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 segment of the circle.

Answer 1

824.99. formula is given angle/360multiply by area of circle

Answer 2

$\blue\bigstar$QUESTION:-

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A chord of a circle of radius 15 cm subtends an angle of 60 at the centre find the area of the corresponding minor and major segment of circle

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$\red\bigstar$SOLUTION:-

↬Given:-

$\sf \bullet \: radius \: (r) = 15cm \\ \sf \bullet \: \theta = 60^{0} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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Now, area of minor sector

$\sf \implies \dfrac{ \theta}{{360}^{0} } \times \pi {r}^{2}$

$\implies \sf\dfrac{ {60}^{0} }{ {360}^{0} } \times 3.14 \times {(15)}^{2}$

$\implies \sf \dfrac{1}{6} \times 3.14 \times 225$

$\implies \sf117.75 \: {cm}^{2}$

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$\sf \: Area \: of \: \triangle AOB$

$\sf Draw \: OM\perp AB$

In right triangles OMA and OMB

OA=OB⠀ ⠀⠀| radii of the circle |

OM=OM ⠀ ⠀| common side |

$\therefore \sf \triangle OMA \cong \triangle OMB \: \: \: |rhs \: congruency|$

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And

$\sf \angle AOM =\angle BOM \: \: \: \: |cpct| \\ \sf \implies \dfrac{1}{2}\angle AOB = \dfrac{1}{2} \times (60)^{0} = 30^{0}$

Now , In right triangle OMA

$\sf \: cos {30}^{0} = \dfrac{OM}{OA}$

$\sf \implies \dfrac{ \sqrt{3} }{2} = \dfrac{OM}{15}$

$\implies \sf OM = \dfrac{ 15 \sqrt{3} }{2} cm$

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$\sf \: sin {30}^{0} = \dfrac{AM}{OA}$

$\implies \sf \dfrac{1}{2} = \dfrac{AM}{15}$

$\implies \sf \: AM = \dfrac{15}{2}$

$\implies \sf \: 2AM = 2 \bigg( \dfrac{15}{2} \bigg) = 15$

$\sf\implies AB=15 cm$

Area of triangle AOB

$\implies \sf \dfrac{1}{2} \times AB\times OM$

$\implies \sf \dfrac{1}{2} \times 15 \times \dfrac{15 \sqrt{3} }{2}$

$\implies \sf \dfrac{225 \sqrt{3} }{4}$

$\sf \implies \dfrac{225 \times 1.73}{4}$

$\implies \sf 97.3125 \: cm^{2}$

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Therefore, Area of corresponding minor segment of the circle =area of minor sector -area of triangle AOB

$\: \: \: \: \: \: \: \: \: \: \: \implies \sf117.75 - 97.3125$

$\: \: \: \: \: \: \: \: \: \: \: \sf \implies20.4375 \: cm^{2}$

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And area of the corresponding major segment of the circle

$\: \: \: \: \: \: \: \: \implies \sf \pi \: {r}^{2} - area \: of \: minor \: segment$

$\: \: \: \: \: \: \: \: \implies \sf3.14 \times 15 \times 15 - 20.4375$

$\: \: \: \: \: \: \: \: \implies \sf706.5 - 20.4375$

$\: \: \: \: \: \: \: \: \implies \sf686.0625 \: cm^{2}$