$PQ = \sqrt{3^2 + 4^2} = 5$. Let length of altitude from $R$ to $h$. Then $\frac{1}{2} \times h \times 5 = 2$.
Step-by-Step Solution
Key Concept: Use the area formula for triangles to find the altitude, then determine how many configurations satisfy both the area constraint and the right angle constraint.
Given that $PQ = 5$ and the triangle has area 2, the altitude from $R$ perpendicular to $PQ$ satisfies $\frac{1}{2} \times 5 \times h = 2$, so $h = \frac{4}{5}$. The maximum possible length of altitude through $R$ with right angle at $R$ of $\triangle PQR$ (without the given area) is equal to $\frac{1}{2}PQ = \frac{5}{2}$. Now $\frac{1}{c} = \frac{1}{2}$. So, 4 triangles are possible.
Correct Answer: 4