7. Equilateral triangle $A B C$ of side length 2 is drawn. Three squares external to the triangle, $A B D E, B C F G$, and $C A H I$, are drawn. What is the area of the smallest triangle that contains these squares?
Solution: The equilateral triangle with sides lying on lines $D G, E H$, and $F I$ has minimal area. (The only other reasonable candidate is the triangle with sides along $D E, F G, H I$, but a quick sketch shows that it is larger.) Let $J, K$, and $L$ be the vertices of this triangle closest to $D, H$, and $F$, respectively. Clearly, $K I=F L=2$. Triangle $F C I$ is a $30^{\circ}-30^{\circ}-120^{\circ}$ triangle, so we can calculate the length of $F I$ as $2 \sqrt{3}$, making the side length of $\triangle J K L$ $4+2 \sqrt{3}$, and its area $12+7 \sqrt{3}$.