(a) The area of sector ABC is given by \(\frac{1}{2} r^2 \theta\). Substituting \(\theta = \frac{1}{6} \pi\), the sector area is \(\frac{1}{2} r^2 \times \frac{1}{6} \pi = \frac{1}{12} \pi r^2\).

The triangle ABD has base AD and height BD. Using trigonometry, \(BD = r \sin \frac{\pi}{6} = \frac{1}{2} r\) and \(AD = r \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} r\).

The area of triangle ABD is \(\frac{1}{2} \times \frac{1}{2} r \times \frac{\sqrt{3}}{2} r = \frac{\sqrt{3}}{8} r^2\).

Therefore, the area of BCD is \(\frac{1}{12} \pi r^2 - \frac{\sqrt{3}}{8} r^2\).

(b) Given BD =\( \frac{\sqrt{3}}{2} r\), use trigonometry to find angle BAC: \(\angle BAC = \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3}\).

Then, \(AD = r \cos \frac{\pi}{3} = \frac{1}{2} r\) and \(CD = \frac{1}{2} r\).

The length of arc BC is \(r \times \frac{\pi}{3}\).

Thus, the perimeter of BCD is \(\frac{\sqrt{3}}{2} r + \frac{1}{2} r + \frac{\pi}{3} r\).