A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the cir

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1. Step-by-step explanation:

   ✿Question:–

   A chord of a circle of radius 12 cm

   subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)

   ✿Answer:–

   [tex]Solution:

   \\ Radius, \: r \: = \: 12 \: cm

   \\ Now, \\ draw \: a \: perpendicular \: OD \: on \: chord \\ AB \: and \: it \: will \: bisect \: chord \: AB.

   \\ So, AD = DB

   [/tex]

   [tex] \\ Now, \\ the \: area \: of \: the \: minor \: sector \: = \: (θ/360°) \: × \: πr {}^{2}

   \\ = \: (120/360) \: × \: (22/7) \: × \: 12 {}^{2}

   \\ = 150.72 cm2

   [/tex]

   [tex] \\ Consider \: the \: ΔAOB,

   \\ ∠ OAB \: = \: 180°- \: (90°+60°) \: = \: 30°

   \\ Now, \\ cos 30° \: = \: AD/OA

   \\ √3/2 \: = \: AD/12

   \\ Or, \\ AD \: = \: 6√3 cm

   [/tex]

   [tex] \\ We \: know \: OD \: bisects \: AB. \\ So,

   \\ AB \: = 2×AD \: = \: 12√3 \: cm

   \\ Now, \: \\ sin 30° \: = \: OD/OA

   [/tex]

   [tex] \\ Or, \: ½ \: = \: OD/12

   \\ ∴ OD \: = \: 6 cm

   \\ So, \\ the \: area \: of \: ΔAOB \: = \: ½ × base × height

   [/tex]

   [tex] \\ Here, \\ base \: = AB \: = \: 12√3 \: and

   \\ Height \: = \: OD \: = \: 6

   \\ So, \\ area \: of \: ΔAOB \: = \: ½×12√3×6 \: = \: 36√3 \: cm \: = \: 62.28 cm2

   [/tex]

   [tex] \\ ∴ \: Area \: of \: the \: corresponding \: Minor \: segment \\ = Area \: of \: the \: Minor \: sector \: – \: Area \: of \: ΔAOB

   \\ = 150.72 cm2 \: – \: 62.28 cm2 \: = \: 88.44 cm2

   [/tex]

   [tex]Hence, \: the \: answer \: is \: 88.44 \: cm {}^{2} [/tex]

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