Hi topspin360!

This is a tricky problem to do algebraically because of the various configurations for the triangle (is it right, isosceles, equilateral, nothing??). So maybe we just start making up some numbers that work. We are only asked about perimeters, so as long as we make the sides of the triangle "legal" (no side is longer than the sum of the other 2 sides), then we're good.

(1) The length of the longest side of T is twice the length of a side S.
To be lazy here, I picked a well known right triangle: 3-4-5, perimeter=12. This means the side of the square is 1/2 of 5, or 2.5, perimeter=10. In this case, the perimeter of the triangle is LARGER than the perimeter of the square. Now I have to test something where the perimeter of the triangle would be LESS than the perimeter of the square....maybe if I make the sides of the triangle really small, and if I make the "longest" side as short as possible, or the same length as the other sides? What about an equilateral with all sides 1, perimeter=3. The square would have sides .5, perimeter 2...still smaller!! So let's think theory... the smallest "long" side for a triangle would be one where it was the same length as the other sides (equilateral), so if the triangle has side lengths t, perimeter 3t, the square would have sides .5T, perimeter 2t...always smaller than 3t!! So it looks like we were able to prove that under this constraint, the perimeter of the triangle is ALWAYS larger than that of the square! SUFFICIENT

(2) T is isoceles.
So here, we don't have ANY relationship between the triangle and the square so we can make up whatever we want. Let's make the triangle 1-1-1 (because ALL equilateral triangles are also isosceles), and the square 2-2-2-2... Here the square has a LARGER perimeter than the triangle. But swap them, make the triangle 2-2-2 and the square 1-1-1-1... Now the square is only 4 while the triangle is 6, so the triangle has a LARGER perimeter. NOT SUFFICIENT

Therefore, the answer is A

Hope this helps!

Whit