jmc

Note that $1.5^2+2^2=2.5^2,$ so $△PED$ has a right angle at $P.$ (Alternatively, you could note that $(1.5,2,2.5)$ is half of the Pythagorean triple $(3,4,5).$) Since the centroid $P$ divides medians $AD$ and $CE$ in the ratio $2:1,$ we have $CP=2\cdot EP=2\cdot 1.5=3$ and $AP=2\cdot DP=2\cdot 2=4.$ Then quadrilateral $AEDC$ consists of four right triangles; we can then compute its area as $$[AEDC]=\frac{1}{2}(4\cdot 1.5+2\cdot 1.5+3\cdot 2+4\cdot 3)=\boxed{13.5}.$$