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# I'm starting a new thread to collect some traps that are

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22 Dec 2006, 06:02
1
KUDOS
Thanks Blissful. Best of lucks in your endeavers too!
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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22 Dec 2006, 14:49
cute_smiley wrote:
i think that (1) insuf

but for the second it cannot be concluded that two angles of the triangle are equal. It can be that y=z+x (all angles are different) from this we can conclude that x+y+z = 180 i.e. y+y=180, y=90 so there is right triangle. But this is not enough because either side can have 90degrees angle so we need first statement: 12^2+15^=k^2, or 15^-12^2=k^2 and so on untill we will get the perimeter which will be the multiple of nine.
Am i right?

I agree with you smiley - we cannot conclude that the two angels are the same but we do still need the first statement so I'm getting answer as C. However, when working through the problem I cant seem to get a suitable numerical answer.

We know that there is an angle of 90 degrees so if the triangle were an isosceles triangle then the sides would have to be in the ratio x: x: x*(sqrt2) which would not allow for sides of 12 and 15 or a perimeter with a multiple of 9.

From 1) we know the perimeter is a multiple of 9. The only configuration of sides with a 90 degree internal angle that would allow this is for the sides to be in the ratio 3:4:5 or 9:12:15 - giving a perimeter of 36.

I understood till the part that says one of the angles is 90. But why cant the other two angles be (35, 55) or (50, 40) .. as they still sum to 90 but not necessarily a special triangle . Please let me know if i am missing something.
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24 Dec 2006, 08:54
trap

(1) The sides of Q have the same length.
(2) The diagonals of Q have the same length.
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24 Dec 2006, 09:14
(1) The quadrilateral could be a square or a rhomboid. Insuff => B or C or E.

(2) Square or rectangle or trapezoid. Insuff => C or E.

(1&2) Only viable shape is the square. Suff => C.

To determine the shape of a quadrilateral you need 3 of its angles.

93.santosh wrote:
trap

(1) The sides of Q have the same length.
(2) The diagonals of Q have the same length.
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03 Jan 2007, 12:33
cute_smiley wrote:
i think that (1) insuf

but for the second it cannot be concluded that two angles of the triangle are equal. It can be that y=z+x (all angles are different) from this we can conclude that x+y+z = 180 i.e. y+y=180, y=90 so there is right triangle. But this is not enough because either side can have 90degrees angle so we need first statement: 12^2+15^=k^2, or 15^-12^2=k^2 and so on untill we will get the perimeter which will be the multiple of nine.
Am i right?

I agree with you smiley - we cannot conclude that the two angels are the same but we do still need the first statement so I'm getting answer as C. However, when working through the problem I cant seem to get a suitable numerical answer.

We know that there is an angle of 90 degrees so if the triangle were an isosceles triangle then the sides would have to be in the ratio x: x: x*(sqrt2) which would not allow for sides of 12 and 15 or a perimeter with a multiple of 9.

From 1) we know the perimeter is a multiple of 9. The only configuration of sides with a 90 degree internal angle that would allow this is for the sides to be in the ratio 3:4:5 or 9:12:15 - giving a perimeter of 36.

Since in the question it asks for the longest side. That means 12 and 15 are not the longest ones. So for right angle triangle k being the hypotenuse (longest side in the right triangle) k^2 = 12^2 + 15^2= 369
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12 Jan 2007, 05:26
All the traps were indeed very good eye openers.
Thanks a lot for your contributions.
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Sanil George

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16 Mar 2007, 06:56
startrek has a good point here!

Since the question is clearly asking for the longest side, it means hypotenuse.

Thanks.
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18 Mar 2007, 06:43
Summer 3 :

The question asks for longest side... but the longest side can be 15 or K.

If you assume 15 to be hypotenuse, then K = 9

But K can also be the longest side, in which case K = 19.2

B is not sufficient.
18 Mar 2007, 06:43

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