Question

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

Answer 1

Radius (r) of circle = 21 cm
Angle subtended by the given arc = 60°

(i) Length of an arc of a sector of angle θ =\(\frac {\theta }{360^{\degree}} \times 2 \pi r\)

Length of arc ACB =\(\frac{60°}{360 °} \times 2 \times \frac{22}7 \times 21\)
= \(\frac{1}{6} \times 2 \times {22} \times 3\)
= 22 cm

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 (ii) Area of sector OACB =  \(\frac{60°}{360 °} \times \pi r^2\)

 = \(\frac{1}{6}  \times \frac{22}7 \times 21\times 21\)
= \(231 cm ^2\)

In ΔOAB, 

∠OAB = ∠OBA (As OA = OB) 
∠OAB + ∠AOB + ∠OBA = 180°
 2∠OAB + 60° = 180°
 ∠OAB = 60° 
Therefore, ΔOAB is an equilateral triangle.

 Area of ΔOAB = \(\frac{ \sqrt3 }{4} \times (Side) ^2\)

= \(\frac{ \sqrt3 }{4} \times (22) ^2 = \frac{441 \sqrt 3}{4} \, cm^2\)

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(iii) Area of segment ACB = Area of sector OACB - Area of ΔOAB
                                          = \((231 - \frac{441 \sqrt3}{4})\, cm^2\) 

Answer 2

Given: a radius \((r) = 21\)cm and an arc angle \((\theta) = 60°\)

Step 1: Find the length of the arc

Length of the arc L: 
\(L = \frac{2\pi r \theta}{360^\circ}\)
\(L = \frac{2 \times \pi \times 21 \times 60^\circ}{360^\circ}\)

Using \(\pi = \frac{22}{7}​:\)
\(L = \frac{2 \times \frac{22}{7} \times 21 \times 60}{360}\)

\(L = \frac{2 \times 22 \times 21 \times 60}{7 \times 360}\)

\(L = \frac{27720}{2520}L = 22 \text{ cm}\)

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Step 2: Find the area of the sector

Area of the sector A: 
\(A = \frac{\theta}{360^\circ} \times \pi r^2\)

\(A = \frac{60^\circ}{360^\circ} \times \pi \times 21^2\)

\(A = \frac{1}{6} \times \pi \times 441\)

Using\(\pi = \frac{22}{7}​:\)

\(A = \frac{1}{6} \times \frac{22}{7} \times 441\)

\(A = \frac{22 \times 441}{42}\)

\(A = \frac{9702}{42}\)

\(A = 231 \text{ cm}^2\)

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Step 3: Find the area of the segment

The area of the segment APB = Area of sector OAPB - Area of triangle OAB

In \(\triangle OAB\), since two sides are equal (radii of the circle), the angles opposite these sides are also equal.

Let \(\angle OAB = \angle OBA = x.\)

Using the sum of angles property of triangles: \(\angle AOB + \angle OAB + \angle OBA = 180^\circ\)
\(60^\circ + x + x = 180^\circ\)
\(60^\circ + 2x = 180^\circ\)
\(2x = 120^\circ\)
\(x = 60^\circ\)

So,\(\angle OAB = \angle OBA = 60^\circ.\)
 Therefore,\(\triangle OAB\) is an equilateral triangle.

Area of an equilateral triangle: 
\(\text{Area} = \frac{\sqrt{3}}{4} ( \text{side} )^2\)

\(\text{Area} = \frac{\sqrt{3}}{4} \times (21)^2\)

\(\text{Area} = \frac{\sqrt{3}}{4} \times 441\)
\(\text{Area} = 110.25\sqrt{3} \approx 190.95 \text{ cm}^2\)

Area of the segment: 
\(\text{Area of the segment} = \text{Area of sector} - \text{Area of triangle}\)
\(\text{Area of the segment} = 231 - 190.95\)
\(\text{Area of the segment} = 40.05 \text{ cm}^2\)

Therefore:

• Length of the arc = 22 cm
• Area of the sector = \(231\ cm^2\)
• Area of the segment = \(40.05\  cm^2\)