1.8. In an isosceles triangle, the heights drawn to the base and to the lateral side are equal to 10 and 12 cm, respectively. Find the length of the base.
1.8. In $\triangle A B C$ we have $A B=B C$, $A E \perp B C, B D=10$ cm and $A E=12$ cm (Fig. 1.7). Let $A C=x, A B=B C=y$. Right triangles $A E C$ and $B D C$ are similar (angle $C$ is common); therefore, $B C: A C=B D: A E$, or $y: x=10: 12=5: 6$. Applying the Pythagorean theorem (1.13) to $\triangle B D C$, we have $B C^{2}=B D^{2}+D C^{2}$, i.e.,
$y^{2}=100+\frac{x^{2}}{4}$. Thus, we arrive at the system of equations

$$
\left\{\begin{array}{l}
\frac{y}{x}=\frac{5}{6} \\
y^{2}=100+\frac{x^{2}}{4}
\end{array}\right.
$$

solving which we get $x=15$.

Answer: 15 cm.