ample 3 : Find the area of the segment AYBshown in Fig. 12.9, if radiu

1.	ample 3 : Find the area of the segment AYBshown in Fig. 12.9, if radius of the circle is 21 cm and120
Answer» In ∆AOB, draw a perpendicular line from O which intersect AB at M. In ∆AOM, angle AMO = 90angle OAM = 30cos 30 = AM/AO√3/2 = AM/21AM = 21×√3/2AB = 2(AM)=2(21×√3/2)=21√3 OM^2 = AO^2-AM^2=21^2-(21√3/2)^2=441-330.51=110.48OM =√110.48OM =10.51 OM = 10.51cm Area of ∆AOM = 1/2 AB × OM=1/2 ×21√3 ×10.51=191.14cm^2 Area of sector AOBY = 120πr^2/360=120×21×21×22/2520=462cm^2 Area of segment AYB = Area of sector OAYB -Area of∆OAB=462-191.14=270.86 Area of segment AYB is 270.86cm^2._______________

ample 3 : Find the area of the segment AYBshown in Fig. 12.9, if radius of the circle is 21 cm and120

Answer»

In ∆AOB, draw a perpendicular line from O which intersect AB at M.

In ∆AOM, angle AMO = 90angle OAM = 30cos 30 = AM/AO√3/2 = AM/21AM = 21×√3/2AB = 2(AM)=2(21×√3/2)=21√3

OM^2 = AO^2-AM^2=21^2-(21√3/2)^2=441-330.51=110.48OM =√110.48OM =10.51

OM = 10.51cm

Area of ∆AOM = 1/2 AB × OM=1/2 ×21√3 ×10.51=191.14cm^2

Area of sector AOBY = 120πr^2/360=120×21×21×22/2520=462cm^2

Area of segment AYB = Area of sector OAYB -Area of∆OAB=462-191.14=270.86

Area of segment AYB is 270.86cm^2._______________