, 2025

In the first diagram shown, we label the vertices of the polygon
and the length $ST=c$, since $ST=QR$.

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.

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).