ABC is an isosceles triangle. BDE, EFG, GHI and IJC are four equal isosceles triangles inside ABC triangle. D and F,F and H,H and J are connected by circular arcs. The angle ABC is 30 degrees and BE is 1m. What is the area of the shaded region ?

Since $$\angle ABC=30^o$$ and $$\triangle DBE$$ is an isosceles triangle, $$\angle BDE=120^o$$

Applying the cosine rule in $$\triangle BDE$$,

$$BE^2=BD^2+DE^2-2\times BD\times DEcos(120)$$

$$1=2BD^2+BD^2$$   [since $$BD=DE$$]

$$BD=\frac{1}{\sqrt 3}$$

Thus, $$BD=DE=EF=FG=GH=HI=IJ=JC=\frac{1}{\sqrt 3}$$

Area of $$\triangle BDE=\frac{1}{2}\times BD\times DE\times sin(120)=\frac{1}{2}\times \frac{1}{\sqrt 3}\times \frac{1}{\sqrt 3}\times \frac{\sqrt 3}{2}=\frac{1}{4\sqrt 3}$$

The total area of all smaller isosceles triangles = $$4\times \frac{1}{4\sqrt 3}=\frac{1}{\sqrt 3} m^2$$

$$\angle DEF=\angle FGH = \angle HIJ=120^o$$

Thus, the total area of circular arcs = $$\pi\times (\frac{1}{\sqrt 3})^2=\frac{\pi}{3} m^2$$

Now, side $$BC=BE+EG+GI+IC=4 m$$

Using the cosine rule in $$\triangle ABC$$, we get $$AB=AC=\frac{4}{\sqrt 3}$$

Thus, Area of $$\triangle ABC=\frac{1}{2}\times \frac{4}{\sqrt 3}\times \frac{4}{\sqrt 3}\times sin(120)=\frac{4}{\sqrt 3} m^2$$

Thus, the area of the shaded region = Area of $$\triangle ABC$$ - Area of smaller isosceles triangles - Area of the circular sections

=$$\frac{4}{\sqrt 3}-\frac{1}{\sqrt 3}-\frac{\pi}{3}=\sqrt 3 - \frac{\pi}{3}$$

Hence, the answer is option C.