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by vinay1983 » Wed Oct 09, 2013 3:16 am
I don't know how to makes changes in the figure.But here it is

See the triangle POQ
Op=4
OQ=3
So as you can see POQ forms a right angle triangle then PQ becomes "5"

Imagine a triangle at the extreme end with point X as the last imaginary point, then PX is 3 and XR is 4, then PR is 5 again.

So now take either PQ or PR as the base and find the area

1/2 * b * h

1/2 * 5*5
25/2
12.5

Hope I am correct
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by GMATGuruNY » Wed Oct 09, 2013 3:17 am
In the rectangular coordinate system below, the area of triangular region PQR is

12.5
14
10√2
16
25
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The area of the rectangle drawn around ∆PQR = 7*4 = 28.
Since ∆PQR takes up less than half the rectangle, ∆PQR < 14.

The correct answer is A.
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by vinay1983 » Wed Oct 09, 2013 3:44 am
GMATGuruNY wrote:
In the rectangular coordinate system below, the area of triangular region PQR is

12.5
14
10√2
16
25
Image

The area of the rectangle drawn around ∆PQR = 7*4 = 28.
Since ∆PQR takes up less than half the rectangle, ∆PQR < 14.

The correct answer is A.
Thanks Mitch, this is what i wanted to do, but was unsure. I could have saved atleast 1 min on this question and moved on on DS!
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by [email protected] » Wed Oct 09, 2013 10:29 am
Hi pareekbharat86,

There's a great tactic that's worth remembering on graphing questions: any diagonal line on a graph is part of a right triangle (that you can draw and use to figure out the length of the line). That tactic works perfectly on this prompt (as you can see from the various explanations). You'd be surprised how often that one "math move" can be used to help you solve graphing questions, so keep in mind as you continue to study.

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