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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). "

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.

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.

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