# Three girls Reshma, Salma and Mandip …

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Preeti Dabral 1 year, 5 months ago

In {tex}\Delta \mathrm { NOR }{/tex} and {tex}\Delta \mathrm { NOM }{/tex}

ON = ON |Common

{tex}\angle \mathrm { NOR } = \angle \mathrm { NOM }{/tex} {tex}| \because{/tex} Equal chords of a circle subtend equal

angle at the centre

OR = OM |Radii of a circle

{tex}\therefore \triangle \mathrm { NOR } \cong \Delta \mathrm { NOM }{/tex} [SAS Rule]

{tex}\therefore \angle O N R = \angle O N M{/tex} [c.p.c.t]

and NR = NM [c.p.c.t.]

But {tex}\angle O N R + \angle O N M = 180 ^ { \circ }{/tex} |Linear Pair Axiom

{tex}\therefore \angle O \mathrm { NR } = \angle \mathrm { O } \mathrm { NM } = 90 ^ { \circ }{/tex}

{tex}\triangle{/tex} ON is the perpendicular bisector of RM,

Draw bisector SN of {tex}\angle \mathrm { R } \mathrm { SM }{/tex} to intersect the chord RM in N.

In {tex}\Delta \mathrm { RSN }{/tex} and {tex}\Delta \mathrm { MSN }{/tex}

RS = MS (= 6 cm each)

SN = SN [Common]

{tex}\angle R S N = \angle M S N{/tex} [By construction]

{tex}\therefore \Delta R S N \cong \Delta \mathrm { NSN }{/tex} [SAS Rule]

{tex}\therefore \angle R N S = \angle M N S{/tex} [c.p.c.t]

and RN = MN [c.p.c.t]

But {tex}\angle \mathrm { RNS } + \angle \mathrm { MNS } = 180 ^ { \circ }{/tex} |Linear Pair Axion

{tex}\therefore \angle R N S = \angle M N S = 90 ^ { \circ }{/tex}

{tex}\therefore \mathrm { SN }{/tex} is the perpendicular bisector of RM and therefore passes through O when produced.

Let ON = x m

Then SN = (5 - x) m

In right triangle ONR,

x

^{2}+ RN^{2 }= 5^{2}, ------ (1) |By Pythagoras theoremIn right triangle SNR,

(5-x)

^{2}+ RN^{2}= 6^{2}---- (2) |By Pythagoras theoremFrom (1),

RN

^{2}= 5^{2}- x^{2}From (2),

RN

^{2}= 6^{2}- (5 - x)^{2}Equating the two values of RN

^{2}, we get5

^{2}- x^{2}= 6^{2}- (5 -x)^{2}{tex}\Rightarrow 25 - x ^ { 2 } = 36 - ( 25 - 10 x + x ) ^ { 2 }{/tex} {tex}\Rightarrow 25 - x ^ { 2 } = 36 - 25 + 10 x - x ^ { 2 }{/tex}

{tex}\Rightarrow 25 - x ^ { 2 } = 11 + 10 x - x ^ { 2 }{/tex}{tex}\Rightarrow 25 - 11 = 10 x{/tex}

{tex}\Rightarrow{/tex} 14 = 10 x {tex}\Rightarrow{/tex}10x = 14

{tex}\Rightarrow x = \frac { 14 } { 10 } = 1.4{/tex}

Putting x = 1.4 in (1), we get

(1.4)

^{2}+ RN^{2}= 5^{2}{tex}\Rightarrow R N ^ { 2 } = 5 ^ { 2 } - ( 1.4 ) ^ { 2 }{/tex} {tex}\Rightarrow \mathrm { RN } ^ { 2 } = 25 - 1.96{/tex}

{tex}\Rightarrow \mathrm { RN } ^ { 2 } = 23.04 \Rightarrow \mathrm { RN } = \sqrt { 23.04 }{/tex}

{tex}\Rightarrow{/tex} RN = 4.8

{tex}\therefore{/tex} RM = 2 RN = {tex}2 \times 4.8 \mathrm { m } = 9.6 \mathrm { m }{/tex}

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

0Thank You