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
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.
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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