A chord of a circle of radius 10 cm. subtends a right angle at the centre. Find the area of the corresponding :

(use π = 3.14)

i. Minor segment

ii. Major segment

i. Let Major segment A₁ minor segment be A₂

∠A = 90°

area of sector ACD = 78.5cm²

Pythagoras

AC² = CD² + AD²

AC² = 10² + 10²

AC² = 200

AC = 10

Height of triangle =

Cos 45° =

AE =

Area of minor segment = Area of sector ACD – Area ΔACD

= 78.5 –10

= 78.5-50

= 28.5 cm²

ii. Major segment = Area of circle – minor segment

= π × r² –minor segment

= π × 10² –28.5

= 314-28.5

= 285.5cm²