Question

A Chord PQ of a circle with radius 15 cm subtends an angle of 120° with centre
of the circle. Find the area of minor as well as major segment.​

Answer 1

$\bf\large\underline\green{SOLUTION:-}$

Given :

Radius ( r ) = 15 cm

angle at the centre ( x ) = 60°

i ) Area of the sector ( OPAQ )

= ( x/360 ) × πr²

= ( 60/360 ) × πr²

= (1/6 ) × 3.14 × 15²

= 117.75 cm² --( 1 )

ii ) Area of the equilateral triangle OPQ

= ( √3/4 ) × side²

= (1.73/4) × 15²

= 0.4325 × 225

= 97.3125 cm²----( 2 )

iii ) Area of the circle ( A ) = πr²

= 3.14 × 15²

= 706.5 cm² ---( 3 )

iii ) Area of the minor segment ( PAQ )

= ( 1 ) - ( 2 )

= 117.75 - 97.3125

= 20.4375 cm² ---( 4 )

iv ) Area of the major segment

= ( 3 ) - ( 4 )

= 706.5 cm² -20.4375 cm²

= 686.0625cm²

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Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

Given :

Radius ( r ) = 15 cm

angle at the centre ( x ) = 60°

i ) Area of the sector ( OPAQ )

= ( x/360 ) × πr²

= ( 60/360 ) × πr²

= (1/6 ) × 3.14 × 15²

= 117.75 cm² --( 1 )

ii ) Area of the equilateral triangle OPQ

= ( √3/4 ) × side²

= (1.73/4) × 15²

= 0.4325 × 225

= 97.3125 cm²----( 2 )

iii ) Area of the circle ( A ) = πr²

= 3.14 × 15²

= 706.5 cm² ---( 3 )

iii ) Area of the minor segment ( PAQ )

= ( 1 ) - ( 2 )

= 117.75 - 97.3125

= 20.4375 cm² ---( 4 )

iv ) Area of the major segment

= ( 3 ) - ( 4 )

= 706.5 cm² -20.4375 cm²

= 686.0625cm²