Question

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

Answer 1

Let Ishita at $A$,

ISha at $b$ and Nisha at $C$.

$\therefore AB=BC=24cm$.......(given)

and R$OA=OC=20m$ (radii)

Let $OD=xm$

$\therefore BD=OB-OD=(20-x)m$

Let $AC = (2y)m

Hence, $AD=\left(y\right)m$

In $\delta ODA$, $O{D}^2+D{A}^2=O{A}^2$

$x^2+y^2=(20{)}^2$ .......By Pythagoras.

$\Rightarrow x^2+y^2=400$

In $\Delta ADB,A{D}^2+B{D}^2=A{B}^2$

$\Rightarrow y^2+(20-x{)}^2=(24{)}^2$

$\therefore y^2+400-40x+x^2=576$

$\therefore (x^2+y^2)+400-576=40x$

$\therefore 400+400-576=40x$

$\Rightarrow x=\frac{224}{40}=5.6m$

Substituting this value in

$x^2+y^2=400$ we get

$(5.6{)}^2+y^2=400$

$\therefore 31.36+y^2=400$

$\Rightarrow y^2=400-31.36$

$\Rightarrow y^2=368.64$

$\Rightarrow y=19.2m$

Now, $AC=2y=2\times 19.2m=38.4m$

Hence, the distance between Ishita and Nisha $=38.4m$