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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 36090=270
Area of the sector making angle 270
=(270360)×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
=(90360)×πr2 cm2
=(14)×102π=25πcm2
=25×3.14 cm2
=78.5 cm2
Area of the minor segment = Area of the sector making angle 90Area of ΔAOB
=78.5cm250cm2=28.5cm2

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