1. A n s w e r: $\frac{1}{4}$.

F i r s t s o l u t i o n. Let the vertices of the square be denoted as $A, B, C, D$, such that $L$ lies on $A B$, $M$ lies on $B C$, $N$ lies on $C D$, and $K$ lies on $A D$. Denote the intersection point of $K M$ and $L N$ as $O$. Let the lengths of segments $A L$ and $A K$ be $x$ and $y$ respectively. Then, by the Pythagorean theorem, $1-x-y=\sqrt{x^{2}+y^{2}}$. Squaring both sides, we get $1=2 x+2 y-2 x y$. Since $A D\|L N\| B C$ and $A B\|K M\| C D$, it follows that $O L=y$ and $O K=x$. Therefore, $O M=1-x$ and $O N=1-y$. This means $N C=1-x$ and $M C=1-y$. Thus, the area of triangle $N M C$ is $\frac{1}{2} N C \cdot M C=\frac{1}{2}(1-x)(1-y)=$ $=\frac{1}{2}(1-x-y+x y)=\frac{1}{4}$.
The second solution. Let's denote the vertices of the square as in the first solution. Consider the circle $\omega$ inscribed in the square. Let $K^{\prime}, L^{\prime}$ be the points of tangency of the circle with the sides $A D$ and $A B$, respectively. These are the midpoints of the sides $A D$ and $A B$. From the fact that the perimeter of triangle $A K L$ is twice the length of segment $A K^{\prime}$, we conclude that the circle $\omega$ is an excircle for triangle $A K L$. Using the formula expressing the area of a triangle in terms of the semiperimeter, the length of a side, and the radius of the excircle touching that side, we have that $S_{A K L}=\frac{1}{2}\left(\frac{1}{2}-K L\right)$. Therefore,

$$
\begin{aligned}
S_{M N C}=\frac{1}{2} S_{M O N C}=\frac{1}{2}\left(1-S_{M K A B}\right. & \left.-S_{N L A D}+2 S_{A K L}\right)= \\
& =\frac{1}{2}\left(1-A K-A L+\frac{1}{2}-K L\right)=\frac{1}{4} .
\end{aligned}
$$