5. Rectangles $A B N F$ and $C M K D$ are overlapped such that their common part is rectangle $M N F K$. It is given that $B N=8$ cm and $K D=9$ cm. Find the area of rectangle $M N F K$, if the area of $A B M K$ is 25 cm $^{2}$, and the area of $C D F N$ is $32 \mathrm{~cm}^{2}$.

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Solution. Let $AB = x$ cm, then the area of rectangle $ABNF$ is $8x \, \text{cm}^2$, and the area of rectangle $MNFK$ is $8x - 25$ cm$^2$. On the other hand, the area of rectangle $CMKD$ is $9x$ cm$^2$, and the area of rectangle $MNFK$ is $9x - 32$ cm$^2$. The area of rectangle $MNFK$ has been found in two ways, so we can set up the equation $8x - 25 = 9x - 32$, from which $x = 7$, so the area of rectangle $MNFK$ is $8 \cdot 7 - 25 = 31 \, \text{cm}^2$. Answer: $31 \, \text{cm}^2$.