$\mathrm{P}\mathrm{Q}=\sqrt{3^2+4^2}=5$
Let, length of altitude through $\mathrm{R}$ is $h$

then $\frac{1}{2}\times \mathrm{h}\times 5=2$

$\Rightarrow \mathrm{h}=\frac{4}{5}$
Maximum possible length of altitude through $\mathrm{R}$ with right angle at $\mathrm{R}$ of $\mathrm{\Delta }\mathrm{P}\mathrm{Q}\mathrm{R}$ (without the given area) is equal to $\frac{1}{2}\mathrm{P}\mathrm{Q}=\frac{5}{2}$
Now, $\frac{4}{5}<\frac{5}{2}$
So, $4$ triangles are possible