2013 AMC 10B Problem 16

1. Look at triangle $PED$ first, since we know all three of its sides: $1.5$, $2$, $2.5$. Doubling them gives $3, 4, 5$, a right triangle. Check: $1.5^2 + 2^2 = 2.25 + 4 = 6.25 = 2.5^2$. So the angle at $P$ is $90^\circ$, which means the medians $AD$ and $CE$ cross at right angles.
2. $P$ is where two medians meet, so it is the centroid. The centroid cuts each median into two pieces, and the piece touching the corner is twice as long as the piece touching the midpoint. So
$$AP = 2 \cdot PD = 4, \qquad CP = 2 \cdot PE = 3.$$
3. The segments $AD$ and $CE$ are the diagonals of $AEDC$, and they cross at $P$. They cut $AEDC$ into four right triangles, all with their right angle at $P$. Each has area half the product of its two legs:

1. Add them up: $3 + 1.5 + 3 + 6 = 13.5$.

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

Two facts do all the work. Three given lengths that fit the Pythagorean theorem reveal a hidden right angle, and the centroid's $2$ to $1$ split turns the short pieces into full medians. Whenever the diagonals of a four-sided shape cross at right angles, splitting it into four right triangles makes the area easy.

A four-sided shape whose diagonals cross at right angles has area equal to half the product of the diagonals (the four right triangles above add up to exactly that). Here $AD = 4 + 2 = 6$ and $CE = 3 + 1.5 = 4.5$, so the area is $\frac{1}{2} \cdot 6 \cdot 4.5 = 13.5$.

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 $\boxed{\textbf{(B)}\ 13.5}$.