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A circle is inscribed in equilateral triangle ABC such that

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

Gmat Gurus help needed....

usman
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Re: Geometry [#permalink] New post 01 Oct 2006, 17:30
Is the answer (b) -> 3*underroot(3) - pi
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 [#permalink] New post 02 Oct 2006, 20:56
Draw perpendicular bisectors from each vertex to the other side of triangle


For each of these bisectors lenght is sqrt(6^2-3^2) = 3sqrt(3)

All three perpendicualr bisectors meet at the center of circle and divide themselves in ratio of 2:1

Thus the radius of circle is sqrt(3)

Area of Eq. triangle is {sqrt(3)/4}* 6^2 = 9sqrt(3) and area of circle is 3*pi

Now Triangle -circle area is 9sqrt(3) - 3*pi

to find area of what is asked divide this by 3 since by symmetry of the problems we have 3 such equal areas

Answewr is 3*sqrt(3) -pi
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 [#permalink] New post 03 Oct 2006, 01:42
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.....
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 [#permalink] New post 03 Oct 2006, 04:28
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
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 [#permalink] New post 08 Oct 2006, 11:01
thanks jainan....much appreciated
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 [#permalink] New post 08 Oct 2006, 11:19
Simple way is

1/3( area of triangle- area of circle)

1/3( 9sqrt(3)- 3 pi)

3sqrt(3)-pi
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