Question

A chord AB of a circle, of radius 14 cm makes an angle of 60° at the centre of the circle. Find the area of the minor segment of the circle. Also, find the area of the major segment of the circle. `("Use"  π = 22/7)`

Answer 1

Given:

Radius r = 14 cm, central angle θ = 60°, `π = 22/7`.

Use: area of minor segment = area of sector – area of triangle.

Step-wise calculation:

1. Area of sector OAB

r² = 14² = 196

Sector area = `(θ/360)·πr^2`

= `(60/360)·(22/7)·196`

`(22/7)·196 = 22·28 = 616` and `60/360 = 1/6`

Sector area = `616·(1/6) = 308/3 cm^2 ≈ 102.6667  cm^2`

2. Area of triangle OAB (isosceles triangle with vertex angle 60°)

Triangle area = `(1/2) r^2 sin θ = (1/2)·196·sin 60^circ`

`sin 60^circ = √3/2`

⇒ Triangle area = `98·(sqrt(3)/2)`

= `49sqrt(3)  cm^2`

Numeric: `sqrt(3) ≈ 1.732`

⇒ Triangle area ≈ 49·1.732 = 84.87 cm²

3. Area of minor segment

Minor = Sector – Triangle

= `308/3 - 49sqrt(3)  cm^2`

Decimal ≈ 102.6667 – 84.87 = 17.80 cm² (rounded to 2 d.p.)

4. Area of the whole circle

Circle area = `πr^2 = (22/7)·196`

= 616 cm²

5. Area of major segment

Major = circle – minor

= `616 - (308/3 - 49sqrt(3))`

Exact form = `616 - 308/3 + 49sqrt(3)`

= `1540/3 + 49sqrt(3)  cm^2`

Decimal ≈ 616 – 17.80 = 598.20 cm² (rounded to 2 d.p.)

Area of the minor segment = `308/3 - 49sqrt(3)  cm^2 ≈ 17.80  cm^2`.

Area of the major segment = `1540/3 + 49sqrt(3)  cm^2 ≈ 598.20  cm^2`.