BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

PS-Circles + Area (OG-11 PS#206)

Expert replies
by haidgmat » Sun Jul 18, 2010 5:47 pm
Hello guys, It's been a while since I took any Geometry. Can anyone explain this solution in simple words?

206. In a circle, PQ is parallel to diameter OR and the length of OR is 18. Angle ORP equals 35 degrees. What is the length of arc PQ?

A. 2Π

B. 9Π/4

C. 7Π/2

D. 9Π/2

E. 3Π

OA-A
Join the discussion
Source: — Problem Solving |

by Rahul@gurome » Sun Jul 18, 2010 6:59 pm
Solution
Let T be the centre of the circle.
Join both P and Q to T.
Let angle PTQ be x.
Or length of arc PQ is (x/360)*2*pi*r = (x/360)*pi*18.
We need to calculate x.
Now since PT and TR are radius, they are same in value.
So triangle PTR is isosceles.
So angle TPR is also 35 degrees (since angle ORP or angle TRP is given as 35 degrees).
So angle PTR is 180 - (35 + 35) = 110.
Since OR is parallel to PQ, angle ORP = angle RPQ = 35 degrees.
Now angle QTR is twice angle QPR (since angle subtended by an arc at centre is double the angle the same arc subtends at any other point on the circumference).
So angle QTR is 70 degrees.
So angle PTQ = x = 110 - 70 = 40 degrees.
So length of arc PQ is (40/360)*pi*18 = 2*pi.

The correct answer is (A).
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Join the discussion

by GMATGuruNY » Sun Jul 18, 2010 7:28 pm
The circumference of the circle is 18pi. We need to determine what fraction of the circumference is arc PQ. The proportionality of circles tells us that:

(degree measurement of arc)/360 = (length of arc)/circumference

The attached drawing shows that the degree measurement of arc PQ is 40 degrees.

40/360 = 1/9
1/9 = (PQ)/18pi
PQ = 2pi.

The correct answer is A.
Attachments
arc question.pdf
(58.44 KiB) Downloaded 162 times
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by haidgmat » Mon Jul 19, 2010 6:07 pm
Thanks Rahul@gurome and GMATGuruNY! The drawing really helped btw!!!

In the real test, what difficulty level would this problem be considered as ?
Join the discussion