Magda cut out two identical isosceles triangles, each with a perimeter of $100 \mathrm{~cm}$. First, she formed a quadrilateral by placing the triangles together along their legs. Then, she formed a quadrilateral by placing the triangles together along their bases. In the first case, the quadrilateral she obtained had a perimeter that was $4 \mathrm{~cm}$ shorter than in the second case.

Determine the lengths of the sides of the cut-out triangles.

(E. Semerádová)
![](https://cdn.mathpix.com/cropped/2024_04_17_698df76b4228de5fb9e6g-3.jpg?height=236&width=1020&top_left_y=1378&top_left_x=521)

Hint. Was the base or the leg of the triangle longer? And by how much?
The perimeter of the first quadrilateral consists of two sides and two bases, while the perimeter of the second quadrilateral consists of four sides. Since the first quadrilateral had a perimeter $4 \mathrm{~cm}$ shorter than the second, the base must have been $2 \mathrm{~cm}$ shorter than the side.

The perimeter of each triangle consists of one base and two sides, which corresponds to three sides minus $2 \mathrm{~cm}$. At the same time, each triangle had a perimeter of $100 \mathrm{~cm}$. Therefore, the sum of the lengths of the three sides was $102 \mathrm{~cm}$.

The sides of the triangles measured $34 \mathrm{~cm}(102: 3=34)$ and the base $32 \mathrm{~cm}(34-2=32)$.

Note. If we denote the length of the base by $z$ and the length of the side by $r$, then the previous considerations can be written as follows (units cm are not specified in the following):

The difference in the perimeters of the quadrilaterals was

$$
4 r-2 r-2 z=2 r-2 z=4
$$

thus $r-z=2$ or $z=r-2$. The perimeter of each triangle was

$$
2 r+z=3 r-2=100,
$$

thus $3 r=102$. From this, we get $r=102: 3=34$ and $z=34-2=32$.

Triangles with such sides are indeed possible, as the triangle inequalities are satisfied $(34+32>34$ and $34+34>32)$.