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In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ? (1) The coordinates of point P are (6,8) (2) The coordinates of point Q are (13,0) Attachment:
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Last edited by Bunuel on 11 Sep 2017, 22:33, edited 3 times in total.
Edited the question and added the diagram.



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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient. (2) The coordinates of point Q are (13,0) > we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient. Answer: A. Hope it's clear. Attachment:
Triangle.png [ 15.88 KiB  Viewed 17920 times ]
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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thanks Bunuel. stmt 1 shows us that the smaller triangle is 1/2*6*8 and the other triangle must be larger than the smaller one. adding the two triangles together you get a number larger than 48.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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09 Jul 2013, 15:37
Bunuel wrote: In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient. (2) The coordinates of point Q are (13,0) > we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient. Answer: A. Hope it's clear. Hi Bunuel, Can thehre be a case when OS= 5.5, SQ=6.5? Then OQ= 12 ....
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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09 Jul 2013, 15:47
Mountain14 wrote: Bunuel wrote: In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient. (2) The coordinates of point Q are (13,0) > we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient. Answer: A. Hope it's clear. Hi Bunuel, Can thehre be a case when OS= 5.5, SQ=6.5? Then OQ= 12 .... No, that's not possible. We know that the coordinates of point P are ( 6,8). PS is altitude, thus the coordinates of point S are ( 6,0), so OS=6. Hope it' clear.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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13 Jul 2013, 01:05
Hi,
Could you explain the following :
Since OP<PQ then OS<SQ and the base OQ is more than 12
Thanks.



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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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12 Oct 2013, 09:03
Hi Bunnel,
From 1 and 2
\(\frac{1}{2}*OS*PS\) =\(\frac{1}{2}*6*8\) = 24 \(\frac{1}{2}*SQ*PS\) =\(\frac{1}{2}*7*8\) = 28
1 + 2 = 52
Whats wrong here?



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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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12 Oct 2013, 10:12



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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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12 Oct 2013, 19:44
Bunuel wrote: GMAT40 wrote: Hi Bunnel,
From 1 and 2
\(\frac{1}{2}*OS*PS\) =\(\frac{1}{2}*6*8\) = 24 \(\frac{1}{2}*SQ*PS\) =\(\frac{1}{2}*7*8\) = 28
1 + 2 = 52
Whats wrong here? Nothing wrong but what are you trying to say? Hi, The answer is A but dont we need both the statements to conclude this



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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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21 Oct 2013, 12:27
Bunuel wrote: Attachment: Triangle.png In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient. (2) The coordinates of point Q are (13,0) > we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient. Answer: A. Hope it's clear. I didn't understand how are we assuming that OS is equal 6? and how are we concluding t either that OS=QS?Can you pls explain?



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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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21 Oct 2013, 21:36
Punyata wrote: Bunuel wrote: Attachment: Triangle.png In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient. (2) The coordinates of point Q are (13,0) > we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient. Answer: A. Hope it's clear. I didn't understand how are we assuming that OS is equal 6? and how are we concluding t either that OS=QS?Can you pls explain? Please read the thread: intherectangularcoordinatesystemaboveifoppqis129092.html#p1244697The coordinates of point P, which is just above S, are ( 6,8), thus the coordinates of point S are ( 6, 0).
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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20 Nov 2013, 22:40
Hi Bunuel, Option (2) Isn't it possible to get the answer, if OPQ>48, from... 1) OP<PQ 2) a^2+b^2=13^2 .... (a=OP, b=PQ) I think there is no option of (a,b) to make ab>96 when OP<PQ because a^2+b^2=13^2 and ab=96 can not have any common answers. a^2 +(96/a)^2 = 169 ...........ab=96 a^4169*a^2+8716=0 A^2169A+9216=0................A=a^2 169^2 4*9216 = 6303............b^24ac <0 (ax^2+bx+c=0) Thanks. Bunuel wrote: Attachment: Triangle.png In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient. (2) The coordinates of point Q are (13,0) > we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient. Answer: A. Hope it's clear.



