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In the \(xy-\)plane, point \(O\) is located at the origin,

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by swerve » Tue Aug 13, 2019 10:40 am

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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 \(q=3\)

The OA is B

Source: Princeton Review
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Source: — Data Sufficiency |

by Jay@ManhattanReview » Tue Aug 20, 2019 11:20 pm
swerve wrote: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 \(q=3\)

The OA is B

Source: Princeton Review
Let's take each statement one by one.

1) \(r=7\)

No information about the height of the triangle. Insufficient.

2) \(p=4\) and \(q=3\)

Let's find out the maximum area of ∆ABO. See the image below.

Image

Drop a perpendicular from vertex A on X-axis. The height of the ∆ABO = AA' = 3 and OA' = 4. Thus, OA = √(3^2 + 4^2) = √25 = 5. Since it is given that \(AO > AB\), AB < 5. Again, in the rightangles ∆AA'B, we would have A'B < 4.

Thus, the base of ∆ABO = 4 + <4 = <8

Area of ∆ABO < 1/2 * 3 * 8 => < 12.

Thus, the area of triangular region \(ABO\) less than 12. The answer is Yes. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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by Ian Stewart » Wed Aug 21, 2019 7:03 am
This is just an inferior copy of an official question (inferior because it doesn't include a diagram, which the GMAT would include, and because it's needlessly wordy) :

https://gmatclub.com/forum/in-the-recta ... 29092.html
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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