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In the figure, point P and Q lie on the circle with center [#permalink]
 03 Sep 2007, 09:07
In the figure, point P and Q lie on the circle with center O. What is the value of S?
A) 1/2
B)1
C) sqrt 2
D) sqrt 3
e) sqrt2 / 2
If you draw a line perpendicular from Point P and Q to x-axis, you can see that the two triangle are congruent. My answer is D. The OA is B. Why is that?
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Note: GMAT diagrams are not drawn to scale. You cannot assume you have congruent triangles.
Recognize that this is a 45-45-90 triangle with sides of length 2 and hypotenuse 2sqrt(2).
So QO = s^2 + t^2 = 4
PQ = sqrt(s+sqrt3)^2 + (t-1)^2 = 2sqrt2
2s(sqrt3) = 2t
s(sqrt3) = t
So QO = s^2 + 3s^2 = 4
s^2 = 1
s = +/- 1. (-1 is invalid as Q lies on the positive x plane).
So s = 1
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How did you figure out that
Recognize that this is a 45-45-90 triangle with sides of length 2 and hypotenuse 2sqrt(2).
So QO = s^2 + t^2 = 4
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thanks ywilfred!!
i see what you mean, i believe both triangles are congruent triangles, with 30, 60 and 90. but the triangle from the left has its 30C angle attached to the centre O and the triangle on the left has its 60C angle attached to centre O.
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hey, plz answer beatgmat's Q.
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From the left we have a triangle: 1-sqrt3-2, so radius = 2.
Triangle in the middle is a right triangle with the legs = radius = 2
From the right we have a triangle 1-sqrt3-2, but the leg sqrt3 is situating opposite 60 degree angle and is a vertical line. So S coordinate of this point = 1
Last edited by Whatever on 03 Sep 2007, 21:57, edited 4 times in total.
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somethingbetter wrote: hey, plz answer beatgmat's Q.
PO=OQ=radius
A triangle that has two sides equal has two adjacent angles equal, since PO=OQ; angle QPR=anglePQO.
The angle between PO and QO is 90, leaving 90 {180 -90} degrees for the other two EQUAL angles.
Thus 90/2 is 45 degrees each angle
45:45:90
sum of angles of any tiangle is 180.
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IrinaOK wrote: somethingbetter wrote: hey, plz answer beatgmat's Q. PO=OQ=radius A triangle that has two sides equal has two adjacent angles equal, since PO=OQ; angle QPR=anglePQO. The angle between PO and QO is 90, leaving 90 {180 -90} degrees for the other two EQUAL angles. Thus 90/2 is 45 degrees each angle 45:45:90 sum of angles of any tiangle is 180. Don't be tricked by the figure:
"Thus 90/2 is 45 degrees each angle" this is incorrect.
It takes be awhile to see this, if you re-draw the graph yourself and have the up-side-down triangle slightly tited to your left.
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mbunny wrote: IrinaOK wrote: somethingbetter wrote: hey, plz answer beatgmat's Q. PO=OQ=radius A triangle that has two sides equal has two adjacent angles equal, since PO=OQ; angle QPR=anglePQO. The angle between PO and QO is 90, leaving 90 {180 -90} degrees for the other two EQUAL angles. Thus 90/2 is 45 degrees each angle 45:45:90 sum of angles of any tiangle is 180. Don't be tricked by the figure: "Thus 90/2 is 45 degrees each angle" this is incorrect. It takes be awhile to see this, if you re-draw the graph yourself and have the up-side-down triangle slightly tited to your left. Yeah, of course,
The the upper part of coordinate system is devided into 30,90 and 60 degrees. and the triagles are not symmetric.
I was saying that the triangle that PQO {that is the one we get when we connect P and Q} is 90:45:45
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IrinaOK wrote: mbunny wrote: IrinaOK wrote: somethingbetter wrote: hey, plz answer beatgmat's Q. PO=OQ=radius A triangle that has two sides equal has two adjacent angles equal, since PO=OQ; angle QPR=anglePQO. The angle between PO and QO is 90, leaving 90 {180 -90} degrees for the other two EQUAL angles. Thus 90/2 is 45 degrees each angle 45:45:90 sum of angles of any tiangle is 180. Don't be tricked by the figure: "Thus 90/2 is 45 degrees each angle" this is incorrect. It takes be awhile to see this, if you re-draw the graph yourself and have the up-side-down triangle slightly tited to your left. Yeah, of course, The the upper part of coordinate system is devided into 30,90 and 60 degrees. and the triagles are not symmetric. I was saying that the triangle that PQO {that is the one we get when we connect P and Q} is 90:45:45I agree that that triangle is a 45:45:90.
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beatgmat wrote: How did you figure out that
Recognize that this is a 45-45-90 triangle with sides of length 2 and hypotenuse 2sqrt(2).
So QO = s^2 + t^2 = 4
Here you go...
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Ok, I got to the point where now I can see there are two 45 45 90 triangles, how did you get to the 45 45 length of root (3) and 1?
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