19. For the convex quadrilateral $A B C D$, the lengths of the sides $A B, B C, C D$, and $D A$ are known, in order, to be $4, 9, 6$, and $11 \, \text{cm}$. Indicated with $E$ and $F$ the midpoints of the sides $A B$ and $C D$, it is also known that the area of the quadrilateral $B E D F$ is $18 \, \text{cm}^2$. By how many $\text{cm}^2$ is the area of the quadrilateral $A B C D$?
(A) 27
(B) 36
(C) 24
(D) more than 50
(E) a value between 40 and 50
Problem 19. The answer is $(\mathbf{B})$.

Draw the diagonal $B D$: the area of the quadrilateral $A B C D$ is equal to the sum of the areas of triangles $A D B$ and $D B C$, while the area of the quadrilateral $B E D F$ is equal to the sum of the areas of triangles $E D B$ and $D B F$. Moreover, triangle $A D B$ has an area twice that of triangle $E D B$ (being the union of triangles $A D E$ and $E D B$, which are congruent, as they have the same height and bases of equal length $A E=E B$). Similarly, triangle $D B C$ has an area twice that of triangle $D B F$ (since $B D F$ and $B F C$ are also congruent). The area of the quadrilateral $A B C D$ is thus equal to twice the area of $B E D F$, which is $36 \mathrm{~cm}^{2}$. 

(Problem proposed by S. Pelizzola)