Question

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 distance between Ishita and Nisha?​

Answer 1

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Answer 2

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Solution:

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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$^{2}$ = OD$^{2}$ + DA$^{2}$

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$\implies$ 20$^{2}$ = x$^{2}$ + y$^{2}$

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$\implies$ x$^{2}$ + y$^{2}$ = 400 ----- (1)

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In $\triangle$ ADB

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AB$^{2}$ = AD$^{2}$ + BD$^{2}$

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

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$\implies$ y$^{2}$ + (20 - x)$^{2}$ = 576

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$\implies$ y$^{2}$ + 400 - 40x + x$^{2}$ = 576

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$\implies$ 40x = 224

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$\implies$ x = 224/40 = 5.6 m

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From equation 1

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

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$\implies$ 31.36 + y$^{2}$ = 400

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$\implies$ y$^{2}$ = 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