Usually, the formula for the area of a parallelogram is derived by cutting out a right triangle from the parallelogram and moving it to the other side of the (right) trapezoid to get a rectangle with the same base, height, and area.
However, sometimes, such a right triangle could not be cut out, particularly when the left endpoint of one of the bases of a parallelogram oriented with horizontal bases is further to the right than the right endpoint of the other base. Is it still true that $A=bh$ for such a parallelogram, where $A$, $b$, and $h$ are the area, base, and height respectively? In this case, to calculate the height, one must extend both of the bases.