### Instruction: In triangle $ABC$, medians $AD$ and $CE$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$.  What is the area of $AEDC$?### Response: I notice that the medians of a triangle divide it into six smaller triangles of equal area.
So, to find the area of $AEDC$, I only need to find the area of one of those triangles and multiply by four.
I choose to focus on triangle $DEP$, since I know all its sides.
I can use Heron's formula to find the area of a triangle given its sides.
Heron's formula says that the area of a triangle with sides $a, b, c$ is $\sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semiperimeter, or half the sum of the sides.
In this case, $a=1.5, b=2, c=2.5$, so $s=\frac{1.5+2+2.5}{2}=3$.
Plugging these values into Heron's formula, I get $\sqrt{3(3-1.5)(3-2)(3-2.5)}=\sqrt{3(1.5)(1)(0.5)}=\sqrt{2.25}=1.5$.
So the area of triangle $DEP$ is 1.5 square units.
Multiplying by four, I get the area of $AEDC$ is 6 square units.