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by aneesh.kg » Mon Apr 30, 2012 10:55 am
Join P and Q to O. Mark C as the centre of the circle.
Given: r = 9

Since PQ is parallel to OR,
angle QOR = angle PRO = 35

angle POQ = angle POR - angle QOR = 20
angle PCQ = 40 (Property of circle: the angle subtended by an arc is twice at the centre than at the circumference)

Thus,
Length of the required arc = (40/360)*2(pi)r
= (1/9)*18*pi
= 2(pi)

(A) is the answer.
This is a good problem but definitely not very difficult because there are many more ways of solving it if you are comfortable with Geometry.
Last edited by aneesh.kg on Mon Apr 30, 2012 11:01 am, edited 2 times in total.
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by bobdylan » Mon Apr 30, 2012 10:59 am
I think the answer is 7 pi/2. the circunference of the circle is 18 pi. The central angle is 70 degrees, because OR is parallel to PQ. So you divide the central angle by 360, and you get 7/36. Then you multiple this to the circunference, to get 7 pi/2.
But I am not sure this is the right answer.
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by bobdylan » Mon Apr 30, 2012 11:07 am
Aneesh, can you please explain better how you got the angle 20. Thanks!
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by aneesh.kg » Mon Apr 30, 2012 11:14 am
Alright.
As I said, first join P to O and Q to O. Since PQ and OR are two parallel chords, triangles PRO and QOR are congruent. So, angle QOR is also 35.
angle POR = 90 - 35 = 55 (because traingle OPR is right-angled)
Now,
angle POR - angle QOR = angle POR = 20
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by Brent@GMATPrepNow » Mon Apr 30, 2012 11:24 am
Here's another approach:
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by aneesh.kg » Mon Apr 30, 2012 11:30 am
Alternate method:

Mark the centre of the circle as C.
The minor arc subtends an angle of 35 at R. So, it must subtend an angle double of it at the centre C.
angle PCO = 70
Similarly, on the other side, angle QCR = 70

angle PCQ = 180 - (angle PCO + angle QCR) = 180 - 140 = 40

arc length = (40/360)*2(pi)r = 2(pi)

(A) is the answer.

Notice that we've seen three methods already to solve this problem. That's why this cannot be categorised as a very difficult problem.
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by bobdylan » Mon Apr 30, 2012 11:49 am
Aneesh, can you please explain better how you got the angle 20. Thanks!
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by sathishkumarjva9888 » Mon Apr 30, 2012 6:43 pm
Thanks Aneesh and Brent. :) You made me understand the central angle concept in a better way. :)
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