In odot(O, 5.6),overline{OA} and overline{OB} | vedclass

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Question

(C) Given that in $\odot(O, 5.6)$,$\overline{OA} \perp \overline{OB}$.Thus,$OA = OB = 5.6 	ext{ cm}$ and $m\angle AOB = 90^\circ$.The area of the min

Vedclass · Accepted Answer

$15.68$

Vedclass · Answer

(C) Given that in $\odot(O, 5.6)$,$\overline{OA} \perp \overline{OB}$.Thus,$OA = OB = 5.6 	ext{ cm}$ and $m\angle AOB = 90^\circ$.The area of the minor sector formed by minor $\widehat{AB}$ is given by $\frac{	heta}{360^\circ} 	imes \pi r^2$.The area of the corresponding minor segment is given by $(	ext{Area of minor sector}) - (	ext{Area of } \Delta OAB)$.Therefore,the difference between the area of the minor sector and the area of the minor segment is equal to the area of $\Delta OAB$.Area of $\Delta OAB = \frac{1}{2} 	imes OA 	imes OB = \frac{1}{2} 	imes 5.6 	imes 5.6 = 15.68 	ext{ cm}^2$.

In $\odot(O, 5.6)$,$\overline{OA}$ and $\overline{OB}$ are radii perpendicular to each other. Then,the difference between the area of the minor sector formed by minor $\widehat{AB}$ and the corresponding minor segment is $\ldots \ldots \ldots \ldots cm^2$.

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