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Manager
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In the figure, point P and Q lie on the circle with center [#permalink]
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03 Sep 2007, 09:07
This topic is locked. If you want to discuss this question please repost it in the respective forum. 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 xaxis, 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 454590 triangle with sides of length 2 and hypotenuse 2sqrt(2).
So QO = s^2 + t^2 = 4
PQ = sqrt(s+sqrt3)^2 + (t1)^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



Manager
Joined: 29 Aug 2007
Posts: 110

How did you figure out that
Recognize that this is a 454590 triangle with sides of length 2 and hypotenuse 2sqrt(2).
So QO = s^2 + t^2 = 4



Manager
Joined: 11 Jul 2006
Posts: 68

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.



Manager
Joined: 06 Jul 2007
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hey, plz answer beatgmat's Q.



Manager
Joined: 03 Sep 2006
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From the left we have a triangle: 1sqrt32, so radius = 2.
Triangle in the middle is a right triangle with the legs = radius = 2
From the right we have a triangle 1sqrt32, 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.



Director
Joined: 22 Aug 2007
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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.



Manager
Joined: 11 Jul 2006
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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 redraw the graph yourself and have the upsidedown triangle slightly tited to your left.



Director
Joined: 22 Aug 2007
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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 redraw the graph yourself and have the upsidedown 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



Manager
Joined: 11 Jul 2006
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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 redraw the graph yourself and have the upsidedown 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
I agree that that triangle is a 45:45:90.



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

beatgmat wrote: How did you figure out that
Recognize that this is a 454590 triangle with sides of length 2 and hypotenuse 2sqrt(2).
So QO = s^2 + t^2 = 4
Here you go...
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Director
Joined: 10 Feb 2006
Posts: 657

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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