(a) The equal base angles are \[t=\angle CAB=\frac{\pi-1.8}{2}.\] The circle radius is \[R=AC=12\sin0.9\approx9.39992.\] Since \(AD=R\) and \(AB=6\), \(BD=R-6\). The boundary comprises BC, BD and arc CD: \[P=6+(R-6)+Rt=R(1+t)\approx15.7\text{ cm}.\]

(b) Subtract triangle ABC from sector ADC: \[A=\frac12R^2t-\frac12(6)(6)\sin1.8\approx12.1\text{ cm}^2.\] Unrounded values of \(R\) and \(t\) are used.