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# DS_geometry_coordinate

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29 Sep 2007, 12:57
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29 Sep 2007, 13:58
Area of ABC = 1/2*a^2*2/a = a
is a>1
(1) testing for various B
B=60; A= 60;=> a >1
B=30; A= 75;=> a >1

SUFF

(2) Perimeter or triangle 2/a (*Sqrt(a^6+1)) + 2/a > 4/a
=> Sqrt(a^6+1) > 1
=> a > 1 ..SUFF

Ans D
--------
EDITED for (1)

Last edited by Juaz on 29 Sep 2007, 14:30, edited 2 times in total.
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29 Sep 2007, 14:08
the answer is (A)

since the area of ABC = 1/a*a^2 = a

and since ABC < 90 ---> and half ABC < 45.

than plugging a=1 will yield a 45 degrees triangle ---> 45:45:90 ---> 1:1:sqrt(2).

any other value for a > 1 will yield a triangle with half ABC > 45 or ABC > 90.

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29 Sep 2007, 14:44
KillerSquirrel wrote:
the answer is (A)

since the area of ABC = 1/a*a^2 = a

and since ABC <90> and half ABC <45> 45:45:90 ---> 1:1:sqrt(2).

any other value for a > 1 will yield a triangle with half ABC > 45 or ABC > 90.

KS, do you have any links to fundamentals of this info ? I am not sure how you determined that "plugging a=1 will yield a 45 degrees triangle ---> 45:45:90 ---> 1:1:sqrt(2). "
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29 Sep 2007, 14:52
pmenon wrote:
KillerSquirrel wrote:
the answer is (A)

since the area of ABC = 1/a*a^2 = a

and since ABC <90> and half ABC <45> 45:45:90 ---> 1:1:sqrt(2).

any other value for a > 1 will yield a triangle with half ABC > 45 or ABC > 90.

KS, do you have any links to fundamentals of this info ? I am not sure how you determined that "plugging a=1 will yield a 45 degrees triangle ---> 45:45:90 ---> 1:1:sqrt(2). "

I really had hard time with this problem and I'm not sure that (A) is correct.

But since you have in the diagram a triangle with the height of a^2 and one side is 1/a than 1^2 = 1 and 1/1 = 1. The only way this can be (1:1:sqrt(2)) is if the triangle has 45:45:90 degrees ratio.

see OG 11 ---> page 130.

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01 Oct 2007, 01:09
Area of the triangle = 1/2 (2/a) (a^2) = a

So the area of triangle will be greater than 1 if a > 1.

St1:
If we set ABC to 90 degrees (slightly out of range), then we will have a = 1. So if ABC is less than 90 degrees, a > 1 and area of triangle > 1. Sufficient.

St2:
Perimeter > 4/a

AB = BC = sqrt(a^6+1)/a
Perimeter = 2sqrt(a^6 + 1)/a + 2/a

If a = 1, then perimeter > 4/a and area of triangle = 1.
If a = 2, then perimeter > 4/a and area of triangle = 2.
Insufficient.

Ans A
01 Oct 2007, 01:09
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# DS_geometry_coordinate

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