10.027. Each of the three equal circles of radius $r$ touches the other two. Find the area of the triangle formed by the common external tangents to these circles.
## Solution.

Triangle $A B C$ is equilateral (Fig. 10.27); therefore, its area is $\frac{a^{2} \sqrt{3}}{4}$, where $a$ is the side of the triangle. Draw $O K \perp A C$. Since $\angle O A K=30^{\circ}$, then $A O=2 r, A K=r \sqrt{3}$, hence $a=2 r \sqrt{3}+2 r=2 r(\sqrt{3}+1)$. Consequently, $S=\frac{4 r^{2}(3+2 \sqrt{3}+1) \sqrt{3}}{4}=2 r^{2}(2 \sqrt{3}+3)$.

Answer: $2 r^{2}(2 \sqrt{3}+3)$