In a circle with radius 42 cm,a minor arc su | vedclass

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Question

(A) Given: Radius $r = 42 \ cm$,Central angle $	heta = 60^{\circ}$.$1$. Area of minor sector = $\frac{	heta}{360^{\circ}} 	imes \pi r^2 = \frac{60}

Vedclass · Accepted Answer

Area of minor sector = $924 \ cm^2$,Area of minor segment = $161.07 \ cm^2$

Vedclass · Answer

(A) Given: Radius $r = 42 \ cm$,Central angle $	heta = 60^{\circ}$.$1$. Area of minor sector = $\frac{	heta}{360^{\circ}} 	imes \pi r^2 = \frac{60}{360} 	imes \frac{22}{7} 	imes 42 	imes 42 = \frac{1}{6} 	imes 22 	imes 6 	imes 42 = 22 	imes 42 = 924 \ cm^2$.$2$. Area of minor segment = Area of minor sector - Area of $	riangle OAB$.Area of $	riangle OAB = \frac{1}{2} r^2 \sin(	heta) = \frac{1}{2} 	imes 42 	imes 42 	imes \sin(60^{\circ}) = 882 	imes \frac{\sqrt{3}}{2} = 441 	imes 1.73 = 762.93 \ cm^2$.Area of minor segment = $924 - 762.93 = 161.07 \ cm^2$.

In a circle with radius $42 \ cm$,a minor arc subtends an angle of $60^{\circ}$ at the centre. Find the area of the minor sector and the minor segment corresponding to this arc. (Use $\sqrt{3} = 1.73$)

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