gmat620 wrote:
A Circle inscribed in a equilateral triangle ABC so that point D lies on the circle and on line segment AC and point E lies on circle and on line segment AB. If line segment AB=6, what is the area of the figure created by line segments AD, AE and minot Arc DE.
A. 3√3 - (9/4) π
B. 3√3 - π
C. 6√3 - 3 π
D. 9√3 - 3 π
5. cannot be determined
To find the area of that region we need to learn three things
First the area of an equilateral triangle is s^2 sqrt (3) / 4 therefore area will be 9sqrt (3)
Next, radius of an inscribed circle will be '6'* sqrt (3) / 6 , where '6' is the side of the equilateral triangle
Therefore, radius will be sqrt (3) and thus the area of the circle will be 3 (pi)
Finally, since the circle divides the equilateral triangle evenly then Area of region ADE = (Area of equilateral triangle - Area of circle)/3 = 3 sqrt (3) - pi
Answer is hence B
Hope it helps
Let me know if you have any doubts OK?
Cheers!
J