An isosceles triangle has a base measuring $10 \mathrm{~cm}$ and two equal sides of $13 \mathrm{~cm}$. Is it possible to change the base of the triangle and obtain another isosceles triangle with the same area?

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-06.jpg?height=244&width=242&top_left_y=366&top_left_x=1424)

10
In an isosceles triangle, the height relative to the base coincides with the median. By drawing this height, we obtain two right triangles with legs measuring \( h \) and 5, and hypotenuse 13. By the Pythagorean Theorem, we have:

\[
h^{2}+5^{2}=13^{2} \Rightarrow h^{2}=13^{2}-5^{2}=144 \Rightarrow h=\sqrt{144}=12
\]

Thus, the area of the triangle is \( A=\frac{b \times h}{2}=\frac{10 \times 12}{2}=60 \mathrm{~cm}^{2} \).

Now, let's "glue" the two right triangles along the side measuring 5, obtaining an isosceles triangle with base \( 12+12=24 \mathrm{~m} \), sides of \( 13 \mathrm{~cm} \), and the height relative to the base equal to \( 5 \mathrm{~cm} \). Therefore, this new isosceles triangle also has an area equal to \( \frac{24 \times 5}{2}=60 \mathrm{~cm}^{2} \).

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-24.jpg?height=464&width=200&top_left_y=722&top_left_x=1518)