Three friends Amit, Vinay and Sukrit are playing a game by standing on a circle of radius 5 m drawn in a park. Amit throws a ball to Vinay, Vinay to Sukrit, Sukrit to Amit. If the distance between Amit and Vinay and between Vinay and Sukrit is 6 m each, what is the distance between Amit and Sukrit?

Let the center of the circular park be O and positions of Amit, Vinay and Sukrit be A, V and S respectively.

The radius of the circle is 5 m.

AV = VS = 6 m

Draw VM ⊥ AS.

In an isosceles triangle, the perpendicular from a vertex between equal sides bisects the opposite side.

∴ AM = MS

Thus,

Let OM = x m and MS = y m

VM = OV - OM = (5 - x) m.

In right-angled triangle VMS,

⇒ VS² = VM² + MS²

⇒ 6² = (5 - x)² + y²

⇒ 36 = (5 - x)² + y²

⇒ y² = 36 - (5 - x)² ….(1)

In right-angled triangle OMS,

⇒ OS² = OM² + MS²

⇒ 5² = x² + y²

⇒ y² = 25 - x²….(2)

From (1) and (2), we get :

⇒ 25 - x² = 36 - (5 - x)²

⇒ 25 - x² = 36 - (25 - 10x + x²)

⇒ 25 - x² = 36 - 25 + 10x - x²

⇒ 25 - x² = 36 - 25 + 10x - x²

⇒ 25 = 11 + 10x

⇒ 10x = 14

⇒ x = 1.4

Substituting value of x in equation (2):

⇒ y² = 25 - (1.4)²

⇒ y² = 23.04

⇒ y = $\sqrt{23.04}$ = 4.8 m

From figure,

⇒ AS = MS + AM

⇒ AS = 2MS

⇒ AS = 2(4.8) = 9.6 m

Hence, the distance between Amit and Sukrit is 9.6 m.