2014 AMC 10A Problem 13

We can find the areas of all the individual pieces and then add them up together. The area of the center equilateral triangle is $$\dfrac{1^2 \sqrt{3}}{4} = \dfrac{\sqrt{3}}{4}.$$ We have that the areas of all the squares is $$3 \cdot 1^2 = 3.$$ We also have that $$\angle EAF = 360^{\circ} - 60^{\circ} - 2 \cdot 90^{\circ}$$$$= 120^{\circ}.$$ Also, $AE=AF=1$ and $\angle EAF=120^\circ$. Dropping the altitude from $A$ shows that $EF=\sqrt3$ and the altitude is $\frac12$, so $[EAF]=\frac{\sqrt3}{4}$. The other two outer triangles have the same area. Thus their combined area is $\frac{3\sqrt3}{4}$. The total area is then $$\dfrac{\sqrt{3}}{4} + \dfrac{3\sqrt{3}}{4} + 3 = 3 + \sqrt{3}.$$ Thus, C is the correct answer.