Question

Find the area of the minor segment of a circle whose sector makes an angle of 120°at the centre.The radius of the circle is 21cm.(take £=22/7)

Answer 1

Answer:

]Area of segment AYB is 270.86cm^2.

Step-by-step explanation:

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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.

Answer 2

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