A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?

$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3}$
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](https://artofproblemsolving.com/wiki/index.php/Triangle) $AFB$ and $DFB$ share an [altitude](https://artofproblemsolving.com/wiki/index.php/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 $x+y = \frac{15}{2}+\frac{21}{2} = \boxed{\textbf{(D) }18}$.