Find the area of the segment AYB shown in fig., if radius of the circle is 21 cm and angle AOB = 120 degree. ( use pie = 22/7 )
Pls solve this question in a easy and short way
Answers
Step-by-step explanation:
In ∆AOB, draw a perpendicular line from O which intersect AB at M.
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 270.86.
Hope it helps u !
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: