A chord of a circle of radius 12 cm subtends an angle of 120 $°$ at the center. Find the area of the corresponding minor segment of the circle. (use π=3.14, $\sqrt{3}$ =1.73)

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88.44 cm2

87 cm2

89.44 cm2

86.74 cm2

answer is A.

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Detailed Solution

Area of minor segment = \frac{θ}{360 °} \times π r^{2} - \frac{1}{2} r^{2} \times sinθ

= \frac{120 °}{360} \times 3.14 \times 12^{2} - \frac{1}{2} \times 12^{2} \times \sin \sin (120 °)

= \frac{1}{3} \times 3.14 \times 12 \times 12 - \frac{1}{2} \times 12 \times 12 \times \frac{\sqrt{3}}{2}

= 150.72 - 62.28

= 88.44 c m^{2}

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