A chord of a circle of radius 14 cm subtends a right angle at the centre. Find the area of the minor segment and the area of the major segment. \(\left(\pi = \dfrac{22}{7}\right)\)

Answer: Minor segment = 56 cm²; major segment = 560 cm²

Let the chord be AB of the circle with centre O; ∠AOB = 90°, r = 14 cm.

Area of sector O-AB \(= \dfrac{90}{360} \times \dfrac{22}{7} \times 14 \times 14 = 154\) cm².

Area of Δ AOB \(= \dfrac{1}{2} \times OA \times OB = \dfrac{1}{2} \times 14 \times 14 = 98\) cm² (right angle at O).

Area of minor segment = 154 − 98 = 56 cm².

Area of circle \(= \pi r^2 = \dfrac{22}{7} \times 14 \times 14 = 616\) cm².

Area of major segment = 616 − 56 = 560 cm².