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:
- Radius of the circle $r = 14 \, \text{cm}$
- Angle subtended by the chord at the center $\theta = 90^\circ$

1. Area of the sector formed by the chord:
   $\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times \pi \times 14^2 = \frac{1}{4} \times \pi \times 196 = 49\pi \, \text{sq cm}$
2. Area of the triangle formed by the chord and the center of the circle: The two radii and the chord form an isosceles triangle with a vertex angle of $90^\circ$. The area of the triangle is given by: $\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 14 = 98 \, \text{sq cm}$
3. Area of the minor segment:
   $\text{Area of minor segment} = \text{Area of sector} - \text{Area of triangle} = 49\pi - 98$ Approximating $\pi = 3.14$: $\text{Area of minor segment} \approx 49 \times 3.14 - 98 = 153.86 - 98 = 55.86 \, \text{sq cm}$
4. Area of the major segment:
   $\text{Area of major segment} = \text{Total area of the circle} - \text{Area of minor segment}$ $\text{Area of major segment} = \pi r^2 - \text{Area of minor segment} = 3.14 \times 14^2 - 55.86 = 615.44 - 55.86 = 559.58 \, \text{sq cm}$

Thus, the area of the minor segment is approximately $55.86 \, \text{sq cm}$, and the area of the major segment is approximately $559.58 \, \text{sq cm}$.