Question

# Question 4 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)

Solution

## Radius of the circle = 10 cm Major sector is making 360∘−90∘=270∘ Area of the sector making angle 270∘ =(270∘360∘)×πr2 cm2 =(34)×102π=75πcm2 =75×3.14cm2=235.5cm2 ∴ Area of the major sector =235.5cm2 Height of ΔAOB=OA=10 cm Base of ΔAOB=OB=10 cm Area of ΔAOB=12×OA×OB =12×10×10=50cm2 Minor sector is making 90∘  Area of the sector making angle 90∘ =(90∘360∘)×πr2 cm2 =(14)×102π=25πcm2 =25×3.14 cm2 =78.5 cm2 Area of the minor segment = Area of the sector making angle 90∘−Area of ΔAOB =78.5cm2−50cm2=28.5cm2

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