if the radius of the circle is 21 cm and AOB is 120 then the area of segment AYB is
Answers
Answer:
Step-by-step explanation:
Area of segment = Area of sector -
area of ∆AOB
= (120°/360°) × π × 21×21 - [(1/2)×(21)^2
× (sin120°)
= 1/3 × 22/7 × 21 × 21 - 441√3/4
= 22 × 21 - 190.95
= 271.04 (approx)
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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: