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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 27 Sep 2008, 16:36
A circle is inscribed in equilateral triangle ABC such that point D lies on circle and on line segment AC , point E lies on circle and line segment AB , and point F lies on circle and on line segment BC. If line segment AB=6 , what is the area of figure created by line segments AD , AE and minor arc DE.


how to get value of OE or OD ( O is center of circle)
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Re: Good Trigo Question [#permalink] New post 27 Sep 2008, 17:05
the equation for an inscribed circle radius for an equilateral triangle is r=a (sqrt3 / 6) I know that most people don't know that (I had to look it up).

So you take 6 * sqrt3 / 6, which equals sqrt3.
the formula for area of a circle is pi r squared, so it is 3.14* sqrt3 squared (3), so the area of the circle is 9.42.
There will inherently be an inscribed equilateral triangle inside this circle, with sides 3,3,3 using the side bisectors D,E, and F as the points.
the area of this triangle can be found by halving the height*base. So, you make half the triangle into a right triangle with a side 1.5 and the hypotenuse of 3 (a 30-60-90 triangle), which means that the height of the triangle is 1.5*sqrt3 (the other side).
so 1.5sqrt3 times 3 divided by 2 is 2.25sqrt3, the area of the triangle, which is 3.8971.
so the area of the circle minus the area of the triangle gives you the aggregate area of the three minor arc sections, which is 9.42-3.8971, 5.5229.
so the area of one arc section is 1.84097,
take the area of the triangle ADE (which is the same as DEF) 3.8971 minus the area of the arc section, 1.84097, so the area of AD,AE, and the minor arc DE is....

2.05613
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Re: Good Trigo Question [#permalink] New post 27 Sep 2008, 17:06
as for getting the O, I'm not sure. someone plz figure it out and post it bc I really want to know now lol.
another way to remember the inscribed circle radius is it is 1/3 the altitude, and a circumscribed radius is 1/2 the altitude.
These kind of formulas are crucial for the GMAT, although I have never seen a geometry problem quite as difficult as this one in prep or on the GMAT.
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Re: Good Trigo Question [#permalink] New post 28 Sep 2008, 02:48
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Unable to draw the figure here.

Since, ABC is an equilateral triangle, point D, E or F will divide AC, AB or BC into two halves.

Now, if O is the center of circle and we focus on triangle OBF, then BF = 3 and since OVF is a 30-60-90 triangle, we know OF = BF/sqrt3 = sqrt3.

Now, we can find out the area of circle, subtract it from the area of triangle and divide the result by 3 to get the result.
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Re: Good Trigo Question [#permalink] New post 28 Sep 2008, 06:46
ok so because OB is bisecting angle EBF, you can conclude that angle OBF is 30 degrees, and that angle OFB is 90, so that OF=x, BF=sqrt3*x, and OB is 2x, with OF being the radius of the circle and BF=3. that makes sense, I think my way is quicker tho.
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Re: Good Trigo Question [#permalink] New post 28 Sep 2008, 08:18
scthakur wrote:
Unable to draw the figure here.

Since, ABC is an equilateral triangle, point D, E or F will divide AC, AB or BC into two halves.

Now, if O is the center of circle and we focus on triangle OBF, then BF = 3 and since OVF is a 30-60-90 triangle, we know OF = BF/sqrt3 = sqrt3.

Now, we can find out the area of circle, subtract it from the area of triangle and divide the result by 3 to get the result.



+1

I got until the part of BF=3. How ever, I failed to recognize that Triangle OBF is 30-60-90 triangle.

Are of equilateral triangle is s ^2 X sqrt (3) /4 leaves us with 9 sqrt (3)

Area of circle is 3pi

Reqd area = (9 sqrt (3) - 3 pi ) /3

Thanks
Re: Good Trigo Question   [#permalink] 28 Sep 2008, 08:18
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