An isosceles triangle has a base of $10 \mathrm{~cm}$ and two equal sides measuring $13 \mathrm{~cm}$. Is it possible to cut this triangle into two other triangles in such a way that, by joining these triangles in another manner, we obtain another isosceles triangle (obviously with the same area)?

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-032.jpg?height=251&width=192&top_left_y=1325&top_left_x=1640)
In an isosceles triangle, the height relative to the base coincides with the median. By drawing this height in the given triangle, with a base of 10, we obtain two right triangles with legs measuring 5 and $h$ and hypotenuse 13. By the Pythagorean Theorem, we have $h^{2}+5^{2}=13^{2}$, hence $h^{2}=13^{2}-5^{2}=144 \text{ and}$, therefore, $h=\sqrt{144}=12$. Thus, the area of the given triangle is

$$
A=\frac{b \times h}{2}=\frac{10 \times 12}{2}=60 \text{~cm}^{2}
$$

Now, by "gluing" the two right triangles along the leg measuring 5, we obtain an isosceles triangle with a base of $12+12=24 \text{~cm}$, sides of $13 \text{~cm}$, and a height relative to the base equal to $5 \text{~cm}$. Thus, this new isosceles triangle also has an area equal to $\frac{24 \times 5}{2}=60 \text{~cm}^{2}$.