December 09, 2018 December 09, 2018 07:00 AM PST 09:00 AM PST Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT. December 10, 2018 December 10, 2018 10:00 PM PST 11:00 PM PST Practice the one most important Quant section  Integer properties, and rapidly improve your skills.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 12 Feb 2011
Posts: 87

Re: In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
11 Feb 2016, 18:59
gmatprav wrote: Bingo! central angle theorem is the rock star! Thanks for the valuable Math book Bunuel!
One other thing I missed in this problem is it asks for the minor arc length PQ. NOT length of PQ.
Follow up question. Is it possible to find length of the chord PQ with the information given in the question? If so what is the length of PQ? Nice question, I am actually trying to figure this. However, I am able to think of a solution for chord length only if the inscribed angle is changed from 35 to 30, making an equilateral triangle. Bunuel  Can you please help confirm if my understanding is correct, i.e. can we find the exact chord length with the values presented in the question (without changing 35 to 30)? Thanks!



Retired Moderator
Joined: 18 Sep 2014
Posts: 1121
Location: India

Re: In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
16 Feb 2016, 07:01
Dienekes wrote: gmatprav wrote: Bingo! central angle theorem is the rock star! Thanks for the valuable Math book Bunuel!
One other thing I missed in this problem is it asks for the minor arc length PQ. NOT length of PQ.
Follow up question. Is it possible to find length of the chord PQ with the information given in the question? If so what is the length of PQ? Nice question, I am actually trying to figure this. However, I am able to think of a solution for chord length only if the inscribed angle is changed from 35 to 30, making an equilateral triangle. Bunuel  Can you please help confirm if my understanding is correct, i.e. can we find the exact chord length with the values presented in the question (without changing 35 to 30)? Thanks! DienekesYou don't have to change any value. I hope below solution helps! Attachment:
1.png [ 26.87 KiB  Viewed 2370 times ]



CEO
Joined: 20 Mar 2014
Posts: 2633
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
16 Feb 2016, 07:42
Nevernevergiveup wrote: Dienekes wrote: gmatprav wrote: Bingo! central angle theorem is the rock star! Thanks for the valuable Math book Bunuel!
One other thing I missed in this problem is it asks for the minor arc length PQ. NOT length of PQ.
Follow up question. Is it possible to find length of the chord PQ with the information given in the question? If so what is the length of PQ? Nice question, I am actually trying to figure this. However, I am able to think of a solution for chord length only if the inscribed angle is changed from 35 to 30, making an equilateral triangle. Bunuel  Can you please help confirm if my understanding is correct, i.e. can we find the exact chord length with the values presented in the question (without changing 35 to 30)? Thanks! DienekesYou don't have to change any value. I hope below solution helps! Attachment: The attachment 1.png is no longer available Another solution not involving trigonometry (if you do know it, youll end up saving a lot of time!). Attachment:
21616 102254 AM.jpg [ 17.72 KiB  Viewed 2615 times ]
Refer to the image for a description of the points and angles. Based on your assumption, \(\angle{PRO} = 30\). As \(\angle{PAO}\) is the central angle to\(\angle{PRO}\) , \(\angle{PAO} = 2*30=60\) > \(\angle {PAR} = 18060=120\) Also, in triangle PAO, PA=OA (as both are the radii of the circle) > Triangle PAO is an equilateral triangle >\(\angle{POA} = 60 = \angle {APO}\) Now, as OR  PQ >\(\angle {PRO} = \angle {RPQ} = 30\) Now consider triangle PAR such that AR=AP=radii > (from \(\angle {PAR} = 120\)) you get \(\angle {APR} = \angle {ARP} = 30\) Finally, in triangle PAQ, \(\angle {APQ} = \angle {APR}+\angle{RPQ} = 30+30 = 60\) and as AP=AQ=radii > triangle APQ is another equilateral triangle > \(\angle {PAQ} = 60\) or Alternately, you can see that as PQOR > \(\angle {PAR }+\angle {APQ} = 180\) (supplementary angles). You have already calculated \(\angle {PAR} = 120\) > \(\angle {APQ} = 180120=60\) Hence length of PQ = r*theta , where theta = 60 degrees in radians = \(60*2*\pi/360\) =\(\pi / 3\) > length minor arc PQ = \(9*\pi/3 = 3\pi\) Hope this helps. Do note that the above solution also stands for 35 degrees as I have not used any particular property of equilateral triangles that you couldnt have applied to 35 degrees or any other angle for that matter. I used property of parallel lines, triangles in general and relations of circle and inscribed central angles.



