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.

Construction: 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

Given: Let Reshma, Salma and Mandip be donated by points R,S & M resp. Given radius = 5 m

& RS = 6m, SM =6m

To find: Length of RM

Construction: Join OR, OS, OM Let RM intersect OS at X

Solution:In A ORS & A OMS

OR = OM

OS = OS

(Both radius)

(Common)

RS = SM (Given)

A ORSA OMS (SSS congruency)

: 2 ROS = 2 MOS (CPCT) ... (1)

In A ORX & A OMX

OR = OM

2 ROX = 2 MOX

OX = OX

A ORXA OMX

.. RX = MX

r = 5 m

6 m

R

X

M

6 m

S

(Both radius)

(From (1): 2 ROS = 2 MOS )

(Common)

(SAS congruency)

(CPCT) ... (2)

Since RX = MX

So, OX bisects chord RM,

.. OX I RM

Now,

Let OX = X,

Line drawn through centre of circle to bisect a chord is perpendicular to the chord

So, XS = OS - OX = 5-x

In A ORX

Applying Pythagoras theorem

OR² = OX² + RX²

5² = x² + RX²

25 = x² + RX²

25 - x² = RX²

RX²= 25-x²

In A XRS

Applying Pythagoras theorem

RS² = XS² + RX²

6²=(5-x)² + RX²

3652+x²2(5)(x) + RX²

3625+ x²10x + RX²

36-25-x²10x = RX²

11-x² + 10x RX²

RX² = 11-x² + 10x ...(4)

From (3) & (4)

25-x2²11-x² + 10x

25-11-x² + x² = 10x

14 = 10x

10x = 14

14

X = 10

X = 1.4

Putting value of x in (3)

RX²= 25-x²

RX²= 25-(1.4)2

RX225-1.96

RX² = 23.04

RX = √23.04

RX = 4.8

Therefore,

RM = 2RX

= 2 × 4.8

= 9.6 m

.. Distance between Reshma and Mandip is 9.6 m

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