In a circle with centre O,overline{OA} and ov | vedclass

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Question

(A) The measure of the angle of the minor sector is $	heta = 90^{\circ}$ (since $OA \perp OB$).Perimeter of a minor sector = Length of the minor arc

Vedclass · Accepted Answer

$126$

Vedclass · Answer

(A) The measure of the angle of the minor sector is $	heta = 90^{\circ}$ (since $OA \perp OB$).Perimeter of a minor sector = Length of the minor arc $+ 2 	imes$ (radius).$75 = \frac{\pi r 	heta}{180} + 2r$$75 = \frac{22}{7} 	imes \frac{r 	imes 90}{180} + 2r$$75 = r \left( \frac{11}{7} + 2 ight)$$75 = \frac{25}{7} r$$r = \frac{75 	imes 7}{25} = 21 \, cm$.Area of minor sector $OACB = \frac{\pi r^2 	heta}{360} = \frac{22}{7} 	imes \frac{21 	imes 21 	imes 90}{360} = 346.5 \, cm^2$.In $\Delta OAB$,$m\angle O = 90^{\circ}$,so Area of $\Delta OAB = \frac{1}{2} 	imes OA 	imes OB = \frac{1}{2} 	imes 21 	imes 21 = 220.5 \, cm^2$.Area of minor segment = Area of minor sector $OACB - 	ext{Area of } \Delta OAB = 346.5 - 220.5 = 126 \, cm^2$.

In a circle with centre $O$,$\overline{OA}$ and $\overline{OB}$ are radii perpendicular to each other. The perimeter of the sector formed by these radii is $75 \, cm$. Find the area of the corresponding minor segment. (in $cm^2$)

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