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?

Given:

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.

The distance between Ishita and Isha and between Isha and Nisha is $24\ m$ each.
To do:

We have to find the distance between Ishita and Nisha.

Solution:

The distance between Isha and Ishita and Ishita and Nisha is equal.

Therefore,

Let $RS = SM = 24\ m$

They are equidistant from the centre.

In right angled triangle $ORL$,

$OL^2 = OR^2 - RL^2$

$=20^{2}-(\frac{24}{2})^{2}$

$=400-144$

$=256$

$=16^{2}$

$\Rightarrow \mathrm{OL}=16 \mathrm{~cm}$

Let $\mathrm{OL} \perp \mathrm{RS}$

$OS$ bisects $RM$

Let $\mathrm{RK}=x$

Therefore,

$ar \Delta \mathrm{ORS}=\frac{1}{2}$ base $\times$ altitude

$=\frac{1}{2} \times 24 \times 16 \mathrm{~m}^{2}$

$=192 \mathrm{~m}^{2}$

$a r \Delta \mathrm{ORS}=\frac{1}{2} \times \mathrm{OS} \times \mathrm{RK}$

$=\frac{1}{2} \times 20 \times x$

$=10 x$

This implies,

$10 x=192$

$x=\frac{192}{10}$

$=19.2 \mathrm{~m}$

$\mathrm{RM}=2 \times \mathrm{RK}=2x$

$=2 \times 19.2 \mathrm{~m}$

$=38.4 \mathrm{~m}$

Hence the distance between Ishita and Nisha is $38.4\ m$. 

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Updated on: 10-Oct-2022

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