Question

a Find the lengths of the arcs cut off from a circle of radius 12 cmn by a
chord 12 cm long. Also, find the area of the minor segment.
(Take n = 3.14 and 13 = 1.73
​

Answer 1

Answer:

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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