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Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn on a park. The distance between Reshma and Salma and between Salma and Mandip is 6m each. What is the distance between Reshma and Mandip?

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a

4.8m

b

9.6m

c

2.4m

d

7.2m

answer is B.

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Detailed Solution

Given that,
Reshma R, Mandip M and Salma S stand on the circle of radius 5m.
Reshma and Salma are 6m apart and also Salma and Mandip are 6m apart.
So, we get RS = 6m, SM = 6m
4453.jfif

4453.jfif

In ΔORS and ΔOMS
OR = OM (both radius)
OS = OS (common)
RS = SM (given)
ΔQRS ≅ ΔOMS [by side-side-side congruence rule]
∴ ∠ROS = ∠MOS (by CPCT) ⋯⋯⋯⋯⋯(1)
In ΔORX and ΔOMX
OR = OM (both radius)
∠ROX = ∠MOX (using i as ∠ROS=∠MOS)
OX = OX
So, ΔORX ≅ ΔOMX ( by SAS congruence rule)
∴ RX = MX (by CPCT) ⋯⋯⋯⋯⋯(2) Since, RX = MX. So, OX bisect chord RM.
∴ OX ⊥ RM as we know a line through the center of a circle to bisect a chord is perpendicular to the chord.
Now, let OX as t. So,
⇒ XS = OS – OX = 5 − t
In triangle ORX, we apply Pythagoras theorem.

{OR}^{2}

=

{OX}^{2}

+

{RX}^{2}

5^{2}

=

t^{2}

+

{RX}^{2}

25 −

t^{2}

=

{RX}^{2}

⋯⋯⋯⋯⋯(3)
Now, applying Pythagoras theorem on triangle XRS, we get:

{RS}^{2}

=

{XS}^{2}

+

{RX}^{2}

6^{2}

=

{(5 - t)}^{2}

+

{RX}^{2}

{RX}^{2}

= 11 −

t^{2}

+ 10t ⋯⋯⋯⋯⋯(4)
From (3) and (4) we get:
25 −

t^{2}

= 11 −

t^{2}

+ 10t
25 – 11 = −

t^{2}

+

t^{2}

+ 10t 14 = 10t
t =

\frac{14}{10}

= 1.4
Now, using t = 1.4 in equation (3) we get:

{RX}^{2}

= 25 −

t^{2}

{RX}^{2}

= 25 –

{(1.4)}^{2}

RX =

\sqrt{25 - 1.96}

So, RX = 4.8
Now, RM = 2RX
So, ∴ RM = 2×4.8 = 9.6m
Therefore, distance between Reshma and Mandip is 9.6m
So, Correct option is 2): 9.6 m

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