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Triangle in XP plane

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Triangle in XP plane

by sugmomo » Wed Jun 16, 2010 4:51 am
Hi.. can u please give a detailed explanation for the below prob...

Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x and y coordinates of P, Q and R are to be integers that satisfy the inequalities -4 <= x <= 5 and 6 <= y <= 16. How many different triangles with these properties could be constructed?

(A) 110
(B) 1,100
(C) 9,900
(D) 10,000
(E) 12,100
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by kvcpk » Wed Jun 16, 2010 6:17 am
is the OA C?
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by kvcpk » Wed Jun 16, 2010 6:24 am
In the XY plane draw the picture for -4 <= x <= 5 and 6 <= y <= 16
you will get a rectangluar region.

given that coordinates of p,q,r are integers. So x can take 10 possible values(-4 to 5) and y can take 11 possible values(6 to 16).

P can take ant of those points. So P can take 10(11) = 110 points

PR is parallel to x axis. SO PR is horizontal. R will have the same Y value as P. and so can have 9 different values. other than P.

PQ has to be vertical because angle QPR is 90. So Q has the same x value as P and can have 10 other values apart from P itself.

therefore total number of triangles = 110(10)(9) = 9900 triangles.

I think this is one of the difficult questions. It took me some time to analyze it. Not sure if it couldbe on GMAT.
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