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# Equilateral Triangle ABC is inscribed in a circle. The arc

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Equilateral Triangle ABC is inscribed in a circle. The arc [#permalink]

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20 May 2007, 14:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Equilateral Triangle ABC is inscribed in a circle. The arc length of ABC is 24. What is the diameter of the circle?

A. 5
B. 8
C. 11
D. 15
E. 18
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Re: GMAT Prep: Triangle Inscribed in Circle [#permalink]

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20 May 2007, 15:38
skinsvt wrote:
Equilateral Triangle ABC is inscribed in a circle. The arc length of ABC is 24. What is the diameter of the circle?

A. 5
B. 8
C. 11
D. 15
E. 18

C = 2pi*r
Arc ABC = 2/3(C) = 2/3(2pi*r) = (4pi*r)/3
(4pi*r)/3 = 24
72=(4pi*r)
18=pi*r = 18=3.14(r) r = 5.7; d = 5.7(2) = approx 11; My answer is C
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20 May 2007, 15:58
Thanks..I'm able to follow the equations. The one place I'm confused is at the beginning. How did you get 2/3(C)? Is the 2/3 a default formula?[/b]
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20 May 2007, 20:01
skinsvt wrote:
Thanks..I'm able to follow the equations. The one place I'm confused is at the beginning. How did you get 2/3(C)? Is the 2/3 a default formula?[/b]

when you have a square inscribed inside a circle it will divide the circumface into 4 equal diffrent segments. the same goes for a triangle inscibed inside a circale it will divide the circumface into 3 equal segments.

knowing that 2 segments equal 24 we can find the circumface of the circale.

24*3/2 = 36

2*pi*r = 36

2*r = diameter = 36/pi

~ 11
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22 May 2007, 15:54
I think I follow...so for a triangle..if a triangle is inscribed into a circle the we have 3 equal parts. Since ABC=24 we have 2/3 = 24.

So If we have ABC equal to 50 would we have 2/3= 50 or a different number.

Also when we have a square. We would have 4 equal parts. And let's assume that Arc ABCD = 24 would we have 3/4 = 24 or is the ratio different for a square?

Thanks
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22 May 2007, 18:19
A triangle only divides a circle into three equal parts if it is an equilateral triangle, because all sides and all angles are equal.

Then, what KillerSquirrel said...
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22 May 2007, 18:29
So I guess the key here is to know that it's an equilateral triangle. Had this not been an equilateral triangle could we have solved it?

Also, how would we have solved this problem had it been a square inscribed inside of a circle with the same facts..Would that be possible?

Sorry for playing devils advocate..just want to make sure I full understand.

Thanks for your help. Couldn't have learned as much math without this forum.
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22 May 2007, 20:37
Awsome explanation Killer squirrel. One question. Why do we only know 2 of the sides? When u say length ABC is that only 2 sides? Like literally A to B then B to C, which would kinda look like this:

^

(sorry that is not bigger).

If this is true, then I get the whole damn problem and understand why u had to do 3/2 for 36. Although since its equilateral not sure if u have to.

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22 May 2007, 21:00
good stuff....I was going for the complicated method of using the r*(theta) = arc length
the proped method didnt occur to me...now I will remember...thanks
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22 May 2007, 22:33
skinsvt & gmatblackbelt, hello

Yes ! in order to solve the given problem, the triangle has to be equilateral ! For a square it's the same princeple , instead of dividing the circle circumface into three segments it will divided it into four smaller then 12 segments (of 9).

please note that it has to be a square ! a rectangular won't do the trick !

thanks all

22 May 2007, 22:33
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