In the figure below, choose point $D$ on $\overline{BC}$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ABD$? [asy]draw((0,0)--(4,0)--(0,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,0), ESE); label("$C$", (0, 3), N); label("$3$", (0, 1.5), W); label("$4$", (2, 0), S); label("$5$", (2, 1.5), NE);[/asy]
Because $\overline{BD} + \overline{CD} = 5,$ we can see that when we draw a line from point $B$ to imaginary point $D$ that line applies to both triangles. Let us say that $x$ is that line. Perimeter of $\triangle{ABD}$ would be $\overline{AD} + 4 + x$, while the perimeter of $\triangle{ACD}$ would be $\overline{AD} + 3 + (5 - x)$. Notice that we can find $x$ from these two equations by setting them equal and then canceling $\overline{AD}$. We find that $x = 2$, and because the height of the triangles is the same, the ratio of the areas is $2:3$, so that means that the area of $\triangle ABD = \frac{2 \cdot 6}{5} = \fbox{\frac{12}{5}}$.