Question

A chord of a circle of radius 14 cm subtends an angle of 90° at the centre. Find the area of the corresponding minor and major segments of the circle.

Answer 1

Given:
- Circle radius \( r = 14 \, \text{cm} \)
- Central angle subtended by the chord \( \theta = 90^\circ \)

1. Sector Area:
   \[ \text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times \pi \times 14^2 = \frac{1}{4} \times \pi \times 196 = 49\pi \, \text{sq cm} \]
2. Triangle Area: Formed by the chord and center, this is an isosceles triangle with a \( 90^\circ \) vertex angle.
   \[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 14 = 98 \, \text{sq cm} \]
3. Minor Segment Area:
   \[ \text{Area of minor segment} = \text{Area of sector} - \text{Area of triangle} = 49\pi - 98 \] Approximating \( \pi \approx 3.14 \): \[ \text{Area of minor segment} \approx 49 \times 3.14 - 98 = 153.86 - 98 = 55.86 \, \text{sq cm} \]
4. Major Segment Area:
   \[ \text{Area of major segment} = \text{Total circle area} - \text{Area of minor segment} \] \[ \text{Area of major segment} = \pi r^2 - \text{Area of minor segment} = 3.14 \times 14^2 - 55.86 = 615.44 - 55.86 = 559.58 \, \text{sq cm} \]

The minor segment area is approximately \( 55.86 \, \text{sq cm} \), and the major segment area is approximately \( 559.58 \, \text{sq cm} \).