, 2025

In the first diagram shown, we label the vertices of the polygon
and the length $ST=c$, since $ST=QR$. [[IMAGE0]] Next, we extend $UT$ by a length equal to $SR$, and we extend $QR$ by a length equal to $ST$, as shown in the second diagram. [[IMAGE1]] Each of the angles in the polygon is a right angle, and so these two extended line segments are perpendicular to each other and will meet at a point that we label $V$. That is, $STVR$ is a rectangle with $TV=SR=b$ and $RV=ST=c$. Each of the following expressions is equal to the perimeter of the original polygon $$\begin{align*} &PQ+QR+SR+ST+TU+PU\\ =\ & PQ+QR+ST+SR+TU+PU\ (\text{reordering the lengths})\\ =\ & PQ+QR+RV+TV+TU+PU\ (\text{since }RV=ST \text{ and } TV=SR)\\ =\ & PQ+QV+UV+PU\ (\text{since }QR+RV=QV \text{ and } TV+TU=UV)\end{align*}$$ which is the perimeter of $PQVU$. Each of the angles in $PQVU$ is a right angle, and $PQ=PU$, and thus $PQVU$ is a square. Since $PQ=UV=UT+TV=a+b$, and $PU=QV=QR+RV=c+c=2c$, then $a+b=2c$. Summarizing, the perimeter of the original polygon is equal to the perimeter of square $PQVU$, and each side length of square $PQVU$ can be expressed as $a+b$ or as $2c$ since $a+b=2c$. If each of the $4$ side lengths is expressed as $a+b$, then the perimeter of $PQVU$ (and thus the perimeter of the original polygon), is equal to $(a+b)+(a+b)+(a+b)+(a+b)=4a+4b$. If $3$ side lengths are expressed as $a+b$ and $1$ side length is expressed as $2c$, then the perimeter is $(a+b)+(a+b)+(a+b)+(2c)=3a+3b+2c$. If $2$ side lengths are expressed as $a+b$ and $2$ side lengths are expressed as $2c$, then the perimeter is $(a+b)+(a+b)+(2c)+(2c)=2a+2b+4c$. If $1$ side length is expressed as $a+b$ and $3$ side lengths are expressed as $2c$, then the perimeter is $(a+b)+(2c)+(2c)+(2c)=a+b+6c$. Finally, if all $4$ sides lengths are expressed as $2c$, the perimeter is $(2c)+(2c)+(2c)+(2c)=8c$. Of the expressions given, $a+b+7c$ remains, and since $a+b+7c=2c+7c=9c$
is not equal to the perimeter, then the correct answer is (B).