101
A chord of a circle of radius 14 cm subtends a right angle at the centre. Find the area of the corresponding (i) minor segment (ii) major segment.
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Solution: (i) Area of minor segment = $\frac{90}{360} \times \frac{22}{7} \times 14 \times 14 - \frac{1}{2} \times 14 \times 14$ (1 Mark)
$= 56$ cm$^2$ (1 Mark)
(ii) Area of major segment = $\frac{22}{7} \times 14 \times 14 - 56$ (1/2 Mark)
$= 560$ cm$^2$ (1/2 Mark)
$= 56$ cm$^2$ (1 Mark)
(ii) Area of major segment = $\frac{22}{7} \times 14 \times 14 - 56$ (1/2 Mark)
$= 560$ cm$^2$ (1/2 Mark)