10.247. The bases of two equilateral triangles with sides $a$ and $3a$ lie on the same straight line. The triangles are located on opposite sides of the line and do not have any common points, and the distance between the nearest ends of their bases is $2a$. Find the distance between the vertices of the triangles that do not belong to the given line.
Solution.

Let in the equilateral triangles $A B C$ and $D F E$ (Fig.10.54) $A C=a$, $D E=3 a, C D=2 a$. Since $\angle B C D=\angle F D C=120^{\circ}$, then $B C \| D F, K$ is the intersection point of the line $D F$ and the line passing through point $B$ parallel to $A C$. Therefore, $B K D C$ is a parallelogram, $B K=C D=2 a$, $K D=B C=a, K F=K D+D F=4 a, \angle B K D=\angle C D F=120^{\circ}$. From $\triangle B K F:$

$$
\begin{aligned}
& B F=\sqrt{B K^{2}+K F^{2}-2 B K \cdot K F \cos \angle B K F}= \\
& =\sqrt{4 a^{2}+16 a^{2}-2 \cdot 2 a \cdot 4 a \cos 120^{\circ}}=2 a \sqrt{7}
\end{aligned}
$$

Answer: $2 a \sqrt{7}$.