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In the circle above, PQ is parallel to diameter OR, and OR

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

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New post 10 Jun 2008, 16:07
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Attachment:
GMATCircle.jpg
GMATCircle.jpg [ 20.23 KiB | Viewed 23006 times ]

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) \(3\pi\)

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-circle-above-pq-is-parallel-to-diameter-or-93977.html

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Re: PS - Geometry  [#permalink]

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New post 10 Jun 2008, 17:53
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jallenmorris wrote:
Attachment:
GMATCircle.jpg


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) \(3\pi\)


hey you have to realize that the PQ and OR are parallel..having said that angle P will also be 35..

ok so now that you this..you draw another line from the center to P, to make a issocles triangle.. this angle will be 35 as well..now knowing this ..you yet again draw another line from the center to Q..and form an issoceles triangle..each angle will be 70 70..

so then the angle at the center O will be 40..
arc length will be 1/9*18pi..i.e 2pi
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Re: PS - Geometry  [#permalink]

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New post 10 Jun 2008, 20:02
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jallenmorris wrote:
Attachment:
The attachment GMATCircle.jpg is no longer available


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:
GMATCircleAnswer.jpg
GMATCircleAnswer.jpg [ 20.99 KiB | Viewed 22861 times ]


Thanks (kudos) for the explanation. In the image above, I realize now how to solve it from your explanation.

Angle RPQ's measure is 35 degrees because of the rule that opposite angles created by a line through 2 parallel lines will be equal. (ORP and RPQ being the opposite angles created).

Then the red line (seen in the image) is drawn creating an isosceles triangle out of PRC (C being the Center of the circle). Because 2 of the 3 sides are the same length, we know it's isosceles triangle and angle CRP = CPR.

Step 2 draw the blue line. THis creates the 2nd isosceles triangle out of CPQ. Because angle RPQ is 35 degrees, and angle CPR is 35, we know that angle CPQ is 70 degrees, and because CPQ is isosceles, CQP is also 70 degrees. This makes PCQ = 40 degrees.

Because the angle is 40 degrees, that means that the arc takes up 40/360 of the circumference of the circle. This is 1/9. The stem tells us the diameter is 18, and the formula for the circumference is d\(\pi\), so we have 18\(\pi\) * 1/9, or 2\(\pi\).
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Re: PS - Geometry  [#permalink]

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New post 15 Jul 2008, 13:11
pi*r - ((2*2*35)/180)*pi*r = pi*r*2/9 = pi*9*2/9=2*pi -> A
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Re: PS - Geometry  [#permalink]

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New post 15 Jul 2008, 13:48
excellent! thank you. I couldn't solve it in 2 mins though.
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Re: PS - Geometry  [#permalink]

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New post 15 Jul 2008, 13:58
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It's really not as complicated as J Allen's explanation....

Given: PQ is parallel to diameter OR
1) PRO = QPR = 35
2) Arc PRO = Arc QPR = 70 (35x2)

Now, a circle = 360 degree
Arc PQ = 360- Arc OR - Arc PO - Arc QR = 350-180-70-70=40

Here is the tricky part: circumference = 2*pi*r=pi*d (this is where I fell and got the wrong answer when I was practicing!)

Circumference=18pi
Arc PQ = 18pi*40/360=2pi

Yup, totally doable under 2 mins....and you can get it correctly if you don't fall into the trap of the 2*pi*r thing like I did.... Fortunately 4*pi was not one of the options and it forced me to review my steps.
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Re: PS - Geometry  [#permalink]

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New post 15 Jul 2008, 14:18
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fatb,

You mistake my long explanation as complicating the question. You forget that people come here to learn how to do these problems. Just because you can solve it in 30 seconds, or explain it in 3 lines doesn't mean it's easy for someone to understand your explanation. I go overboard on my explanations to make sure that ANYONE can understand the method I used so they can learn the concept. It doesn't do any good or improve this community, if all of the explanations are so short that no one can understand them.

In order to help people learn, we have to simplify things...once they get it, there are certainly steps that can be done without writing it out, or consciously thinking about that step, but until people get to that point...they need all of the steps which it appears you think complicate the explanation.

fatb wrote:
It's really not as complicated as J Allen's explanation....

