Math, asked by niveditatuli1836, 9 months ago

The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of the other chord from the centre?

Answers

Answered by nikitasingh79
4

Given: The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre.

 

To find : The distance of the other chord from the centre.

Solution :  

Let the distance of smaller chord AB from centre of circle , OM = 4 cm

MB = AB/2 = 6/2  

MB = 3 cm

 

In ∆OMB ,  By using Pythagoras theorem ,  

OM² + MB² = OB²

4² + 3² = OB²

16 + 9 = OB²

25 = OB²

OB = √25

OB = 5 cm

 

In OND,

OD = OB = 5 cm  

[Radii of same circle]

ND = CD/2 = 8/2

ND = 4 cm

By using Pythagoras theorem ,  

ON² + ND² = OD²

ON² + 4² = 5²

ON² + 16 = 25

ON² = 25 - 16

ON² = 9

ON = √9

ON = 3 cm

Hence , the distance of bigger chord from the circle is 3 cm.

 

HOPE THIS ANSWER WILL HELP YOU…..

 

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Attachments:
Answered by Anonymous
2

Answer:

Step-by-step explanation:

MB = 3 cm

 

In ∆OMB ,  By using Pythagoras theorem ,  

OM² + MB² = OB²

4² + 3² = OB²

16 + 9 = OB²

25 = OB²

OB = √25

OB = 5 cm

 

In OND,

OD = OB = 5 cm  

[Radii of same circle]

ND = CD/2 = 8/2

ND = 4 cm

By using Pythagoras theorem ,  

ON² + ND² = OD²

ON² + 4² = 5²

ON² + 16 = 25

ON² = 25 - 16

ON² = 9

ON = √9

ON = 3 cm

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