Problem 7. Let's introduce the notation as shown in the figure. By extending the lateral sides, denote their point of intersection as $E$. The point of intersection of the diagonals is called $F$. Right triangles $B F C$ and $D F A$ are similar, and if the legs of the first are denoted as $x$ and $y$, then the corresponding legs of the second are $4 x$ and $4 y$.

The height of the trapezoid $h$ is composed of the heights of triangles $B F C$ and $A F D$:

$$
h=\frac{B F \cdot F C}{1}+\frac{A F \cdot F D}{4}=x y+\frac{4 x \cdot 4 y}{4}=5 x y
$$

The area of the trapezoid is $\frac{15}{16}$ of the area of triangle $A E D$. We get:

$$
\frac{1}{2} A C \cdot B D=\frac{15}{16} \cdot \frac{1}{2} \cdot A E \cdot E D \cdot \sin 60^{\circ}=\frac{15 \sqrt{3}}{64} \cdot \frac{4}{3} A B \cdot \frac{4}{3} C D=\frac{15 \sqrt{3}}{36} A B \cdot C D
$$

from which[^0]

$$
\frac{25}{2} x y=\frac{15 \sqrt{3}}{36} \sqrt{x^{2}+16 y^{2}} \cdot \sqrt{y^{2}+16 x^{2}}
$$

Considering that $x^{2}+y^{2}=1$, we find: $\frac{30}{\sqrt{3}} x y=\sqrt{1+15 y^{2}} \cdot \sqrt{1+15 x^{2}}$.

Performing obvious transformations:

$$
300 x^{2} y^{2}=1+15\left(x^{2}+y^{2}\right)+225 x^{2} y^{2} ; \quad 75 x^{2} y^{2}=16 ; \quad 5 x y=\frac{4}{\sqrt{3}}
$$
Answer: $\frac{4}{\sqrt{3}}$.