## Book: RS Aggarwal - Mathematics

### Chapter: 18. Area of Circle, Sector and Segment

#### Subject: Maths - Class 10th

##### Q. No. 15 of Exercise 18B

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15
##### Find the lengths of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long. Also, find the area of the minor segment. [Take π = 3.14 and √3 = 1.73.]

Consider the figure shown above.

In this, the triangle AOB is an equilateral triangle as all the sides are equal; therefore, it is obvious that the central angle of the sector is 60 degrees. Now by simply applying the formula of length of an arc, we can easily calculate the length of arc of the sector AOB.

Given Radius of circle = R = 12 cm,

Length of chord AB = 12 cm

Central angle = θ = 60° ( ∆AOB is an equilateral triangle)

Where R = radius of the circle and θ = central angle of the sector

Put the values of R and θ in equation 1

= 2×3.14×2

= 12.56 cm

Now, Length of major arc = 2πR – Length of minor arc

Length of major arc = 2π(12) – 12.56 (put π = 3.14)

Length of major arc = (2×3.14×12) – 12.56

Length of major arc = 75.36 – 12.56

Length of major arc = 62.8 cm

Now, Area of minor segment = Area of sector – Area of triangle eqn1

= 75.36 cm2 eqn2

Area of triangle = 1.73×36

Area of triangle = 62.28 cm2 eqn3

Put the values of equation 2 and 3 in equation 1,

Area of minor segment = 75.36 – 62.28

= 13.08 cm2

Length of major arc is 62.8 cm and of minor arc is 12.56 cm and area of minor segment is 13.08 cm2.

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