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

A 3√3- 9π/4
B 3√3- π
C 6√3- π
D 9√3- 3π
E It cannot be determined from the information given.

Here is a drawing:


Region ADE = (∆ABC - circle)/3.

∆ABC:
Area of an equilateral triangle = (s²√3)/4.
Thus, the area of ∆ABC = (6²√3)/4 = 9√3.

Circle:
The sides of a 30-60-90 triangle in the following ratio:
x : x√3 : 2x.
In the 30-60-90 triangle shown above, x√3=3, so x=√3.
x=√3 is also the RADIUS of the circle.
Thus, the area of the inscribed circle = πr² = π(√3)² = 3π.

Region ADE:
Area = (∆ABC - circle)/3 = (9√3 - 3π)/3 = 3√3 - π.

The correct answer is B.

An alternate approach is to BALLPARK.
The area of ∆ABC = 9√3 ≈ (9)(1.7) ≈ 15.
Region ADE seems to constitute less than 1/4 of ∆ABC.
Thus, region ADE ≈ 3.
Only B yields a value close to 3:
3√3 - π ≈ (3)(1.7) - 3 = 5.1 - 3 = 2.1.