Three girls Reshma,Salma and Mandip are playi | vedclass

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(D) Let $O$ be the centre of the circle. Given $OR = OM = OS = 5 \, m$ (radii of the circle) and $RS = SM = 6 \, m$.In quadrilateral $ORSM$,$OR = OM =

Vedclass · Accepted Answer

$9.6$

Vedclass · Answer

(D) Let $O$ be the centre of the circle. Given $OR = OM = OS = 5 \, m$ (radii of the circle) and $RS = SM = 6 \, m$.In quadrilateral $ORSM$,$OR = OM = 5 \, m$ and $RS = SM = 6 \, m$.Therefore,quadrilateral $ORSM$ is a kite.Thus,its diagonal $OS$ bisects the diagonal $RM$ at right angles at point $K$.Therefore,$\angle RKO = 90^{\circ}$ and $K$ is the midpoint of $RM$,so $RM = 2 RK$.Draw $OL \perp RS$. Since $OL$ is perpendicular from the centre to the chord $RS$,$L$ is the midpoint of $RS$.Therefore,$RL = \frac{1}{2} RS = \frac{1}{2} 	imes 6 = 3 \, m$.In right-angled $\Delta RLO$,by Pythagoras theorem:$RO^2 = OL^2 + RL^2$$5^2 = OL^2 + 3^2$$25 = OL^2 + 9$$OL^2 = 16 \implies OL = 4 \, m$.Now,the area of $\Delta ROS$ can be calculated in two ways:Area $= \frac{1}{2} 	imes RS 	imes OL = \frac{1}{2} 	imes 6 	imes 4 = 12 \, m^2$.Also,Area $= \frac{1}{2} 	imes OS 	imes RK = \frac{1}{2} 	imes 5 	imes RK$.Equating the two areas: $\frac{1}{2} 	imes 5 	imes RK = 12$$5 	imes RK = 24 \implies RK = 4.8 \, m$.Since $RM = 2 RK$,$RM = 2 	imes 4.8 = 9.6 \, m$.Thus,the distance between Reshma and Mandip is $9.6 \, m$.

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