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Please Explain - Coordinate Geometry Problem

This question is from the GMATPrep software. I picked the "obvious" but apparently wrong answer (sqrt 3). The official answer is 1. Can someone please demystify this? Many thanks!

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draw 2 rectangles...
one rectangle has a line that joins p to X-axis and and the other line p to y-axis

you will get a rectangle with sides squareroot of 3 and 1..

now draw another rectangle joining Q to X-xis and Y-axis...
find the diagonal of the first rectangle.......sides squareroot of 3 and 1..
the diagonal will be 2...
2 is the radius of the circle ...
so the second figure should also have a 2 radius as diagonal...
Now find the angles made by the diagonals on each rectangle...
tan 30 degrees=1/root 3
so the angles on the diameter would be 30,60,30,60.....
now tan 60=root 3/1

so the height becomes root 3..BUT the x co-ordinate values remains 1..

p.s:even if O isn't the 0,0 the distance will still be root 3 and 1.....from the center of the circle.............
phew!

Please does anyone have small solution for this..I will really be happy to learn it..........

hope it helps!

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Originally Posted by ak7
This question is from the GMATPrep software. I picked the "obvious" but apparently wrong answer (sqrt 3). The official answer is 1. Can someone please demystify this? Many thanks!
I just want to confirm whether O means origin?

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I also assumed that 0 meant origin.

I drew the triangles you describe as well, but I didn't go as far as trig, since I don't remember it from high school and the geometry questions are "supposed" to be solvable without it. Here's what I don't get: if we draw these triangles, and we assume that the height of the rectangles are equal, then why aren't the lengths of the rectangles equal?

There is nothing to suggest that, if we draw two lines down from the two points to the X-axis, we would not get 2 perfectly symmetrical triangles, or rectangles.

If they are symmetrical, then shouldn't s equal sqrt(3)?
And if they are not symmetrical, and O is not the origin, then how do you get a solution at all?

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If O can be assumed as origin, then It becomes simple. The value of s is 1

use two equations
PQ^2 = PO^2 + OQ^2
PO^2 = OQ^2 ( because PO and OQ are each equal to radius)

Due to the difficult of writing the entire equations , I am unable to show you the solution , but ultimately, s equals 1

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This is a very tricky question. I missed it too, but my girlfriend is a chemistry major and she just explained it to me. (Not that it has anything to do with chemistry)

First draw one rectangle on the left side of the Y-axis.

Here is how:
One line should join Point P to X-axis and the other line should join P to y-axis.

Next, Draw the diagonal of this rectangle and you have a 30-60-90 triangle.

Which is in the ratio of 1:2:Root 3

In the 30, 60, 90 degree triangle:

The 30 degree angle faces toward: 1

The 60 degree angle faces towards: root 3

& the 90 degree angle faces towards: 2

Make sure you fill these angles in with the appropriate degrees or you might miss what I’m saying

Now GMAT Prep Tells us that their is a central angle that has 90 degrees (Thus the square drawn smack dab in the middle)

We also know that the angle below this central angle (on the left side of the Y axis in the bottom left corner) is a 30 degree angle. (You drew it in right?)

And we know that there are 180 degrees in a line. So 30 + 90 + X = 180. (Where X is the mirror image of our 30 degree angle on the right side of the Y axis)

X = 60

So the mirror image of our 30 degree angle is actually a 60 degree angle on the right side of the Y axis. (I know, I thought it was 30 too).

Be sure to fill in this 60 degree angle!!!

So now, you need to draw a new rectangle on the right side of the Y-axis.

Here is how:
One line should join Point Q to X-axis and the other line should join Q to y-axis.

Draw the diagonal of this rectangle and once again you have a 30-60-90 triangle. Which is in the ratio of 1:2:Root 3

Now remember what we talked about before:

In the 30, 60, 90 degree triangle:

The 30 degree angle faces toward: 1

The 60 degree angle faces towards: root 3

& the 90 degree angle faces towards: 2

Therefore, S = 1. (I know this problem is lame.)

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Thanks a lot Jaybird! It's been haunting me since I got it wrong (for some reason, I assumed the diagram was to scale and that 90 degree angle was composed of two 45-degree angles).

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awesome explaination jaybird - you rock!!!

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this is simple question. Use the following concepts.
1. y = mx is the equaltion of line passing through origin.
2. for perpendicular lines m1 X m2 = -1
3. and equation of a circle with center at origin is x ^ + y ^2 = r ^ 2.

Let me know is something is not clear.

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