Let $M$ be on segment$ BC$ of $\vartriangle ABC$ so that $AM = 3$, $BM = 4$, and $CM = 5$. Find the largest possible area of $\vartriangle ABC$.
1. Given that $M$ is on segment $BC$ of $\triangle ABC$ such that $AM = 3$, $BM = 4$, and $CM = 5$, we need to find the largest possible area of $\triangle ABC$.
2. To maximize the area of $\triangle ABC$, we should consider $AM$ as the altitude from $A$ to $BC$. This is because the area of a triangle is given by:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   Here, the base is $BC$ and the height is $AM$.
3. The length of $BC$ can be calculated as:
   \[
   BC = BM + CM = 4 + 5 = 9
   \]
4. Since $AM$ is perpendicular to $BC$ and $AM = 3$, the area of $\triangle ABC$ is:
   \[
   \text{Area} = \frac{1}{2} \times BC \times AM = \frac{1}{2} \times 9 \times 3 = \frac{27}{2}
   \]

The final answer is $\boxed{\frac{27}{2}}$