In the xy-plane, point O is located at the origin, point A has coordin

Question

In the xy-plane, point O is located at the origin, point A has coordinates (p,q), and point B has coordinates (r,0). If p, q, and r are all positive values and AO > AB, is the area of triangular region ABO less than 12 ?

(1) r = 7

(2) p = 4 and q = 3

(1) r = 7

(2) p = 4 and q = 3

saurabh87 · Answer

I am getting D.

St 2 is correct for obvious reasons.
St 1 Given r=7
Also, AO>AB

^1/2> ^1/2
P^2+Q^2>(P-R)^2+Q^2
P^2>(P-R)^2
P^2>P^2+R^2-2PR
2PR>R^2
2P>R
2P>7
P>3.5
Area(AOB)= 1/2b.h
=1/2X7XP
since P>3.5
Area(AOB)>12.25
Thus A is sufficient.

Let me know what mistake I made..

abhi1693 · Answer

Saurabh I think you considered the height of the triangle in incorrectly

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abhi1693 · Answer

By using Heron's Formula
Area of triangle is 1/2(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))
Consider statement 1 which is not enough
Now consider statement 2 which still does not gives us all three coordinates of vertices.
Thus, C is the answer.

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mbabubble · Answer

comprehensive soln anyone folks?

shashanksagar · Answer

Ans should be B.
The most important clue that we have is AO>AB
This restricts how far r can go.
AO from coordinates will be 5 units in length. So maximum value of AB < 5.
And all points are in first quadrant only so maximum value of r can be 8 (excluding 8).
So maximum area of tringle could be
1/2 * 8 * 3 = 12 (excluding 12)
Let me know if its not clear or if I am wrong somewhere.

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shashanksagar · Answer

To add to my solution above,
Algebraically we can show how r should be less than 8.
We know that length of a line between two coordinates is equal to sqrt of ( (x1-x2)^2 + (y1-y2)^2 )
So from AO > AB we get:
AO^2 > AB^2 (since both sides are positive )
So (4-0)^2 + (3-0)^2 > (r-4)^2 + (3-0)^2
Simplifying :
(r-4)^2 < 16
This gives us 0 < r < 8

Let me know if its not clear or if I am wrong somewhere.

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mvictor · Answer

oh wow...took 3 minutes just to make sure i don't need 1...
a truly 700 level question

umabharatigudipalli · Answer

Hey! Why is it excluding 8 here? maximum value of r can be 8 rite? area can also be 12, which is the maximum. Pls throw some light here!

Thanks,
Uma

ENGRTOMBA2018 · Answer

Detailed solution. We can clearly see the following things from the question statement: 1. All p,q,r >0 2. As AO > AB p^2 + q^2 > (p-r)^2 + q^2 Giving you, 2*p*r > r^2 2 cases possible --> Case 1: 2pr-r^2 > 0 ---> r>0 and 2p>r or Case 2: 2pr-r^2 >0 ---> r<0 and 2p 0. Hence ignore this case. Thus the only possible set of values are r>0 and 2p>r 3. Area of Triangle ABO = 0.5*r*q The question asks, is 0.5*r*q<12 or is r*q<24 Per Statement 1: No information on q. NOT SUFFICIENT. Per Statement 2: q=3 and p=2 --> 2p>r --> r<8 Thus, the value of r*q MUST BE <24 as q=3 and r<8. By any combination of values of r and q, you will NEVER get a value >= 24. Hence, this statement is sufficient. Hope this helps. The question is very straightforward if you break it down into recognizable chunks.

TheMechanic · Answer

While I understood the algebraic approach to deduce that r should be <8. The length of the 3rd side should be: greater than 1 and less than 9. Is there any other way to deduce that r<8?

Bunuel : Please throw some light here.

fskilnik · Answer

{S_{\Delta ABO}} = \frac{{r \cdot q}}{2}\,\,\mathop < \limits^? \,\,12\,\,\,\,\,\, \Leftrightarrow \,\,\,\boxed{\,\,r \cdot q\,\,\mathop < \limits^? \,\,24\,\,}

https://preview.ibb.co/hapUzf/19-Nov18-4j.gif

\left( 1 \right)\,\,r = 7\,\,\,\left\{ \matrix{
 \,{\rm{Take}}\,\,q = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr 
 \,{\rm{Take}}\,\,q = 4\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.

\left( 2 \right)\,\,\left( {p,q} \right) = \left( {4,3} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,r \cdot q < 8 \cdot 3\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle

\left( * \right)\,\,4 = p > {r \over 2}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,r < 8

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

fskilnik · Answer

Hi,  TheMechanic !

I will "intrude" to help. I hope Bunuel and yourself accept my apologies, if needed.

The perpendicular bisector of line segment OB is the set of all points (in the plane) that are equidistant of O (the origin) and B (given).

Hence AO > AB occurs if, and only if, A is "nearer" B, i.e., "to the right" of the perpendicular bisector mentioned, as shown in my figure (post above).

That´s why we must have p greater than r/2 , therefore (in sttm 2)), r/2 less than 4, i.e., r less than 8.

I hope you got things clearer now.

Regards,
Fabio.

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