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655-705 Level|   Coordinate Plane|   Geometry|                  
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In the rectangular coordinate system above, if OP < PQ, is the area of 07 Aug 2018, 06:45
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Bunuel

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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Hi Bunuel,

how do we know that OS<SQ? I mean how do we deduce that from OP<OQ
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In the rectangular coordinate system above, if OP < PQ, is the area of 30 Aug 2018, 12:41
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eybrj2

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)

Show Spoiler
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\(?\,\,\,:\,\,\,{S_\Delta }\,\,\mathop > \limits^? \,\,48\)

\(\left( 1 \right)\,\,{S_\Delta }\,\,\mathop > \limits^{{\text{see}}\,\,{\text{our}}\,\,{\text{figure !}}} \,\,{S_{\Delta OPQ\,dash}} = \frac{{\left( {6 + 6} \right) \cdot 8}}{2} = 48\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)

\(\left( 2 \right)\,\,{\text{Geometric}}\,\,{\text{Bifurcation}}\)

Please note that:

> The first figure (on the right) is "the same" as the one we used in our first statement (see our figure), in which we answered <YES>
> The second figure (on the right-below) shows that the area-focus may be as small as we which (but still positive, of course) , hence <NO>

This solution follows the notations and rationale taught in the GMATH method.
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30Ago18_7q.gif
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In the rectangular coordinate system above, if OP < PQ, is the area of 09 Dec 2018, 04:44
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Its such a good trap question !!!

How one can be aware of such trap questions ?
Is practice and seeing the pattern is the solution ?
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In the rectangular coordinate system above, if OP < PQ, is the area of 09 Dec 2018, 04:55
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ENEM
Bunuel

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.

Show Spoiler
Attachment:
Triangle.png

Hi Bunuel,

how do we know that OS<SQ? I mean how do we deduce that from OP<OQ
Show more


From the question we have OP < OQ .....(a)
also from Pythagoras theorem we have \(OP^2\) = \(PS^2\) +\(OS^2\) and \(OQ^2\) = \(PS^2\) +\(SQ^2\)
Putting these value in equation a we have
\(PS^2\) +\(OS^2\) < \(PS^2\) +\(SQ^2\)
=> OS<SQ

Hope its clear now
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In the rectangular coordinate system above, if OP < PQ, is the area of 22 Jul 2019, 18:45
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More simplified version of what Bunuel explained.

We MUST test OP = PQ because under this constraint we can see that the base equals 12 and thus the area =1/2 *12*8=48, but since OP is actually less than PQ, the base must be larger (can deduce through inequality theorem).
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In the rectangular coordinate system above, if OP < PQ, is the area of Updated on: 12 Jul 2020, 07:19
Originally posted by jabhatta2 on 12 Jul 2020, 07:16.
Last edited by jabhatta2 on 12 Jul 2020, 07:19, edited 1 time in total.
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Bunuel

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.

Show Spoiler
Attachment:
Triangle.png
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Hi Bunuel VeritasKarishma - can i make the following general take-away ?

In any isosceles triangle
- Right angle isosceles like 45-45-90 or
- Non right angle isosceles triangle like 20-20-140

The perpendicular from the vertex to the base line (the height) in both types of isosceles triangles divides the base into two equal lengths ?
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In the rectangular coordinate system above, if OP < PQ, is the area of 16 Jul 2020, 00:17
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Bunuel

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.

Show Spoiler
Attachment:
Triangle.png

Hi Bunuel VeritasKarishma - can i make the following general take-away ?

In any isosceles triangle
- Right angle isosceles like 45-45-90 or
- Non right angle isosceles triangle like 20-20-140

The perpendicular from the vertex to the base line (the height) in both types of isosceles triangles divides the base into two equal lengths ?
Show more

Yes, but only if the base is the non-equal side.
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In the rectangular coordinate system above, if OP < PQ, is the area of 01 Mar 2021, 16:49
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Bunuel

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.

Show Spoiler
Attachment:
Triangle.png
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This is clever, but I am wondering why my approach was wrong?

Given P(6,8), we can calculate the distance OP --> √(6-0)^2 + (8-0)^2 --> OP = 10.

Since PQ > OP, we can do some test values

PQ = 11 --> Area = 44
PQ = 13 --> Area = 104/2 = 52

Insufficient.

Just wondering why that's not an equally valid approach?
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In the rectangular coordinate system above, if OP < PQ, is the area of 02 Mar 2021, 00:09
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Bunuel

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.

Show Spoiler
Attachment:
Triangle.png

This is clever, but I am wondering why my approach was wrong?

Given P(6,8), we can calculate the distance OP --> √(6-0)^2 + (8-0)^2 --> OP = 10.

Since PQ > OP, we can do some test values

PQ = 11 --> Area = 44
PQ = 13 --> Area = 104/2 = 52

Insufficient.

Just wondering why that's not an equally valid approach?
Show more

Not sure how you calculated the areas but in both cases the area is different and greater than 48.
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In the rectangular coordinate system above, if OP < PQ, is the area of 02 Mar 2021, 00:15
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Bunuel
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Bunuel

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.

