Question

in a circle of radius 21 cm an arc subtends an angle of 60 degree at the centre. find the area of the sector formed by the arc ​

Answer 1

Given:

• Radius of Circle, r = 21 cm
• Angle subtended by arc, $\theta$ = 60°

⠀⠀⠀⠀⠀⠀⠀

To find:

• Area of sector?

⠀⠀⠀⠀⠀⠀⠀

Solution:

⠀⠀⠀⠀⠀⠀⠀

$\setlength{\unitlength}{1.2mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(25,30){\line(5, - 4){16}}\put(25,30){\circle*{1}}\put(33,25){\sf{21 cm}}\put(25,30){\line( - 5, - 4){16}}\qbezier(22,27.5)(24,24)(28,27.5)\put(10,25){\sf{21 cm}}\put(23,22){\sf{60^\circ }}\end{picture}$

⠀⠀⠀⠀⠀⠀⠀

We know that,

⠀⠀⠀⠀⠀⠀⠀

$\star\;{\boxed{\sf{\purple{Area_{\;(sector)} = \dfrac{ \theta}{360^\circ} \times \pi r^2}}}}$

Putting values,

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf \dfrac{ \cancel{60}}{ \cancel{360}} \times \dfrac{22}{7} \times 21 \times 21$

$:\implies\sf \dfrac{1}{6} \times \dfrac{22}{ \cancel{7}} \times \cancel{21} \times 21$

$:\implies\sf \dfrac{1}{ \cancel{6}} \times 22 \times \cancel{3} \times 21$

$:\implies\sf \dfrac{1}{ \cancel{2}} \times \cancel{22} \times 21$

$:\implies\sf 11 \times 21$

$:\implies{\boxed{\sf{\pink{231\;cm^2}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;the\;area\;of\:sector\;formed\;by\;the\;arc\;is\; \bf{231\;cm^2}.}}}$