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

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Manager
Joined: 30 Jan 2006
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Right triangle PQR is to be constructed in the xy-plane so [#permalink]

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15 Apr 2006, 21:01
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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?
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15 Apr 2006, 21:43
This is a Q. My guess is

(9*10)*(10*11) --> Assuming there is no restriction that x co-ordinate of P must be > x co-ordinate of Q.

For each base and height chosen, there are (5+4)*(16-6) = 9*10 triangles possible.

Also there are:
1) 11 bases (PR) and to choose from (y=6 through y = 16, both included).
2) 10 heights (QR) to choose from (x=-4 through x = 5, both included)

Total = 9*10*10*11?
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Manager
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16 Apr 2006, 16:30
Absolutely correct

Answer is 9900, which you have given above.
Senior Manager
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16 Apr 2006, 17:39
giddi77 wrote:
This is a Q. My guess is

(9*10)*(10*11) --> Assuming there is no restriction that x co-ordinate of P must be > x co-ordinate of Q.

For each base and height chosen, there are (5+4)*(16-6) = 9*10 triangles possible.

Also there are:
1) 11 bases (PR) and to choose from (y=6 through y = 16, both included).
2) 10 heights (QR) to choose from (x=-4 through x = 5, both included)

Total = 9*10*10*11?

Great Solution.
16 Apr 2006, 17:39
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# Right triangle PQR is to be constructed in the xy-plane so

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