Question

find the area of the segment AYB shown in fig., if radius of the circle is 21cm and angle AOB = 120 degree. ( use pie = 22/7 )

Answer 1

☺☺ HERE IS YOUR ANSWER ☺☺
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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 270.86cm^2.
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☺ HOPE IT HELPS YOU ☺

THANKS..

Answer 2

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 270.86cm^2