In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

(i) the length of the arc

(ii) area of the sector formed by the arc

(iii) area of the segment formed by the corresponding chord

(i) We know that,

Length of an arc of sector of angle θ = $\dfrac{θ}{360°} \times$ 2πr

Substituting values we get :

$\text{Length of arc APB} = \dfrac{60°}{360°} \times 2 \times \dfrac{22}{7} \times 21 \\[1em] = \dfrac{1}{6} \times 2 \times 22 \times 3 \\[1em] = 22 \text{ cm}.$

Hence, length of arc = 22 cm.

(ii) We know that,

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

Substituting values we get :

$\Rightarrow \text{Area of sector AOBP} = \dfrac{60°}{360°} \times \dfrac{22}{7} \times 21^2 \\[1em] = \dfrac{1}{6} \times 22 \times 3 \times 21 \\[1em] = 11 \times 21 \\[1em] = 231 \text{ cm}^2.$

Hence, area of sector = 231 cm².

(iii) 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 RHS axiom)

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

In △MOB,

⇒ sin 30° = $\dfrac{MB}{OB}$

⇒ $\dfrac{1}{2} = \dfrac{MB}{r}$

⇒ MB = $\dfrac{r}{2} = \dfrac{21}{2}$ = 10.5 cm

By C.P.C.T.

MA = MB = 10.5 cm

AB = MA + MB = 21 cm.

Since, OA = OB = AB.

∴ △AOB is an equilateral triangle.

We know that,

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

Substituting values we get :

$\text{Area of △AOB} = \dfrac{\sqrt{3}}{4} \times 21^2 \\[1em] = \dfrac{441\sqrt{3}}{4} \text{ cm}^2.$

From figure,

Area of segment APB = Area of sector AOBP - Area of triangle AOB

= $\Big(231 - \dfrac{441\sqrt{3}}{4} \text{ cm}^2\Big).$

Hence, area of segment APB = $\Big(231 - \dfrac{441\sqrt{3}}{4} \text{ cm}^2\Big).$