16. (6 points) As shown in the figure, there is a point $P$ inside quadrilateral $A B C D$ such that the distances from $P$ to the four sides $A B, B C, C D, D A$ are $P E, P F, P M, P N$, all equal to 6 cm. If the perimeter of quadrilateral $A B C D$ is 57 cm, then the area of quadrilateral $A B C D$ is $\qquad$ square cm.
【Analysis】
Connecting $P A, P B, P C, P D$ results in four triangles, $\triangle P A B, \triangle P B C, \triangle P C D, \triangle P D A$. The area of quadrilateral $A B C D$ is equal to the sum of the areas of these four triangles.

【Solution】
The area of quadrilateral $A B C D = S \triangle P A B + S \triangle P B C + S \triangle P C D + S \triangle P D A = \frac{1}{2} A B \times P E + \frac{1}{2} B C \times$
$$
P F + \frac{1}{2} C D \times P M + \frac{1}{2} A D \times P N
$$

Since $P E = P F = P M = P N = 6$ cm, and $A B + B C + C D + A D =$ the perimeter of quadrilateral $A B C D$ is 57 cm,
Therefore, $S$ quadrilateral $A B C D = S \triangle P A B + S \triangle P B C + S \triangle P C D + S \triangle P D A$
$$
\begin{array}{l}
= \frac{1}{2} A B \times P E + \frac{1}{2} B C \times P F + \frac{1}{2} C D \times P M + \frac{1}{2} A D \times P N \\
= \frac{1}{2} \times 6 \times (A B + B C + C D + A D) \\
= \frac{1}{2} \times 6 \times 57 \\
= 171 \text{ (square cm); }
\end{array}
$$

Answer: The area of quadrilateral $A B C D$ is 171 square cm.
Therefore, the answer is: 171.