Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle

Solution:

Draw two parallel chords AB and CD of lengths 5 cm and 11 cm. Let the center of the circle be O. Join one end of each chord to the center.

Draw two perpendiculars OM and ON to AB and CD, respectively, which bisects the chords.

Thus, MB = 2.5 cm and ND = 5.5 cm [The perpendicular drawn from the center of the circle to the chords bisects it.]

Let OM = x and ON = 6 - x

Consider ΔOMB

By Pythagoras theorem,

OM² + MB² = OB²

x² + 2.5² = OB²

x² + 6.25 = OB²..................(1)

Consider ΔOND

By Pythagoras theorem,

ON² + ND² = OD²

(6 - x)² + 5.5² = OD²

36 + x² - 12x + 30.25 = OD²

x² - 12x + 66.25 = OD²............... (2)

OB and OD are the radii of the circle. Therefore OB = OD.

Thus, OB² = OD²

Equating (1) and (2) we get,

x² + 6.25 = x² - 12x + 66.25

12x = 60

x = 5

Substituting the value of x in (1),

OB² = x² + 6.25

OB² = 5² + 6.25

OB² = 31.25

OB = 5.59 (approx.)

Thus, we get the radius of the circle = 5.59 cm.

☛ Check: Class 9 Maths NCERT Solutions Chapter 10

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Video Solution:

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its center. If the distance between AB and CD is 6 cm, find the radius of the circle

Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.6 Question 2

Summary:

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its center, if the distance between AB and CD is 6 cm, the radius of the circle is 5.59 cm.

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