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 distance between Ishita and Isha and between Isha and Nisha is 24m each, what is the distance between Ishita and Nisha.​

Answer 1

Step-by-step explanation:

Let Ishita at A, Isha at B and Nisha at C.

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Therefore, AB = BC = 24 m

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OA = OC = 20 m

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Let OD = x m.

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BD = BO - BD

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BD = (24 - x) m

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Let AD = y m

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In \triangle△ ODA

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By using Pythagoras theorem,

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OA^{2}

2

$= OD^{2} </p><p>2</p><p> + DA^{2} </p><p>2</p><p> </p><p></p><p>⠀</p><p>$

\implies⟹ 20^{2}

2

$= x^{2} </p><p>2</p><p> + y^{2} </p><p>2</p><p> </p><p></p><p>⠀$

\implies⟹ x^{2}

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AB^{2}

$2</p><p> = AD^{2} </p><p>2</p><p> + BD^{2} </p><p>2$

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$\implies⟹ 24^{2} </p><p>2</p><p> = y^{2} </p><p>2</p><p> + (20 - x)^{2} </p><p>2</p><p>$

$\implies⟹ y^{2} </p><p>2</p><p> + (20 - x)^{2} </p><p>2</p><p> = 576</p><p>$

$\implies⟹ y^{2} </p><p>2</p><p> + 400 - 40x + x^{2} </p><p>2</p><p> = 576$

$\implies⟹ 40x = 224$

$\implies⟹ x = 224/40 = 5.6 m$

$From equation 1$

$(5.6)^{2} </p><p>2</p><p> + y^{2} </p><p>2</p><p> = 400$

$\implies⟹ 31.36 + y^{2} </p><p>2$

$= 400$

$\implies⟹ y^{2} 2$

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$= 368.64$

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$\implies⟹ y = 19.2 m$

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$Now,$

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$AC = 2xy = 2 × 19.2$

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$= 38.4 m$

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Hence, the distance between Ishita and Nisha is 38.4 m