21 સેમી ત્રિજ્યાવાળા વર્તુળના 120^{circ} ના ખ | vedclass

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Question

Given that, radius of the circle $(r)=21 \,cm$ and central angle of the sector $\operatorname{AOBA}(	heta)=120^{\circ}$

So, area of the circle $=\

Vedclass · Accepted Answer

$462$

Vedclass · Answer

Given that, radius of the circle $(r)=21 \,cm$ and central angle of the sector $\operatorname{AOBA}(	heta)=120^{\circ}$

So, area of the circle $=\pi r^{2}=\frac{22}{7} 	imes(21)^{2}=\frac{22}{7} 	imes 21 	imes 21$

$=22 	imes 3 	imes 21=1386 \,cm ^{2}$

Now, area of the minor sector $AOEA$ with central angle $120^{\circ}$

$=\frac{\pi r^{2}}{360^{\circ}} 	imes 	heta=\frac{22}{7} 	imes \frac{21 	imes 21}{360^{\circ}} 	imes 120$

$=\frac{22 	imes 3 	imes 21}{3}=22 	imes 21=462 \,cm ^{2}$

$	herefore$ Area of the major sector $A B O A$

$=$ Area of the circle $-$ Area of the sector $AOBA$

$=1386-462=924\, cm ^{2}$

Difference of the areas of a sector $AOBA$ and its corresponding major sector $ABOA$

$=\mid$ Area of major sector $A B O A$ $-$ Area of minor sector $A O B A \mid$

$=|924-462|=462 \,cm ^{2}$

Hence, the required difference of two sectors is $462 \,cm ^{2}$

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