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Manager  B
Joined: 13 Feb 2011
Posts: 81
Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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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 S
Joined: 18 Sep 2014
Posts: 1100
Location: India
Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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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!

Dienekes
You don't have to change any value.
I hope below solution helps!
Attachment: 1.png [ 26.87 KiB | Viewed 2835 times ]
CEO  S
Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44 GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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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!

Dienekes
You 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: 2-16-16 10-22-54 AM.jpg [ 17.72 KiB | Viewed 3083 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$$\angle{PRO}$$ , $$\angle{PAO} = 2*30=60$$ ---> $$\angle {PAR} = 180-60=120$$

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 PQ||OR ---> $$\angle {PAR }+\angle {APQ} = 180$$ (supplementary angles). You have already calculated $$\angle {PAR} = 120$$ ---> $$\angle {APQ} = 180-120=60$$

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  S
Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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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!

Dienekes
You 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 S
Joined: 18 Sep 2014
Posts: 1100
Location: India
Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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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  S
Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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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.
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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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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  S
Status: Come! Fall in Love with Learning!
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Location: India
Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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Attachment: Circle.jpg [ 21.32 KiB | Viewed 2667 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 = 55-35 =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

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In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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1
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.

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Manager  B
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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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VeritasPrepKarishma, chetan2u could you please share your approach to this problem? Thanks.
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In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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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 $$- 2*7\pi/2 = 9\pi - 7\pi = 2\pi$$
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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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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 = 180-35-90=55
angle QOR = 35 ( similar triangles)

so angle POQ = 55-35=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 1341 times ]

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In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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VeritasPrepKarishma 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.
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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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C is the center of circle
OR 18
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 egg-like circle... 15345239420186686756818382894670.jpg [ 1.9 MiB | Viewed 877 times ]

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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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

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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. $$3\pi$$

Attachment:
The attachment Circle.jpg is no longer available

The explanation is on the picture. I color coded the angles with each step to make things easy to follow
Attachments GMATCLUB.png [ 35.28 KiB | Viewed 150 times ] Re: In the circle above, PQ is parallel to diameter OR, and OR has length   [#permalink] 19 Apr 2019, 12:48

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