The area of parallelogram $A B C D$ is $B C$ times the perpendicular distance from $B C$ to $A D$. The area of triangle $\triangle B C M$ is half the base $B C$ times the perpendicular distance from $B C$ to $A D$. Therefore the area of $\triangle B C M$ is half the area of parallelogram $A B C D$; that is, $1234 / 2=617 \mathrm{~cm}^{2}$.

Similarly, the area of $\triangle B C N$ is half the area of parallelogram $B F E C$; that is, $2804 / 2=1402 \mathrm{~cm}^{2}$. Since the area of $\triangle M N C$ is the sum of the two triangular areas, we obtain that its area is $617+1402=2019$ $\mathrm{cm}^{2}$.

Note that the answer is the average of the areas of the two parallelograms.

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In the figure below, parallelograms $A B C D$ and $B F E C$ have areas $1234 \mathrm{~cm}^{2}$ and $2804 \mathrm{~cm}^{2}$, respectively. Points $M$ and $N$ are chosen on sides $A D$ and $F E$, respectively, so that segment $M N$ passes through $B$. Find the area of $\triangle M N C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_3170889cf6798dffe43cg-2.jpg?height=721&width=653&top_left_y=775&top_left_x=736)

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