CEO
Joined: 20 Mar 2014
Posts: 2633
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
16 Feb 2016, 07:44
Nevernevergiveup wrote: Dienekes wrote: gmatprav wrote: Bingo! central angle theorem is the rock star! Thanks for the valuable Math book Bunuel!
One other thing I missed in this problem is it asks for the minor arc length PQ. NOT length of PQ.
Follow up question. Is it possible to find length of the chord PQ with the information given in the question? If so what is the length of PQ? Nice question, I am actually trying to figure this. However, I am able to think of a solution for chord length only if the inscribed angle is changed from 35 to 30, making an equilateral triangle. Bunuel  Can you please help confirm if my understanding is correct, i.e. can we find the exact chord length with the values presented in the question (without changing 35 to 30)? Thanks! DienekesYou don't have to change any value. I hope below solution helps! Attachment: 1.png How would you find sin 20 or cos 70 or sin 70 or cos 20 as they are not the "standard angles" in the limited time in GMAT?



Retired Moderator
Joined: 18 Sep 2014
Posts: 1121
Location: India

Re: In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
16 Feb 2016, 08:19
Engr2012 wrote: How would you find sin 20 or cos 70 or sin 70 or cos 20 as they are not the "standard angles" in the limited time in GMAT?
I just solved to show him the possibility of problem getting solved. I believe since GMAT does not test such complex aspects of mathematics, it will either give us an approximate value or indicate sinθ in answer choices. I think we can find approximate value,as we know that sinθ value varies from 0 to 1 and vice versa for cosθ.



CEO
Joined: 20 Mar 2014
Posts: 2633
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
16 Feb 2016, 08:28
Nevernevergiveup wrote: Engr2012 wrote: How would you find sin 20 or cos 70 or sin 70 or cos 20 as they are not the "standard angles" in the limited time in GMAT?
I just solved to show him the possibility of problem getting solved. I believe since GMAT does not test such complex aspects of mathematics, it will either give us an approximate value or indicate sinθ in answer choices. I think we can find approximate value,as we know that sinθ value varies from 0 to 1 and vice versa for cosθ. See this is what I thought! Trigonometry is NOT required for GMAT and as such GMAT will never provide any options with trigonometric functions in the options. When you do present a solution with trigonometry, make sure that you clarify that people do no need to know trigonometry. Finding sin 20 correctly is tedious to say the least and of the options are close enough, you won't be able to approximate the values. If you do present an alternate solution, make sure to provide a complete solution.



Director
Joined: 12 Nov 2016
Posts: 735
Location: United States
GPA: 2.66

Re: In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
08 Mar 2017, 20:58
Okay Bunuel's explanation in this problem kicked ass I HATE it when I'm getting real into the problem and then I scroll down and see the answer. I had to carefully grab the scrollbar on my page and examine each step in his explanation and was able to figure it out by applying a concept I just learned in a previous problem.



Director
Status: Come! Fall in Love with Learning!
Joined: 05 Jan 2017
Posts: 505
Location: India

Re: In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
09 Mar 2017, 01:42
Attachment:
Circle.jpg [ 21.32 KiB  Viewed 2198 times ]
Join OP and OQ angle OPR = angle PRO = 35 (parallel lines) angle PQO = angle PRO = 35 (angle made on circumference by chord on the same are equal) angle QOR = angle PQO = 35 (parallel lines) since OR is diameter, angle OPR = 90 and hence POR = 180  (90+35) = 55 angle POR = angle POQ +angle QOR or 55 = angle POQ + 35 or angle POQ = 5535 =20 angle made by chord on center = 2* angle made by chord on circumference = 2* angle POQ= 2*20 =40 length of arc PQ = (2*pi/360) *angle made by arc on center*radius = (2*pi/360) *angle made by chord on center*radius = (2*pi/360) *40*9 = 2*pi Option A Hit kudos and visit our page for free GMAT prep question and articles : www.byjus.com/freegmatprep
_________________
GMAT Mentors



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830

In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
26 Mar 2018, 16:34
LM wrote: In the circle above, PQ is parallel to diameter OR, and OR has length 18. What is the length of minor arc PQ? A. \(2\pi\) B. \(\frac{9\pi}{4}\) C. \(\frac{7\pi}{2}\) D. \(\frac{9\pi}{2}\) E. \(3\pi\) Since PQ is parallel to OR, angle RPQ is 35 degrees. Thus, since arc PO is twice angle RPQ, arc PO is 70 degrees and it is thus 70/360 of the circumference. Since arc QR = arc PO, arc QR is also 70/360 of the circumference. We also see that arc OPQR is 180/360 of the circle, so arc PQ is: 180/360  70/360  70/360 = 40/360 = 4/36 = 1/9 of the circumference. Since the diameter is 18, the circumference of the circle is 18π. So arc PQ is 1/9 x 18π = 2π. Answer: A
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Manager
Joined: 10 Sep 2014
Posts: 83
Location: Bangladesh
GPA: 3.5
WE: Project Management (Manufacturing)

