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

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

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

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

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

Merging similar topics.

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

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

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.

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.

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.

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

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?

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