2013 AMC 10B Problem 16

Attempt Problem 16 of the 2013 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 10B solutions, or check the answer key.

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16.

In triangle $\triangle ABC,$ medians $AD$ and $CE$ intersect at $P,$ $PE=1.5,$ $PD=2,$ and $DE=2.5.$ What is the area of $AEDC?$

$13$

$13.5$

$14$

$14.5$

$15$

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 .