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In the circle above, PQ is parallel to diameter OR

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In the circle above, PQ is parallel to diameter OR [#permalink] New post 10 May 2010, 08:37
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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
[Reveal] Spoiler: OA
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Re: PS6 [#permalink] New post 10 May 2010, 15:18
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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.

For more on circle check the circles chapter of Math Book (link in my signature).
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Re: PS6 [#permalink] New post 06 Jun 2010, 13:56
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I hope this helps. The explanation goes as:
-We know angle CBA=35
-AOB=diameter and hence angle ACB=90
-Thus in triangle ACB, angle CAB=55
-Now in triangle AOC, OA=OC=Radius and thus angle ACO=55 and angle AOC=70.
-Similarly we can find angle DOB=70
-Thus angle COD=180-70-70=40
And length of arc calculation is given in the image=2π
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Re: PS6 [#permalink] New post 19 Apr 2011, 18:32
40/180 = x/pi*18/2

x = 2pi

Answr - A
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Re: Geometry Problem [#permalink] New post 08 Jan 2012, 04:46
kotela wrote:
Can anyone please help me in solving this problem.......


arc OP = arc QR = (70 * 18 pi )/360 as angle OPR = angle QPR (35 degree)

angle for arc PQ = 180 - (2 arc QP) = 180 - 140 = 40

length of arc PR = (40 * 18pi)/360 = 2pi
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Re: Geometry Problem [#permalink] New post 11 Jan 2012, 00:21
kp1811 can u explain this step
angle for arc PQ = 180 - ( 2arc QP )

how u got that ? plz update
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Re: Geometry Problem [#permalink] New post 11 Jan 2012, 01:32
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kotela wrote:
Can anyone please help me in solving this problem.......



There are 2-3 methods, lets see one of them:---
1]You are given that OR is diameter=18 (radius =9); therefore arc OR is a semicircle
the semicircle is composed of 3 arc--arc OP + arc PQ+ arc QR {arc PQ = 180- arc OP + arc QR}

2]the lines PQ & OR are parallel , therefore than alternate angles are congruent ( angle PRO = angle=QPR )
but these are inscribed angles and we need the central angles to compute the length of arc given by the formula :
{central angle (teta) /360 * 2*pi*radius}

3] Central angle of arc QR or arc OP =2 * inscribed angle (2*35= 70)
thus arc's QR + OP have central angles 70 +70= 140

4] thus, central angle for arc PQ =180-140=40 ;

length of arc PQ= 40/360 *2* pi* 9 = 2 pi

You may also employ back solving if you are comfortable with that for faster solution
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Re: Geometry Problem [#permalink] New post 11 Jan 2012, 01:46
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pbull78 wrote:
kp1811 can u explain this step
angle for arc PQ = 180 - ( 2arc QP )

how u got that ? plz update


we know arc OP = arc QR so we can write either 180 -(arc OP + arc QR) or 180 - 2OP or 180 - 2QR
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In the circle above, PQ is parallel to diameter OR [#permalink] New post 10 May 2012, 19:40
Here I'm giving an elaborate solution.
Hope it helps.
Let A be the center of the circle. We join the mid point of PQ, let's say S with A. Clearly angle ASQ=90 degree.
Now angle ROQ= angle PRO= 35 degree
As at center any arc subtends double the angle what it subtends at any peripherial point, we can conclude angle RAQ= 70 degree.
Hence angle PAQ= 2* angle SAQ= 2*(90-70)= 40 degree
180 degree= pi radian
40 degree= 2 pi/9 rad
Hence minor arc PQ= 9*2 pi/9 (As radius of the circle=9)
Arc PQ= 2 pi
Ans-(A)
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In the circle above, PQ is parallel to diameter OR, and OR has [#permalink] New post 30 Jul 2012, 15:03
I encountered this problem on one of the tests. It was around my 12th or 13 question. I do admit that this question stumped me and I took some amount of time in solving this question.

I started off with the geometric approach but then quickly got tangled in it. I then shifted to approximating the area and using the process of elimination. I am explaining my approach even though it is unconventional and based only on the situation that I was in.

assumption pi=3.14

The radius is given as (OR/2)=9. Hence the circumference of the circle is 18pi.

Clearly looking at the figure, the length of arc PQ is much smaller than 1/4 the circumference.

18*pi=18*3.14~42+ (read it as something greater than 42)

Hence 1/4 of the circumference is ~ 10+ (read as something greater than 10). The length of Arc PQ is much smaller than 1/4 of the circumference as well.

