imc

The area of the equilateral triangle is $\frac{\sqrt{3}}{4}$. The area of the three squares is $3\times 1=3$. Since $\angle C=360$, $\angle GCH=360-90-90-60=120$. Dropping an altitude from $C$ to $GH$ allows to create a $30-60-90$ triangle since $△GCH$ is isosceles. This means that the height of $△GCH$ is $\frac{1}{2}$ and half the length of $GH$ is $\frac{\sqrt{3}}{2}$. Therefore, the area of each isosceles triangle is $\frac{1}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$. Multiplying by $3$ yields $\frac{3\sqrt{3}}{4}$ for all three isosceles triangles. Therefore, the total area is $3+\frac{\sqrt{3}}{4}+\frac{3\sqrt{3}}{4}=3+\frac{4\sqrt{3}}{4}=3+\sqrt{3}⟹\boxed{3+\sqrt{3}}$.