As shown in the diagram,overline{OA} and over | vedclass

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Question

(A) For the sector $OACB$,the radius $r = 35 	ext{ cm}$ and the measure of the angle $	heta = 90^{\circ}$ (since $\overline{OA} \perp \overline{OB}$

Vedclass · Accepted Answer

$752.5$

Vedclass · Answer

(A) For the sector $OACB$,the radius $r = 35 	ext{ cm}$ and the measure of the angle $	heta = 90^{\circ}$ (since $\overline{OA} \perp \overline{OB}$).Area of sector $OACB = \frac{\pi r^2 	heta}{360^{\circ}}$$= \frac{22}{7} 	imes \frac{35 	imes 35 	imes 90^{\circ}}{360^{\circ}}$$= 962.5 	ext{ cm}^2$In $\Delta ODA$,$\angle O = 90^{\circ}$.Area of $\Delta ODA = \frac{1}{2} 	imes OD 	imes OA$$= \frac{1}{2} 	imes 12 	imes 35$$= 210 	ext{ cm}^2$Area of the shaded region = Area of sector $OACB - 	ext{Area of } \Delta ODA$$= 962.5 - 210$$= 752.5 	ext{ cm}^2$Thus,the area of the shaded region is $752.5 	ext{ cm}^2$.

As shown in the diagram,$\overline{OA}$ and $\overline{OB}$ are two radii of $\odot(O, 35 \text{ cm})$ perpendicular to each other. If $OD = 12 \text{ cm}$,find the area of the shaded region. (in $\text{cm}^2$)

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