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# circle properties handy to solve ths?

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Intern
Joined: 27 Aug 2009
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circle properties handy to solve ths? [#permalink]  07 Sep 2009, 09:14
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Hi,

i m not too convinced with the OA... can i have answers with explanations? thanks
Current Student
Joined: 13 Jul 2009
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Location: Barcelona
Schools: SSE
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Re: circle properties handy to solve ths? [#permalink]  07 Sep 2009, 09:34
Senior Manager
Joined: 18 Aug 2009
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Re: circle properties handy to solve ths? [#permalink]  07 Sep 2009, 09:36
meenal8284 wrote:
Hi,

i m not too convinced with the OA... can i have answers with explanations? thanks

What is the question?
Intern
Joined: 23 Aug 2009
Posts: 35
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Kudos [?]: 7 [0], given: 5

Re: circle properties handy to solve ths? [#permalink]  07 Sep 2009, 10:57
meenal8284 wrote:
Hi,

i m not too convinced with the OA... can i have answers with explanations? thanks

Where is the -1 option for kudos ??? hehe, just kidding.
Intern
Joined: 27 Aug 2009
Posts: 28
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Kudos [?]: 2 [0], given: 1

Re: circle properties handy to solve ths? [#permalink]  07 Sep 2009, 12:31
looks like the file isnt getting uploaded...
pls find one of the questions attached.
thanks
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DS- Circle properties.JPG [ 33.46 KiB | Viewed 909 times ]

Senior Manager
Joined: 18 Aug 2009
Posts: 331
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Kudos [?]: 197 [0], given: 13

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

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
Intern
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Re: circle properties handy to solve ths? [#permalink]  07 Sep 2009, 14:06
thts the OA... but can you pls elaborate on the conclusion that u drew form the st 1...
Senior Manager
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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.
Re: circle properties handy to solve ths?   [#permalink] 07 Sep 2009, 14:27
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