Question

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Answer 1

Question:-

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, 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?

Step by step explanation:-

Given: Let Reshma, Salma and Mandip be denoted by point R S and M respectively.

To find: Length of RM

Construction: Join OR, QS, OM

Let RM intersect OS at X

Solution:

In ∆ORS & ∆OMS

OR = OM (Both radius)

OS = OS (Common)

RS = SM (Given )

ΔORS ≅ ΔOMS (SSS Congruency)

∴ ∠ROS = ∠MOS (CPCT) _______(1)

In ΔORX & ΔOMX

OR = OM ( Both radius)

∠ROX = ∠MOX (From eq (1): ∠ROS = ∠MOS)

OX = OX (Common)

ΔORX ≅ ΔOMX (SAS Congruency)

∴ RX = MX (CPCT) _______(2)

Since RX = MX

So, OX bisects chord RM,

∴ OX ⏊ RM

Now,

Let OX = X

So, XS = OS - OX = S - X

In ΔORX

Applying pythagoras theorem

OR^2 = OX^2 + RX^2

5^2 = x^2 + RX^2

25 = x^2 + RX^2

25 - x^2 = RX^2

RX^2 = 25 - x^2 ___________(3)

In ΔXRS

Applying pythagoras theorem

RS^2 = XS^2 = RX^2

6^2 = (5 - x)^2 + RX^2

36 = 5^2 + x^2 - 2(5)(x) + RX^2

36 = 25 + x^2 - 10x + RX^2

36 - 25 - x^2 + 10x = RX^2

11 - x^2 + 10x = RX^2

RX^2 = 11 - x^2 + 10x __________ (4)

From equation (3) & (4)

25-x^2 = 11 - x^2 + 10x

25 - 11- x^2 + x^2 = 10x

14 = 10x

10x = 14

x = 14 / 10

x = 1.4

Putting value x in equation (3)

RX^2 = 25 - x^2

RX^2 = 25 - (1.4)^2

RX^2 = 25 - 1.96

RX^2 = 23.04

RX = √23.04

RX = 4.8

Therefore,

RM = 2RX

= 2 × 4.8

= 9.6 m.

∴ Distance between Reshma and Mandip is 9.6m

Answer:-

Distance between Reshma and Mandip is 9.6m

• I hope it's help you...☺

Answer 2

hope above answer may be help u

hii

I'm Annie

class 10

from kv school and u