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Diagram - Semicircle with point O as center. Y axis is

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Diagram - Semicircle with point O as center. Y axis is [#permalink] New post 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.

The correct answer that was given is 1.
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 [#permalink] New post 03 May 2006, 18:54
Can you please upload the diagram here?
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Re: GMATPrep - geometry [#permalink] New post 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
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 [#permalink] New post 04 May 2006, 09:21
Attaching the diagram. Ignore my poor drawing skills. :wink:

I don't see why when P (-sqrt3, 1) the x value of Q (or s) should be 1.
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 [#permalink] New post 04 May 2006, 14:01
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 :cry:
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 [#permalink] New post 04 May 2006, 15:34
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
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 [#permalink] New post 04 May 2006, 15:35
pesquadero wrote:
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 :wink:
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 [#permalink] New post 04 May 2006, 21:20
I still don't get it. How do you know it is a 30-60-90. Can you please send the complete explanation
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 [#permalink] New post 04 May 2006, 22:03
Loner wrote:
I still don't get it. How do you know it is a 30-60-90. Can you please send the complete explanation


a 30-60-90 triangle have sides of x, 2x, and x*3^(1/2), in this case, 1, 2, and 3^(1/2), it is just a rule that you have to remember.

And a 45-45-90 triangle have sides of x, x, x*2^(1/2)
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 [#permalink] New post 04 May 2006, 23:03
Gosh! I was missing a minor point. Thanks a ton buddy!
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 [#permalink] New post 06 May 2006, 11:05
Attaching the diagram. Ignore my poor drawing skills. :wink:

I don't see why when P (-sqrt3, 1) the x value of Q (or s) should be 1.
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 [#permalink] New post 06 May 2006, 11:51
Where is this problem from?
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 [#permalink] New post 06 May 2006, 18:12
You can solve it without knowing about the nature of the ▲.

__2
OP = 3 + 1 = 4
__2
OQ = s**2 + t**2
__2
PQ = (s+√3)**2 + (t-1) **2

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.
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  [#permalink] 06 May 2006, 18:12
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Diagram - Semicircle with point O as center. Y axis is

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