设腰长为$x$，底边长为$y$，
则$\left\{\begin{array}{l}{x+\frac{x}{2}=12}\\{y+\frac{x}{2}=9}\end{array}\right.$，或$\left\{\begin{array}{l}{x+\frac{x}{2}=9}\\{y+\frac{x}{2}=12}\end{array}\right.$
解得：$\left\{\begin{array}{l}{x=8}\\{y=5}\end{array}\right.$，或$\left\{\begin{array}{l}{x=6}\\{y=9}\end{array}\right.$，
经检验，都符合三角形的三边关系.
等腰三角形的腰长为$6$或$8$.
故答案为：$6$或$8$.

设腰长为$x$，底边长为$y$，
则$\left\{\begin{array}{l}{x+\frac{x}{2}=12}\\{y+\frac{x}{2}=9}\end{array}\right.$，或$\left\{\begin{array}{l}{x+\frac{x}{2}=9}\\{y+\frac{x}{2}=12}\end{array}\right.$
解得：$\left\{\begin{array}{l}{x=8}\\{y=5}\end{array}\right.$，或$\left\{\begin{array}{l}{x=6}\\{y=9}\end{array}\right.$，
经检验，都符合三角形的三边关系.
等腰三角形的腰长为$6$或$8$.
故答案为：$6$或$8$.