Problem 2. Given is the rectangle $A B C D$ with sides $\overline{A B}=5 \mathrm{~cm}$ and $\overline{B C}=4 \mathrm{~cm}$. Point $M$ lies on side $A B$, and point $N$ lies on side $C D$. Determine the length of segment $M N$ if the perimeters of quadrilaterals $A M N D$ and $M B C N$ are both $14 \mathrm{~cm}$.
Solution. The perimeter of rectangle $ABCD$ is equal to $2(5+4)=18 \mathrm{~cm}$. The sum of the perimeters of quadrilaterals $AMND$ and $MBCN$ is equal to $2 \cdot 14=28 \mathrm{~cm}$.

If we subtract these sums, we will get twice the length of segment $MN$. Therefore,

$$
\overline{MN}=(28-18): 2=5 \mathrm{~cm}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_b353751bdd87cc544500g-09.jpg?height=330&width=397&top_left_y=633&top_left_x=1027)