12. 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 $5, 8, 7$, and $12 \, \text{cm}$. 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 $24 \, \text{cm}^2$. By how many $\text{cm}^2$ is the area of the quadrilateral $A B C D$?
(A) 32
(B) 36
(C) 48
(D) a value between 50 and 60
(E) more than 60
Problem 12. The answer is (C).

Draw the diagonal $BD$: the area of the quadrilateral $ABCD$ is equal to the sum of the areas of triangles $ADB$ and $DBC$, while the area of the quadrilateral $BEDF$ is equal to the sum of the areas of triangles $EDB$ and $DBF$. Moreover, triangle $ADB$ has an area twice that of triangle $EDB$ (being the union of triangles $ADE$ and $EDB$, which are congruent, since they have the same height and bases of equal length $AE=EB$). Similarly, triangle $DBC$ has an area twice that of triangle $DBF$ (since $BDF$ and $BFC$ are also congruent). The area of the quadrilateral $ABCD$ is thus equal to twice the area of $BEDF$, which is $48 \mathrm{~cm}^{2}$. 

(Problem proposed by S. Pelizzola)