imc

Label the points in the figure as shown below, and draw the segment $CF$. This segment divides the quadrilateral into two triangles, let their areas be $x$ and $y$. Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$. Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$. Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB=3:7$. Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7)=3:7$, which gives us $7x=3(y+7)$. Substituting $y=x+3$ into the second equation we get $7x=3(x+10)$, which solves to $x=\frac{15}{2}$. Then $y=x+3=\frac{15}{2}+3=\frac{21}{2}$, and the total area of the quadrilateral is $x+y=\frac{15}{2}+\frac{21}{2}=\boxed{18}$.