Let the positions ofReshma, Salma and Mandip be represented as A, B and C respectively.
From the question, weknow that AB = BC = 6cm.
So, the radius of thecircle i.e. OA = 5cm
Now, draw aperpendicular BM ⊥ AC.
Since AB = BC, ABC canbe considered as an isosceles triangle. M is mid-point of AC. BM is theperpendicular bisector of AC and thus it passes through the centre of thecircle.
Now,
let AM = y and
OM = x
So, BM will be =(5-x).
By applying Pythagoreantheorem in ΔOAM we get,
OA2 =OM2 +AM2
⇒ 52 =x2 +y2 — (i)
Again, by applyingPythagorean 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 thisequation we get the value of x as
x = 7/5
Substituting the valueof x in equation (i), we get
y2 +(49/25)= 25
⇒ y2 =25 – (49/25)
Solving it we get thevalue of y as
y = 24/5
Thus,
AC = 2×AM
= 2×y
= 2×(24/5) m
AC = 9.6 m
So, the distancebetween Reshma and Mandip is 9.6 m.