Show Spoiler
Attachment:
Triangle.png

This is clever, but I am wondering why my approach was wrong?

Given P(6,8), we can calculate the distance OP --> √(6-0)^2 + (8-0)^2 --> OP = 10.

Since PQ > OP, we can do some test values

PQ = 11 --> Area = 44
PQ = 13 --> Area = 104/2 = 52

Insufficient.

Just wondering why that's not an equally valid approach?

Not sure how you calculated the areas but in both cases the area is different and greater than 48.
Show more

So given a height of 8 ---> 8 x 11 /2 = 88 /2 = 44 <48 and 8 x 13 /2 = 104 / 2 = 52 > 48...two different solutions?
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In the rectangular coordinate system above, if OP < PQ, is the area of 02 Mar 2021, 00:40
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CEdward
Bunuel
CEdward

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 clever, but I am wondering why my approach was wrong?

Given P(6,8), we can calculate the distance OP --> √(6-0)^2 + (8-0)^2 --> OP = 10.

Since PQ > OP, we can do some test values

PQ = 11 --> Area = 44
PQ = 13 --> Area = 104/2 = 52

Insufficient.

Just wondering why that's not an equally valid approach?

Not sure how you calculated the areas but in both cases the area is different and greater than 48.

So given a height of 8 ---> 8 x 11 /2 = 88 /2 = 44 <48 and 8 x 13 /2 = 104 / 2 = 52 > 48...two different solutions?
Show more

The area of a triangle 1/2*height*base = 1/2*PS*OQ.

Height = PS = 8;
Base = OQ. (If PQ = 11, then \(OQ = 6 + \sqrt{57}\) and if PQ = 13, then \(OQ = 6 + \sqrt{105}\)).

You are calculating the area as 1/2*PS*PQ, which is wrong.
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In the rectangular coordinate system above, if OP < PQ, is the area of 24 Aug 2021, 03:38
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Isn't assuming that OQ is a base an improper assumption? brunel
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In the rectangular coordinate system above, if OP < PQ, is the area of 03 Mar 2022, 06:02
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Good Question +1

This question reminds me that GMAT is not only about math, Its MATH + LOGIC skills.

Let's try to analyze this question.

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)

Attachment:
gcpic.JPG
gcpic.JPG [ 19.48 KiB | Viewed 1788 times ]

If you drop a perpendicular from point P to the base OQ and meet at a point S and this line will divide the △ POQ into 2 triangles.

△1 would be a right-angled triangle with base OS and height PS.

By using the coordinates of point P (6,8 ), we can say that side OS = 6 and PS = 8.

Hence, Area of △1 = 1/2 * base * height = 1/2 * 6 * 8 = 24.

Now let us consider △2. It will also be a right-angled triangle with base SQ and height PS.

Since it is given in the question that the hypotenuse PQ > OP,

\(OS^2 + PS^2 = OP^2 \) in △1

\(SQ^2 + PS^2 = PQ^2 \) in △ 2

Comparing both equations, since PQ > OP, we can conclude that the base SQ > OS.

Since the height of 2 triangles, PS is the same. The base of △ 2 is greater than △ 1.

Area of △2 = 1/2 * base SQ * height PS

Area of △1 = 1/2 * base OS * height PS = 24.

Therefore, we conclude that the area of △2 is greater than △ 1 i.e 24.

So the combined area of 2 triangles should be greater than 24+24 i.e 48

This will answer the question stem. Hence, Statement 1 alone is sufficient.

(2) The coordinates of point Q are (13,0)

From the coordinates of Q, we can say that the base OQ=13 but we don't have any idea about the coordinates of P.
Coordinates of P will determine the height of the triangle. Since the height is unknown, we will not be able to figure out if the area of △ OPQ is greater than 48 or not.

Statement 2 alone is not sufficient.

Option A is the correct answer.

Thanks,
Clifin J Francis,
GMAT QUANT SME
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In the rectangular coordinate system above, if OP < PQ, is the area of 08 Jul 2024, 04:52
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won't this question be considered a part of geometry?
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In the rectangular coordinate system above, if OP < PQ, is the area of 18 Jul 2025, 18:27
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I arrived at the same answer. Even if it were to be on the x-axis, it can be (-1,0) and satisfy the conditions and have area<48. Believe the question is incorrect.

@bunnel
germanfellow
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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In the rectangular coordinate system above, if OP < PQ, is the area of 18 Jul 2025, 20:41
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In this question we can't distort the question as in the question it states the word "above" which means question wants us to refer to image as provided and not assume anything else.
Anyway, these geometry questions are not a part of focus edition so it's not wise to put time to practice them.
pmahaja2
I arrived at the same answer. Even if it were to be on the x-axis, it can be (-1,0) and satisfy the conditions and have area<48. Believe the question is incorrect.

@bunnel
germanfellow
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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In the rectangular coordinate system above, if OP < PQ, is the area of 10 Nov 2025, 02:09
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Hi, can such a question be tested in Focus edition?
Bunuel

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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In the rectangular coordinate system above, if OP < PQ, is the area of 10 Nov 2025, 02:44
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soumyab12
Hi, can such a question be tested in Focus edition?

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No. Especially in DS.
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