Question

The perimeter of sector OACB of the circle centred at O and of radius 24, is 73.12 cm.

1. Find the central angle ∠AOB.
2. Find the area of the minor segment ACB. (Use π = 3.14 and `sqrt3` = 1.73)

Answer 1

i. Given, perimeter of OACB = 73.12 cm

Radius = 24 cm

Perimeter of OACB = Length of arc ACB + 2 × r

`73.12 = θ/360 xx 2πr + 2 xx 24`

`73.12 - 48 = θ/360 xx 2 xx 3.14 xx 24`

`25.12 = θ/15 xx 2 xx 3.14`

`(25.12 xx 15)/(2 xx 3.14) = θ`

`376.8/6.28 = θ`

θ = 60°

ii. Area of minor segment ACB = Area of sector − Area of Δ

= `θ/360 xx πr^2 - sqrt3/4 a^2`

= `60/360 xx 3.14 xx 24 xx 24 - 1.73/4 xx 24 xx 24`

= 3.14 × 4 × 24 − 1.73 × 6 × 24

= 301.44 − 249.12

= 52.32 cm²