16. As shown in Figure 1, in $\triangle A B C$, it is known that $A D$ and $C E$ are medians of $\triangle A B C$, and $A D$ intersects $C E$ at point $P$. If $P E=1.5, P D=2, D E=2.5$, then the area of quadrilateral $A E D C$ is ( ).
(A) 13
(B) 13.5
(C) 14
(D) 14.5
(E) 15
16. B.

From the problem, we know that $P$ is the centroid of $\triangle ABC$. Therefore, $AP = 2PD = 4$, and $CP = 2PE = 3$.
Also, $DE$ is the midline of $\triangle ABC$, and $DE = 2.5$, so $AC = 5$.
Notice that, $AC^2 = AP^2 + CP^2$.
Thus, $AD \perp CE$.
Therefore, $S_{\text{quadrilateral } AEDC} = \frac{1}{2} AD \cdot CE = \frac{1}{2} \times 6 \times 4.5 = 13.5$.