A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14 )

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 Solution:
⇨ Given
    r = 10 cm
    AO = OB =10 cm
    θ1 = 360 - 90 = 270° (Major)
    θ2 = 90° (Minor)

⇨(i) Area of Major sector
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14 )
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14 )
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14 )
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14 )
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14 )
⇨(i) Area of Major sector
 
⇨ Area of Minor sector 
⇨ Area of Minor sector
⇨ Area of Minor sector
⇨ Area of Minor sector
⇨ Area of Minor sector
⇨ Area of Minor sector

⇨ Now in △ AOB
    AO = OB = 10 cm
⇨ Area of △ AOB
⇨ Area of △ AOB

⇨(ii) Area of Minor Segment 
Area of Minor sector - Area of △ AOB
= 78.5 - 50 
= 28.5


  Exercise 12.2  
















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