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

Question

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.

gmatmba · Answer

Can you please upload the diagram here?

TeHCM · Answer

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

sideslip · Answer

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.

TeHCM · Answer

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:

Attachments

Circle.jpg [ 7.46 KiB | Viewed 2806 times ]

pesquadero · Answer

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

TeHCM · Answer

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

Loner · Answer

I still don't get it. How do you know it is a 30-60-90. Can you please send the complete explanation

TeHCM · Answer

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)

Loner · Answer

Gosh! I was missing a minor point. Thanks a ton buddy!

sideslip · Answer

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.

shevy · Answer

Where is this problem from?

Zooroopa · Answer

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.

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

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