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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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14 Mar 2012, 14:21
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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)
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Mar 2012, 15:08, edited 2 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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14 Mar 2012, 15:30
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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.

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

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30 Mar 2012, 17:51
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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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28 May 2013, 04:29
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Bumping for review and further discussion.
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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.

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.

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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09 Jul 2013, 15:51
Bunuel wrote:
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.

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.

Yes Got it!... Thanks
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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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13 Jul 2013, 01:54
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.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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30 Sep 2013, 05:28
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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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
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?
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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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13 Oct 2013, 02:41
GMAT40 wrote:
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

To conclude WHAT? The exact area? If so, then yes.
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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.

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.

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?

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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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^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.

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
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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 ?

(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.

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 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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15 Apr 2014, 11:32
Hi Bunuel,

In the above diagram,what if we consider the triangle to be inscribed in a rectangle and prove angle p to be a right angle, in that case, statement 1 and 2 might be sufficient to answer the question if the area of the triangle is greater than 48 right??
Re: In the rectangular coordinate system above, if OP < PQ, is   [#permalink] 15 Apr 2014, 11:32

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