Let the positions of Reshma, Salma and Mandip be represented as A, B and C respectively.

From the question, we know that AB = BC = 6cm.

So, the radius of the circle i.e. OA = 5cm

Now, draw a perpendicular BM ⊥ AC.

Since AB = BC, ABC can be considered as an isosceles triangle. M is mid-point of AC. BM is the perpendicular bisector of AC and thus it passes through the centre of the circle.

Now,

let AM = y and

OM = x

So, BM will be = (5-x).

By applying Pythagorean theorem in ΔOAM we get,

OA² = OM² +AM²

⇒ 5² = x² +y² — (i)

Again, by applying Pythagorean theorem in ΔAMB,

AB² = BM² +AM²

⇒ 6² = (5-x)²+y² — (ii)

Subtracting equation (i) from equation (ii), we get

36-25 = (5-x)² +y² -x²-y²

Now, solving this equation we get the value of x as

x = 7/5

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

y² +(49/25) = 25

⇒ y² = 25 – (49/25)

Solving it we get the value of y as

y = 24/5

Thus,

AC = 2×AM

= 2×y

= 2×(24/5) m

AC = 9.6 m

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