Hi there. I'm happy to help with this.

Prompt: P, Q and S belong to the circumference with center O, R is the intersection of the OS and PQ, and OS is perpendicular to PQ. If OR = 0.6, what is the value of PQ?

What you need to know about this is a Geometry theorem that says:
a) if a radius is perpendicular to a chord, then it bisects the chord
b) if a radius bisects a chord, then it is perpendicular to the chord.

See the diagram I attached. Given that OS is perpendicular to PQ, we know that PR must equal RQ. If we found one, we could find the other. PQ = 2*PR = 2*RQ

Also, notice that OP = OS = OQ = radius. In right triangle ORP, we already know OR = 0.6. If we knew the radius, we would know OP. Then, we would have two sides of a right triangle, and we could use the Pythagorean Theorem to find the third side, PR. Once we knew PR, we could double it to find PQ. Thus, anything that allows us to figure out the radius allows us to answer the question.

Statement #1: RS = 0.4

Well, OS = OR + RS = 0.6 + 0.4 = 1, so we know the radius, so we could figure out PQ. Statement #1, by itself, is sufficient.

Statement #2: The circumference is 2Ï€ long

From the circumference, you can calculate the radius. From the radius, you can calculate PQ. Statement #2, by itself, is sufficient.

Answer = D

Does that make sense?

Here's another geometry DS question, for more practice.
https://gmat.magoosh.com/questions/1016
When you submit your answer to that question, the following page will have the explanation video.

Let me know if you have any further questions.

Mike

rahulvsd wrote:Hi Mike,
Do we actually need to apply the theorem you mentioned at the start of your explanation to this problem?
The question stem states that PQ is perpendicular to OS, therefore we can use Pythagoras theorem to find out RQ and PR independently from the value of the radius, and PR and RQ to get PQ.

Well, here's the thing. What the theorem says is pretty obvious, once you draw in radii OP and OQ ----> you get congruent triangles, and all the parts have to be congruent. It's just a handy time saver --- that way you know, when you calculate PR, you don't have to do any extra work to solve for RQ -- all you have to do is multiply by 2. Yes, you could do the entire problem with no knowledge of that theorem, but knowing the theorem saves you time, and time is by far the most precious commodity on the GMAT. Does that make sense?

Let me know if you have any further questions.

Mike