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Find the area of a segment of a circle of radius 21cm, if the angle made by the arc of the segment has a measure of $60 °$ ?

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45.27 cm2

40.27 cm2

40.8 cm2

44.27 cm2

answer is B.

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Detailed Solution

If the angle is 360 ° the area is π r^{2}

\to If angle is 60 °, Area of sector OAPB = \frac{π r^{2}}{360} \times 60

= \frac{22}{7} \times \frac{21^{2}}{360} \times 60

= 231 c m^{2}

In ∆ AOM, \sin \sin 30 ° = \frac{AM}{OA}

\frac{1}{2} = \frac{AM}{21}

AM = \frac{21}{2} cm

\cos \cos 30 ° = \frac{OM}{OA}

\frac{\sqrt{3}}{2} = \frac{OM}{21}

OM = \frac{21 \sqrt{3}}{2} cm

AB = 2 AM

AB = 2 \times \frac{21}{2}

AB = 21 cm

Area of ∆ AOB = \frac{1}{2} \times AB \times OM

= \frac{1}{2} \times 21 \times \frac{21 \sqrt{3}}{2}

= \frac{441 \sqrt{3}}{4} c m^{2}

Area of Segment = Area of sector - Area Of triangle

= 231 - \frac{441 \sqrt{3}}{4}

= 231 - 190.83

= 40.27 c m^{2}

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