2013 AMC 10B Problem 16

The three given lengths satisfy $1.5^2 + 2^2 = 2.25 + 4 = 6.25 = 2.5^2$, so triangle $PED$ is right-angled at $P$. In other words, the medians $AD$ and $CE$ are perpendicular.

$P$ is the centroid, which cuts each median in a $2 : 1$ ratio measured from the vertex. Hence
$$AP = 2 \cdot PD = 4, \qquad CP = 2 \cdot PE = 3,$$
so $AD = 4 + 2 = 6$ and $CE = 3 + 1.5 = 4.5$.

Quadrilateral $AEDC$ has diagonals $AD$ and $CE$, which are perpendicular. Its area is half the product of the diagonals:
$$\frac{1}{2} \cdot 6 \cdot 4.5 = 13.5.$$

The answer is $\boxed{\textbf{(B)}\ 13.5}$.

Two facts do all the work: the centroid's $2:1$ division converts the given fragments into full medians, and a right-triangle check on the three given numbers reveals the perpendicularity. Any quadrilateral with perpendicular diagonals $d_1, d_2$ has area $\frac12 d_1 d_2$, since the diagonals split it into four right triangles.

Sum the four right triangles around $P$ directly: $\frac12(4 \cdot 3 + 3 \cdot 2 + 2 \cdot 1.5 + 1.5 \cdot 4) = \frac12(12 + 6 + 3 + 6) = 13.5$. Or use $[ABC] = \frac{2}{3} \cdot AD \cdot CE$ for perpendicular medians, giving $[ABC] = 18$, and then $[AEDC] = [ABC] - [BDE] = 18 - \frac14 \cdot 18 = 13.5$ since $DE$ is a midline.

• Using PD and PE as the full medians, or forgetting that the centroid divides each median in a 2:1 ratio.
• Not noticing the $3$-$4$-$5$ triangle and attempting Heron's formula or the law of cosines on triangle $PED$.