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Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?

Circles

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Answer

Let R, S and M be the position of Reshma, Salma and Mandip respectively.

From center O,

Draw OA, perpendicular to chord RS and OB, perpendicular to chord SM.

Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip? NCERT Class 9 Mathematics CBSE Solutions.

We know that,

Perpendicular from the center to the chord, bisects it.

⇒ AR = AS = RS2=62\dfrac{RS}{2} = \dfrac{6}{2} = 3 m

From figure,

⇒ OR = OS = OM = 5 m (Radii of circle)

In ∆ OAR,

By pythagoras theorem,

⇒ OR2 = OA2 + AR2

⇒ (5)2 = OA2 + (3)2

⇒ 25 = OA2 + 9

⇒ OA2 = 25 - 9

⇒ OA2 = 16

⇒ OA = 16\sqrt{16}

⇒ OA = 4 m

By formula,

Area of triangle = 12×\dfrac{1}{2} \times base x height

Area of ∆ ORS = 12\dfrac{1}{2} x OA x RS ……..(1)

From figure,

In ∆ ORS for base OS, RC is the altitude.

Area of ∆ ORS = 12\dfrac{1}{2} x OS x RC ……..(2)

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

12\dfrac{1}{2} x OS x RC = 12\dfrac{1}{2} x OA x RS

12\dfrac{1}{2} x RC x 5 = 12\dfrac{1}{2} x 4 x 6

⇒ RC x 5 = 24

⇒ RC = 245\dfrac{24}{5}

⇒ RC = 4.8 m

As OC is the perpendicular to the chord RM,

∴ OC bisects RM.

⇒ MC = RC

⇒ MC = RC = 4.8 m

⇒ RM = 2 RC = 2 x 4.8 = 9.6 m

Hence, the distance between Reshma and Mandip is 9.6 m.

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