(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)