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In the figure above, O is the center of the circle. If OP [#permalink]
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17 Jan 2004, 23:02
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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
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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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the answer is D actually...anyone have any ideas on why D?



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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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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]
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08 Sep 2013, 00:39
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]
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08 Sep 2013, 01:38
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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]
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08 Sep 2013, 05:55



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Re: In the figure above, O is the center of the circle. If OP [#permalink]
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09 Sep 2013, 05:20
rc197906 wrote: Attachment: Untitled.png 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]
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09 Sep 2013, 06:19
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]
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Re: In the figure above, O is the center of the circle. If OP [#permalink]
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10 Sep 2013, 05:24
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]
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10 Sep 2013, 05:31
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]
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28 Nov 2013, 17:56
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]
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29 Nov 2013, 02:20
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 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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Re: In the figure above, O is the center of the circle. If OP [#permalink]
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04 Dec 2013, 10:52
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.



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