Let R, S and M be the position of Reshma, Salma and Mandip respectively.

From center O,

Draw OA, perpendicular to chord RS and OB, perpendicular to chord SM.

We know that,

Perpendicular from the center to the chord, bisects it.

⇒ AR = AS = $\dfrac{RS}{2} = \dfrac{6}{2}$ = 3 m

From figure,

⇒ OR = OS = OM = 5 m (Radii of circle)

In ∆ OAR,

By pythagoras theorem,

⇒ OR² = OA² + AR²

⇒ (5)² = OA² + (3)²

⇒ 25 = OA² + 9

⇒ OA² = 25 - 9

⇒ OA² = 16

⇒ OA = $\sqrt{16}$

⇒ OA = 4 m

By formula,

Area of triangle = $\dfrac{1}{2} \times$ base x height

Area of ∆ ORS = $\dfrac{1}{2}$ x OA x RS ........(1)

From figure,

In ∆ ORS for base OS, RC is the altitude.

Area of ∆ ORS = $\dfrac{1}{2}$ x OS x RC ........(2)

From equation (1) and (2), we get :

⇒ $\dfrac{1}{2}$ x OS x RC = $\dfrac{1}{2}$ x OA x RS

⇒ $\dfrac{1}{2}$ x RC x 5 = $\dfrac{1}{2}$ x 4 x 6

⇒ RC x 5 = 24

⇒ RC = $\dfrac{24}{5}$

⇒ RC = 4.8 m

As OC is the perpendicular to the chord RM,

∴ OC bisects RM.

⇒ MC = RC

⇒ MC = RC = 4.8 m

⇒ RM = 2 RC = 2 x 4.8 = 9.6 m

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

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