Which of these is equivalent to the sum of the lengths of arc corresponding to the minor and major segment of a circle of radius 12 cm?
Answers
Step-by-step explanation:
given a circle of radius 12 cm.
OA=OB=12 cm= radius of the circle and .Let AB be the chord of 12 cm. hence we get an equilateral triangle OAB inside the circle that means ∠O=∠A=∠B=60
o
to find the length of the arcs (APB and ADB) of circle.
circumference of the circle =2πr=2π×12 length of the arc APB of the circle =
360
2π×12×60
=
360
2π×12×60
=4π=12.56 cm
Now the length of the arc AQB=
360
2π×12×(360−60)
=
360
2π×12×300
=20π
=62.80 cm
Now to find the Area of the minor segment
= Area of the sector ABCA= Area of the sector AOBCA
= Area of the triangle OAB
Area of the sector AOBCA=π×(12)
2
×
360
60
=π×12×2=24π=75.36 cm
2
Area of the triangle OAB =
4
3
×12×12=36
3
=62.28 cm
2
hence area of the minor segment is =(75.36−62.28) cm
2
=13.08 cm
2
Given : a circle of radius 12 cm
To find : sum of lengths of arcs corresponding to minor and major segment of the circle
solution:
sum of lengths of arcs corresponding to minor and major segment of the circle = circumference of the circle
circumference of a circle = 2 π R
R = Radius = 12 cm
=> circumference of the circle = 2 π ( 12)
= 24 π cm
≈ 75.4 cm
sum of lengths of arcs corresponding to minor and major segment of the circle of radius 12 cm is about 75.4 cm
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