Question

Find the area of the segment of a circle of radius 14cm, when the angle of the corresponding Sector is 30°

Answer 1

[ Figure in the attachment ]

Given :- Radius of Circle = 14 cm

The angle of the corresponding sector
( |_ AOB ) :- 30°

Solution :-

$theta \: denoted \: by \: \alpha$

$area \: of \: segment \: = area \: of \: sector \: - \: area \: of \: corresponding \: triangle$

( swipe left )

$area \: of \: segment = \frac{ \alpha }{360} \times \pi \: {r}^{2} - \frac{1}{2} \sin( \alpha ) {r}^{2}$

$= {r}^{2} ( \frac{30}{360} \times \frac{22}{7} - \frac{1}{2} \sin(30) )$

$= 196( \frac{1}{12} \times \frac{22}{7} - \frac{1}{2} \times \frac{1}{2} )$

$= 196( \frac{11}{42} - \frac{1}{4} )$

$= \frac{44 - 42}{168} \times 196$

$= 2.33 {cm}^{2}$

Answer 2

Answer:

2.33 cm²

Step-by-step explanation:

r = 14 cm      θ = 30°

area of major segment = area of major sector - (1/2)r² sinθ

                                        = (θ/360°)×π×r² - (1/2)r²sinθ

                                        = r²[(θ/360°)×π - (1/2)sinθ]

                                        = 14×14×[(30/360)×(22/7) - (1/2)sin30°

                                        = 14×14×[(11/42) - (1/2)(1/2)]

                                        = 14×14×[(11/42) - (1/4)]  

                                        = 14×14×[(22-21)/84]    {LCM of 42 and 4 is 84}

                                        =14×14×(1/84)

                                        =7/3

                                  =2.33 cm²