1. As shown in Figure 1, in $\triangle A B C$, $\angle A=70^{\circ}$, $\angle B=90^{\circ}$, the symmetric point of point $A$ with respect to $B C$ is $A^{\prime}$, the symmetric point of point $B$ with respect to $A C$ is $B^{\prime}$, and the symmetric point of point $C$ with respect to $A B$ is $C^{\prime}$. If the area of $\triangle A B C$ is 1, then the area of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is $\qquad$
-1.3 .
As shown in Figure 6, connect $B^{\prime} B$, and extend it to intersect $C^{\prime} A^{\prime}$ at point $D$, and intersect $A C$ at point $E$. Given $C^{\prime} B=B C$, $A^{\prime} B=B A, A C \perp A^{\prime} C^{\prime}$, and $B B^{\prime} \perp A C, B^{\prime} E=B E$, we get $B^{\prime} D=3 B E$. Therefore,
$$
\begin{array}{c}
S_{\triangle A^{\prime} B C}=\frac{1}{2} B^{\prime} D \cdot A^{\prime} C^{\prime} \\
=3 \times \frac{1}{2} B E \cdot A C=3 S_{\triangle A B C}=3
\end{array}
$$