Math, asked by Kunjeet, 9 months ago

A chrod of a cirle of a radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segment of the circle.
 \rm \: use \: \pi = 3.14 \: and \:  \sqrt{3}  = 1.73

Answers

Answered by falgunivaidya23
3

Answer:

minor segment = 20.38 ≈ 20.4 cm²

major segment = area of circle - area of minor segment

                         = 706.86 - 20.4

                         = 686.46 cm²

Answered by Anonymous
102

 \huge{\underline {\underline{\boxed{\red{Solution:-}}}}}

 \rm \: r = 15 \: cm

 \theta \:  = 60 \degree

 \rm \: Area \: of \: the \: minor \: sector

  \:  \:  \:  \:  \:  \: \rm =  \frac{ \theta}{360 \degree} \times \pi r {}^{2}  \\

  \: \:  \:  \:  \:  \:  \rm =  \frac{60 \degree}{360 \degree} \times 3.14 \times (15) {}^{2}  \\

 \:  \:  \:  \:  \:  \:  \:  \rm = 117.75 \: cm {}^{2}

 \bf Area \: of \triangle \: AOB

  \: \:  \:  \:  \rm =  \frac{1}{2} \: r {}^{2}  \: sin  \: \theta \\

 \:  \:  \:  \:  \:  \:  \:  \:  \rm =  \frac12(15) {}^{2}  \: sin \: 60 \degree \\

 \:  \:  \:  \:  \:  \rm =  \frac12 \times 15 \times 15  \: \times \frac{\sqrt3}{2} \\

 \:  \:  \:  \:  \:  \rm =  \frac{225 \times  \sqrt{3 \: } }{4} \\

 \:  \:  \:  \:  \:  \:  \:  \rm =  \frac{225 \times 1.73}{4} \\

\implies \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \sf  [∵ \sqrt{3}  = 1.73(given)]

 \:  \:  \:  \:  \:  \:  \rm = 97.3125 \: cm {}^{2}

 \therefore \:  \:  \:  \:  \:  \rm Area \: of \: the \: corresponding \: minor \: segment \:  \\  \rm \:  \:  \:  \:  \:  \:  \:  \:  \:  \: of \: the \: circle \:

\:  \rm = \:  Area \: of \: the \: minor \: sector - Area \: of \triangle \: AOB

 \:  \:  \:  \:  \:  \rm =  \: 117.75 - 97.3125 = 20.4375 \: cm {}^{2}

 \:  \:  \rm \: and, \: area \: of \: the \: corresponding \: major \: segment \: of \\  \:  \:  \:  \:  \:  \rm the \: circle

 \:  \:  \:  \:  \:  \rm = Area \: of \: the \: circle \: i.e. \: \pi  r {}^{2}   \\  \rm \:   \:  \:  \:  \:  \:  \: -  \: Area \: of \: the \: corresponding \: minor \: segment \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \rm of \: the \: circle

 \:  \:  \:  \:  \:  \:  \:  \rm =  \:  \: 3.14 \times 15 \times 15 - 20.4375

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \rm = 706.5 - 20..4375

 \:  \:  \: \:  \:   \:  \:  \:  \:  \sf \red {= 686.0625 \: cm {}^{2} }

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