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 $\sqrt{3}$ = 1.73)

The circle of radius 15 cm with its chord AB subtending an angle of 60° at the centre is shown in the figure below:

In triangle OAB,

OM ⊥ AB

In triangle OAM and OMB,

∠OMA = ∠OMB = 90°

OA = OB (Radius of same circle)

OM = OM (Common)

∴ △OAM ≅ △OMB (By SAS axiom)

∴ ∠MOB = ∠MOA = $\dfrac{60°}{2}$ = 30°. [By C.P.C.T.]

In △MOB,

⇒ sin 30° = $\dfrac{MB}{OB}$

⇒ $\dfrac{1}{2} = \dfrac{MB}{OB}$

⇒ MB = $\dfrac{OB}{2} = \dfrac{15}{2}$ = 7.5 cm

By C.P.C.T.

MA = MB = 7.5 cm

AB = MA + MB = 15 cm.

Since, OA = OB = AB.

∴ △AOB is an equilateral triangle.

By formula,

Area of equilateral triangle = $\dfrac{\sqrt{3}}{4} \times$ (side)²

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Area of minor segment APB = Area of sector BOAP - Area of triangle AOB

Substituting values we get :

$\text{Area of minor segment APB} = \dfrac{60°}{360°} \times 3.14 \times 15^2 - \dfrac{\sqrt{3}}{4} \times 15^2 \\[1em] = \dfrac{1}{6} \times 3.14 \times 225 - \dfrac{225 \times 1.73}{4} \\[1em] = 117.75 - 97.3125 \\[1em] = 20.4375 \text{ cm}^2.$

Area of circle = πr²

= 3.14 × 15²

= 3.14 × 225

= 706.5 cm².

From figure,

Area of major segment AQB = Area of circle - Area of minor segment APB

= 706.5 - 20.4375 = 686.0625 cm².

Hence, area of minor segment = 20.4375 cm² and area of major segment = 686.0625 cm².