Question

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)

Answer 1

$o \: is \: the \: centre \: of \: a \: circle \: \\ ab \: is \: a \: chord \\ axb \: is \: a \: major \: arc \\ oa = ob = radius = 15 \\ arc \: axb \: subtends \: an \: angle \: 60 \: deg \\ area \: of \: sector \: aob = 60 \div 360 \times \pi \times r {}^{2} \\ 60 \div 360 \times 3.14 \times (15) {}^{2} = 117.75cm {}^{2} \\ area \: of \: minor \: segment \: (area \: of \: shadedregion) = area \: of \: sector \: aob - area \: of \: traingle \: aob \\ by \: trigonmetry \\ ac = 15sin30 \\ oc = 15 \: cos \: 30 \\ and \: ab \: = 2ac \\ ab = 2 \times 15sin30 = 15cm \\ oc \: = 15cos \: 30 = 15 \times \sqrt{} 3 \div 2 = 15 \times 1.73 \div 2 = 12.975 = 97.3125cm {}^{2} \\ area \: of \: triangle \: aob = 0.5 \times 15 \times 12.975 = 97.3125cm {}^{2} \\ area \: of \: minor \: segment \: (area \: of \: shaded \: region \: ) = 117.75 - 97.3125 = 20.4375cm {}^{2} \\ area \: of \: major \: segment \: = area \: of \: circle \: - area \: of \: minor \: segment \: \\ = (3.14 \times 15 \times 15) - 20.4375 \\ = 686.0625cm {}^{2}$

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