Minor arc AB in the diagram is 30 and the radius of the circle is 6. Find the area of circular segment AB.
+2668 Oh wait, never mind, I think I know how to solve it.
The circular segment is equal to the area of the region \(ABO-\triangle{ABO}\).
We know the circular region's area is \({30 \over360} \times 36\pi =3\pi\).
\(\triangle{ABO}\) is isosceles, with degrees of 30, 70, 70, and 2 sides with 6. Using trig, we find that the area of the triangle is 9.
Thus, the area of the circular region is: \(\color{brown}\boxed{3\pi - 9}\)
