Question

8. (a) Calculate the area of the shaded portion in Fig 2.1 (a sector of radius 8cm with angle 54 degrees). (b) A tin of milk has radius 6.5cm and height 13cm. Find, correct to 2 decimal places: i. Total surface area of the tin ii. Volume of liquid in litres that will fill the tin (pi=22/7). (12 marks)

Accepted Answer

(a) Area of sector = 54/360 x 22/7 x 8^2 = 30.171cm^2. Area of triangle = 1/2 x 8^2 x sin(54) = 25.8885cm^2. Area of segment (shaded portion) = 30.171-25.8885 = 4.2825cm^2.
(b)(i) Total surface area = 2 pi r(h+r) = 2 x 22/7 x 6.5 x (13+6.5) = 796.71cm^2.
(ii) Volume = pi r^2 h = 22/7 x 6.5^2 x 13 ~= 1726.21cm^3 = 1.73 litres (since 1 litre=1000cm^3).

GCE Mathematics 2025 Theory — Question 8

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