In triangle $ABC$, medians $AD$ and $CE$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$. What is the area of $AEDC$? $\textbf{(A)}13\qquad$ $\textbf{(B)}13.5\qquad$ $\textbf{(C)}14\qquad$ $\textbf{(D)}14.5\qquad$ $\textbf{(E)}15\qquad$
Solution
$1.5^2+2^2=2.5^2$, so $\triangle PDE$ has a right angle at $P$, i.e. $AD\perp CE$. Since $P$ is the centroid, $CE=3PE=4.5$ and $AD=3PD=6$.
$[AEDC]=\frac12\cdot 4.5\cdot 6=13.5$, by theorem 16.
