Question

An arc of a circle of radius 21 cm subtends an angle of 60° at the centre. Find the area of the sector formed by the arc and the area of the segment formed by the corresponding chord.

Hint:

The area of a sector is calculated using the formula \( A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 \), and the area of a segment is found by subtracting the area of the triangle from the sector's area.

Answer 1

The given data is:
- Radius of the circle, \( r = 21 \, \text{cm} \),
- Angle subtended by the arc, \( \theta = 60^\circ \).
Step 1: The formula for the area of a sector of a circle is: \[ A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2. \] Substitute the given values: \[ A_{\text{sector}} = \frac{60^\circ}{360^\circ} \times \pi \times (21)^2 = \frac{1}{6} \times \pi \times 441 = 73.5 \pi \, \text{cm}^2. \] Thus, the area of the sector is: \[ A_{\text{sector}} = 73.5 \pi \, \text{cm}^2 \approx 230.94 \, \text{cm}^2. \]
Step 2: To find the area of the segment, we need to subtract the area of the triangle formed by the radius lines and the chord from the area of the sector. The area of the triangle is given by the formula: \[ A_{\text{triangle}} = \frac{1}{2} r^2 \sin \theta. \] Substitute the values: \[ A_{\text{triangle}} = \frac{1}{2} \times (21)^2 \times \sin 60^\circ = \frac{1}{2} \times 441 \times \frac{\sqrt{3}}{2} = \frac{441 \sqrt{3}}{4} \approx 190.57 \, \text{cm}^2. \]
Step 3: Now, find the area of the segment by subtracting the area of the triangle from the area of the sector: \[ A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = 73.5 \pi - 190.57 \approx 230.94 - 190.57 = 40.37 \, \text{cm}^2. \]
Conclusion: The area of the sector is approximately \( 230.94 \, \text{cm}^2 \) and the area of the segment is approximately \( 40.37 \, \text{cm}^2 \).