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

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

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.

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.

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.

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.

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)



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

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,
Is there any other way(as fast as the given solution) to solve this without drawing the perpendicular from P to X-axis?

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).

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.

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.

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.

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

Hi Bunuel,

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