Answer :

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 m

In right ΔOPA,

OP

^{2}+ AP

^{2}= OA

^{2}

OP

^{2}+ (12)

^{2}= (20)

^{2}

OP

^{2}= 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 = x OP x AB

= x 16 x 24 = 192 sq m

Also area of ΔOAB = x OB x AR

Hence, x OB x AR = 192

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

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