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Diagram - Semicircle with point O as center. Y axis is [#permalink]
03 May 2006, 18:06
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
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 1251 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. _________________