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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
⇨ Area of Minor sector
⇨ Now in △ AOB
AO = OB = 10 cm
⇨ Area of △ AOB
⇨(ii) Area of Minor Segment
= Area of Minor sector - Area of △ AOB
= 78.5 - 50
= 28.5
Exercise 12.2