Question

Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius $20m$ 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 $24m$ each, what is the distance between Ishita and Nisha?

Answer 1

Since, Distance between Isha and Ishita and Ishita and Nisha is same.

$\therefore RS=SM=24m$

and $OS=20m$ [Since, radius of the circle]

Now, in $△ORS$

Let $RS=a=24m$

$OS=b=20m$

and $OR=c=20m$

$S=\frac{a+b+c}{2}=\frac{20m+20m+24m}{2}$

$\Rightarrow S=32m$

Area of triangle $△=\sqrt{S(S-a)(S-b)(S-c)}$

$\Rightarrow △=\sqrt{32m(32m-24m)(32m-20m)(32m-20m)}$

$\Rightarrow △=\sqrt{32m\left(8m\right)\left(12m\right)\left(12m\right)}$

$\Rightarrow △=\sqrt{4\times 8\times 8\times 12\times 12}{m}^2$

$\Rightarrow △=2\times 8\times 12m^2$

$\Rightarrow △=192m^2$...(i)

In radius $OS$ perpendicularly bisects the chord $RM$.

$\Rightarrow OK\perp RM$ and $RK\perp OS$.

Let $RK=x$

Now, $ar\left(\Delta ORS\right)=\frac{1}{2}\times \text{base}\times \text{altitude}$

$=\frac{1}{2}\times OS\times RK$

$\Rightarrow \frac{1}{2}\times 20\times x=10x$

$\therefore 10x=192m^2\Rightarrow x=\frac{192}{10}=19.2m$ [From (i)]

Now $RM=2\times RK=2x$

$=2\times 19.2m=38.4m$

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