1. From a rectangular sheet of paper with dimensions $54 \mathrm{~cm}$ and $15 \mathrm{~cm}$, Marko cuts out the largest possible square. From the remaining piece of paper, he again cuts out the largest possible square and continues the process until the entire sheet is cut into squares, i.e., until the remaining piece of paper is also a square. What is the sum of the perimeters of all the squares thus obtained?
## Solution.

From a rectangular piece of paper with dimensions $54 \mathrm{~cm}$ and $15 \mathrm{~cm}$, Marko first cuts out three squares with a side length of $15 \mathrm{~cm}$.

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After that, from the rectangular piece of paper with dimensions $54-3 \cdot 15=9$, i.e., $9 \mathrm{~cm}$ and $15 \mathrm{~cm}$, Marko cuts out one square with a side length of $9 \mathrm{~cm}$.

After that, from the rectangular piece of paper with dimensions $15-9=6$,

i.e., $6 \mathrm{~cm}$ and $9 \mathrm{~cm}$, Marko cuts out one square with a side length of $6 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_e5ca9b3e8971fb159cf9g-1.jpg?height=200&width=282&top_left_y=1913&top_left_x=1595)

After that, from the rectangular piece of paper with dimensions $9-6=3$,

i.e., $3 \mathrm{~cm}$ and $6 \mathrm{~cm}$, Marko cuts out two squares with a side length of $3 \mathrm{~cm}$.

The sum of the perimeters of all the squares is $3 \cdot 60+36+24+2 \cdot 12=180+36+24+24=264 \mathrm{~cm}$.