Example 3 : Find the area of the segment AYB
shown in Fig. 12.9.if radius of the circle is 21 cm and
21 cm
120°
ZAOB = 120° (Use T =
21 cm
Fig. 12.9
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Answer:
In ∆AOM, angle AMO = 90
angle OAM = 30
cos 30 = AM/AO
√3/2 = AM/21
AM = 21×√3/2
AB = 2(AM)
=2(21×√3/2)
=21√3
OM^2 = AO^2-AM^2
=21^2-(21√3/2)^2
=441-330.51
=110.48
OM =√110.48
OM =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
Step-by-step explanation:
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