Question

A chord of a circle of radius 21 cm subtends a right angle at the centre. Find the area of the corresponding minor segment.

Hint:

Area of a segment = (Area of sector) – (Area of triangle). For a 90° angle, the sector is one-fourth of the circle.

Answer 1

Step 1: Given data.
Radius \( r = 21 \) cm, central angle \( \theta = 90° \).

Step 2: Area of the sector.
\[ \text{Area of sector} = \frac{\theta}{360} \times \pi r^2 = \frac{90}{360} \times \frac{22}{7} \times 21^2 \] \[ = \frac{1}{4} \times \frac{22}{7} \times 441 = \frac{22 \times 63}{4} = 346.5 \text{ cm}^2 \]
Step 3: Area of the triangle formed by the two radii.
\[ \text{Area of triangle} = \frac{1}{2} r^2 \sin\theta = \frac{1}{2} \times 21^2 \times \sin90° = \frac{1}{2} \times 441 = 220.5 \text{ cm}^2 \]
Step 4: Area of minor segment.
\[ \text{Area of segment} = \text{Area of sector} - \text{Area of triangle} \] \[ = 346.5 - 220.5 = 126 \text{ cm}^2 \] Step 5: Conclusion.
Hence, the area of the minor segment is \( \boxed{126\ \text{cm}^2} \).