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Director  Joined: 03 Sep 2006
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In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 56% (02:48) correct 44% (02:52) wrong based on 702 sessions

HideShow timer Statistics 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: Circle.jpg [ 20.23 KiB | Viewed 65657 times ]
Math Expert V
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In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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27 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 $$\angle PCO=2\angle PRO=70$$.

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

$$\angle PCQ=180-( \angle PCO+ \angle QCR)=180-70-70=40$$.

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

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, and OR has length  [#permalink]

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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
Attachments IMAG0432.jpg [ 344.09 KiB | Viewed 53726 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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40/180 = x/pi*18/2

x = 2pi

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

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kotela wrote:

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
Intern  Joined: 16 Dec 2011
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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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kp1811 can u explain this step
angle for arc PQ = 180 - ( 2arc QP )

how u got that ? plz update
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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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kotela wrote:

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

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

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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)
Intern  Joined: 14 Jun 2012
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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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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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Intern  Joined: 11 Jul 2012
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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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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, and OR has length  [#permalink]

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Ousmane wrote:
Bunuel how do we know that OR is the diameter of the circle?

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

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Bunuel 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. $$3\pi$$

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

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

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fozzzy wrote:
Bunuel 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. $$3\pi$$

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

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 $$\angle PCQ = 180 - (\angle PCO + \angle QCR) = 180 - 70 - 70 = 40$$.
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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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< 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, and OR has length  [#permalink]

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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, and OR has length  [#permalink]

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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: 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 $$PQ=\frac{40}{360}*circumference=\frac{2\pi{r}}{9}=2\pi$$

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

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

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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: 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 $$PQ=\frac{40}{360}*circumference=\frac{2\pi{r}}{9}=2\pi$$

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: Hope it's clear.

Attachment: Untitled.png [ 7.15 KiB | Viewed 47690 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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Bunuel - how do you know the red angle is twice the blue and not vice versa? <rpq is not inscribed in <cqr
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Re: In the circle above, PQ is parallel to diameter OR, and OR has length  [#permalink]

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bankerboy30 wrote:
Bunuel - how do you know the red angle is twice the blue and not vice versa? <rpq is not inscribed in <cqr

Blue angle is an inscribed angle.
Red angle is a central angle.

The measure of inscribed angle is always half the measure of the central angle.

Check here for more: math-circles-87957.html
_________________ Re: In the circle above, PQ is parallel to diameter OR, and OR has length   [#permalink] 12 Oct 2014, 15:04

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