(A) \(\frac{3}{\sqrt{2}}\)
(B) 3
(C) \(3\sqrt{3}\)
(D) 9
(E) \(9\sqrt{3}\)
Kudos for a correct solution.Attachment:
2015-06-16_1428.png
This is a 30-60-90 triangle and we are told that the circumference is \(6\pi{\sqrt{3}}\). Set \(6\pi{\sqrt{3}}\) = \(2(\pi)r\) giving you r = \(3{\sqrt{3}}\)
You know the diameter, line AC equal 2r or \(6{\sqrt{3}}\)
If you remember your 30-60-90 angles, you can calculate the 60 degree angle, \(3{\sqrt{3}}\)*\({\sqrt{3}}\) BC = 9