Question

In a circle of radius 21 cm an are subteends an a hole of 60° at the center.Find the length of arc and area of sector formed the arc. Solution

Answer 1

Answer :

$\large\boxed{{\star\:\:\bf{Length\:of\:the\:Arc\:=\:22\:cm\:and\:Area\:of\:the\:Sector\:=\:231\:cm^2}\:\:\star}}$

Explanation :

Figure :–

$\setlength{\unitlength}{25} \begin{picture}(6,6) \put(2,2){\circle{14}}\put(2,2){\circle*{0.09}}\qbezier(2,2)(2.4,2.5)(2.5,2.6)\qbezier(2,2)(2.4,1.3)(2.4,1.35)\qbezier(2.15,1.7)(2.5,2)(2.2,2.2)\put(2.4,1.8){ \bf{{60}^{\circ} } }\put(1,0.5){ \bf r = 21 cm }\put(2.2,1.8){\vector(-1,-2){0.5}}\end{picture}$

Given :–

• Radius = 21 cm

• Angle subtended on the circle = 60°

• $\sf{\pi\:=\:\dfrac{22}{7} }$

To Find :–

• (i) Length of the arc of the Sector .
• (ii) Area of the Sector .

Formulae Applied :–

• $\boxed{\bf{\star\:\:Length\:of\:the\:Arc\:=\:\dfrac{\theta}{360}\:\times\:2\pi r \:\:\star}}$

• $\boxed{\bf{\star\:\:Area\:of\:the\:Sector\:=\:\dfrac{\theta}{360}\:\times\:\pi r^2 \:\:\star}}$

Solution :–

(i) We have , r = 21 cm and θ = 60° .

  Putting these Values in the Length of the Arc Formula :-

$\rightarrow\sf{Length\:of\:the\:Arc\:=\:\dfrac{60}{360}\:\times\:2\:\times\:\dfrac{22}{7} \:\times\:21}$

$\rightarrow\sf{Length\:of\:the\:Arc\:=\:\dfrac{1}{6}\:\times\:2\:\times\:22\:\times\:3}$

$\rightarrow\sf{Length\:of\:the\:Arc\:=\:\dfrac{1}{2}\:\times\:2\:\times\:22}$

$\rightarrow\bf{Length\:of\:the\:Arc\:=\:22\:cm}$

(ii) We have ,  r = 21 cm and θ = 60° .

   Putting these Values in the Area of the Sector Formula :-

$\rightarrow\sf{Area\:of\:the\:Sector\:=\:\dfrac{60}{360}\:\times\:\dfrac{22}{7} \:\times\:(21)^2}$

$\rightarrow\sf{Area\:of\:the\:Sector\:=\:\dfrac{1}{6}\:\times\:\dfrac{22}{7} \:\times\:441}$

$\rightarrow\sf{Area\:of\:the\:Sector\:=\:\dfrac{1}{3}\:\times\:11\:\times\:63}$

$\rightarrow\sf{Area\:of\:the\:Sector\:=\:11\:\times\:21}$

$\rightarrow\bf{Area\:of\:the\:Sector\:=\:231\:cm^2}$