imc

Introduce a coordinate system where $Q=(0,0)$, $P=(2,0)$ and $R=(0,2)$. In this coordinate system we have $M=(2,1)$, and the line $QM$ has the equation $2y-x=0$. As the line $RN$ is orthogonal to $QM$, it must have the equation $y+2x+q=0$ for some suitable constant $q$. As this line contains the point $R=(0,2)$, we have $q=-2$. Substituting $x=2y$ into $y+2x-2=0$, we get $y=\frac{2}{5}$, and then $x=\frac{4}{5}$. We can note that in $△RNQ$ $x$ is the height from $N$ onto $RQ$, hence its area is $[RNQ]=\frac{x\cdot RQ}{2}=\frac{2x}{2}=x=\frac{4}{5}$, and therefore the answer is $3[RNQ]+4=3\cdot \frac{4}{5}+4=\boxed{\frac{32}{5}}$.