In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=\(12\sqrt{3}\) , what is the radius of the circle?A. 12
B. 18
C. 24
D. 36
E. 72
Good question. It tests three
must know properties for the GMAT:
1. A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.According to the above, we have that ABC is a right triangle, with a right angle at B.
2. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.According to the above, central angle AOB is twice inscribed angle ACB (notice that both subtend the same arc AB). Hence since AOB=120 degrees (third of 360 degrees), then ACB=60 degrees. So, we have that triangle ABC is 30°-60°-90° right triangle.
3. In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\).Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).
Thus, \(AB:AC=\sqrt{3}: 2\) --> \(\frac{12\sqrt{3}}{AC}=\frac{\sqrt{3}}{2}\) --> AC=diameter=24 --> radius=12.
Answer: A.
For more on triangles check:
math-triangles-87197.htmlFor more on circles check:
math-circles-87957.htmlHope it helps.
I'm having trouble visualizing your 2nd point; could you draw a rough illustration of it? I'm familiar with the theorem, great illustration here:
, but I can't visualize it in this problem. I got the answer correct as the sqrt(3) tipped me off to a likely 30-60-90, but I'd like to learn a bit more from this problem