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# In the rectangular coordinate system above, if OQ >= PQ, is

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In the rectangular coordinate system above, if OQ >= PQ, is  [#permalink]

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Updated on: 27 Sep 2013, 09:44
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In the rectangular coordinate system above, if OQ >= PQ, is the area of region OPQ less than 30?

(1) The coordinates of point P are (6,0).
(2) The coordinates of point Q are (3,8)

Originally posted by PrateekDua on 27 Sep 2013, 09:15.
Last edited by Bunuel on 27 Sep 2013, 09:44, edited 1 time in total.
Renamed the topic and edited the question.
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In the rectangular coordinate system above, if OQ >= PQ, is  [#permalink]

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27 Sep 2013, 09:57
2

In the rectangular coordinate system above, if OQ >= PQ, is the area of region OPQ less than 30?

(1) The coordinates of point P are (6,0). We know the length of the base (OP=6) but know nothing about the height (QS), which may be 1 or 100, so the area may or may not be less than 30. Not sufficient.

(2) The coordinates of point Q are (3,8).

Now, if OQ were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SP and the area would be: $$\frac{1}{2}*base*height=\frac{1}{2}(OS+SP)*QS=\frac{1}{2}*(3+3)*8=24$$. Since $$OQ\geq{PQ}$$ then $$OS\geq{SP}$$ and the base OP is less than or equal to 6, which makes the area less than or equal 24. Sufficient.

Similar question to practice: in-the-rectangular-coordinate-system-above-if-op-pq-is-129092.html

Hope it's clear.

Attachment:

Triangle.png [ 19.83 KiB | Viewed 3042 times ]

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Re: In the rectangular coordinate system above, if OQ >= PQ, is  [#permalink]

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27 Sep 2013, 10:07
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PrateekDua wrote:
Attachment:
Untitled.jpg
In the rectangular coordinate system above, if OQ >= PQ, is the area of region OPQ less than 30?

(1) The coordinates of point P are (6,0)
(2) The coordinates of point Q are (3,8)

I'm happy to help.

Statement #1: The coordinates of point P are (6,0)
The base = 6, but the height could be anything, so the area could be anything. This statement, alone and by itself, is not sufficient.

Statement #2: The coordinates of point Q are (3,8)
This is tricky. Segment OQ has some length --- we could find this length from the Pythagorean theorem (3^2 + 8^2, then take a square root), but the exact number is not important. The requirement OQ >= PQ allows for two cases --- let's look at them:
Case #1: Suppose OQ = PQ --- then this would be an isosceles triangle, and when PQ came down 8 vertical units from point Q, it would have to go over three horizontal units to (6, 0) --- base = 6, height = 8, area = 12
Case #2: Suppose OQ > PQ --- then PQ would have to be steeper, and come down the 8 vertical units while going fewer than 3 horizontal units to the right. This would make the base less than 6, so the area would be less than 12.
Either way, we have a definitive answer to the prompt. This statement, alone and by itself, is sufficient.

Does this make sense?
Mike
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Re: In the rectangular coordinate system above, if OQ >= PQ, is  [#permalink]

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27 Sep 2013, 23:47
Thanks Bunuel and Mike, the explanation you provided was way easier than the one I read.

Kudos to both of you!
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Re: In the rectangular coordinate system above, if OQ >= PQ, is  [#permalink]

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19 Jun 2015, 18:43
I guess , the coordinates of point O is (0,0) should be given in the stem, unless otherwise could be stated
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Re: In the rectangular coordinate system above, if OQ >= PQ, is  [#permalink]

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20 Jun 2015, 06:48
mikemcgarry wrote:
PrateekDua wrote:
Attachment:
Untitled.jpg
In the rectangular coordinate system above, if OQ >= PQ, is the area of region OPQ less than 30?

(1) The coordinates of point P are (6,0)
(2) The coordinates of point Q are (3,8)

I'm happy to help.

