2000 AMC 10 Problem 5

Since $MN$ is a midsegment, $MN = \tfrac12 AB,$ which is fixed. The base $AB$ and the height from $P$ to $AB$ are both constant as $P$ slides along the parallel line, so the area of $\triangle PAB$ does not change. The trapezoid $ABNM$ is the triangle minus $\triangle PMN,$ both of whose areas are constant, so its area does not change either. Only the perimeter changes, since $PA$ and $PB$ vary as $P$ moves. So exactly one quantity changes. Thus, the correct answer is B.