Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?
Let A, B and C represent the positions of Reshma, Salma and Mandip respectively.
AB = 6cm and BC = 6cm.
Radius OA = 5cm
BM ⊥ AC is drawn.
ABC is an isosceles triangle as AB = BC, M is mid-point of AC. BM is perpendicular bisector of AC and thus it passes through the centre of the circle.
Let AM = y and OM = x then BM = (5-x).
Applying Pythagoras theorem in ΔOAM,
OA2 = OM2 + AM2
⇒ 52 = x2 + y2 --- (i)
Applying Pythagoras theorem in ΔAMB,
AB2 = BM2 + AM2
⇒ 62 = (5-x)2 + y2 --- (ii)
Subtracting (i) from (ii), we get
36 - 25 = (5-x)2 - x2 -
⇒ 11 = 25 - 10x
⇒ 10x = 14 ⇒ x= 7/5
Substituting the value of x in (i), we get
y2 + 49/25 = 25
⇒ y2 = 25 - 49/25
⇒ y2 = (625 - 49)/25
⇒ y2 = 576/25
⇒ y = 24/5
Thus,
AC = 2×AM = 2×y = 2×(24/5) m = 48/5 m = 9.6 m
Distance between Reshma and Mandip is 9.6 m.