Statement #1: The coordinates of point P are (6,0)
The base = 6, but the height could be anything, so the area could be anything. This statement, alone and by itself, is not sufficient.

Statement #2: The coordinates of point Q are (3,8)
This is tricky. Segment OQ has some length --- we could find this length from the Pythagorean theorem (3^2 + 8^2, then take a square root), but the exact number is not important. The requirement OQ >= PQ allows for two cases --- let's look at them:
Case #1: Suppose OQ = PQ --- then this would be an isosceles triangle, and when PQ came down 8 vertical units from point Q, it would have to go over three horizontal units to (6, 0) --- base = 6, height = 8, area = 12
Case #2: Suppose OQ > PQ --- then PQ would have to be steeper, and come down the 8 vertical units while going fewer than 3 horizontal units to the right. This would make the base less than 6, so the area would be less than 12.
Either way, we have a definitive answer to the prompt. This statement, alone and by itself, is sufficient.

Does this make sense?
Mike

Hi Mike ,

I could not understand ur explanation for statement B.
Are we assuming here points O and P are fixed , and we moving point Q in order to consider scenario of OQ>PQ ??
and how can be assume point P to be fixed ?? O is fixed since it is origin
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In the rectangular coordinate system above, if OQ >= PQ, is  [#permalink]

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20 Jun 2015, 08:21
mikemcgarry wrote:
PrateekDua wrote:
Attachment:
Untitled.jpg
In the rectangular coordinate system above, if OQ >= PQ, is the area of region OPQ less than 30?

(1) The coordinates of point P are (6,0)
(2) The coordinates of point Q are (3,8)

I'm happy to help.

Statement #1: The coordinates of point P are (6,0)
The base = 6, but the height could be anything, so the area could be anything. This statement, alone and by itself, is not sufficient.

Statement #2: The coordinates of point Q are (3,8)
This is tricky. Segment OQ has some length --- we could find this length from the Pythagorean theorem (3^2 + 8^2, then take a square root), but the exact number is not important. The requirement OQ >= PQ allows for two cases --- let's look at them:
Case #1: Suppose OQ = PQ --- then this would be an isosceles triangle, and when PQ came down 8 vertical units from point Q, it would have to go over three horizontal units to (6, 0) --- base = 6, height = 8, area = 12
Case #2: Suppose OQ > PQ --- then PQ would have to be steeper, and come down the 8 vertical units while going fewer than 3 horizontal units to the right. This would make the base less than 6, so the area would be less than 12.
Either way, we have a definitive answer to the prompt. This statement, alone and by itself, is sufficient.

Does this make sense?
Mike

Hi Mike ,

I could not understand ur explanation for statement B.
Are we assuming here points O and P are fixed , and we moving point Q in order to consider scenario of OQ>PQ ??
and how can be assume point P to be fixed ?? O is fixed since it is origin

Statement 2: The coordinates of point Q are (3,8)

i.e. X co-ordinate of Q is 3 and Y-co-ordinate is 8 i.e. OQ= $$\sqrt{8^2 + 3^2} = \sqrt{73}$$

BUT Since, OQ >= PQ i.e. PQ <$$\sqrt{73}$$ But in Triangle PQS since QS=8 therefore PS<3

i.e. OP = OS + PS = 3 + (<3) i.e. OP <6

i.e. Area of Triangle OPQ = (1/2)*OP*QS = (1/2)*(<6)*(8)

i.e. Area of Triangle OPQ < 24 Which answers the Questions with Certainty

Hence, SUFFICIENT
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Re: In the rectangular coordinate system above, if OQ >= PQ, is  [#permalink]

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22 Jun 2015, 10:18
Thanks Mike and Bunuel
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Re: In the rectangular coordinate system above, if OQ >= PQ, is  [#permalink]

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18 Aug 2018, 22:42
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Re: In the rectangular coordinate system above, if OQ >= PQ, is &nbs [#permalink] 18 Aug 2018, 22:42
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