આકૃતિમાં ચતુષ્કોણ ABCD નાં શિરોબિંદુઓ A, B, C | vedclass

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Question

Given that, radius of each $\operatorname{arc}(r)=21 \,cm$

Area of sector with $\angle A=\frac{\angle A}{360^{\circ}} 	imes \pi r^{2}=\frac{\angle

Vedclass · Accepted Answer

$1396$

Vedclass · Answer

Given that, radius of each $\operatorname{arc}(r)=21 \,cm$

Area of sector with $\angle A=\frac{\angle A}{360^{\circ}} 	imes \pi r^{2}=\frac{\angle A}{360^{\circ}} 	imes \pi 	imes(21)^{2}\, cm ^{2}$

[$\because$ area of any sector with central angle $	heta$ and radius $\left.r=\frac{\pi^{2}}{360^{\circ}} 	imes 	hetaight]$

Area of sector with $\angle B=\frac{\angle B}{360^{\circ}} 	imes \pi r^{2}=\frac{\angle B}{360^{\circ}} 	imes \pi 	imes(21)^{2}\, cm ^{2}$

Area of sector with $\angle C=\frac{\angle C}{360^{\circ}} 	imes \pi r^{2}=\frac{\angle C}{360^{\circ}} 	imes \pi 	imes(21)^{2} \,cm ^{2}$

and area of sector with $\angle D=\frac{\angle D}{360^{\circ}} 	imes \pi r^{2}=\frac{\angle D}{360^{\circ}} 	imes \pi 	imes(21)^{2} \,cm ^{2}$

Therefore, sum of the areas (in $cm ^{2}$ ) of the four sectors.

$=\frac{\angle A}{360^{\circ}} 	imes \pi 	imes(21)^{2}+\frac{\angle B}{360^{\circ}} 	imes \pi 	imes(21)^{2}+\frac{\angle C}{360^{\circ}} 	imes \pi 	imes(21)^{2}+\frac{\angle D}{360^{\circ}} 	imes \pi 	imes(21)^{2}$

$=\frac{(\angle A+\angle B+\angle C+\angle D)}{360^{\circ}} 	imes \pi 	imes(21)^{2}$ [$\because$ sum of all interior angles in any quadrilateral $\left.=360^{\circ}ight]$

$=22 	imes 3 	imes 21=1386\, cm ^{2}$

Hence, required area of the shade region is $1386\, cm ^{2}$

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