Find the area of the segment if radius is 21cm and angle AOB=120°
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The area of a circle(360°) of radius r is=πr²
Now...area of a sector of 120° is
=πr²×120°/360=
=(πr²)/3
Now ...r=21cm
={22×21×21}/(3×7)
=22×3×7 cm²
=462 cm²
area of the triangle ∆AOB. is
=[(21×21×sin120°)/2] cm^2
=(441√3)/4=190.96 cm^2
there area of the segment....
=(462-190.96)cm²
=271.04 cm²
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Answer:
271.05 cm^2
Step-by-step explanation:
r = 21cm, length of chordAB =2r sin(A/2)
AB=2×21×(sin120/2)= 42×√3/2 = 21√3
area of the sector OAB
=120/360 * 22/7 *(21)^2=462cm^2
height of isosceles ∆ OAB
h = √[(21)^2 - {(21√3)^2/4}]
= 21/2 cm
area of ∆ OAB =1/2 * 21√3 * 21/2
= 190.95
area of the segment
= 462 - 190.95= 271.05 cm^2
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