Two circles with centers A and A' and having radius = 2 cm
\(\angle BAC = 120^\circ \, \text{(since ABA' forms an equilateral triangle)} \, \text{and length of an arc} = \frac{\theta}{360^\circ} \times 2\pi r\)
\(\text{Perimeter of shaded portion} = 2 \times \left( \text{length of arc BC} \right)\)
\(= 2 \times \left( \frac{120^\circ}{360^\circ} \times 2\pi r \right)\)
\(= 2 \times \left( \frac{1}{3} \times 2\pi \times 2 \right)\)
\(= 2 \times \frac{4\pi}{3} = \frac{8\pi}{3} \)
The correct answer is (A) : (8/3) π