## Solution.

Let $A L=a, L B=b$ and $L K=c$ (Fig. R3 103). Then the area of the given square $S_{1}=(a+b)^{2}=a^{2}+2 a b+b^{2}=c^{2}+2 a b$. Since $\angle B L K=60^{\circ}$, then $\angle B K L=30^{\circ}$, from which $c=2 b, a=b \sqrt{3}$ and, therefore, $S_{1}=4 b^{2}+2 b^{2} \sqrt{3}$. The area of the inscribed square $S_{2}=c^{2}=4 b^{2}$. Thus,

$$
\frac{S_{2}}{S_{1}}=\frac{4 b^{2}}{4 b^{2}+2 b^{2} \cdot \sqrt{3}}=\frac{2}{2+\sqrt{3}}=4-2 \sqrt{3}
$$

Answer: $4-2 \sqrt{3}$.
10.107. A square is inscribed in another square, with its vertices lying on the sides of the first square, and its sides forming angles of $60^{\circ}$ with the sides of the first square. What fraction of the area of the given square is the area of the inscribed square?