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 distance between Ishita and Nisha?

Answer 1

Let A be the position of Ishita, B be the position of Isha and C be the position of Nisha

Given,

AB = BC = 24 m 

OA = OB = OC = 20 m 

[Radii of circle]

Draw perpendiculars OP and OQ on AB and BC respectively

AP = PB = 12 min 

right ΔOPA,

OP² + AP² = OA²

OP² + (12)² = (20)²

OP² = 256 sq m

Therefore, 

OP = 16 m

From the figure, 

OABC is a kite since OA = OC and AB = BC.

Recall that the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.

Therefore, 

∠ARB = 90° and AR = RC

Area of ΔOAB,

= \(\frac{1}{2}\) x OP x AB

= \(\frac{1}{2}\) x 16 x 24

= 192 sq m

Also,

Area of ΔOAB = \(\frac{1}{2}\) x OB x AR

Hence,

 \(\frac{1}{2}\) x OB x AR = 192

 \(\frac{1}{2}\) x 20 x AR = 192

Therefore, 

AR = 19.2 m 

But,

AC = 2AR 

= 2(19.2) = 38.4 m

Thus, 

the distance between Ishita and Nisha is 38.4 m