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

☆ Diagram :

$\setlength{\unitlength}{1.2mm}\begin{picture} \thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\qbezier(30,26)(25,18)(20,26)\put(25,30){\line(5, - 4){16}}\put(25,30){\circle*{1}}\put(24,32){\sf\large{O}}\put(15,40){\sf\large{Major Sector}}\put(5,14){\sf\large{A}}\put(25,30){\line(- 5, -4){16}}\put(43,14){\sf\large{B}}\put(14,16){\sf\large{Minor Sector}}\put(22,19){\sf\large{ {60}^{ \circ} }}\end{picture}$

Answer :

$:\implies \sf Area \: of \: sector = \dfrac{\theta}{360^{\circ}} \times \pi r^2$

$:\implies \sf Area \: of \: sector = \dfrac{ {60}^{ \circ} }{360^{\circ}} \times \pi r^2$

$:\implies \sf Area \: of \: sector = \dfrac{ {60}^{ \circ} }{360^{\circ}} \times \dfrac{22}{7} \times 21 \times 21$

$:\implies \sf Area \: of \: sector = \dfrac{ 6 }{36} \times 22 \times 3 \times 21$

$:\implies \sf Area \: of \: sector = \dfrac{ 1 }{6} \times 22 \times 3 \times 21$

$:\implies \sf Area \: of \: sector = \dfrac{1}{6} \times 66 \times 21$

$:\implies \sf Area \: of \: sector = 11 \times 21$

$:\implies \underline{ \boxed{ \sf Area \: of \: sector = 231 \: {cm}^{2} }}$

Therefore,Area of sector is 231 cm².