7. A line is parallel to side $\mathrm{BC}$ of the known $\triangle \mathrm{ABC}$, intersects the other two sides at $D$ and $E$, and makes the area of $\triangle BDE$ equal to the constant $\mathrm{k}^{2}$. What is the relationship between $\mathrm{k}^{2}$ and the area of $\triangle \mathrm{ABC}$ for the problem to be solvable, and how many solutions are there?
7. Let S denote the area of $\triangle ABC$ (Fig. 7), and set $\frac{AD}{AB}=x$. Then the area of $\triangle ADE$ is equal to $x^{2} S$, and the area of $\triangle ABE$ is equal to $\mathrm{J} \mathbf{x S}$ (this conclusion is not difficult to obtain - poem difficulty). According to the conditions of the problem, we can obtain the following equation:
$$
x S - x^{2} S = k^{2} \text{.}
$$

Solving it, we get:
$$
x = \frac{1 \pm \sqrt{1 - \frac{4 k^{2}}{S}}}{2} .
$$

Therefore, when $S \geqslant 4 k^{2}$, the problem is solvable; and if $S > 4 k^{2}$, there are two solutions; if $S = 4 k^{2}$, there is one solution.