Find the area of the major sector of a circle with radius 42 cm and of angle 30°.
(Use π = $\frac{22}{7}$

5082

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Solution

The correct option is A 5082
Given sector is OAPB,

$A r e a o f t h e m i n o r s e c t o r$
$= \frac{θ}{360^{\circ}} \times π r^{2}$
$= \frac{30^{\circ}}{360^{\circ}} \times π \times r^{2}$
$= \frac{1}{12} \times π r^{2}$
$= \frac{π r^{2}}{12}$

$A r e a o f t h e m a j o r s e c t o r = A r e a o f t h e c i r c l e - A r e a o f t h e m i n o r s e c t o r$
$A r e a o f t h e m a j o r s e c t o r = (π r^{2} - \frac{π r^{2}}{12}) c m^{2}$
$A r e a o f t h e m a j o r s e c t o r = (\frac{12 π r^{2} - π r^{2}}{12}) c m^{2}$
$A r e a o f t h e m a j o r s e c t o r = (\frac{11 π r^{2}}{12}) c m^{2}$

Now, using $π = \frac{22}{7}$
$A r e a o f t h e m a j o r s e c t o r = (\frac{11 \times 22 \times 42 \times 42}{12 \times 7}) c m^{2}$
$A r e a o f t h e m a j o r s e c t o r = 11 \times 11 \times 42 c m^{2}$
$A r e a o f t h e m a j o r s e c t o r = 121 \times 42 c m^{2}$
$A r e a o f t h e m a j o r s e c t o r = 5082 c m^{2}$

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