Question

19) The central angle of a sector of circle of
area 9n sq.cm is 60', the perimeter of the
sector is
Α) π
B) 3+1 C) 6+ 1
D) 6​

Answer 1

Given:

✰ Area of the central angle of a sector of circle = 9π cm²

✰ The central angle of a sector of circle = 60°

To find

☯ The perimeter of a sector of circle

Solution:

As we know the area of the circle is 9π cm² so we will find out radius first. After calculating radius, we will find the perimeter of a sector of circle by using formula.

Area of circle = πr²

➩ 9π = πr²

Now, π on R.H.S ( right hand side ) gets cancel with π on L.H.S ( left hand side )

➩ 9 = r²

➩ r = √9

➩ r = 3

Now, let's find out the perimeter of a sector of circle by using formula.

$\sf{ \underline{ \boxed{ \sf{Perimeter \: of \: sector = \dfrac{2\pi r \theta}{360 \degree} + 2r}}}}$

$\implies \sf{ \dfrac{2 \times \pi \times 3 \times 60 \degree}{360 \degree} + 2 \times 3}$

$\implies \sf{ \dfrac{2 \times \pi \times 3 \times \cancel{60} \degree}{ \cancel{360 \degree}_{6 \degree}} + 2 \times 3}$

$\implies \sf{ \dfrac{2 \times \pi \times 3 \times 1 \degree }{6 \degree} + 2 \times 3}$

$\implies \sf{ \dfrac{6 \degree \times \pi }{6 \degree} + 2 \times 3}$

$\implies \sf{ \dfrac{ \cancel{6 \degree} \times \pi }{ \cancel{6 \degree}} + 2 \times 3}$

$\implies \sf {\pi + 2 \times 3}$

$\implies \sf {\pi + 6}$

∴ The perimeter of a sector of circle = $\green{ \underline{ \boxed{ \sf{ \pi + 6}}}}$

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Answer 2

Answer

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• ➽ 6 + π cm${\boxed{\green{\checkmark{}}}}$

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$\orange{\sf{\star\;Step\;-\;by\;-\;step\;explanation:}}$

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• ➽ The central angle of a sector of circle of area 9 Pi square centimetre is 60 degree the perimeter of the sector is

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Area of circle = π Radius²   = 9π

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• ➽Radius² = 9

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• ➽ radius = 3  cm

• ➽ Circumference = 2πR  = 2π*3 = 6π cm

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• ➽ Length of Arc = (60/360)*6π = π cm

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$\green{\sf{\star\;Perimeter\; of\; Arc \;=\; Length \;of\; arc\; +\; radius\; + \;Radius}}$

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• ➽ = π + 3 + 3

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$\pink{\sf{\star\;=  \;6\; + \;π \;cm}}$

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