Question

Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandeep, Mandeep to Reshma. If the distance between Reshma and Salma and between Salma and Mandeep is 6m each, what is the distance between Reshma and Mandeep?
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Answer 1

${ \large{ \sf{ \underbrace{\underline{\bigstar \:Solution :-}}}}}$

$\textbf{Let R, S and M represent the position of Reshma, Salma and Mandeep respectively.}$

$\bold{\boxed{Clearly\: △\:RSM\: is\: an\: isosceles\: triangle\: as}}$

$\bold\pink{\fbox{\sf{RS\:=\: SM\: = \:6m}}}$

$\therefore\textbf{Join OS which intersect RM at A.}$

$\bold\pink{\fbox{\sf{In \:△\:ROS \:and\: △\:MOS}}}$

$\underline{\bf{OR = OM}}$ (Radii of the same circle)

$\underline{\bf{OS = OS}}$(Common)

$\underline{\bf{RS = SM}}$ (Each 6cm)

∴ △ROS ≅ △MOS (By SSS congruence criterion)

∴ ∠RSO = ∠MSO (CPCT)

In △RAS and △MAS

AS = AS (Common)

∴ ∠RSA = ∠MSA (Since, ∠RSO = ∠MSO)

RS = MS (Given)

∴ △RAS ≅ △MAS (By SAS congruence criterion)

∴ ∠RAS = ∠MAS (CPCT)

Since, ∠RAS + ∠MAS = 180° (Linear pair)

⇒ ∠RAS = ∠MAS = 90°

Let OA = x m ⇒ AS = (5 - x)m

$\textit{In \:right \:triangle\: RAS}$

$\sf{RS2 = RA2 + AS2}$

$\sf{⇒62 = RA2 + (5 - x)2}$

$\sf{⇒RA2 = 62 - (5 - x)2 .(i)}$

$\sf\color{pink}{In\: right\: triangle\: RAO}$

RO2 = RA2 + OA2}[/tex]

$\sf{⇒52 = RA2 + x2}$

$\sf{⇒RA2 = 52 - x2 .(ii)}$

From Equation 1 and 2 we have :-

$\sf{62 - (5-x)2 = 52 - x2}$

$\sf{62 - 52 = (5-x)2 - x2}$

$\sf{36 - 25 = 25 + x2 - 10x - x2}$

$\sf{11 = 25 - 10x ⇒ 10x = 14 ⇒ x = 1.4m}$

$\bold\color{red}{\boxed{From \:equation\:(ii), \:we \:have}}$

$\sf{RA2 = 52 -(1.4)2 = 25 - 1.96}$

$\sf{RA2 = 23.04 ⇒ RA = √23.04}$

$\sf{RA = 4.8m}$

$\sf\color{green}As\:the\: perpendicular\:from \;the\: centre \:of \:a \:circle\: bisects\: the\: chord. .$

$\tt{∴ RM = 2RA}$

$\tt{RM = 2 x 4.8 = 9.6m}$

$\sf\color{indigo}Hence,\: distance\: between\: Reshma \:and\: Mandeep/: is\: 9.6m.$

Thankyou :)

Answer 2

Let  O be the centre of the circle. A, B and C represent the positions of Reshma, Salma

and Mandip.

$\tt \: given \: that$

• AB = 6cm
• BC = 6cm.
• Radius OA = 5cm (given)

$\tt \: construction \:$

Draw BM ⊥ AC

$\tt \: solution$

ABC is an isosceles triangle as AB = BC, M is

mid-point of AC. BM is perpendicular bisector of AC and thus it passes through the centre of the

circle.

Let,

• AM = y
• OM = x
• BM = (5-x).

$In \:ΔOAM,$

OA²=OM²+AM² ( by Pythagoras theorem)

5²=x²+ y²—(i)

$In\: ΔAMB$

AB²=BM² +AM² (by Pythagoras theorem)

6²= (5-x)²+y² — (ii)

Subtracting (i) from (ii),

36 – 25 = (5-x)² -x² 

11 = 25x²– 2×5×x) - x²

11= 25+x²-10x - x²

11= 25-10x

10x = 14

x= 7/5

Substituting the value of x in (i), we get

y²+ 49/25 = 25

y² =25 – 49/25

y² =(625 – 49)/25

y²=576/25

y = 24/5

Thus,

AC = 2×AM

AC = 2×y

AC= 2×(24/5) m

= 48/5 m = 9.6 m

Hence, the Distance between Reshma and Mandip is 9.6 m.

_____________________________

$\sf \: @WildCat7083$