Given: PQ is parallel to diameter OR
1) PRO = QPR = 35
2) Arc PRO = Arc QPR = 70 (35x2)

Now, a circle = 360 degree
Arc PQ = 360- Arc OR - Arc PO - Arc QR = 350-180-70-70=40

Here is the tricky part: circumference = 2*pi*r=pi*d (this is where I fell and got the wrong answer when I was practicing!)

Circumference=18pi
Arc PQ = 18pi*40/360=2pi

Yup, totally doable under 2 mins....and you can get it correctly if you don't fall into the trap of the 2*pi*r thing like I did.... Fortunately 4*pi was not one of the options and it forced me to review my steps.

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Re: PS - Geometry  [#permalink]

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New post 15 Jul 2008, 18:11
I think the only underlying assumption that was made in my explanation was that people should know the following: the angle formed by any three points on a circle is the half of the measure of the arc subtended.
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Re: PS - Geometry  [#permalink]

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New post 16 Jul 2008, 18:18
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heres how I did it.

angle PRO = 35 so for arc PO central angle will be 70 and thus PO is 70/360 of the circumference which is 18*Pi = 7/2*pi

Since lines PQ and OR are parallel, QR must also be 70/360 * 18 * pi = 7/2*pi

The semicircle in the bottom part will be 1/2 *18 * pi = 9*pi

length of arc PQ = circumference - lenght of arc PO - length of arc QR - length of semicircle
= 18*pi - 7/2* pi - 7/2*pi - 9*pi
= pi ( 18 - 16)
= 2*pi

Thus A
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Re: PS - Geometry  [#permalink]

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New post 16 Aug 2009, 16:18
Guys major confusion!

We know that minor arcs are always twice their inscribed angles right?
So,minor arc PO=70=minor arc QR.
minor arc PQ=180-70-70=40.So we have minor arc PQ=40.
Same logic gives us Angle PCQ(C is the centre of the circle) as 20 deg.!!!
Applying formula for length of an arc=2pi(radius)(angle)/360
=2pi(9)(20)/360=pi :(
What did I do wrong?
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Re: PS - Geometry  [#permalink]

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New post 16 Aug 2009, 18:45
The problem you're having is that at first you use INSCRIBED angles, so you know 35 degrees, so minor arc PO = 70. This is correct. It also means that QR is 70 which is right. Then you use that 180-70-70 = 40 to get the degree measure of PQ. The problem you have here is that 40 is not the INSCRIBED angle. It is THE angle of the arc. The INSCRIBED angle is not the angle formed with center. The inscribed angle must be one end of the diameter to some point on the circle. We double that angle measurement to get the angle measured from point on the cirle to CENTER then to Point on the circle. We already have the 40 degree measurement for PQ. We don't need to find the INSCRIBED angle for this arc.

Go to the formula you already have and use 40 instead of 20 and you get \(2pi\) which is the correct answer.
tejal777 wrote:
Guys major confusion!

We know that minor arcs are always twice their inscribed angles right?
So,minor arc PO=70=minor arc QR.
minor arc PQ=180-70-70=40.So we have minor arc PQ=40.
Same logic gives us Angle PCQ(C is the centre of the circle) as 20 deg.!!!
Applying formula for length of an arc=2pi(radius)(angle)/360
=2pi(9)(20)/360=pi :(
What did I do wrong?

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Re: PS - Geometry  [#permalink]

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New post 16 Aug 2009, 19:58
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Tejal 777.. you are all most at it but minor mistake


You found arc PQ = 40 Degree (in terms of length )
360 degree = 2*Pi*r = 2*pi*9
40 degree = (2*pi*r*9/360 ) *40 = 2Pi
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Re: PS - Geometry  [#permalink]

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New post 17 Aug 2009, 01:51
ok ok..i'm getting there..ALMOST understood..
So lemme see..
we need the measurement of minor arc PQ.I got the measurement of that but in terms of length which is 40.but the answer choices do not indicate it so we have to go furthur.Great.

