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

Let the three girls Reshma, Salma and Mandip are standing on the circle of radius $5$ m at points $R,S$ and $M$ respectively.

Given: $RS=SM=6$ m

We know that center $O$ lies on the bisector of $\angle RSM$ i.e. ray $OS$

Let $OA\perp$ chord $RM$

We know the perpendicular from center to the chord, bisects the chord.

Hence $RA=AM$

In $△SAR$, $\angle A={90}^o$

$S{R}^2=S{A}^2+R{A}^2$ [By Pythagoras theorem]

$\Rightarrow 36=S{A}^2+R{A}^2$

$\Rightarrow R{A}^2=36-S{A}^2$ ... (1)

In the $△OAR$,

$R{O}^2=A{O}^2+R{A}^2$ [By Pythagoras theorem]

$\Rightarrow 25=(OS-AS{)}^2+R{A}^2$

$\Rightarrow R{A}^2=25-(OS-AS{)}^2$

$R{A}^2=25-(5-AS{)}^2$ ... (2)

From (1) and (2) , we get

$\Rightarrow 36-S{A}^2=25-(5-SA{)}^2$

$\Rightarrow 11-S{A}^2+(5-SA{)}^2=0$

$\Rightarrow 11-S{A}^2+25-10SA+A{S}^2=0$

$\Rightarrow 10SA=36$

$\Rightarrow SA=3.6$
Putting $SA=3.6$ in (1), we get

$R{A}^2=3.6-(3.6{)}^2$

$=36-12.96$

$=36-12.96$

$=23.04$ m

$\therefore RA=4.8$ m

$\Rightarrow RM=2RA=2\times 4.8=9.6$ m

Hence, the distance between Reshma and Mandip $=9.6$ m