Math, asked by ravishah49, 3 days ago

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Answers

Answered by SarcasticKarma
3

Answer:

 \:\\ \large{\underline{\underline{\green{Firstly,\;let's\;understand\;the\;concept\;used\;:-}}}}

Here the concept of Areas of Sectors has been used. We see that we are given the area of Sector. Now we need to find angle θ subtended by the sector. We can simply apply the values we got into the equations we formed.

Let's do it !!

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★ Formula Used :-

 \:\\ \large{\boxed{\sf{\red{Area\;\:of\;\:Sector\;\:=\;\bf{\blue{\dfrac{\pi r^{2} \theta}{360^{\circ}}}}}}}}

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★ Question :-

If the area of a sector of a circle of radius 6 cm is 9 pie cm square then the angle subtended at the centre of the circle is ?

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★ Solution :-

  • » Radius of the circle = r = 6 cm
  • » Area of sector = 9π cm²

• Let the angle subtended by the sector at the centre of the circle be θ.

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For the Value of θ :-

 \: \\ \qquad \large{\sf{:\Longrightarrow\;\;\: Area\;\:of\;\:Sector\;\:=\;\bf{\dfrac{\pi r^{2} \theta}{360^{\circ}}}}}

 \: \\ \qquad \large{\sf{:\Longrightarrow\;\;\: 9 \pi \;\:cm^{2}\;\:=\;\bf{\dfrac{\pi (6)^{2}\:\times\: \theta}{360^{\circ}}}}}

 \: \\ \qquad \large{\sf{:\Longrightarrow\;\;\: \theta \;\: = \; \: \bf{\dfrac{9\;\times\;\cancel{\pi}\;\times\;360^{\circ}}{\:\cancel{\pi}\:\times\:6\;\times\;6} \: \: = \: \: \underline{\underline{90^{\circ}}}}}}

 \: \\ \large{\underline{\underline{\rm{Thus,\;the\;angle\;formed\;by\;sector\;at\;centre\;is\;\;\boxed{\bf{90^{\circ}}}}}}}

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 \: \\ \large{\underbrace{\underbrace{\sf{More\;Formulas\;to\;know\;:-}}}}

 \: \\ \;\;\sf{\leadsto\;\;\; Area\;subtended\;by\;a\;chord\;=\;\dfrac{\pi r^{2} \theta}{360^{\circ}} \;-\; \dfrac{1}{2}\:r^{2}\:sin\theta}

 \: \\ \leadsto\;\; \sf{Length\;of\;an\;Arc\;=\;\dfrac{2 \pi r\theta}{360^{\circ}}}

Answered by EvilExtinction
5

Answer:

\huge{\tt{\red{}\green{A}\purple{N}\pink{S}\blue{W}\orange{E}\red{R}}}:)

» Radius of the circle = r = 6 cm

» Area of sector = 9π cm²

• Let the angle subtended by the sector at the centre of the circle be θ.

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For the Value of θ :-</p><p>

\begin{gathered} \: \\ \qquad \large{\sf{:\Longrightarrow\;\;\: Area\;\:of\;\:Sector\;\:=\;\bf{\dfrac{\pi r^{2} \theta}{360^{\circ}}}}}\end{gathered}</p><p>

Step-by-step explanation:

</p><p>\begin{gathered} \: \\ \large{\underbrace{\underbrace{\sf{More\;Formulas\;to\;know\;:-}}}}\end{gathered}</p><p>

⇝Areasubtendedbyachord=360∘πr2θ−21r2sinθ</p><p></p><p>[tex]</p><p>\begin{gathered} \: \\ \leadsto\;\; \sf{Length\;of\;an\;Arc\;=\;\dfrac{2 \pi r\theta}{360^{\circ}}}\end{gathered}

⇝LengthofanArc=360∘2πrθ</p><p>

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