Area of shaded region

[Ashujain]

In the figure in the attachment, BC is tangent to circle A at point B, BC is parallel to ED, and BF (not shown) is equal to 2. If minor arc BF is one-sixth of the circumference of circle A, what is the area of the shaded region? (I have replaced 'pie' with 3.14 in the answer options)
A) 2 - 2*3.14/3
B) 18/3.14^2 - 3*3.14
C) 3.14/3
D) 2
E) 2 - 3.14/3

OA: E

[gmat_and_me]

ARC BF = 1/6-th => BAF = 60 degree => BAF is an equilateral triangle.
=> radius of the circle = 2

Now area of the shaded portion = Area of triangle ACE - Area of sector
AFEA

Area of triangle ACE = 1/2 * base * height = 1/2 * 2 * 2 (since AE = 2
and AB = 2) = 2

Area of sector AFEA = 1/12-th of area of the circle = 1/12 * 3.14 * (2)^2
= 1/3 * (3.14) [ 1/12-th because angle FAE = 30 degree ]

So area of shaded region = 2 - 3.14/3

HTH

In the figure in the attachment, BC is tangent to circle A at point B, BC is parallel to ED, and BF (not shown) is equal to 2. If minor arc BF is one-sixth of the circumference of circle A, what is the area of the shaded region? (I have replaced 'pie' with 3.14 in the answer options)
A) 2 - 2*3.14/3
B) 18/3.14^2 - 3*3.14
C) 3.14/3
D) 2
E) 2 - 3.14/3

OA: E

[Anurag@Gurome]

As BF is 1/6 of the circumference, measure of angle BAF is 360°/6 = 60°
In triangle ABF, AB = AF and angle BAF = 60°
Hence, ABF is an equilateral triangle.
As length of the cord BF = 2, AB = AE = radius of the circle = 2

Now, as BC and AE are parallel, height of the triangle ACE = AB
Area of the triangle ACE = (base)*(height)/2 = (AE)*(AB)/2 = (2)*(2)/2 = 2

Now, BC is a tangent to the circle at B and parallel to AE, measure of angle BAE is 90°
Hence, measure of angle FAE is (90° - 60°) = 30°
Therefore, area of the sector AFE = (πr²)*(30/360) = πr²/12 = π(2)²/12 = 4π/12 = π/3

Hence, are of the shaded region = (Area of the triangle ACE - Area of the sector AFE) = (2 - π/3)

The correct answer is E.