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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]

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03 May 2006, 19:06
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.

The correct answer that was given is 1.

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03 May 2006, 19:54

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03 May 2006, 21:42
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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04 May 2006, 10:21
Attaching the diagram. Ignore my poor drawing skills.

I don't see why when P (-sqrt3, 1) the x value of Q (or s) should be 1.
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GF8U2170.jpg [ 44.78 KiB | Viewed 1443 times ]

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04 May 2006, 15:01

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
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Circle.jpg [ 7.46 KiB | Viewed 1386 times ]

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04 May 2006, 16:34
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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04 May 2006, 16:35
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
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04 May 2006, 22:20
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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04 May 2006, 23:03
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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05 May 2006, 00:03
Gosh! I was missing a minor point. Thanks a ton buddy!

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06 May 2006, 12:05
Attaching the diagram. Ignore my poor drawing skills.

I don't see why when P (-sqrt3, 1) the x value of Q (or s) should be 1.
Attachments

GF8U2170.jpg [ 44.78 KiB | Viewed 1408 times ]

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06 May 2006, 12:51
Where is this problem from?

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06 May 2006, 19:12
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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Thanks,
Zooroopa

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06 May 2006, 19:12
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