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

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Right triangle PQR is to be constructed in the xy-plane so [#permalink] New post 10 Aug 2008, 04:05
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63% (02:25) correct 36% (01:47) wrong based on 5 sessions
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

OPEN DISCUSSION OF THIS QUESTION IS HERE: right-triangle-pqr-is-to-be-constructed-in-the-xy-plane-so-71597.html
[Reveal] Spoiler: OA
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Re: PS: Right triangle PQR [#permalink] New post 10 Aug 2008, 05:02
Total possible Orientations for Triangle = 4

Total possible values for x = 9*10/2
Total possible values for y = 10*11/2

Total possible values of xy = 9*10/2 * 10*11/2 = 9900/4

As there are 4 possible orientations so total possible triangle = 4*9900/4 = 9900

This is an OG problem, so you can refer its solution as well.

Please find the attached document to see the possible orientations.
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Re: PS: Right triangle PQR [#permalink] New post 10 Aug 2008, 05:12
abhijit_sen wrote:
Total possible Orientations for Triangle = 4

Total possible values for x = 9*10/2
Total possible values for y = 10*11/2

Total possible values of xy = 9*10/2 * 10*11/2 = 9900/4

As there are 4 possible orientations so total possible triangle = 4*9900/4 = 9900

This is an OG problem, so you can refer its solution as well.

Please find the attached document to see the possible orientations.


ca you explain why x=9*10/2
:shock: :roll: probability of understanding probability is less :?
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Re: PS: Right triangle PQR [#permalink] New post 10 Aug 2008, 05:32
spriya wrote:
ca you explain why x=9*10/2
:shock: :roll: 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.
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Re: PS: Right triangle PQR [#permalink] New post 10 Aug 2008, 06:47
abhijit_sen wrote:
spriya wrote:
ca you explain why x=9*10/2
:shock: :roll: 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
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Re: PS: Right triangle PQR [#permalink] New post 10 Aug 2008, 08:05
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)

total number of triangles = 100*9*11 = 9900
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Re: PS: Right triangle PQR [#permalink] New post 10 Aug 2008, 09:19
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

So, total combination = 110 * 90 = 9900
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Re: PS: Right triangle PQR [#permalink] New post 10 Aug 2008, 09:23
judokan wrote:
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

Answer = 11*10*10*9 = 9900
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Re: PS: Right triangle PQR [#permalink] New post 24 May 2011, 07:25
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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Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink] New post 24 Dec 2012, 01:06
how would you solve this in under 2 minutes?
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Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink] New post 24 Dec 2012, 01:10
fozzzy wrote:
how would you solve this in under 2 minutes?


Check here: right-triangle-pqr-is-to-be-constructed-in-the-xy-plane-so-71597.html

Similar questions to practice:
triangle-abc-will-be-constructed-in-a-xy-plane-according-to-132573.html
right-triangle-pqr-is-to-be-constructed-in-the-xy-plane-so-88380.html
right-triangle-abc-is-to-be-drawn-in-the-xy-plane-so-that-88958.html
a-right-triangle-abc-has-to-be-constructed-in-the-xy-plane-94644.html
a-right-triangle-abc-has-to-be-constructed-in-the-xy-plane-100675.html
right-triangle-rst-can-be-constructed-in-the-xy-plane-such-137129.html

Hope it helps.
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Re: Right triangle PQR is to be constructed in the xy-plane so   [#permalink] 24 Dec 2012, 01:10
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