Question

find the area of a sector of a circle with diameter 28m and the angle of the sector is 60​

Answer 1

Answer:

The area of the sector of the circle is 102.67 m² ( Approx. ).

Step-by-step-explanation:

We have given that,

Diameter of circle = 28 m

Central angle of circle $\sf\:(\:\theta\:)\:=\:60^{\circ}$

We have to find the area of the sector of the circle.

Now, we know that,

$\displaystyle{\pink{\sf\:Area\:of\:sector\:=\:\dfrac{\theta}{360}\:\times\:\pi\:r^2}\sf\:\:\:-\:-\:[\:Formula\:]}$

$\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{\theta}{360}\:\times\:\pi\:\times\:\left(\:\dfrac{d}{2}\:\right)^2\:\:\:-\:-\:\left[\:\because\:r\:=\:\dfrac{d}{2}\:\right]}$

$\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\cancel{\dfrac{60}{360}}\:\times\:\dfrac{22}{7}\:\times\:\left(\:\dfrac{28}{2}\:\right)^2}$

$\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{1}{6}\:\times\:\dfrac{22}{\cancel{7}}\:\times\:\dfrac{\cancel{28}}{2}\:\times\:\dfrac{28}{2}}$

$\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{1}{6}\:\times\:22\:\times\:\cancel{\dfrac{4}{2}}\:\times\:\dfrac{28}{2}}$

$\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{1}{6}\:\times\:22\:\times\:2\:\times\:\cancel{\dfrac{28}{2}}}$

$\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{1}{6}\:\times\:22\:\times\:2\:\times\:14}$

$\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{22\:\times\:\cancel{2}\:\times\:14}{\cancel{6}}}$

$\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{22\:\times\:14}{3}}$

$\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\cancel{\dfrac{308}{3}}}$

$\displaystyle{\implies\sf\:Area\:of\:sector\:=\:102.666}$

$\displaystyle{\implies\boxed{\red{\sf\:Area\:of\:sector\:\approx\:102.67\:m^2}}}$

Answer 2

Answer:

area of the sector of the circle is 102.67 m² ( Approx. ).

Step-by-step-explanation:

We have given that,

Diameter of circle = 28 m

Central angle of circle \sf\:(\:\theta\:)\:=\:60^{\circ}(θ)=60

∘

We have to find the area of the sector of the circle.

Now, we know that,

\displaystyle{\pink{\sf\:Area\:of\:sector\:=\:\dfrac{\theta}{360}\:\times\:\pi\:r^2}\sf\:\:\:-\:-\:[\:Formula\:]}Areaofsector=

360

θ

×πr

2

−−[Formula]

\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{\theta}{360}\:\times\:\pi\:\times\:(\:\dfrac{d}{2}\:)^2\:\:\:-\:-\:[\:\because\:r\:=\:\dfrac{d}{2}\:]}⟹Areaofsector=

360

θ

×π×(

2

d

)

2

−−[∵r=

2

d

]

\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\cancel{\dfrac{60}{360}}\:\times\:\dfrac{22}{7}\:\times\:(\:\dfrac{28}{2}\:)^2}⟹Areaofsector=

360

60

×

7

22

×(

2

28

)

2

\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{1}{6}\:\times\:\dfrac{22}{\cancel{7}}\:\times\:\dfrac{\cancel{28}}{2}\:\times\:\dfrac{28}{2}}⟹Areaofsector=

6

1

×

7

22

×

2

28

×

2

28

\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{1}{6}\:\times\:22\:\times\:\cancel{\dfrac{4}{2}}\:\times\:\dfrac{28}{2}}⟹Areaofsector=

6

1

×22×

2

4

×

2

28

\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{1}{6}\:\times\:22\:\times\:2\:\times\:\cancel{\dfrac{28}{2}}}⟹Areaofsector=

6

1

×22×2×

2

28

\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{1}{6}\:\times\:22\:\times\:2\:\times\:14}⟹Areaofsector=

6

1

×22×2×14

\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{22\:\times\:\cancel{2}\:\times\:14}{\cancel{6}}}⟹Areaofsector=

6

22×

2

×14

\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\dfrac{22\:\times\:14}{3}}⟹Areaofsector=

3

22×14

\displaystyle{\implies\sf\:Area\:of\:sector\:=\:\cancel{\dfrac{308}{3}}}⟹Areaofsector=

3

308

\displaystyle{\implies\sf\:Area\:of\:sector\:=\:102.666}⟹Areaofsector=102.666

\displaystyle{\implies\boxed{\red{\sf\:Area\:of\:sector\:\approx\:102.67\:m^2}}}⟹

Areaofsector≈102.67m

2