1. Let $\triangle A B C$ have interior angles $\angle A, \angle B, \angle C$ with opposite sides of lengths $a, b, c$ respectively. If $c=1$, and the area of $\triangle A B C$ is equal to $\frac{a^{2}+b^{2}-1}{4}$, then the maximum value of the area of $\triangle A B C$ is $\qquad$
- 1. $\frac{\sqrt{2}+1}{4}$.

By the cosine theorem and the area formula, we obtain
$$
\angle C=45^{\circ} \text {. }
$$

Thus $1=a^{2}+b^{2}-\sqrt{2} a b \geqslant(2-\sqrt{2}) a b$
$$
\Rightarrow S_{\triangle A B C}=\frac{\sqrt{2}}{4} a b \leqslant \frac{\sqrt{2}+1}{4} \text {, }
$$

with equality holding if and only if $a=b=\sqrt{\frac{2+\sqrt{2}}{2}}$.