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# GEOMETRY-gmatprep qs

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Senior Manager
Joined: 05 Oct 2008
Posts: 267

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16 Jan 2009, 08:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

attached. sorry don't know how to copy the image as jpeg.
would appreciate if someone who knows can put it as an image in his reply. Thanks.
Attachments

Geometry-Minor Arc PQ.JPG [ 25.61 KiB | Viewed 1425 times ]

prepqs2.doc [59 KiB]

Manager
Joined: 04 Jan 2009
Posts: 237
Re: GEOMETRY-gmatprep qs [#permalink]

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16 Jan 2009, 09:44
2
KUDOS
study wrote:
attached. sorry don't know how to copy the image as jpeg.
would appreciate if someone who knows can put it as an image in his reply. Thanks.

I don't know either. Is the answer one you chose or OA?
Anyway, here is a little bit cumbersome way to solve it:
Let C be the center of the circle. cr=co = 9.
<crp=<rpq=35.
Further because of isosceles triangle,
cpr=rpq=35
Thus, <cpq=70
Angle submitted by arc on center=<pcq=40.
Length of the minor arc = pi*18*40/360=2*pi
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tusharvk

Senior Manager
Joined: 05 Oct 2008
Posts: 267
Re: GEOMETRY-gmatprep qs [#permalink]

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16 Jan 2009, 10:30
Thanks!

OA is A, which is 2pi.
SVP
Joined: 29 Aug 2007
Posts: 2467
Re: GEOMETRY-gmatprep qs [#permalink]

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16 Jan 2009, 11:00
tusharvk wrote:
study wrote:
attached. sorry don't know how to copy the image as jpeg.
would appreciate if someone who knows can put it as an image in his reply. Thanks.

I don't know either. Is the answer one you chose or OA?
Anyway, here is a little bit cumbersome way to solve it:
Let C be the center of the circle. cr=co = 9.
<crp=<rpq=35.
Further because of isosceles triangle,
cpr=rpq=35
Thus, <cpq=70
Angle submitted by arc on center=<pcq=40.
Length of the minor arc = pi*18*40/360=2*pi

Fully agree with PQ = 2 Pi R (40/360) = 2 (pi)

A.
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Intern
Joined: 16 Jan 2009
Posts: 14
Re: GEOMETRY-gmatprep qs [#permalink]

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16 Jan 2009, 12:51
yup

2pi

PQ = (40/360) 2 pi R
PQ = ( 40/360 ) 2 pi . 9
PQ = 2pi

Re: GEOMETRY-gmatprep qs   [#permalink] 16 Jan 2009, 12:51
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# GEOMETRY-gmatprep qs

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