Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Diagram - Semicircle with point O as center. Y axis is [#permalink]
03 May 2006, 18:06

Diagram - Semicircle with point O as center. Y axis is bisecting the semicircle and x axis is the diameter of the circle (passes though point O). Central angle of 90 degrees yields points P (-srt3, 1) and Q (s, t) on the circle.

What is the value of s?

I guessed +srt3 because the diagram looked symetrical about the y axis.

Re: GMATPrep - geometry [#permalink]
03 May 2006, 20:42

If I'm not mistaken, if P and Q are 'on' the circle formed by the 90 degrees, then P(-srt3, 1), isn't 1 the radius? Thus, Q should have an x value of 1 as well _________________

Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...

Sorry my bad. I had a different 90 degree angle in mind. The answer is right.

The points formed may not be symmetrical. The triangles formed are identical triangles and these are 30:60:90 triangles so the sides are 1:3^(1/2):2 with the radius being 2

Please execuse my drawing, my drawing is pretty elementary but you get the point

Attachments

Circle.jpg [ 7.46 KiB | Viewed 1075 times ]

_________________

Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...

Actually, I took a different approach. Please let me know if there is any error in my method.

The distance from (sqrt3,1) to the origin would also be the radius of the circle.
Therefore,
length of line OP = srt((-srt(3)-0}^2 + (1-0)^2)]
= srt(3+1) = srt4
= 2

Actually, I took a different approach. Please let me know if there is any error in my method.

The distance from (sqrt3,1) to the origin would also be the radius of the circle. Therefore, length of line OP = srt((-srt(3)-0}^2 + (1-0)^2)] = srt(3+1) = srt4 = 2

Yes, that's how I knew it was a 30:60:90 triangle _________________

Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...

Apply the Pythagorean rules and equate the sum of the two sides to the hypotenuse. You would get s = +/-1. Since s is above the x-axis you should have s = 1. _________________