The perimeter of a sector of a circle of radius r cm and of central angle $θ$ (in degrees) is .................

2 r (\frac{π θ}{360^{o}} + 2)

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2 r (\frac{π θ}{360^{o}} + 1)

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2 r (\frac{π θ}{180^{o}} + 1)

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2 r (\frac{π θ}{1800^{o}} + 2)

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Solution

The correct option is C $2 r (\frac{π θ}{360^{o}} + 1)$
$G i v e n - A O B i s a s e c t o r o f r a d i u s = r a n d t h e c e n t r a l a n g l e = θ . T o f i n d o u t - t h e p e r i m e t e r o f t h e s e c t o r A O B = ? S o l u t i o n - T h e a r c l e n g t h A B = \frac{θ}{360^{o}} \times 2 \times π \times r = \frac{π θ}{360^{o}} \times 2 r . N o w t h e p e r i m e t e r o f t h e s e c t o r = O A + O B + a r c A B = r + r + \frac{θ}{360^{o}} \times 2 π r = 2 r (\frac{π θ}{360^{o}} + 1) . A n s - O p t i o n B .$

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Similar questions

The area of a sector of angle $θ^{\circ}$ of a circle with radius $R$ is

(a) $\frac{2 π R θ}{180}$

(b) $\frac{π R^{2} θ}{180}$

(d) $\frac{π R^{2} θ}{360}$

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The length of an arc of a sector of angle $θ^{\circ}$ of a circle with radius R is

(a) $\frac{2 π R θ}{180}$

(b) $\frac{2 π R θ}{360}$

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Figure shows a sector of a circle, centre O, containing an angle $θ^{\circ}$ . Prove that:

(i) Perimeter of the shaded region is $r (t a n θ + s e c θ + \frac{π θ}{180} - 1)$

(ii) Area of the shaded region is $\frac{r^{2}}{2} (t a n θ - \frac{π θ}{180})$

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