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In the figure above, O is the center of the circle. If OP

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In the figure above, O is the center of the circle. If OP [#permalink] New post 17 Jan 2004, 22:02
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In the figure above, O is the center of the circle. If OP and QS are perpendicular and bisect each other, what is the length of OS?

(1) The length of PR is \sqrt{3}
(2) The length of QS is 6
[Reveal] Spoiler: OA
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 [#permalink] New post 17 Jan 2004, 23:05
A? We need to find OS which is equal to OP as both are radius. from 1 we can find the OS
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 [#permalink] New post 19 Jan 2004, 11:12
the answer is D actually...anyone have any ideas on why D?
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 [#permalink] New post 19 Jan 2004, 13:49
Rakesh explains why 1 is sufficient. Here's why 2 is sufficient:


The length of QS is 6.

Hence, QR and RS are each 3.

Call OR and PR BOTH "x".

(Since OR and PR combine to form a radius, just like OS)

Call OS "2x."

Since we have a right triangle, we can establish that X^2 +3^2=(2x)^2.

Solve this and you see that x= sqrt 3, and 2x (in this case line OS) is 2*sqrt3.

I think
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 [#permalink] New post 19 Jan 2004, 16:51
Stoolfi ur explanation for statement (2) is good. I dont know why I miss out those things. Probably due to time constrains. Anyways from (1) we are given that PR= sqrt 3, we know that PR=OR (Bisects), so OP will be 2sqrt3 and as both OP and OS are radius, OP=OS. So OS= 2sqrt3
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In the figure above, O is the center of the circle. If OP an [#permalink] New post 07 Sep 2013, 23:39
In the figure above, O is the center of the circle. If OP and QS are perpendicular and bisect each other. What is the length of OS?

1) The length of PR is \sqrt{3}
2) The length of QS is 6

Please provide detailed explanations!
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Re: In the figure above, O is the center of the circle. If OP an [#permalink] New post 08 Sep 2013, 00:38
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fozzzy wrote:
In the figure above, O is the center of the circle. If OP and QS are perpendicular and bisect each other. What is the length of OS?

1) The length of PR is \sqrt{3}
2) The length of QS is 6

Please provide detailed explanations!


In the given figure,OS=OP=radius "r"
and PR=RO=r/2
QR=RS,angleORS=90 degrees
We need find length of OS, which is radius of the circle

From stmt 1)PR= [square_root]3 and PR=r/2 hence radius=2*[square_root]3, hence 1 alone is sufficient
From stmt 2)QS=6 => RS=3 (OP is bisector of QS,R is the midpoint of QS)
OS=radius=r
and OR=r/2 (R is the midpoint of OP, OP being the radius of the circle)
r^2=(r/2)^2+3^2
r=2*[square_root]3
Hence stmt 2 alone is sufficient

Ans is D
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Re: In the figure above, O is the center of the circle. If OP an [#permalink] New post 08 Sep 2013, 04:55
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fozzzy wrote:
In the figure above, O is the center of the circle. If OP and QS are perpendicular and bisect each other. What is the length of OS?

1) The length of PR is \sqrt{3}
2) The length of QS is 6

Please provide detailed explanations!


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Re: In the figure above, O is the center of the circle. If OP [#permalink] New post 09 Sep 2013, 04:20
rc197906 wrote:
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In the figure above, O is the center of the circle. If OP and QS are perpendicular and bisect each other, what is the length of OS?

(1) The length of PR is \sqrt{3}
(2) The length of QS is 6


Hi,

St 1 is obvious so lets not dwell on that.

From St 2 we get QS =6 -----> QR=RS=3

Let OR=x since it is being bisected by QS then PR is also x
Also OQ= 2x since OP=OS as radius of circle
Now in triangle OQR we have 3^2+x^2= 4X^2

9=3 x^2
x=\sqrt{3}

Hence OQ= 2\sqrt{3}

Ans is D
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Re: In the figure above, O is the center of the circle. If OP [#permalink] New post 09 Sep 2013, 05:19
Is there any property tested here? I'm really clueless when it comes to geometry...
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Re: In the figure above, O is the center of the circle. If OP [#permalink] New post 10 Sep 2013, 02:19
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fozzzy wrote:
Is there any property tested here? I'm really clueless when it comes to geometry...

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In the figure above, O is the center of the circle. If OP and QS are perpendicular and bisect each other. What is the length of OS?

Notice that OS is the radius of the circle.

(1) The length of PR is \sqrt{3}. PR is half of the radius, so the radius is twice of that. Sufficient.

(2) The length of QS is 6. This one tests pythagorean theorem. Triangle ORS is a right triangle. We know that RS=QS/2=3 and we know that the other two sides are r (OS) and r/2 (OR) --> r^2=(r/2)^2+3^2 --> r=2\sqrt{3}. Sufficient.

Answer: D.

Hope this helps.
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Re: In the figure above, O is the center of the circle. If OP [#permalink] New post 10 Sep 2013, 04:24
So for such questions never rely on the figure. In this particular question, the figure is misleading.
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Re: In the figure above, O is the center of the circle. If OP [#permalink] New post 10 Sep 2013, 04:31
Expert's post
fozzzy wrote:
So for such questions never rely on the figure. In this particular question, the figure is misleading.


Why it's misleading?

OG13, page 150:
Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

OG13, page 272:
A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2).
Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight.
You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees.
All figures lie in a plane unless otherwise indicated.

Hope it helps.
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Re: In the figure above, O is the center of the circle. If OP [#permalink] New post 28 Nov 2013, 16:56
Bunuel wrote:
fozzzy wrote:
So for such questions never rely on the figure. In this particular question, the figure is misleading.


Why it's misleading?

OG13, page 150:
Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

OG13, page 272:
A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2).
Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight.
You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees.
All figures lie in a plane unless otherwise indicated.

Hope it helps.


Hi Bunuel, question for you
How do we know that PR = OR from the fact that the chord is perpendicular to the radius? I mean what's the property here if you will
Thanks
Cheers
J :)
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Re: In the figure above, O is the center of the circle. If OP [#permalink] New post 29 Nov 2013, 01:20
Expert's post
jlgdr wrote:
Bunuel wrote:
fozzzy wrote:
So for such questions never rely on the figure. In this particular question, the figure is misleading.


Why it's misleading?

OG13, page 150:
Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

OG13, page 272:
A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2).
Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight.
You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees.
All figures lie in a plane unless otherwise indicated.

Hope it helps.


Hi Bunuel, question for you
How do we know that PR = OR from the fact that the chord is perpendicular to the radius? I mean what's the property here if you will
Thanks
Cheers
J :)

Image
We are told that OP and QS are perpendicular and bisect each other, which means that PR = OR and QR = RS.

Hope it's clear.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In the figure above, O is the center of the circle. If OP [#permalink] New post 04 Dec 2013, 09:52
From statement 2.) it seems like we know that triangle ORS is a 30:60:90. How do we know this?

I think I got it but I want to be sure:

We know OP = OS because they are both radiuses. QS bisects OP so OR (where R is the midpoint of OP) is equal to 1/2 OS. We can say that OS = 2x and OR = x. From statement 2 we know that QS = 6 and RS therefore must = 3 because it is bisected by OP. We can now set up Pythagorean equation a^2 + b^2 = c^2 --> x^2 + 3^2 = 2x^2 --> x^2 + 9 = 2x^2 --> Subtract x^2 from both sides --> x^2 = 9 --> x = 3. If x = 3, then 2x (i.e. line OS) = 6.
Re: In the figure above, O is the center of the circle. If OP   [#permalink] 04 Dec 2013, 09:52
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