Ab 8 cm and cd = 6 cm are two parallel chords on the opposite side of the centre of a circle

Solution:

The perpendicular drawn from the center of the circle to the chords bisects it. 

Draw two parallel chords AB and CD of lengths 6 cm and 8 cm. Let the circle's center be O. Join one end of each chord to the center. Draw 2 perpendiculars OM and ON to AB and CD, respectively, which bisects the chords.

AB = 6 cm CD = 8 cm MB = 3 cm ND = 4 cm

Given OM = 4 cm and let ON = x cm Consider ΔOMB

By Pythagoras theorem,

OM² + MB² = OB²

4² + 3² = OB²

OB² = 25

OB = 5 cm

OB and OD are the radii of the circle.

Therefore OD = OB = 5 cm.

Consider ΔOND

By Pythagoras theorem,

ON² + ND² = OD²

x² + 4² = 5²

x² = 25 - 16

x² = 9

x = 3

The distance of the chord CD from the center is 3 cm.

☛ Check: NCERT Solutions for Class 9 Maths Chapter 10

Video Solution:

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

Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.6 Question 3

Summary:

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 center, we have found that the distance of the other chord CD from the center is 3 cm.

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