A trapez has parallel sides of lengths $20 \mathrm{~cm}$ and $17 \mathrm{~cm}$, a height of $2 \mathrm{~cm}$, and one of its legs forms a $45^{\circ}$ angle with the longer base. We want to divide the trapezoid into congruent pieces. For this, we can use segments whose endpoints lie on the parallel sides. Is there such a division?
It is easy to see that the given trapezoid $T=ABCD$ is not isosceles. Let $AB=20 \text{ cm}$ be the longer base. The feet of the perpendiculars from $C$ and $D$ to $AB$ are denoted by $T_1$ and $T_2$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_512639c36a306da67017g-1.jpg?height=358&width=1575&top_left_y=244&top_left_x=253)

Since one of the legs, for example $AD$, forms a $45^\circ$ angle with the base, the right triangle $ADT_2$ is isosceles, and thus $AT_2 = DT_2 = 2 \text{ cm}$. Therefore, $T_1B = 1 \text{ cm}$. All angles in the division must be equal to the angles of $\angle DAT_2$ and $\angle CBT_1$. Therefore, we translate $CT_1B$ so that $DT_2$ and $CT_1$ coincide, and try to divide $T$ into triangles congruent to $\triangle ADB$. The next triangle fits $\triangle ADB'$ rotated $180^\circ$ relative to $\triangle ADB'$, and so on. However, since one of the sides of a small triangle lying on the parallel sides of the trapezoid is $3 \text{ cm}$, and the lengths of the bases of the trapezoid are not multiples of 3, the division is certainly not possible. Let's try with another shape.

We have already seen that the dividing segments must be parallel to one of the legs of the trapezoid. If we start from $AD$, the first dividing segment will be parallel to $BC$ and should not pass through $D$.

![](https://cdn.mathpix.com/cropped/2024_05_02_512639c36a306da67017g-1.jpg?height=356&width=1586&top_left_y=1020&top_left_x=256)

The next dividing segment will be parallel to $AD$ and so on. It is obvious that the resulting pieces will be trapezoids, the difference in the lengths of their bases will be $3 \text{ cm}$, and since $T$ is not symmetric, their number will be odd.

Let $2k+1$ be the number of small trapezoids and $x > 0$ be the length of the shorter parallel side. Then for the longer base of $T$, the segment of length $x+3$ will be measured $(k+1)$ times, and the segment of length $x$ will be measured $k$ times, that is,

\[
(k+1)(x+3) + kx = 20
\]

![](https://cdn.mathpix.com/cropped/2024_05_02_512639c36a306da67017g-1.jpg?height=454&width=1488&top_left_y=1686&top_left_x=297)

Figure 3

from which

\[
x = \frac{17 - 3k}{2k + 1}, \quad \text{and since } x > 0, \quad k < \frac{17}{3}
\]

The possible values of $k$ and the corresponding values of $x$ are:

| $k$ | 1 | 2 | 3 | 4 | 5 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $x$ | $\frac{14}{3}$ | $\frac{11}{5}$ | $\frac{8}{7}$ | $\frac{5}{9}$ | $\frac{2}{11}$ |

These divisions indeed exist, for example, for $k=1$, $x=4 \frac{2}{3}$.