A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.

(Use π = 3.14 and $\sqrt{3}$ = 1.73)

Let AB be the chord subtending angle 120° at the center.

Draw OM ⊥ AB.

In triangle OAM and OMB,

∠OMA = ∠OMB = 90°

OA = OB (Radius of same circle)

OM = OM (Common)

∴ △OAM ≅ △OMB (By RHS axiom)

∴ ∠MOB = ∠MOA = $\dfrac{120°}{2}$ = 60°. [By C.P.C.T.]

In △MOB,

$\Rightarrow \text{sin 60°} = \dfrac{MB}{OB} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{MB}{OB} \\[1em] \Rightarrow MB = \dfrac{OB\sqrt{3}}{2} \\[1em] \Rightarrow MB = \dfrac{12 \times 1.73}{2} = 10.38 \text{ cm} \\[1em] \Rightarrow \text{tan 60°} = \dfrac{MB}{MO} \\[1em] \Rightarrow \sqrt{3} = \dfrac{10.38}{MO} \\[1em] \Rightarrow MO = \dfrac{10.38}{\sqrt{3}} = \dfrac{10.38}{1.73} = 6 \text{ cm}.$

By C.P.C.T.

MA = MB = 10.38 cm

AB = MA + MB = 20.76 cm.

We know that,

Area of triangle = $\dfrac{1}{2} \times \text{ base} \times \text{height}$

Substituting values we get :

Area of triangle OAB = $\dfrac{1}{2} \times AB \times OM$

= $\dfrac{1}{2} \times 20.76 \times 6$ = 62.28 cm².

We know that,

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,

$\Rightarrow \text{Area of minor segment APB} = \dfrac{120°}{360°} \times 3.14 \times 12^2 - 62.28 \\[1em] = \dfrac{1}{3} \times 3.14 \times 144 - 62.28 \\[1em] = 150.72 - 62.28 \\[1em] = 88.44 \text{ cm}^2.$

Hence, area of corresponding segment of the circle = 88.44 cm².