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0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
easy one;)
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mirhaque wrote: easy one  Cordinate of A is less than 3 so Agree with B.
but do not think it is easy.
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1) A < 3
So A can be 1 or 2 (assuing integers) Max area < 15.
Sufficient.
2)triangle is right
Not sufficient. Can't tell.
(A) - statement 1 is sufficient
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"B".....
state 1....A can be -12 or 2....so different areas > or < 15...insuff
state 2.....there is only 2 points where the lines will be meet at right angle in Y axis(#2)...suff
Last edited by banerjeea_98 on 24 Mar 2005, 19:05, edited 1 time in total.
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Director
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banerjeea_98 wrote: "B".....
state 1....A can be -12 or 2....so different areas > or < 15...insuff
state 2.....there is only 1 point where the lines will be meet at right angle in Y axis...suff
can u explain the statement 2? i am not too sure.. what u r talking about.
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vprabhala wrote: i thought it should be E..
if the triangle is a right triangle, the hypoteneous is 5 and other two sides (suppose one is height and other is base) have a maximum measurement of Sqrt(5/2). if so, then the maximum area of the triangle could be 1/2[sqrt(5/2)xsqrt(5/2)], which equal 12.5. this is the maximum possible are of the triangle with the given information. therefore, B is suff.
In the meantime, coordinates of A could be negative, not only positive.
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VP
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I pick (E)
There is nothing that tells us that the hypotenuse is 5 so why would we assume the longest side of the right triangle is the given side?
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Folaa3 wrote: I pick (E). There is nothing that tells us that the hypotenuse is 5 so why would we assume the longest side of the right triangle is the given side?
Folaa3, Note that third vertex of the triangle have coordinates (x, y) as (0, A). value of x = 0 means the value of lies somewhere on the verticle line and B says the triangle is right angle triangle. if you construct the right angle triangle with these constrains, you will find the hypoteneous, base and height, then the area, which is not more than 12.5, of the triangle.
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MA wrote: Folaa3 wrote: I pick (E). There is nothing that tells us that the hypotenuse is 5 so why would we assume the longest side of the right triangle is the given side? Folaa3, Note that third vertex of the triangle have coordinates (x, y) as (0, A). value of x = 0 means the value of lies somewhere on the verticle line and B says the triangle is right angle triangle. if you construct the right angle triangle with these constrains, you will find the hypoteneous, base and height, then the area, which is not more than 12.5, of the triangle. I don't get it MA.. if the value or x in the third vertex = 0, A (which is y) could be anywhere along the y-axis......ah shoooot! I got it...so no matter where A lies on the y-axis, both sides will be equal so it only makes sense that 5 be the hypotenus since a right angle triangle can only have 1 hypotheneous...Got it. Thanks MA
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vp, in ur dia....if u see where the A,0 meets with either side of the axis u wud see that these 2 lines can meet in only 2 points on the yaxis and also be perpendicular ro each other, points r (0,2) and (0,-2)....A can be only these coordinates, so once we know A we know the area, whatever that may be.....If u r interested in calculating the A coordinate, try to think in terms of 2 lines meeting at (0,A) at penpendicular...u will see it can be #2
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its E A could be -18 or 1 or anything less than 3
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banerjeea_98 wrote: "B".....
state 1....A can be -12 or 2....so different areas > or < 15...insuff
state 2.....there is only 2 points where the lines will be meet at right angle in Y axis(#2)...suff
Right on the mark. You made it simple but not any simpler 
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