Ans: C
Since $BC=CD=DE$ and $\angle BCD = 90^\circ$, then $BCDE$ is a square.
Since $BC=CD=DE$ and $\angle BCD = 90^\circ$, then $BCDE$ is a square.
Since $\angle BAE = 90^\circ$, then $A$ is the intersection point of the diagonals. Therefore, the area of the pentagon $ABCDE$
$\begin{array}{cl}
= & \dfrac{3}{4} \times \text{the area of the square $BCDE$} \\
= & \dfrac{3}{4} \times 16^2 \\
= & 192\text{ cm}^2
\end{array}$
