Let's color the sides of the pieces as follows: the longest side green, the side parallel to it blue, the side perpendicular to them yellow, and the remaining side red; the lengths of the corresponding sides will be denoted briefly as $z, m$, ž, and č. Further, let's mark the selected "vertices" of the individual pieces as in the following image:

![](https://cdn.mathpix.com/cropped/2024_04_17_f803abc88728297ab2efg-3.jpg?height=757&width=674&top_left_y=495&top_left_x=725)

The perimeter of the shape is composed of:

- 3 blue segments $B C, H I$, and $K L$, whose lengths are $m$,
- 2 green segments $A B$ and $J K$, whose lengths are $z$,
- 4 yellow segments $C D, D E, N O$, and $O A$, whose lengths are $\check{z}$,

![](https://cdn.mathpix.com/cropped/2024_04_17_f803abc88728297ab2efg-3.jpg?height=45&width=1099&top_left_y=1531&top_left_x=290)

- 2 identical segments $E F$ and $M N$ and 1 segment $G H$, whose lengths are currently unknown.

The lengths of segments $M N$ and $E F$ together with the green side $E R$ and the blue side $R N$ form the segment $M F$, which is composed of two green sides. In other words,

$$
|E F| + |M N| = z + z - (z + m) = z - m \text{.}
$$

The length of the blue side $K G$ can be expressed as the sum of the lengths of the yellow side $K H$ and the segment $G H$, i.e.,

$$
|G H| = m - \check{z} \text{.}
$$

In total, the perimeter of the shape is

$$
3 m + 2 z + 4 \check{z} + 3 \check{c} + (z - m) + (m - \check{z}) = 3 m + 3 z + 3 \check{z} + 3 \check{c} = 3(m + z + \check{z} + \check{c}) \text{.}
$$

We have thus obtained three times the perimeter of one quadrilateral piece, so the perimeter of the entire shape is

$$
3 \cdot 17 = 51(\mathrm{~cm}) \text{.}
$$

Note. If we first move some parts of the shape so that the perimeter remains unchanged, some considerations can be simplified, see for example the following image.

![](https://cdn.mathpix.com/cropped/2024_04_17_f803abc88728297ab2efg-4.jpg?height=452&width=854&top_left_y=228&top_left_x=635)

Another solution. Separate the "chimney" from the "roof" and the "roof" from the "walls" as shown in the image.

![](https://cdn.mathpix.com/cropped/2024_04_17_f803abc88728297ab2efg-4.jpg?height=777&width=574&top_left_y=908&top_left_x=775)

The sum of the perimeters of these three shapes can be easily determined (we will use the notation $m, z, \check{z}, \check{c}$ as above):

$$
5 m + 5 z + 5 \check{z} + 3 \check{c} \text{.}
$$

The given sum is greater than the perimeter of the original shape by two lengths $\check{z}$, which was caused by separating the "chimney," and by two sums of lengths $m + z$, which was caused by separating the "walls." The perimeter of the original shape is therefore

$$
(5 m + 5 z + 5 \check{z} + 3 \check{c}) - \check{z} - \check{z} - (m + z) - (m + z) = 3 m + 3 z + 3 \check{z} + 3 \check{c} \text{.}
$$

We see that the original shape has a perimeter three times larger than that of the quadrilateral piece, i.e., $3 \cdot 17 = 51(\mathrm{~cm})$.
In the following figure, a shape is composed of seven identical quadrilateral pieces of a construction set. What is the perimeter of this shape if the perimeter of one quadrilateral piece is $17 \mathrm{~cm}$?

(K. Pazourek)

![](https://cdn.mathpix.com/cropped/2024_04_17_f803abc88728297ab2efg-2.jpg?height=700&width=568&top_left_y=1877&top_left_x=775)