Re: In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
02 Apr 2018, 21:38
VeritasPrepKarishma, chetan2u could you please share your approach to this problem? Thanks.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8649
Location: Pune, India

In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
03 Apr 2018, 01:44
sadikabid27 wrote: VeritasPrepKarishma, chetan2u could you please share your approach to this problem? Thanks. My approach here isn't very different from what others have posted. When I read parallel lines and see a transversal PR between them, I know that the alternate interior angles PRO and RPQ will be equal i.e. both will be 35 degrees. Then the angles subtended by their respective arcs PO and QR at the centre will be twice of 35 that is 70 degrees (by central angle theorem) So length of arcs PO = QR = \((70/360)*18\pi = 7\pi/2\) Length of PQ = Length of semi circle \( 2*7\pi/2 = 9\pi  7\pi = 2\pi\)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Math Expert
Joined: 02 Aug 2009
Posts: 7095

Re: In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
03 Apr 2018, 02:39
sadikabid27 wrote: VeritasPrepKarishma, chetan2u could you please share your approach to this problem? Thanks. SEE THE ATTACHED FIGURE.. Triangle POR is right angled triangle, so angle POR = 1803590=55 angle QOR = 35 ( similar triangles) so angle POQ = 5535=20.. this angle is the angle subtended by the arc from circumference, so from centre the angle to the arc will be TWICE = 20*2=40 entire 360 = entire circumference = 2*pi*r = 2*pi*9 = 18pi so 40 degree will give 40/360 of circumference = \(\frac{40}{360}*18*pi=2*pi\)
Attachments
Untitled100.png [ 7.36 KiB  Viewed 875 times ]
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Manager
Joined: 10 Sep 2014
Posts: 83
Location: Bangladesh
GPA: 3.5
WE: Project Management (Manufacturing)

In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
04 Apr 2018, 06:24
VeritasPrepKarishma wrote: sadikabid27 wrote: VeritasPrepKarishma, chetan2u could you please share your approach to this problem? Thanks. My approach here isn't very different from what others have posted. When I read parallel lines and see a transversal PR between them, I know that the alternate interior angles PRO and RPQ will be equal i.e. both will be 35 degrees. Then the angles subtended by their respective arcs PO and QR at the centre will be twice of 35 that is 70 degrees (by central angle theorem) So length of arcs PO = QR = \((70/360)*18\pi = 7\pi/2\) Length of PQ = Length of semi circle \( 2*7\pi/2 = 9\pi  7\pi = 2\pi\) VeritasPrepKarishma please explain this part "Length of PQ = Length of semi circle \( 2*7\pi/2 = 9\pi  7\pi = 2\pi\)" why length of PQ would be Length of semi circle \( 2*7\pi/2 and why we are multiplying 7\pi/2\) with 2 ? Thanks.



Manager
Joined: 02 Aug 2017
Posts: 62
Concentration: Strategy, Nonprofit
GPA: 3.71

Re: In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
17 Aug 2018, 08:40
C is the center of circle OR 18 Hence radius OC=CR=9 Join PC, and consider triangle PCR PC=CR Hence, angle CPR = angle CRP = 35 PQ and OR are parallel,hence angle PRC = angle RPQ = 35 Similarly consider triangle CPQ PC=CQ Hence angle CPQ = angle CQP = 70 Angle PCQ=40 Hence arc length PQ =(40/360)2*Pi*9 = 2*Pi Posted from my mobile device
Attachments
File comment: Please pardon my egglike circle...
15345239420186686756818382894670.jpg [ 1.9 MiB  Viewed 410 times ]
_________________
Everything is in flux, nothing stays still
MGMAT1 :590 Q42 V30 (07/07/18) VERITAS :660 Q48 V33 (16/07/18) GMATPREP1 :690 Q46 V36 (22/07/18) GMATPREP2 :740 Q51 V39 (06/08/18) ECONOMIST :740 Q49 V44 (11/08/18) KAPLAN :690 Q49 V36 (17/08/18) PRINCETON :690 Q48 V38 (26/08/18) MGMAT2 :720 Q43 V45 (02/09/18)



Manager
Joined: 07 Feb 2017
Posts: 188

Re: In the circle above, PQ is parallel to diameter OR, and OR has length
[#permalink]
Show Tags
17 Aug 2018, 09:20
Is there way to solve without Central Angle Theorem?




Re: In the circle above, PQ is parallel to diameter OR, and OR has length &nbs
[#permalink]
17 Aug 2018, 09:20



Go to page
Previous
1 2
[ 35 posts ]



