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In the rectangular coordinate system above, if OP < PQ, is

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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:
Triangle.png
Triangle.png [ 12.98 KiB | Viewed 21780 times ]

Originally posted by eybrj2 on 14 Mar 2012, 15:21.
Last edited by Bunuel on 11 Sep 2017, 23:33, 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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New post 14 Mar 2012, 16:30
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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.

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

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New post 30 Mar 2012, 18:51
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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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New post 09 Jul 2013, 16:37
Bunuel wrote:
Image
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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New post 09 Jul 2013, 16:47
Mountain14 wrote:
Bunuel wrote:
Image
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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New post 13 Jul 2013, 02:05
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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New post 13 Jul 2013, 02:54
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Re: In the rectangular coordinate system above, if OP < PQ, is  [#permalink]

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New post 30 Sep 2013, 06:28
Hi Bunuel,
Is there any other way(as fast as the given solution) to solve this without drawing the perpendicular from P to X-axis?
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Re: In the rectangular coordinate system above, if OP < PQ, is  [#permalink]

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New post 21 Oct 2013, 13:27
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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New post 21 Oct 2013, 22:36
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: in-the-rectangular-coordinate-system-above-if-op-pq-is-129092.html#p1244697

The 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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New post 20 Nov 2013, 23:40
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^4-169*a^2+8716=0
A^2-169A+9216=0................A=a^2
169^2 -4*9216 = -6303............b^2-4ac <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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New post 21 Nov 2013, 02:59
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^4-169*a^2+8716=0
A^2-169A+9216=0................A=a^2
169^2 -4*9216 = -6303............b^2-4ac <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 ?
Image

(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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New post 22 Nov 2013, 00:29
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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 co-ordinates 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= 110.25-64 = 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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New post 24 Sep 2014, 10:50
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 x-axis. 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 x-axis 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 x-axis 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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Screen Shot 2014-09-24 at 12.45.28 PM.png
Screen Shot 2014-09-24 at 12.45.28 PM.png [ 46.79 KiB | Viewed 8191 times ]

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

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New post 24 Jun 2015, 18:06
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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New post 25 Jun 2015, 04:20
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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 x-coordinate of Q as anything less than 12.
Hope that clears your doubt.
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Re: In the rectangular coordinate system above, if OP < PQ, is  [#permalink]

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New post 17 Jul 2015, 06:46
So, we have OP<PQ, Area of OPQ = (PS*OQ) / 2 > 48 --> PS*OQ > 96 ?

1) P (6,8) means --> Heght PS=8, OS=6 then we can find OP^2=8^2+6^2=100, OP=10
Since PQ>OP --> SQ>6 (SQ^2=PQ^2 (let's say 10,1) - 64 means SQ>6) --> OQ>12 -> 8*OQ(>12) > 96 SUFFICIENT
2) We need the height to be able to calculate the area - INSUFFICIENT
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Re: In the rectangular coordinate system above, if OP < PQ, is  [#permalink]

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New post 05 Oct 2015, 03:57
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,

Could you explain why OS < SQ? is there a theory/concept behind it?
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Re: In the rectangular coordinate system above, if OP < PQ, is  [#permalink]

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New post 05 Oct 2015, 05:37
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harishbiyani8888 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,

Could you explain why OS < SQ? is there a theory/concept behind it?


Image

PS^2 + OS^2 = OP^2
PS^2 + SQ^2 = PQ^2

Since we are given that OP < PQ, then OS < SQ.

Hope it's clear.
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In the rectangular coordinate system above, if OP < PQ, is  [#permalink]

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New post 11 Sep 2017, 23:59
eybrj2 wrote:
Image
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:
Triangle.png


Let us say OP=PQ
1) P= 6,8
OP =10
PQ=10 we have assumed
=> Q = 12,0

A= 0.5 x 8 x 12 =48
But it is given OP < PQ => A<48 DEFINITE NO
Sufficient
2) Q= 13,0
Insufficient as it will depend upon height ie y coordinate of P
if y coordinate of p < 96/13 then only above false => if y coord of P > 96/13 above is true
we dont know y coord of P => INSUFFICIENT


ANSWER IS A
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In the rectangular coordinate system above, if OP < PQ, is   [#permalink] 11 Sep 2017, 23:59

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