In the rectangular coordinate system above, if OP

Question

https://gmatclub.com/forum/download/file.php?id=16829 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) Triangle.png

Bunuel · Accepted Answer

https://gmatclub.com/forum/download/file.php?id=16833 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 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. Triangle.png

calreg11 · Answer

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.

HarveyS · Answer

Hi Bunuel,

Can thehre be a case when OS= 5.5, SQ=6.5? Then OQ= 12 ....

Bunuel · Answer

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.

Kriti2013 · Answer

Hi,

Could you explain the following :

Since OP<PQ then OS<SQ and the base OQ is more than 12

Thanks.

Bunuel · Answer

Since OS=6 and OS<SQ, then SQ>6, thus OQ=OS+SQ=6+(more than 6)=(more than 12).

Hope it's clear.

bagdbmba · Answer

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?

AIMGMAT770 · Answer

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?

Bunuel · Answer

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).

ohora · Answer

Hi Bunuel, Option (2) Isn't it possible to get the answer, if OPQ>48, from... 1) OP96 when OP

Bunuel · Answer

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.

WoundedTiger · Answer

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

germanfellow · Answer

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.

robinpallickal · Answer

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.

PrepTap · Answer

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.

BrainLab · Answer

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

harishbiyani8888 · Answer

Hi Bunuel,

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

Bunuel · Answer

https://gmatclub.com/forum/download/file.php?id=16833

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.

sahilvijay · Answer

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

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

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

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