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govinam
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Challenge Problem: Coordinate Plane Triangle 19 Aug 2007, 13:38
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Please explain the attatched. I don't understand the official answer. Thanks for the help.



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bkk145
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Challenge Problem: Coordinate Plane Triangle 19 Aug 2007, 14:17
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govinam
Please explain the attatched. I don't understand the official answer. Thanks for the help.
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D for me.

Area = (1/2) * (2/a) * (a^2) = a
Want to know if Area > 1, so same as if
a>1?

(1) If you plug in a =1, then you get 45,45,90 triangle. If a<1, then angle ABC will be greater than 90. If a>1, then it will be smaller than 90. Thus, a will always be great than 1.
SUFFICIENT.

(2) Perimeter = (2/a) + 2*sqrt((1/a^2) + a^4)
(2/a) + 2*sqrt((1/a^2) + a^4) > 4/a
Solving that, you get a^4 > 1
This can only be true when a>1
SUFFICIENT.
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govinam
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Challenge Problem: Coordinate Plane Triangle 19 Aug 2007, 14:57
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answer is supposed to be A.
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bkk145
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Challenge Problem: Coordinate Plane Triangle 19 Aug 2007, 15:03
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govinam
answer is supposed to be A.
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I was wrong. I solved the problem wrong in the (2) part. Here is where I got wrong and the correct answer:

(2/a) + 2* sqrt ((1/a^2) + a^4) > 4/a
=> 2* sqrt ((1/a^2) + a^4) > 2/a
=> sqrt ((1/a^2) + a^4) > 1/a
=> 1/a^2 + a^4 > 1/a^2
=> a^4 > 0
(This is where I got wrong...I mistakenly thought a^4 > 1, duh!)
So yes, you can't tell from here since a can pretty much be anything.
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Fig
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Challenge Problem: Coordinate Plane Triangle 19 Aug 2007, 15:15
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govinam
Please explain the attatched. I don't understand the official answer. Thanks for the help.

D for me.

Area = (1/2) * (2/a) * (a^2) = a
Want to know if Area > 1, so same as if
a>1?

(1) If you plug in a =1, then you get 45,45,90 triangle. If a<1, then angle ABC will be greater than 90. If a>1, then it will be smaller than 90. Thus, a will always be great than 1.
SUFFICIENT.

(2) Perimeter = (2/a) + 2*sqrt((1/a^2) + a^4)
(2/a) + 2*sqrt((1/a^2) + a^4) > 4/a
Solving that, you get a^4 > 1
This can only be true when a>1
SUFFICIENT.
Show more


(A) For me :)

Same reasoning for Stat 1. With area = |a| for me :)

But for 2:
(2/a) + 2*sqrt((1/a^2) + a^4) > 4/a
<=> 2*sqrt((1/a^2) + a^4) > 2/a
<=> sqrt((1/a^2) + a^4) > 1/a
=> ( sqrt((1/a^2) + a^4) ) ^2 > (1/a) ^2 >>> as a could be consider > 0 (the figure is perfectly the same if a > 0 or a < 0) and x^2 is increasing when x is increasing and x > 0.
<=> |1/a^2 + a^4| > 1/a^2
<=> 1 + a^6 > 1 as a is never considered equal to 0
<=> a^6 > 0
=> a != 0

We cannot conclude on the area > 1 ?
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Challenge Problem: Coordinate Plane Triangle 19 Aug 2007, 15:16
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govinam
answer is supposed to be A.

I was wrong. I solved the problem wrong in the (2) part. Here is where I got wrong and the correct answer:

(2/a) + 2* sqrt ((1/a^2) + a^4) > 4/a
=> 2* sqrt ((1/a^2) + a^4) > 2/a
=> sqrt ((1/a^2) + a^4) > 1/a
=> 1/a^2 + a^4 > 1/a^2
=> a^4 > 0
(This is where I got wrong...I mistakenly thought a^4 > 1, duh!)
So yes, you can't tell from here since a can pretty much be anything.
Show more


Ok ;) :)



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!