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Re: Coordinate System and Geometry [#permalink]
19 Nov 2012, 22:30
Expert's post
are you sure about \(\sqrt{3}\) as the answer. As per my calculations i got \(\sqrt{2}\) as the answer. If I am correct, then I will surely post my answer. _________________
Re: Coordinate System and Geometry [#permalink]
19 Nov 2012, 22:47
Expert's post
Step 1: Measure the radius. That will come out to be 2. Step 2: Observe the diagram. Since both OP and OQ are the radii, their measure will be 2 only. Step 3: Draw a line parallel to x axis, from P to Q. Step 4: Angle OQP and angle OPQ= 45 In the diagram attached, the angles marked X are 45 and 45 degrees. Step 5: Draw a perpendicular from Q to x axis. The point at which it meets the X axis, mark it as A. Step 5: The triangle OPA is a 45:45: 90 degree tiangle. This triangle has the property that the hypotenuese:side 1:side 2 will be in the ration \(\sqrt{2}\):1:1. Step 6: Since side 1(OA) and side 2(AQ) are equal, therefore let them be x. These will be equal to \(2/\sqrt{2}\). Step 7: The length of side 1 will come out to be \(2/\sqrt{2}\) or simply \(\sqrt{2}\).
Please let me know what my mistake, if my solution is wrong.
Re: Coordinate System and Geometry [#permalink]
19 Nov 2012, 23:51
Marcab wrote:
Step 1: Measure the radius. That will come out to be 2. Step 2: Observe the diagram. Since both OP and OQ are the radii, their measure will be 2 only. Step 3: Draw a line parallel to x axis, from P to Q. Step 4: Angle OQP and angle OPQ= 45 In the diagram attached, the angles marked X are 45 and 45 degrees. Step 5: Draw a perpendicular from Q to x axis. The point at which it meets the X axis, mark it as A. Step 5: The triangle OPA is a 45:45: 90 degree tiangle. This triangle has the property that the hypotenuese:side 1:side 2 will be in the ration \(\sqrt{2}\):1:1. Step 6: Since side 1(OA) and side 2(AQ) are equal, therefore let them be x. These will be equal to \(2/\sqrt{2}\). Step 7: The length of side 1 will come out to be \(2/\sqrt{2}\) or simply \(\sqrt{2}\).
Please let me know what my mistake, if my solution is wrong.
your solution is wrong in Step 3. PQ is parallel to x axis is an assumption in the solution and has not been stated in question. Figures are not drawn to scale. _________________
Re: Plane Geometry, Semicircle from GMATPrep [#permalink]
10 Dec 2013, 05:34
Bunuel wrote:
zz0vlb wrote:
I need someone like Bunuel or walker to look at my statement and tell if this is true.
Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.
In the figure above, points P and Q lie on the circle with center O. What is the value of s?
Yes you are right.
Point Q in I quadrant \((1, \sqrt{3})\); Point P in II quadrant \((-\sqrt{3}, 1)\); Point T in III quadrant \((-1, -\sqrt{3})\) (OT perpendicular to OP); Point R in IV quadrant \((\sqrt{3},-1)\) (OR perpendicular to OQ).
Is it the case of reflection in Y axis ? I doubt that but still want to clarify..
Re: Plane Geometry, Semicircle from GMATPrep [#permalink]
10 Dec 2013, 05:40
Expert's post
ygdrasil24 wrote:
Bunuel wrote:
zz0vlb wrote:
I need someone like Bunuel or walker to look at my statement and tell if this is true.
Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.
In the figure above, points P and Q lie on the circle with center O. What is the value of s?
Yes you are right.
Point Q in I quadrant \((1, \sqrt{3})\); Point P in II quadrant \((-\sqrt{3}, 1)\); Point T in III quadrant \((-1, -\sqrt{3})\) (OT perpendicular to OP); Point R in IV quadrant \((\sqrt{3},-1)\) (OR perpendicular to OQ).
Is it the case of reflection in Y axis ? I doubt that but still want to clarify..
No. \((-\sqrt{3}, 1)\) is not a reflection of \((1, \sqrt{3})\) around the Y-axis. The reflection of \((1, \sqrt{3})\) around the Y-axis is \((-1, \sqrt{3})\). _________________
Re: In the figure above, points P and Q lie on the circle with [#permalink]
11 Dec 2013, 11:51
In the figure above, points P and Q lie on the circle with center O. What is the value of s?
This problem took be a bit to conceptualize but I got it as soon as I realized that the graph I was looking at wasn't drawn completely accurately.
If you are to accurately graph the point -√3,1 on a piece of graph paper, you see that the angle it forms with regards to the origin (and the x plane) is less than 45 degrees. Because of this, and the fact that both points lie on the circle, and that the two radii form a 90 degree angle, Q must be higher on the circle than P. Upon looking at this graph, I thought that the x and y coordinates of Q would be the same (except positive) which is, I am sure, the intent of the test makers. The answer is simply the positive reciprocal of point P.
Re: In the figure above, points P and Q lie on the circle with [#permalink]
12 Dec 2014, 22:19
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: In the figure above, points P and Q lie on the circle with [#permalink]
24 Jul 2015, 05:28
Bunuel wrote:
zz0vlb wrote:
I need someone like Bunuel or walker to look at my statement and tell if this is true.
Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.
In the figure above, points P and Q lie on the circle with center O. What is the value of s?
Yes you are right.
Point Q in I quadrant \((1, \sqrt{3})\); Point P in II quadrant \((-\sqrt{3}, 1)\); Point T in III quadrant \((-1, -\sqrt{3})\) (OT perpendicular to OP); Point R in IV quadrant \((\sqrt{3},-1)\) (OR perpendicular to OQ).
Hi Bunuel
Cant we do this by applying 45-45-90 getting distance between O and Q as 1/sqroot 3? As OP=OQ (radii)? but with this I get t/s=1/sqroot3. Pls advise
Re: In the figure above, points P and Q lie on the circle with [#permalink]
24 Jul 2015, 06:10
sinhap07 wrote:
Bunuel wrote:
zz0vlb wrote:
I need someone like Bunuel or walker to look at my statement and tell if this is true.
Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.
In the figure above, points P and Q lie on the circle with center O. What is the value of s?
Yes you are right.
Point Q in I quadrant \((1, \sqrt{3})\); Point P in II quadrant \((-\sqrt{3}, 1)\); Point T in III quadrant \((-1, -\sqrt{3})\) (OT perpendicular to OP); Point R in IV quadrant \((\sqrt{3},-1)\) (OR perpendicular to OQ).
Hi Bunuel
Cant we do this by applying 45-45-90 getting distance between O and Q as 1/sqroot 3? As OP=OQ (radii)? but with this I get t/s=1/sqroot3. Pls advise
Let me try to answer this.
\(\angle{POA} = 30\) by following steps:
In triangle POA, OA = \(\sqrt{3}\), AP = 1, thus this triangle is a 30-60-90 (as the sides are in the ratio \(1:\sqrt{3}:2\)) triangle (either remember this or use trigonometry to figure it out!).
Now, per your question, triangle OPQ is isosceles with 45-45-90 triangle and sides in the ratio \(1:1:\sqrt{2}\)
Thus, PQ should be = \(\sqrt{2}\) and not 1/sqroot 3
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