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A circle is inscribed in equilateral triangle ABC such that [#permalink]
01 Oct 2006, 10:43

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1. 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 cirlce and on line segment BC. If line segment AB = 6, what is the area of the figure created by line segments AD, AE and the minor arc DE?

a) 3*underroot(3) - 9/4pi
b) 3*underroot(3) - pi
c) 6*underroot(3) - pi
d) 9*underroot(3) - 3pi
e)cannot be determined from the info given

Great explanation jainan but can you explain this point that you made....
"All three perpendicualr bisectors meet at the center of circle and divide themselves in ratio of 2:1"

Can you please explain this point....
Thanks for much.....

Since the triangle formed by the radius of the circle (perpendicular),

half the base of original equilateral triangle (base) and distance from one of the vertices to the center of circle (hypotenuse) form a 30-60-90 triangle and we also know that sum of base and hypotenuse of that triangle by symmetry is 3sqrt(3), thus you can find hypotenuse = 2sqrt(3) and perpendicualr (radius) = sqrt(3).

If you draw the figure accurately, you can see the symmetry