Solution: Triangle $A B C$ has area $\sqrt{3}$, and each of the three squares has area 4 . The three remaining regions are congruent, so just consider the one that includes vertex $B$. Triangle $K B N$ has two sides of length 2 and an angle of $120^{\circ}$ between them, to bisecting that angle we get two halves of an equilateral triangle of side length 2 , so the area is again $\sqrt{3}$. The remaining region is an equilateral triangle of side length $2 \sqrt{3}$, so its area is $(2 \sqrt{3})^{2} \sqrt{3} / 4=3 \sqrt{3}$. Therefore the area of $A^{\prime} B^{\prime} C^{\prime}$ is $\sqrt{3}+3 \cdot 4+3 \cdot \sqrt{3}+3 \cdot 3 \sqrt{3}=\mathbf{1 2}+\mathbf{1 3} \sqrt{\mathbf{3}}$

Note that this problem is still solvable, but much harder, if the first triangle is not equilateral.
Squares $A B K L, B C M N, C A O P$ are drawn externally on the sides of a triangle $A B C$. The line segments $K L, M N, O P$, when extended, form a triangle $A^{\prime} B^{\prime} C^{\prime}$. Find the area of $A^{\prime} B^{\prime} C^{\prime}$ if $A B C$ is an equilateral triangle of side length 2 .