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

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10 Aug 2008, 04:05

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68% (02:03) correct
33% (02:05) wrong based on 79 sessions

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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 < or= x < or = 5 and 6 < or= y < or = 16. How many different triangles with these properties could be constructed?

ca you explain why x=9*10/2 probability of understanding probability is less

x can have 9 possible with 1 unit length. x can have 8 possible with 2 unit length. x can have 7 possible with 3 unit length. . . . x can have 1 possible with 9 unit length.

ca you explain why x=9*10/2 probability of understanding probability is less

x can have 9 possible with 1 unit length. x can have 8 possible with 2 unit length. x can have 7 possible with 3 unit length. . . . x can have 1 possible with 9 unit length.

Similarly y will have its own case.

Hope this helps.

I got to work hard in probability _________________

number of ways point P can be selected = 100 ( x can be anything from 4 to 5 (including) and y can be anything from 6 to 16(including)) number of ways point R can be selected = 9 (the y co-orodinate will be same as P and we can choses x co-ordinate in 9 different ways) number of ways point Q can be selected = 11 (the x co-orodinate will be same as P and we can choses y co-ordinate in 11 different ways)

For the rt triange PQR, we will have same Y-corodinate value for point R and P and same X-coordinate value for point Q and P.

So Total possible values for X = ( number of ways X can be selected for point R) * ( number ways X can be selected for point Q and P ) = 10 * 9 ( as Q and R will have same X-coordinate and will be different than that of R ) =90

Similarly, total number of ways for selecting Y = 11* 10 =110

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 < or= x < or = 5 and 6 < or= y < or = 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

P can be chosen from 11*10 points and Q can be chosen from 10 points and R from 9 points

I understand the solution given, but doesn't that solution count multiple identical triangles as "different"? That is, we are making a right triangle, so we should only need to worry about the lengths of the two legs. Given the possible x coordinates, there should be 9 possible lengths for PR (1..9). Given the possible y coordinates, there should be 10 possible lengths for PQ (1..10).

The solution provided doesn't give us the number of unique right triangles with integer lengths but instead the total number of unique triangles * the total possible arrangements in the given area for each.

Am I just way off and the wording of the question should have made me understand that we're also looking for unique positions of congruent triangle?

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