15.11. It is given that $\triangle A B C \sim \triangle D E F$. If the area of $\triangle A B C$ is $\frac{3}{2}$ times that of $\triangle D E F$ and $A B=B C=A C=2$, what is the perimeter of $\triangle D E F ?$
15.11. $2 \sqrt{6}$
We first define a notation. Let $(X Y Z)$ denote the area of $\triangle X Y Z$.
$$
\begin{array}{c}
\left(\frac{A B}{D E}\right)^{2}=\frac{(A B C)}{(D E F)}=\frac{\frac{3}{2}(D E F)}{(D E F)}=\frac{3}{2} \Longrightarrow \frac{A B}{D E}=\sqrt{\frac{3}{2}} \\
D E=\sqrt{\frac{2}{3}} A B=\frac{2 \sqrt{6}}{3}=E F=D F \\
\text { perimeter of } \triangle D E F=3 \cdot \frac{2 \sqrt{6}}{3}=2 \sqrt{6}
\end{array}
$$