2013 AMC 10B Problem 16 — Solution | LIVE by Po-Shen Loh

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2013 AMC 10B Problem 16

Below is the professionally curated solution for Problem 16 of the 2013 AMC 10B, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:centroid median (geometry)Pythagorean Triple area

Difficulty rating: 1600

16.

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

$13$

$13.5$

$14$

$14.5$

$15$

Solution:

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 .

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