smc

Construct the midpoints $E=(4,0)$ and $F=(10,0)$ and triangle $△EMF$ as in the diagram, where $M$ is the center of square $PQRS$. Also construct points $G$ and $H$ as in the diagram so that $BG\parallel PQ$ and $CH\parallel QR$. Observe that $△AGB\sim △CHD$ while $PQRS$ being a square implies that $GB=CH$. Furthermore, $CD=6=3\cdot AB$, so $△CHD$ is 3 times bigger than $△AGB$. Therefore, $HD=3\cdot GB=3\cdot HC$. In other words, the longer leg is 3 times the shorter leg in any triangle similar to $△AGB$. Let $K$ be the foot of the perpendicular from $M$ to $EF$, and let $x=EK$. Triangles $△EKM$ and $△MKF$, being similar to $△AGB$, also have legs in a 1:3 ratio, therefore, $MK=3x$ and $KF=9x$, so $10x=EF=6$. It follows that $EK=0.6$ and $MK=1.8$, so the coordinates of $M$ are $(4+0.6,1.8)=(4.6,1.8)$ and so our answer is $4.6+1.8=6.4=$ $\boxed{\frac{32}{5}}$.