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

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Question

The measure of the angle of the minor sector

$	heta=90(\because OA \perp \overline{ OB })$

Perimeter of a minor sector

$=$ Length of the min

Vedclass · Accepted Answer

$126$

Vedclass · Answer

The measure of the angle of the minor sector

$	heta=90(\because OA \perp \overline{ OB })$

Perimeter of a minor sector

$=$ Length of the minor arc $+2$ (radius)

$	herefore 75=\frac{\pi r 	heta}{180}+2 r$

$	herefore 75=\frac{22}{7} 	imes \frac{r 	imes 90}{180}+2 r$

$	herefore 75=r\left(\frac{11}{7}+2ight)$

$	herefore 75=\frac{25}{7} r$

$	herefore 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, $m \angle O=90$

$	herefore$ 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 $\overline{ AB } \cup \widehat{ ACB }$

$=$ Area of minor sector OACB - Area of $\Delta$ OAB

$=346.5-220.5$

$=126 cm ^{2}$

Thus, the area of minor segment $\overline{ AB } \cup \widehat{ ACB }$ is $126 cm ^{2}$.

In a circle with centre $O, \overline{O A}$ and $\overline{O B}$ 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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