2013 AMC 10B Problem 16

Since $PE:PD:DE=1.5:2:2.5$ $=3:4:5$, triangle $DPE$ is right at $P$. Thus medians $AD$ and $CE$ are perpendicular. The centroid divides each median in a $2:1$ ratio, so $CE=3\cdot PE=4.5$ and $AD=3\cdot PD=6$. Quadrilateral $AEDC$ has perpendicular diagonals $AD$ and $CE$, so its area is $\frac12(AD)(CE)=\frac12\cdot6\cdot4.5=13.5$. Thus, the correct answer is B .