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(a) The area of a smaller circle is equal to the area of a sector of a larger circle with central angle $120^{\circ}$. The radii of the smaller and larger circles are '$r$' and '$R$' respectively. Find $r : R$.
OR
(b) In the given figure, $O$ is the centre of a circle of radius $7 \text{ cm}$. $AB$ is a chord of the circle. Find the perimeter of the shaded region.
OR
(b) In the given figure, $O$ is the centre of a circle of radius $7 \text{ cm}$. $AB$ is a chord of the circle. Find the perimeter of the shaded region.
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Solution:(a) A.T.Q.
$\pi r^2 = \frac{120}{360} \pi R^2$ [1 mark]
$\frac{r^2}{R^2} = \frac{1}{3}$ [1/2 mark]
$r : R = 1 : \sqrt{3}$ [1/2 mark]
OR
(b) $\Delta AOB$ is an equilateral triangle as $\angle AOB = 60^{\circ}$
$\therefore AB = 7 \text{ cm}$ [1/2 mark]
Length of minor arc $AB = \frac{60}{360} \times 2 \times \frac{22}{7} \times 7 = \frac{22}{3}$ [1/2 mark]
$\therefore$ Perimeter of the shaded region $= \frac{22}{3} + 7$ [1/2 mark]
$= \frac{43}{3} \text{ cm}$ or $14.33 \text{ cm}$ [1/2 mark]
$\pi r^2 = \frac{120}{360} \pi R^2$ [1 mark]
$\frac{r^2}{R^2} = \frac{1}{3}$ [1/2 mark]
$r : R = 1 : \sqrt{3}$ [1/2 mark]
OR
(b) $\Delta AOB$ is an equilateral triangle as $\angle AOB = 60^{\circ}$
$\therefore AB = 7 \text{ cm}$ [1/2 mark]
Length of minor arc $AB = \frac{60}{360} \times 2 \times \frac{22}{7} \times 7 = \frac{22}{3}$ [1/2 mark]
$\therefore$ Perimeter of the shaded region $= \frac{22}{3} + 7$ [1/2 mark]
$= \frac{43}{3} \text{ cm}$ or $14.33 \text{ cm}$ [1/2 mark]