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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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21 Nov 2013, 01:59
ohora wrote: Hi Bunuel, Option (2) Isn't it possible to get the answer, if OPQ>48, from... 1) OP<PQ 2) a^2+b^2=13^2 .... (a=OP, b=PQ)I think there is no option of (a,b) to make ab>96 when OP<PQ because a^2+b^2=13^2 and ab=96 can not have any common answers. a^2 +(96/a)^2 = 169 ...........ab=96 a^4169*a^2+8716=0 A^2169A+9216=0................A=a^2 169^2 4*9216 = 6303............b^24ac <0 (ax^2+bx+c=0) Thanks. Bunuel wrote: Attachment: Triangle.png In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient. (2) The coordinates of point Q are (13,0) > we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient. Answer: A. Hope it's clear. This is not correct because we are not given that OPQ is a right triangle, thus you cannot write OP^2 + PQ^2 = OQ^2. Hope it's clear.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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21 Nov 2013, 23:29
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eybrj2 wrote: Attachment: Triangle.png In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ? (1) The coordinates of point P are (6,8) (2) The coordinates of point Q are (13,0) From St 1, we have the coordinates of P (6,8) ie. Pt S (6,0) and Pt O (0,0) Now let's take SQ=x so basically the questions becomes is 1/2 (6+x)*8>48 or (6+x)>12 or x>6 We know OP =10 so PQ>10.....If now we take PQ= 10.5 and apply Pythagorus (10.5)^2= 8^2+x^2 > x^2= 110.2564 = 46.25 ie x =6.8 which is more than 6 and therefore sufficient Option B,C AND E ruled out. From St 2, we know only OQ =13 but don't know the height so not sufficient. D ruled out so Ans is A
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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24 Sep 2014, 09:50
Although I understand the discussed solution I come to to a different solution due to a different interpretation of the question. Maybe some of you have a similar opinion... In my opinion, the question at no point requires point Q to lie on the xaxis. The question only requires OP<PQ. Although the graph and question may be interpreted in a way that leads you to think that Q lies on the xaxis they certainly don't state this requirement explicitly. If OP has a length of 10 and PQ has a length of 10+ but Q is located not on the xaxis but rather close to the origin the area of OPQ may be less than 48 despite having P located at (6,8) I attached a graph of my example. I know that the solutions manual offers the same solution as the forum agrees upon here. Is the solution manual wrong? Please point out any mistakes of mine.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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24 Jun 2015, 17:06
Bunuel wrote: Kriti2013 wrote: Hi,
Could you explain the following :
Since OP<PQ then OS<SQ and the base OQ is more than 12
Thanks. Since OS=6 and OS<SQ, then SQ>6, thus OQ=OS+SQ=6+(more than 6)=(more than 12). Hope it's clear. Hi Bunuel, This is my first post. Can you please explain how we got OS<SQ? When I solved, I took the smallest value of Q possible i.e. Q (7,0) and got Area= 1/2 *7*6 = 21 so concluded that both (i) and (ii) are required.



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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]
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25 Jun 2015, 03:20
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robinpallickal wrote: Bunuel wrote: Kriti2013 wrote: Hi,
Could you explain the following :
Since OP<PQ then OS<SQ and the base OQ is more than 12
Thanks. Since OS=6 and OS<SQ, then SQ>6, thus OQ=OS+SQ=6+(more than 6)=(more than 12). Hope it's clear. Hi Bunuel, This is my first post. Can you please explain how we got OS<SQ? When I solved, I took the smallest value of Q possible i.e. Q (7,0) and got Area= 1/2 *7*6 = 21 so concluded that both (i) and (ii) are required. Hi robinpallickal You are probably overlooking the condition given in the question that OP < PQ If \(OP < PQ\) > \(\sqrt{OS^2 + PS^2} < \sqrt{SQ^2 + PS^2}\) > \(OS < SQ\) (as OS and SQ are both positive) Therefore, you can't take the xcoordinate of Q as anything less than 12. Hope that clears your doubt.




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