Question

AB and CD are two parallel chords and O is the centre. If AB=24cm, CD=18cm and radius = 15cm , find the distance between the chords if they are on the opposite sides of the
centre.​

Answer 1

Answer :

Ratio of the time periods of the pendulum is 3 : 1.

Explanation :

Given :.

Ratio of the time periods of the pendulum, \sf{L_{1} : L_{2} = 9 : 1}L

1

:L

2

=9:1

To find :

Ratio of the time periods of the pendulum, \sf{t_{1} : t_{2} = ?}t

1

:t

2

=?

Knowledge required :

Formula for time period of a pendulum :

\boxed{\sf{T = 2\pi\sqrt{\dfrac{L}{g}}}}

T=2π

g

L

Where,

T = Time period of the pendulum.

2π = Constant

L = Length of the pendulum

g = Acceleration due to gravity

Solution :

Let the length of the two simple pendulum in terms of x as 9x and 1x.

So first let us find the time period of both the pendulums , individually :

Time period of the pendulum with Length of 9x.

By using the formula for time period of a pendulum and substituting the values in it, we get :

\begin{gathered}:\implies \sf{T = 2\pi\sqrt{\dfrac{L}{g}}} \\ \\ :\implies \sf{T_{1} = 2\pi\sqrt{\dfrac{9}{g}}} \\ \\ \boxed{\therefore \sf{T_{1} = 2\pi\sqrt{\dfrac{9}{g}}}} \\ \\ \end{gathered}

:⟹T=2π

g

L

:⟹T

1

=2π

g

9

∴T

1

=2π

g

9

Time period of the pendulum with Length of 1x.

By using the formula for time period of a pendulum and substituting the values in it, we get :

\begin{gathered}:\implies \sf{T = 2\pi\sqrt{\dfrac{L}{g}}} \\ \\ :\implies \sf{T_{2} = 2\pi\sqrt{\dfrac{1}{g}}} \\ \\ \boxed{\therefore \sf{T_{2} = 2\pi\sqrt{\dfrac{9}{g}}}} \\ \\ \end{gathered}

:⟹T=2π

g

L

:⟹T

2

=2π

g

1

∴T

2

=2π

g

9

Now let's find out the ratio of the time periods.

\begin{gathered}:\implies \sf{T_{1} : T_{2} = \dfrac{2\pi\sqrt{\dfrac{9}{g}}}{2\pi\sqrt{\dfrac{1}{g}}}} \\ \\ \\ :\implies \sf{T_{1} : T_{2} = \dfrac{\sqrt{\dfrac{9}{g}}}{\sqrt{\dfrac{1}{g}}}} \\ \\ \\ :\implies \sf{T_{1}^{2} : T_{2}^{2} = \dfrac{\dfrac{9}{g}}{\dfrac{1}{g}}} \\ \\ \\ :\implies \sf{T_{1}^{2} : T_{2}^{2} = \dfrac{9}{g} \times \dfrac{g}{1}} \\ \\ \\ :\implies \sf{T_{1}^{2} : T_{2}^{2} = \dfrac{9}{1}} \\ \\ \\ :\implies \sf{T_{1} : T_{2} = \sqrt{\dfrac{9}{1}}} \\ \\ \\ :\implies \sf{T_{1} : T_{2} = \dfrac{3}{1}} \\ \\ \\ \boxed{\therefore \sf{T_{1} : T_{2} = 3 : 1}} \\ \\ \\ \end{gathered}

:⟹T

1

:T

2

=

2π

g

1

2π

g

9

:⟹T

1

:T

2

=

g

1

g

9

:⟹T

1

2

:T

2

2

=

g

1

g

9

:⟹T

1

2

:T

2

2

=

g

9

×

1

g

:⟹T

1

2

:T

2

2

=

1

9

:⟹T

1

:T

2

=

1

9

:⟹T

1

:T

2

=

1

3

∴T

1

:T

2

=3:1

Therefore,

Ratio of the time periods of the pendulum, \sf{t_{1} : t_{2} = 3 : 1}t

1

:t

2

=3:1

Answer 2

Answer:

In the figure, chords AB∥CD

O is the centre of the circle

Radius of the Circle = 15 cm

Length of AB = 24 cm and CD = 18 cm

Join OA and OC

AM = MB = 24/2 = 12 cm

Similarly ON⊥CD

CN = ND = 18/2 = 9 cm

In right △AMO

OA

2

=OM

2

+AM

2

OM

2

=OA

2

−AM

2

OM

2

=15

2

+12

2

=225−144

OM=

81

=9

Similarly in right △CNO

OC

2

=CN

2

+ON

2

(15)

2

=(9)

2

+ON

2

225=81+ON

2

ON

2

=225−81

ON=12cm

Now MN=OM+ON=9+12=21cm

solution