NOW,the formula we have,
Length of an arc=(circumference) (INSCRIBED ANGLE)/360

If we put length=18 pi * 40/360 we are getting the answer..but what exactly did we do..?!how are we inserting 40 for the value of the inscribed angle?
Waht a mess..
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Re: PS - Geometry  [#permalink]

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New post 17 Aug 2009, 04:32
40 is not the value of an inscribed angle. In this problem, ORP is an example of an inscribed angle. I'll refer to the center as C, so it won't be confusing.

Since we know ORP is 35 degrees. Arc OP is 70 degrees, which means that the angle of OCP (where C is the center) measures 70 degrees because it's double 35, and 35 is the measure of the inscribed angle. Because we know that PQ and OR are parallel, Arc OP and arc QR will be the same length. These 2 arcs take up the only space other than what we're trying to find, so you did the 180-70-70=40 for arc PQ. This is where the label of Inscribed was misapplied. 40 is the measure of arc PQ, not an inscribed angle. When we find the measure of an arc, we are always measuring the angle formed by the start of the arc (here P) to the center back out to the end point of the arc (here Q). This is NOT an inscribed angle. This is like a slice of pizza. Because we don't have an inscribed angle with value of 40, we do not need to figure out anything else with regard to an inscribed angle. We now have a slice of pizza that is 40/360 of the total size of the circle. So take \(\frac{2*pi*r*(angle-of-arc)}{360}\)

You're getting caught up in what the 40 degrees actually is.

tejal777 wrote:
ok ok..i'm getting there..ALMOST understood..
So lemme see..
we need the measurement of minor arc PQ.I got the measurement of that but in terms of length which is 40.but the answer choices do not indicate it so we have to go furthur.Great.

NOW,the formula we have,
Length of an arc=(circumference) (INSCRIBED ANGLE)/360

If we put length=18 pi * 40/360 we are getting the answer..but what exactly did we do..?!how are we inserting 40 for the value of the inscribed angle?
Waht a mess..

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Re: PS - Geometry  [#permalink]

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New post 30 May 2011, 19:07
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A good question and uses very basic concepts though hides it well at first look of question:
a) An angle drawn from intersects of lines from circle's perimeter is half of angle drawn from intersects of lines from circles's center for the same ARC
b) straight line has angle of 180
c) Arch of circle will be angle from center/ 360 * 2pi* r

So assume circle center is C and draw two lines CP & CQ, so angle OCP is 70 (as it would double of angle ORP) ALSO similarly angle RCQ is 70 (Using concept a)

Using concept b) if OCP + RCQ = 140 then angle PCQ = 40 and it is at the center of circle

using concept c) 40/360 * 2* pi * 18/2 = 2 pi so answer is A
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Re: PS - Geometry  [#permalink]

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New post 24 Sep 2011, 20:40
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central angle = 2*inscribed angle

let C be the center of the circle.

given ANGLE PRC = 35

this angle PRC is nothing but inscribed angle of arc OP.

=> OP's central angle = 70

also given PQ || OR

=> ANGLE QPR = 35 (AS PRC is 35)

this angle QPR is nothing but inscribed angle of arc QR.

=> QR's central angle = 70.


Length of arc PQ = (PQ's Central angle/360)*(2*pi*r)

= ((180-70-70)/360)*(2*pi*9)

=2pi

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

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New post 22 Jul 2013, 02:54
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Good explanations. Just wanted to sum up that in such problems involving length of arc, the angle swiped by the arc can be found either by using CENTER ANGLE theorem, or by using the property of isosceles triangle, whereever 2 sides of a triangle is the radii.
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Re: In the circle above, PQ is parallel to diameter OR, and OR  [#permalink]

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New post 22 Jul 2013, 02:58
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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. \(3\pi\)

The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

Let C be the center of the circle.

According to the central angle theorem above <PCO=2<PRO=70.

As PQ is parallel to OR, then <QPR=<PRO=35. Again, according to the central angle theorem above <QCR=2<QPR=70.

<PCQ=180-(<PCO+<QCR)=180-70-70=40.

Minor arc \(PQ=\frac{40}{360}*circumference=\frac{2\pi{r}}{9}=2\pi\)

Answer: A.

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-circle-above-pq-is-parallel-to-diameter-or-93977.html
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Re: In the circle above, PQ is parallel to diameter OR, and OR  [#permalink]

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