smc

The unit square is of area 1, so the five unit squares have area 5. Therefore the shaded space must occupy 2.5. The missing unit square is of area 1, and if reconstituted the original triangle would be of area 3.5. It can then be inferred: $(3-c)\ast 3=7$. $3-c=\frac{7}{3}$, so $3-\frac{7}{3}=c$. $3-\frac{7}{3}=\frac{9-7}{3}=\frac{2}{3}$. $\boxed{\frac{2}{3}}$.