આકૃતિમાં P ,Q અને R ને કેન્દ્ર લઈ 14 સેમીની ત | vedclass

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Question

Given that, radii of each arc ( $r$ ) $=14\, cm$

Now, area of the sector with central $\angle P=\frac{\angle P}{360^{\circ}} 	imes \pi r^{2}$

$

Vedclass · Accepted Answer

$308$

Vedclass · Answer

Given that, radii of each arc ( $r$ ) $=14\, cm$

Now, area of the sector with central $\angle P=\frac{\angle P}{360^{\circ}} 	imes \pi r^{2}$

$=\frac{\angle P}{360^{\circ}} 	imes \pi 	imes(14)^{2} \,cm ^{2}$

[$\because$ area of any sector with central angle $	heta$ and radius $\left.r=\frac{\pi^{2}}{360^{\circ}} 	imes 	hetaight]$

Area of the sector with central angle $=\frac{\angle Q}{360^{\circ}} 	imes \pi r^{2}=\frac{\angle Q}{360^{\circ}} 	imes \pi 	imes(14)^{2}\,cm ^{2}$

and area of the sector with central angle $R=\frac{\angle R}{360^{\circ}} 	imes \pi r^{2}=\frac{\angle R}{360^{\circ}} 	imes \pi 	imes(14)^{2} \,cm ^{2}$

Therefore, sum of the areas (in $cm ^{2}$ ) of three sectors

$=\frac{\angle P}{360^{\circ}} 	imes \pi 	imes(14)^{2}+\frac{\angle 	heta}{360^{\circ}} 	imes \pi 	imes(14)^{2}+\frac{\angle R}{360^{\circ}} 	imes \pi 	imes(14)^{2}$

$=\frac{\angle P+\angle Q+\angle R}{360} 	imes 196 	imes \pi=\frac{180^{\circ}}{360^{\circ}} 	imes 196 \pi \,cm ^{2}$

[since, sum of all interior angles in any triangle is $\left.180^{\circ}ight]$

$=98 \pi \,cm ^{2}=98 	imes \frac{22}{7}$

$=14 	imes 22=308 \,cm ^{2}$

Hence, the required area of the shaded region is $308 cm ^{2}$.

આકૃતિમાં $P ,Q$ અને $R$ ને કેન્દ્ર લઈ $14$ સેમીની ત્રિજ્યાનાં ચાપ દોરેલા છે. રેખાંકિત ભાગનું ક્ષેત્રફળ શોધો. (સેમી$^2$ માં)

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