Answer :

i. Let Major segment A_{1} minor segment be A_{2}

∠A = 90°

area of sector ACD = 78.5cm^{2}

Pythagoras

AC^{2} = CD^{2} + AD^{2}

AC^{2} = 10^{2} + 10^{2}

AC^{2} = 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 cm^{2}

ii. Major segment = Area of circle – minor segment

= π × r^{2} –minor segment

= π × 10^{2} –28.5

= 314-28.5

= 285.5cm^{2}

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