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Re: circle properties handy to solve ths? [#permalink]
07 Sep 2009, 12:39

Answer is C

1) Arc AB = 120, so angle AEB = 60 Thus triangle AEB is an equilateral triangle, but we do not know the measure of any side - Not sufficient 2) Segment AB = 2 - Not sufficient

Together sufficient to calculate the perimeter of triangle AEB

Re: circle properties handy to solve ths? [#permalink]
07 Sep 2009, 14:27

meenal8284 wrote:

thts the OA... but can you pls elaborate on the conclusion that u drew form the st 1...

Statement 1: Arc AB = 120. Theorem: The angle subtended by the arc to any point on the circumference of the circle of which it is a part. This angle is always half the central angle. Thus angle AEB = 60. You can refer to this link for more information: http://www.mathopenref.com/arc.html Furthermore, the angle EFB = 90. Therefore triangle AEB is an equilateral triangle. Hope that helps.

gmatclubot

Re: circle properties handy to solve ths?
[#permalink]
07 Sep 2009, 14:27

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