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

9th Class Maths – NCERT Important Questions

Ans:

Step 1: Let’s make a diagram for better understanding of the question:

Here is a circle with center O.

Distance between Reshma (R) and Salma (S) is 6 m and Distance between Salma (S)  and Mandeep (M) is 6 m.

We need to find the distance between R and M.

Step 2: Here, in Δ RSM, RS and SM are 6 m each.

∴ ΔRSM is an isosceles triangle.

If we draw a perpendicular from point S on RM, it will bisect RM and pass through center O.

∴ RP = PM and ∠ SPR = 90⁰

∴ ∠ OPR = 90⁰

We need to find length of RM (distance between R and M)

Let’s consider this distance RM is 2X

Therefore, RP = PM = = X

Step 3: Now in Δ ORP, by Pythagoras Theroem, we get:

PR ² + OP ² = OR ²

∴ X ² + Y ² = 5 ² = 25 …. (i)

Step 4: Similarly, in Δ SRP, by Pythagoras Theroem, we get:

PR ² + PS ² = RS ²

∴ X ² + (5 – Y) ² = 6 ² = 36

∴ X ² + 25 +  Y ² – 10 Y = 36

∴ X ² + Y ² + 25 – 10 Y = 36

Step 5: By substituting values from equation (i), we get:

∴ X ² + Y ² + 25 – 10 Y = 36

∴ (25) + 25 – 10 Y = 36

∴ 50 – 10 Y = 36

∴ 10 Y = 50 – 36 = 14

∴ Y =

Step 6: By substituting value of Y from equation (i), we get:

X ² + Y ² = 25

∴ X ² + = 25

∴ X ² + = 25

∴ X ² = 25 –

∴ X =  = = 4.8 m

Step 7: Here, X is the distance of RP

And length of RM = 2 X = 2 x 4.8 = 9.6 m

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

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