Discussion

Explanation:

Since ∠ABC = 30° and ∆DBE is an isosceles triangle, ∠BDE = 120°

Applying the cosine rule in △BDE.

BE² = BD² + DE² - 2 × BD × DE cos(120)

1 = 2BD² + BD²                [since BD = DE]

BD = $\frac{1}{\sqrt{3}}$

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

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

The total area of all smaller isosceles triangles = 4 × 4$\frac{1}{\sqrt{3}}$ = $\frac{1}{\sqrt{3}}$ m²

∠DEF = ∠FGH = ∠HIJ = 120°
Thus, the total area of circular arcs = π × ($\frac{1}{\sqrt{3}}$)² = $\frac{\pi }{3}$m²

Now, side BC = BE + EG + GI + IC = 4m

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

Thus, Area of △ABC = $\frac{1}{2}$ × $\frac{4}{\sqrt{3}}$ × $\frac{4}{\sqrt{3}}$ × sin(120) = $\frac{4}{\sqrt{3}}$m²

Thus, the area of the shaded region = Area of △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.

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