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A chord of a circle, of radius $14$ cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.
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Sol. $\therefore \triangle OAB$ is an equilateral triangle. (1/2 Mark)
$\Rightarrow AB = 14$ cm (1/2 Mark)
Area of smaller sector = $\frac{60}{360}\times\frac{22}{7}\times 14\times 14$ (1/2 Mark)
= $\frac{308}{3}$ cm² or $102.6$ cm² (approx.) (1 Mark)
Perimeter of smaller segment = $14 + \frac{60}{360}\times 2\times\frac{22}{7}\times 14$ (1/2 Mark)
= $\frac{86}{3}$ cm or $28.6$ cm (approx.)
$\Rightarrow AB = 14$ cm (1/2 Mark)
Area of smaller sector = $\frac{60}{360}\times\frac{22}{7}\times 14\times 14$ (1/2 Mark)
= $\frac{308}{3}$ cm² or $102.6$ cm² (approx.) (1 Mark)
Perimeter of smaller segment = $14 + \frac{60}{360}\times 2\times\frac{22}{7}\times 14$ (1/2 Mark)
= $\frac{86}{3}$ cm or $28.6$ cm (approx.)