A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)

We know that,  formula for the area of the sector of a circle

Area of the sector = $\frac{\theta }{{360}^0}$ $\times$ ${\mathrm{\pi r}}^2$

 Area of the segment  = Area of the sector - Area of the corresponding  triangle

Length of the arc = $\frac{\theta }{{360}^0}$ $\times$ 2$\mathrm{\pi }$r

Here, r = 15 cm, θ = 60⁰

 Area of minor segment APB = Area of sector OAPB - Area of ΔAOB

 Area of major segment AQB = ${\mathrm{\pi r}}^2$ - Area of minor segment APB

(i)Area of the sector = $\frac{\theta }{{360}^0}$ $\times$ ${\mathrm{\pi r}}^2$

Area of the sector OABP= $\frac{\theta }{{360}^0}$ $\times$ ${\mathrm{\pi r}}^2$

= 60°/360° $\times$ 3.14 $\times$ 15$\times$ 15

= 117.75 cm²

In ΔAOB,OA = OB = r

∠OBA = ∠OAB (Angles opposite to the equal sides in a triangle are equal)

∠AOB $+$ ∠OBA $+$ ∠OAB = 180⁰

60° $+$ ∠OAB $+$ ∠OAB = 180⁰

2 ∠OAB = 120⁰

∠OAB = 60⁰

Therefore, ΔAOB is an equilateral triangle because all its angles are equal.

⇒ AB = OA = OB = r

Area of ΔAOB = √3/4 $\times$ (side)²

= √3/4 $\times$ r²

= √3/4 $\times$ (15 )²

= 1.73/4 $\times$ 225

= 97.3125 cm²

(i) Area of minor segment APB = Area of sector OAPB - Area of ΔAOB

= 117.75  - 97.3125

= 20.4375 cm²

(ii)Area of the major segment AQB = Area of the circle - Area of minor segment APB

=  $\mathrm{\pi }$ $\times$ (15)² - 20.4375 

= 3.14 $\times$ 225  - 20.4375 

= 706.5  - 20.4375 

= 686.0625 cm²