(a) In the diagram, O is the centre of the circle radius 3.2cm. If < PRQ = 42°, calculate, correct to two decimal places, the area of the: (i) minor sector ...

From the diagram, \(O\) is the centre, radius \(r = 3.2\ \text{cm}\), and \(\angle PRQ = 42^\circ\) is an angle at the circumference standing on chord \(PQ\).

Angle at the centre. The angle subtended at the centre is twice the angle at the circumference on the same arc:

\[ \angle POQ = 2 \times 42^\circ = 84^\circ \]

(a)(i) Area of minor sector POQ.

\[ A_{\text{sector}} = \frac{\theta}{360^\circ}\,\pi r^2 = \frac{84}{360}\times \pi \times (3.2)^2 \]

\[ = \frac{84}{360}\times \pi \times 10.24 = 0.23333 \times 32.1699 = 7.51\ \text{cm}^2 \]

(a)(ii) Area of the shaded part. The shaded region is the minor segment cut off by chord \(PQ\); it equals the sector minus triangle \(POQ\).

\[ A_{\triangle POQ} = \tfrac{1}{2} r^2 \sin\theta = \tfrac{1}{2}\times 10.24 \times \sin 84^\circ = 5.12 \times 0.99452 = 5.0920\ \text{cm}^2 \]

\[ A_{\text{shaded}} = 7.5063 - 5.0920 = 2.41\ \text{cm}^2 \]

(b) Base radius of the cone. When the sector is rolled into a cone, its arc length becomes the circumference of the base, while the sector radius becomes the slant height.

Arc length of sector:

\[ \ell = \frac{\theta}{360^\circ}\times 2\pi r = \frac{84}{360}\times 2\pi \times 3.2 = 4.6915\ \text{cm} \]

Set equal to base circumference \(2\pi R\):

\[ 2\pi R = 4.6915 \;\Rightarrow\; R = \frac{4.6915}{2\pi} = 0.7467 \approx 0.7\ \text{cm} \]

Answers: (i) \(7.51\ \text{cm}^2\); (ii) \(2.41\ \text{cm}^2\); (b) base radius \(\approx 0.7\ \text{cm}\).