Question

At one end A of a diameter,AB of a circle of radius 5 cm,tangents XAY is drawn to a circle.The length of the chord CD parallel to XY and At a distance 8 cm from A is

Answer 1

HELLO DEAR,

CD=2×MC

given that:-


radius of circle = 5cm=AO=OC

AM=8CM

and

AM=OM+AO


THEN

OM =AM-AO


Put the vlues in this equation given above

We Get

=> OM= (8-5) =3CM


we know that;-

OM is perpendicular to the chord CD.

In∆OCM <OMC=90°

using Pythagoras theorem

we get,

OC²=OM²+MC²

.
${(mc)}^{2} = {(omc)}^{2} - {(om)}^{2} \\ = > {(mc)}^{2} = {5}^{2} - {3}^{2} \\ = > {(mc)}^{2} = 25 - 9 \\ = > {(mc)} = \sqrt{16 } \\ = > {(mc)} = 4cm$

Hence,

CD= 2 ×CM  = 8 cm


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

CD=2×MC

given that:-

radius of circle = 5cm=AO=OC

AM=8CM

and

AM=OM+AO

THEN

OM =AM-AO

Put the vlues in this equation given above

We Get

=> OM= (8-5) =3CM

we know that;-

OM is perpendicular to the chord CD.

In∆OCM <OMC=90°

using Pythagoras theorem

we get,

OC²=OM²+MC²

.

\begin{gathered}{(mc)}^{2} = {(omc)}^{2} - {(om)}^{2} \\ = > {(mc)}^{2} = {5}^{2} - {3}^{2} \\ = > {(mc)}^{2} = 25 - 9 \\ = > {(mc)} = \sqrt{16 } \\ = > {(mc)} = 4cm\end{gathered}

(mc)

2

=(omc)

2

−(om)

2

=>(mc)

2

=5

2

−3

2

=>(mc)

2

=25−9

=>(mc)=

16

=>(mc)=4cm

Hence,

CD= 2 ×CM = 8 cm