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

Given :-

• OR = OS = OM = 5m
• RS = SM = 6m
• RM = ?

$✧\: \rm{By \: theorem \: OP \: \perp \: bisector \: of \: RM}$

$\rm{ \bullet \: \: Let \: OP = xm,} \\ \\ \rm{ \bullet \: \: PS = (5-x) m}$

$★\underline{\bf{ \triangle OPR \: is \: right \: angled \: triangle}}$

$⇝\sf{OR^2=OP^2+RP^2} \\ \\ ⇝ \sf{5^2=X^2+RP^2} \\ \\ ⇝ \sf{25=X^2+RP^2} \\ \\ ⇝ \sf{ \pink{25-X^2=RP^2---(1)}}$

$★\underline{ \bf{\triangle RPS \: is \: right \: angled \: triangle}}$

$⇝\sf{RS^2=SP^2+RP^2} \\ \\ ⇝ \sf{6^2=(5-x)^2+RP^2} \\ \\⇝ \sf{36=5^2-25x+x^2+RP^2} \\ \\ ⇝\sf{36=25-10x+x^2+RP^2} \\ \\ ⇝ \sf{36-25+10x-x^2=RP^2} \\ \\ ⇝ \sf{ \pink{11+10x-x^2=RP^2----(2)}}$

$❒\underline{ \bold{ subtract \: equns \:(1)-(2)}}$

$↦\sf{25-x^2=11+10x-x^2} \\ \\ ↦ \sf{25-11=10x} \\ \\↦ \sf{14=10x} \\ \\ ↦ \sf{14=10x} \\ \\ ↦ \sf{x=\dfrac{14}{10}} \\ \\ ↦\boxed{\frak{ \blue{x=1.4cm}}}$

$❒\underline{ \bold{Substitute \: x \: in \: equation \: 1}}$

$⇝\sf{25-x^2=RP^2} \\ \\ ⇝\sf{25-(1.4)^2=RP^2} \\ \\ ⇝ \sf{25-1.96=RP^2} \\ \\ ⇝ \sf{23.04=RP^2} \\ \\ ⇝\sf{RP=\sqrt{23.04}} \\ \\⇝ \boxed{ \frak{ \orange{RP=4.8cm}}}$

$\underline{ \underline{ \rm{ \red{\therefore \: the \: distance \: between \: reshma \: and \: mandeep \: is \: QRP \: = \: 2(4.8)=9.6m}}}}$