jmc

We first label the trapezoid $ABCD$ as shown in the diagram below. Since $AD$ is the height of the trapezoid, then $AB$ and $DC$ are parallel. The area of the trapezoid is $$\begin{array}{rl}\frac{AD}{2}\times (AB+DC)& =\frac{12}{2}\times (AB+16)\\ & =6\times (AB+16).\end{array}$$ Since the area of the trapezoid is $162,$ we have $6\times (AB+16)=162$ so that $$AB+16=\frac{162}{6}.$$ Solving for $AB$, we obtain $AB=11.$

Construct a perpendicular from $B$ to $E$ on $DC.$ Since $AB$ and $DE$ are parallel, and both $AD$ and $BE$ are perpendicular to $DE,$ we know that $ABED$ is a rectangle. Thus, $DE=AB=11,$ $BE=AD=12,$ and $$\begin{array}{rl}EC& =DC-DE\\ & =16-11\\ & =5.\end{array}$$ Since $\angle BEC=90^{\circ },$ we know that $△BEC$ is a right triangle. Thus by the Pythagorean Theorem, $$\begin{array}{rl}B{C}^2& =B{E}^2+E{C}^2\\ & =12^2+5^2\\ & =169\end{array}$$ so that $BC=13$ (since $BC>0$).

The perimeter of the trapezoid is $$AB+BC+CD+DA=11+13+16+12=\boxed{52}\text{ cm}.$$