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by Sankofi » Sun Oct 05, 2008 6:24 pm
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

Please explain

OA is C
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by stop@800 » Sun Oct 05, 2008 8:56 pm
One point can be chosen in 11 ways
another in 10 ways and last one in 9 ways

so 11*10*9
9900
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by aj5105 » Sun Oct 05, 2008 9:12 pm
can you elaborate ur answer a bit more?
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by mental » Mon Oct 06, 2008 3:57 am
stop@800 wrote:One point can be chosen in 11 ways
another in 10 ways and last one in 9 ways

so 11*10*9
9900
STOP: i suppose there was a typo
11*10*9 = 990

neways, you must have meant (11*10)*(10*9)

y can have 11 integral co-ordinates, out of that two can be chosen in 11*10 ways

x can have 10 integral co-ordinates, so two can be chosen in 10*9 ways

so total triangles = 11*10*10*9 = 9900

THIS QUESTION CAN BE FOUND IN EARLIER POSTS
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by Sankofi » Mon Oct 06, 2008 7:23 am
Thanks. I did a search and was not able to find, sorry.
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