MA wrote:

A circle is inscribed in equilaterel triangle ABC such that point D lies on the circle and on the line segment AC, point E lies on the circle and segment AB, and point F lies on the circle and on the line segment BC. if line segment AB=6, what is the are of the figure created by line segment AD,AE and minor arc DE.

a. 3 sqrt(3) - (9/4) pi

b. 3 sqrt(3) - pi

c. 6 sqrt(3) - pi

d. 9 sqrt(3) - 3 pi

e. it can not be determined from the infomation available.

pls support you answer with explanation............

will clearify if any

B.

The area being asked for is exactly 1/3 of all the area not covered by the circle. So we just need both the area of the circle and the triangle, then subtract, then divide by 3.

The triangle is easy. It's equilateral and the side is 6, so the area is 9root3.

The circle is tougher. If we start in the center and draw a radius to point F, it perpendicularly bisects line BC. then draw a line to point B from the center, and we have a 30-60-90 triangle, with the xroot3 side equalling 3. That means the radius side is x, and solving we get simply, root 3.

That means the area of the circle is 3pi. So the area of everything not in the circle is (9root3-3pi), and the area of just that piece is 1/3 of that, or 3root3 - pi.