Question

do it with explanation​

Answer 1

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

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

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

Answer 2

Answer:

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 !!

_________________________________________________

★ Formula Used :-

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

_________________________________________________

★ 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 ?

_________________________________________________

★ 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.

_________________________________________________

For the Value of θ :-

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

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

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

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

_________________________________________________

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

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

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