Question

$\huge \bf \red{❁Question}$

● The perimeter of sector of a circle is equal to length of the arc of semicircle having the same radius. The angle of this sector is / are :-

(a) $\sf 114⁰32'44"$
(b) $\sf 65⁰27'16"$
(c) $\sf (π-2)rad$
(d) $\sf (π-3)rad$

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⭐ use $\sf\dfrac{\theta}{180}×πr$ as length of arc of the sector.

Step by step explanation

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Answer 1

Answer:

C is correct answer

Step-by-step explanation:

Perimeter of sector of circle = length of arc of semi-circle. θ=1301011or 130.91. Now, we have to convert it into degrees, minutes, seconds. Hence, we found the angle of sector =130.91∘.

Answer 2

Given -

• The perimeter of sector of a circle is equal to length of the arc of semicircle having the same radius.

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To find -

• Angle of the sector.

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Solution -

Firstly,

➝ Let the angle of the sector be θ.

➝ Let the length of the arc be l.

➝ Let the radius of the circle be r.

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We know that,

➝ l = θr

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As we can see in the figure,

➝ Perimeter of sector = r + r + l

➝ $\sf{P_1}$ = 2r + θr

➝ $\sf{P_1}$ = r(2 + θ)

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Since,

➝ Perimeter of semicircle ($\sf{P_2}$) = πr

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It is given that,

$\: \: \: \: \: \: \: \: \: \: \: \: \boxed{\bf{P_1 = P_2}}$

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Therefore,

➝ r(2 + θ) = πr

➝ 2 + θ = π

➝ θ = (π - 2)

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Converting in radians

$\tt:\implies\: \: \: \: \: \: \: \: {\theta = (\pi - 2) \times \dfrac{180^{\circ}}{\pi}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {\theta = \pi \times \dfrac{180^{\circ}}{\pi} - \dfrac{2 \times 180^{\circ}}{\pi}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {\theta = 180^{\circ} - \dfrac{360^{\circ}}{\pi}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {\theta = 180^{\circ} - 114^{\circ} 32' 44''}\: \: \: \: {\bigg\lgroup{\dfrac{360^{\circ}}{\pi} = 114^{\circ} 32' 44''}\bigg\rgroup}$

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$\bf:\implies\: \: \: \: \: \: \: \: {\purple{\theta = 65^{\circ} 27' 16''}}$

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Therefore,

• Angle of sector (θ) = 65°27'16''

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★ $\small\underline{\sf{Hence,\: option\: (b)\: is\: correct.}}$