In a circle of diameter 40cm, …

Given :   a circle of diameter 40cm,the length of a chord is 20 cm,

To Find : the length of the minor arc of the chord

Solution:

Chord Length = 20 cm

Diameter = 40 cm

Perpendicular from center on chord bisect the chord

Hence Half of chord = 10 cm

Radius = 40/2 = 20 cm

Sin ( 1/2 of chord angle )  =  10 /20

=> Sin ( 1/2 of chord angle )  = 1/2

=> Sin ( 1/2 of chord angle )  = Sin 30°

=>  1/2 of chord angle =  30°

=> chord angle =  60°

Minor arc angle = 60°  

or another way to get angle

as Radius = 20 cm

Chord length = 20 cm

Hence it forms an Equilateral triangle

Hence angle formed by chord at center = 60°

Minor arc angle =  60°  

length of the major arc of the chord = (60/360) * 2π * Radius

= ( 1/6 ) * 2π * 10

= 10π/3

= 50 * 3.14/3

=  10.47  cm

Diameter of the circle =40 cm
∴ Radius (r) of the circle = 40​/2 =20 cm
Let AB be a chord (length = 20 cm) of the circle.
In △OAB, OA = OB = Radius of circle = 20 cm
Also, AB = 20 cm
Thus, △OAB is an equilateral triangle.
∴θ=60^∘=π/3​ radian
We know that in a circle of radius r unit, if an arc of length I unit subtends an angle θ radian at the centre, then θ=l​/r
∴π/3 ​= arc AB​/20

⟹arc AB = 20/ 3​π cm