(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 POQ ; (ii) shaded part. (b) If the sector POQ in (a) is used to form the curved surface of a cone with vertex O, calculate the base radius of the cone, correct to one decimal place.
Explanation
(a) (i) Area of minor segment POQ = \(\frac{84}{360} \times \frac{22}{7} \times (3.2)^{2}\) = \(\frac{11 \times 10.24}{15}\) = \(7.5093 cm^{2} \approxeq 7.51 cm^{2}\) (2 decimal places) (ii) Area of \(\Delta\) in POQ = \(\frac{1}{2} \times 3.2 \times 3.2 \times \sin 84°\) = \(\frac{127.296}{25} = 5.0918 cm^{2}\) Area of segment PQ (shaded part) = Area of sector POQ - Area of triangle = \(7.5093 cm^{2} - 5.0918 cm^{2} = 2.4175 cm^{2}\) \(\approxeq 2.42 cm^{2}\) (b) Area of sector POQ = Curved surface area of the cone \(7.5093 cm^{2} = \pi rl\) where r = base radius, L = slant height = 3.2cm \(7.5093 = \frac{22}{7} \times r \times 3.2 = \frac{70.4r}{7}\) \(70.4r = 7.5093 \times 7 \implies r = \frac{7.5093 \times 7}{70.4} = 0.7467 cm\) \(\approxeq 0.7cm\)
