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

Question

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

rakesh1239 · Answer

A? We need to find OS which is equal to OP as both are radius. from 1 we can find the OS

rc197906 · Answer

the answer is D actually...anyone have any ideas on why D?

stoolfi · Answer

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

rakesh1239 · Answer

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

fozzzy · Answer

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!

kusena · Answer

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=  3 and PR=r/2 hence radius=2* 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* 3
Hence stmt 2 alone is sufficient

Ans is D

Bunuel · Answer

Merging similar topics.

WoundedTiger · Answer

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

fozzzy · Answer

Is there any property tested here? I'm really clueless when it comes to geometry...

Bunuel · Answer

https://gmatclub.com/forum/download/file.php?id=21004 
 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.

fozzzy · Answer

So for such questions never rely on the figure. In this particular question, the figure is misleading.

Bunuel · Answer

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.

jlgdr · Answer

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

Bunuel · Answer

https://gmatclub.com/forum/download/file.php?id=21004 
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.

WholeLottaLove · Answer

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.

CEdward · Answer

Totally makes sense. The only issue is that I have taught myself to never assume certain things on DS. How can we be sure that PR is half the radius? What if OS bisects PO at some point other than half? Similarly, how can we be sure that PO bisects OS at exactly the half way point?

I am struggling here with understanding when we should and should not assume something.

I am struggling here with understanding when we should and should not assume something.

Bunuel · Answer

We are told that OP and QS are perpendicular and  bisect  each other. In geometry  bisect  means divide into two exactly equal parts.

CEdward · Answer

Bunuel , this is beyond helpful. Thank you. As a follow-up question, suppose a line seemingly 'bissects' one angle of a triangle in half. Can I assume that the two resulting angles are equal (particularly if there is no mention of 'bisect') or would that be a mistake?

Bunuel · Answer

No, you cannot assume that a line is the bisector if you are not told that.

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

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