overline{ OA } and overline{ OB } are radii o | vedclass

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(A) Given that the radii $\overline{ OA }$ and $\overline{ OB }$ are perpendicular to each other,the central angle $	heta = 90^\circ$.The radius of t

Vedclass · Accepted Answer

$24.64$

Vedclass · Answer

(A) Given that the radii $\overline{ OA }$ and $\overline{ OB }$ are perpendicular to each other,the central angle $	heta = 90^\circ$.The radius of the circle is $r = 5.6 \, cm$.The formula for the area of a minor sector is $	ext{Area} = \frac{\pi r^2 	heta}{360^\circ}$.Substituting the values: $	ext{Area} = \frac{22}{7} 	imes (5.6)^2 	imes \frac{90^\circ}{360^\circ}$.$	ext{Area} = \frac{22}{7} 	imes 31.36 	imes \frac{1}{4}$.$	ext{Area} = \frac{22}{7} 	imes 7.84 = 22 	imes 1.12 = 24.64 \, cm^2$.

$\overline{ OA }$ and $\overline{ OB }$ are radii of a circle perpendicular to each other. If $OA = 5.6 \, cm$,then the area of the minor sector formed by those radii is .......... $cm^2$.

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