Question

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?

Answer 1

Let O be

the centre of the circle. A, B and C represent the positions of Reshma, Salma

and Mandip.

AB = 6cm and BC = 6cm.

Radius OA = 5cm.

(given)

Draw BM ⊥ AC

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).

In ΔOAM,

OA²=OM²+AM² ( by Pythagoras

theorem)

5²=x²+ y²—-----(1)

In ΔAMB,

AB²=BM² +AM²

(by Pythagoras theorem)

6²= (5-x)²+y² — (ii)

Subtracting (i) from (ii),

36 – 25 = (5-x)² -x²

11 = (25+x²– 2×5×x) - x²

11= 25+x²-10x - 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

AC = 2×y

AC= 2×(24/5) m

= 48/5 m = 9.6 m

Hence, the Distance between Reshma and Mandip

is 9.6 m.