Question

Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius $5m$ drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is $6m$ each, what is the distance between Reshma and Mandip?

Answer 1

Draw perpendiculars $OA$ and $OB$ on $RS$ and $SM$ respectively.

$AR=AS=\frac{1}{2}\times 6=3m$

$OR=OS=OM=5m$ (Radii of the circle)

In $\Delta OAR$, we have

$O{A}^2+A{R}^2=O{R}^2$ [Using pythagoras theorem]

$O{A}^2+(3m{)}^2=(5m{)}^2$

$O{A}^2=(25-9)m^2=16m^2$

$OA=4m$

$ORSM$ will be a kite ($OR=OM$ and $RS=SM$). We know that the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.

Let the diagonals of the kite $ORSM$ bisects each other at a point $C.$
$\therefore \angle RCS$ will be of ${90}^{\circ }andRC=CM$

Area of $\Delta ORS=\frac{1}{2}\times OA\times RS=\frac{1}{2}\times 4\times 6=12m^2\dots \dots \left(i\right)$
Also, Area of $\Delta ORS=\frac{1}{2}\times RC\times OS=\frac{1}{2}\times RC\times 5\dots \dots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right),$ we have
$\frac{1}{2}\times RC\times 5=12$
$RC=\frac{12\times 2}{5}=4.8m$
$RC=4.8m$

We have, $RM=RC+CM$
We know that $RC=CM$ (given)
$\therefore$ $RM=2RC=2\left(4.8\right)=9.6m$
Therefore, the distance between Reshma and Mandip is $9.6m.$