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Right triangle PQR is to be constructed in the xy-plane so

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Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

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New post 16 Jul 2015, 10:33
there are 10 points on the x axis from which we can choose any twos and then form the desired PR base.

By the same way ,there are 11 points on the y-axis from which we can choose any twos and then form the desired PQ perpendicular.

We can form 10P2 * 11P2=9900
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Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

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New post 07 Jan 2017, 15:19
This was a very hard question for me.

I know that the set of vertices that should satisfy the inequalities apply to all P,Q and R. I took a look at the options and noticed only C that could be divisible by 3.. since set of all the vertices/3. Hence picked C.
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Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

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New post 17 Mar 2018, 19:36
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Re: Right triangle PQR is to be constructed in the xy-plane so   [#permalink] 17 Mar 2018, 19:36

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Right triangle PQR is to be constructed in the xy-plane so

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