Options C and D can be eliminated since C=~10.5 and D=~13.5. These can be clearly eliminated. Similarly Option E : 3pi ~ very close to 10 and hence I eliminated it.

After this it was purely on the figure given than I eliminated the other choices. The arc PQ looks to be more like closer to half of the 1/4 i.e. ~5+. I really have no justification as to how I deduced it. Just an observation so that I could guess the answer.

Option B [(9pi)/4] comes to be roughly 7- (read as less than 7)

Option A (2pi) comes to ~6.5- (read as less than 6.5) which is closer to mu initial estimate of being 5+.

Hence I chose A.

I am very well aware that on some other day it is very much possible to choose B and to get the question wrong. Also I am not advocating this method in any way :)

Its just that this was an educated guess which I guessed had 50% chance of being right.

The above explanations and diagramatic representations have helped me realize my mistake and should such a problem appear again I am dead sure of the mathematical way to solve it.
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Re: In the circle above, PQ is parallel to diameter OR [#permalink] New post 08 Dec 2012, 01:59
Bunuel how do we know that OR is the diameter of the circle?
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Re: In the circle above, PQ is parallel to diameter OR [#permalink] New post 08 Dec 2012, 03:55
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Re: PS6 [#permalink] New post 22 Dec 2012, 05:09
Bunuel wrote:
Image
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-<QPR=180-70-70=40.

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

Answer: A.

For more on circle check the circles chapter of Math Book (link in my signature).


<PCO AND < QPR ALTERNATE ANGLES? how are those 2 angles equal?
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Re: PS6 [#permalink] New post 22 Dec 2012, 05:35
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fozzzy wrote:
Bunuel wrote:
Image
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-<QPR=180-70-70=40.

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

Answer: A.

For more on circle check the circles chapter of Math Book (link in my signature).


<PCO AND < QPR ALTERNATE ANGLES? how are those 2 angles equal?


It should be <PCQ=180-(<PCO+<QCR)=180-70-70=40.
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In the circle above, PQ is parallel to diameter OR, and OR has l [#permalink] New post 21 May 2013, 00:43
< RPQ = 35 degress

we have 2 inscribed angles so they are 70 degrees

now take one portion of the semicircle and we get 180 - 70 - 70 = 40 degrees

40 / 360 = 1/9th

Minor arc = Pi D

1/9 * 18 Pi = 2 Pi
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Re: In the circle above, PQ is parallel to diameter OR [#permalink] New post 12 Feb 2014, 02:56
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?
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Re: In the circle above, PQ is parallel to diameter OR [#permalink] New post 22 Jun 2014, 23:41
Bunuel wrote:
Image
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.

For more on circle check the circles chapter of Math Book (link in my signature).


Bunnel, don't you think that this question asks for a little too much imagination to be on GMAT?
Just curious, would you rate this as a 750 range question on the test?
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Re: In the circle above, PQ is parallel to diameter OR [#permalink] New post 23 Jun 2014, 00:48
Expert's post
Kconfused wrote:
Bunuel wrote:
Image
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.

For more on circle check the circles chapter of Math Book (link in my signature).


Bunnel, don't you think that this question asks for a little too much imagination to be on GMAT?
Just curious, would you rate this as a 750 range question on the test?


I'd say it's 650 at most.
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Re: In the circle above, PQ is parallel to diameter OR [#permalink] New post 04 Oct 2014, 11:02
Hi Bunuel,
I did not understand how Central angle theorem applied here.

Again, according to the central angle theorem above <QCR=2<QPR=70.




Bunuel wrote:
Image
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.

For more on circle check the circles chapter of Math Book (link in my signature).

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Re: In the circle above, PQ is parallel to diameter OR [#permalink] New post 05 Oct 2014, 01:31
Expert's post
sunita123 wrote:
Hi Bunuel,
I did not understand how Central angle theorem applied here.

Again, according to the central angle theorem above <QCR=2<QPR=70.




Bunuel wrote:
Image
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.

For more on circle check the circles chapter of Math Book (link in my signature).


Red central angle below is twice inscribed blue angle, because they subtend the same arc QR:
Attachment:
Untitled.png
Untitled.png [ 7.15 KiB | Viewed 559 times ]


Hope it's clear.
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Re: In the circle above, PQ is parallel to diameter OR   [#permalink] 05 Oct 2014, 01:31
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