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Fun one... Right triangle PQR is to be constructed in the

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Senior Manager
Joined: 30 Oct 2004
Posts: 286
Followers: 1

Kudos [?]: 12 [0], given: 0

Fun one... Right triangle PQR is to be constructed in the [#permalink]  29 Sep 2005, 21:32
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Fun one...

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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-Vikram

Manager
Joined: 03 Aug 2005
Posts: 134
Followers: 1

Kudos [?]: 2 [0], given: 0

I pick C)9900

We have 10 possibilities for x and 11 for y.

We can proceed by first determining the right angle's vertex and then the length of the legs.

For the vertex, we have 11*10 possibilities.
For the legths we have:
horizontal: 9 possibilities
vertical: 10 possibilities.

Thus 11*10*10*9=9900
VP
Joined: 22 Aug 2005
Posts: 1123
Location: CA
Followers: 1

Kudos [?]: 43 [0], given: 0

C.

x co-ordinates count = 5 - (-4) + 1 = 10 {-4 <= x <= 5}
y co-ordinates count = 16 - 6 + 1 = 11

all combinations for P: 10 * 11
all combinations for R: 9 * 1 (y remains same as that of P as PR is parallel to x-axis)
all combinations for Q: 1 * 10 (x same as P. y other than what is selected for P or R)

total combinations : 10 * 11 * 9 * 1 * 1 * 10 = 9900

the problem becomes simple if you draw tringle on paper.
Senior Manager
Joined: 29 Nov 2004
Posts: 486
Location: Chicago
Followers: 1

Kudos [?]: 9 [0], given: 0

Got C, P and q should have same y value and P and R should have same x vaue so

11*10*9*10 = 9900
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Fear Mediocrity, Respect Ignorance

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Fun one... Right triangle PQR is to be constructed in the

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