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 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:-

$Solution:</p><p></p><p> \\ Radius, \: r \: = \: 12 \: cm</p><p></p><p> \\ Now, \\ draw \: a \: perpendicular \: OD \: on \: chord \\ AB \: and \: it \: will \: bisect \: chord \: AB.</p><p></p><p> \\ So, AD = DB</p><p></p><p>$

$\\ Now, \\ the \: area \: of \: the \: minor \: sector \: = \: (θ/360°) \: × \: πr {}^{2} </p><p></p><p> \\ = \: (120/360) \: × \: (22/7) \: × \: 12 {}^{2} </p><p></p><p> \\ = 150.72 cm2</p><p></p><p>$

$\\ Consider \: the \: ΔAOB,</p><p></p><p> \\ ∠ OAB \: = \: 180°- \: (90°+60°) \: = \: 30°</p><p></p><p> \\ Now, \\ cos 30° \: = \: AD/OA</p><p></p><p> \\ √3/2 \: = \: AD/12</p><p></p><p> \\ Or, \\ AD \: = \: 6√3 cm</p><p></p><p>$

$\\ We \: know \: OD \: bisects \: AB. \\ So,</p><p></p><p> \\ AB \: = 2×AD \: = \: 12√3 \: cm</p><p></p><p> \\ Now, \: \\ sin 30° \: = \: OD/OA</p><p></p><p>$

$\\ Or, \: ½ \: = \: OD/12</p><p></p><p> \\ ∴ OD \: = \: 6 cm</p><p></p><p> \\ So, \\ the \: area \: of \: ΔAOB \: = \: ½ × base × height</p><p></p><p>$

$\\ Here, \\ base \: = AB \: = \: 12√3 \: and</p><p></p><p> \\ Height \: = \: OD \: = \: 6</p><p></p><p> \\ So, \\ area \: of \: ΔAOB \: = \: ½×12√3×6 \: = \: 36√3 \: cm \: = \: 62.28 cm2</p><p></p><p>$

$\\ ∴ \: Area \: of \: the \: corresponding \: Minor \: segment \\ = Area \: of \: the \: Minor \: sector \: – \: Area \: of \: ΔAOB</p><p></p><p> \\ = 150.72 cm2 \: – \: 62.28 cm2  \: = \: 88.44 cm2</p><p></p><p>$

$Hence, \: the \: answer \: is \: 88.44 \: cm {}^{2}$