Question

Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha.

Answer 1

Let R, S and M be the position of Ishita, Isha and Nisha respectively.

Since OA is a perpendicular bisector on RS, so AR = AS = 24/2 = 12 cm 

Radii of circle = OR = OS = OM = 20 cm (Given) 

In ΔOAR: 

By Pythagoras theorem, 

OA² + AR² = OR² 

OA² + 12² = 20² 

OA² = 400 – 144 = 256 

Or OA = 16 m …(1) 

From figure, OABC is a kite since OA = OC and AB = BC. We know that, diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal. 

So in ΔRSM, 

∠RCS = 90° and RC = CM …(2) 

Now, Area of ΔORS = Area of ΔORS 

=>1/2 x OA x RS = 1/2 x RC x OS 

=> OA x RS = RC x OS 

=> 16 x 24 = RC x 20 

=> RC = 19.2 

Since RC = CM (from (2), we have 

RM = 2(19.2) = 38.4 

So, the distance between Ishita and Nisha is 38.4 m.