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# In the figure, point P and Q lie on the circle with center

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Manager
Joined: 11 Jul 2006
Posts: 68
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 re-post 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 x-axis, you can see that the two triangle are congruent. My answer is D. The OA is B. Why is that?
Attachments

Half Circle copy.jpg [ 15.24 KiB | Viewed 1275 times ]

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

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03 Sep 2007, 09:17
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
Manager
Joined: 29 Aug 2007
Posts: 110

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03 Sep 2007, 09:35
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
Manager
Joined: 11 Jul 2006
Posts: 68

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03 Sep 2007, 17:37
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
Posts: 57

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03 Sep 2007, 21:43
Manager
Joined: 03 Sep 2006
Posts: 233

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03 Sep 2007, 21:51
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.
Director
Joined: 22 Aug 2007
Posts: 566

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03 Sep 2007, 21:52
somethingbetter wrote:

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
Posts: 68

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04 Sep 2007, 17:48
IrinaOK wrote:
somethingbetter wrote:

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.
Director
Joined: 22 Aug 2007
Posts: 566

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04 Sep 2007, 17:53
mbunny wrote:
IrinaOK wrote:
somethingbetter wrote:

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
Manager
Joined: 11 Jul 2006
Posts: 68

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04 Sep 2007, 18:06
IrinaOK wrote:
mbunny wrote:
IrinaOK wrote:
somethingbetter wrote:

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

I agree that that triangle is a 45:45:90.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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04 Sep 2007, 20:27
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...
Attachments

tri.jpg [ 29.3 KiB | Viewed 1211 times ]

Director
Joined: 10 Feb 2006
Posts: 657

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18 Oct 2007, 20:20
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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18 Oct 2007, 20:20
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