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# PS Circle and Equilateral Triangle

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Current Student
Joined: 29 Jan 2005
Posts: 5218
PS Circle and Equilateral Triangle [#permalink]

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11 Dec 2005, 04:47
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions

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A: Pi(r)/6
B. Pi(r)/3
C. Pi(r)/2
D. Pi(r)
E. None of these
Attachments

R and B # 21.doc [104.5 KiB]

Director
Joined: 14 Sep 2005
Posts: 985
Location: South Korea

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11 Dec 2005, 04:56
POQ is 60 degree, and so its length should be (1/6)*pi*r

(A)
_________________

Auge um Auge, Zahn um Zahn !

Current Student
Joined: 29 Jan 2005
Posts: 5218

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11 Dec 2005, 05:04
gamjatang wrote:
POQ is 60 degree, and so its length should be (1/6)*pi*r

(A)

That was my guess too, but we are both wrong.
SVP
Joined: 24 Sep 2005
Posts: 1885

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11 Dec 2005, 05:13
gamjatang wrote:
POQ is 60 degree, and so its length should be (1/6)*pi*r

(A)

POQ= 60 => PQ= 60/360 * the perimeter of the circle= 1/6 * 2pi*r
=1/3 pi*r

BTW, Matt, there're some typos in the question AB should be diameter, not radius, right?!
Current Student
Joined: 29 Jan 2005
Posts: 5218

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11 Dec 2005, 05:28
Thanx Laxie. OA is (B). Drawing two additional support lines OP and OQ really simplify this problem. Three equilateral triangles give us an interior angle of 60dg. 60=1/6 or the circum. 1/6*2pi(r)= pi(r)/3
11 Dec 2005, 05:28
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