It has finally dawned on me. The concept you seem to be missing is that for area you need height and if two lines are parallel then the perpendicular height between them is always the same. Think about it.

I know that area = base*height and that the distance between parallel lines is constant, but I guess I wasn't posting like that. It looks like given a parallelogram that is not a rectangle, if the perimeter remains the same and the angles get more equal (closer to a rectangle), the area increases, and if the perimeter remains the same and the angles get less equal, the area decreases. Is that true? Given a fixed perimeter, the greatest four-sided area is a square, which is a special case of a rectangle. Therefore leaving all the sides and therefore the perimeter alone and changing the angles so it is no longer a square will decrease the area.

A parallelogram can be divided into two congruent triangles. By reflecting a triangle over one of its sides, you can make any triangle be half of a parallelogram. In that case, any way of making the perimeter and area of a triangle move in opposite directions also works for a parallelogram that is not a rectangle. For a rectangle, the congruent triangles would be right triangles. The sides of the rectangle would be the legs of a right triangle, which leaves only one possibility for the hypotenuse. If you know two sides of a triangle, that it is not a right triangle, and nothing else about the angles, the third side can be any length that satisfies the inequality that the sum of the smallest two sides is greater than the larger side.

Given two sides of a triangle, you can calculate the range of possibilities for the third side. Can you calculate a range of areas? For a right triangle, the area is half the product of the two smallest sides, and for any other triangle, the area is less than half the product of the two smallest sides. Let's say two sides are 2 and 3. The third side ranges from 1 to 5 excluding 1 and 5. The product of the two smallest sides is less than or equal to 6, so the area must be less than or equal to 3.