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A circle is inscribed in equilaterel triangle ABC such that

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A circle is inscribed in equilaterel triangle ABC such that [#permalink] New post 31 Jan 2005, 13:21
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B
C
D
E

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

a. 3 sqrt(3) - (9/4) pi
b. 3 sqrt(3) - pi
c. 6 sqrt(3) - pi
d. 9 sqrt(3) - 3 pi
e. it can not be determined from the infomation available.

pls support you answer with explanation............

will clearify if any
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 [#permalink] New post 31 Jan 2005, 14:19
I pick "B".

As it is an equilateral triangle the diagonals r perpendicular bisector of opposite sides. Let's say r is the radius of the circle and center of O. So the area that is being asked is :

2*area of triangle AOE - (120)/360 * pi*r2.....(1)

area of AOE = 1/2 * 3 * r (r is also the altotude here)....so total area = 3r

Now know that the diagonals bisect the angles of equilateral....so

we see that r / 3 = tan 30 .....r = (3)^1/2 .....(2)


Therefore from eqn 1 and 2 ....we get

3r - pi*r2 / 3 = 3sqrt(3) - pi
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Re: PS: lengthy Geometry [#permalink] New post 31 Jan 2005, 19:05
MA wrote:
A circle is inscribed in equilaterel triangle ABC such that point D lies on the circle and on the line segment AC, point E lies on the circle and segment AB, and point F lies on the circle and on the line segment BC. if line segment AB=6, what is the are of the figure created by line segment AD,AE and minor arc DE.

a. 3 sqrt(3) - (9/4) pi
b. 3 sqrt(3) - pi
c. 6 sqrt(3) - pi
d. 9 sqrt(3) - 3 pi
e. it can not be determined from the infomation available.

pls support you answer with explanation............

will clearify if any


B.

The area being asked for is exactly 1/3 of all the area not covered by the circle. So we just need both the area of the circle and the triangle, then subtract, then divide by 3.

The triangle is easy. It's equilateral and the side is 6, so the area is 9root3.

The circle is tougher. If we start in the center and draw a radius to point F, it perpendicularly bisects line BC. then draw a line to point B from the center, and we have a 30-60-90 triangle, with the xroot3 side equalling 3. That means the radius side is x, and solving we get simply, root 3.

That means the area of the circle is 3pi. So the area of everything not in the circle is (9root3-3pi), and the area of just that piece is 1/3 of that, or 3root3 - pi.
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 [#permalink] New post 31 Jan 2005, 20:53
Nice explanations. I got the same answer, (B).
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 [#permalink] New post 01 Feb 2005, 02:20
hi guys,

this might help some people.
a website which intorduces the basic of geometry.

http://library.thinkquest.org/20991/geo/ietri.html
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 [#permalink] New post 01 Feb 2005, 02:22
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 [#permalink] New post 01 Feb 2005, 02:25
one more, better
http://www.mathstutor.com/Geometry(ii).html
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 [#permalink] New post 01 Feb 2005, 18:09
thanks Ian and baner for good explanations and mdf for useful likns.

ma
  [#permalink] 01 Feb 2005, 18:09
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