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# 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

OM^{2} + MB^{2} = OB^{2}

x^{2} + 2.5^{2} = OB^{2}

x^{2} + 6.25 = OB^{2}..................(1)

Consider ΔOND

By Pythagoras theorem,

ON^{2} + ND^{2} = OD^{2}

(6 - x)² + 5.5^{2} = OD^{2}

36 + x^{2} - 12x + 30.25 = OD^{2}

x^{2} - 12x + 66.25 = OD^{2}............... (2)

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

Thus, OB^{2} = OD^{2}

Equating (1) and (2) we get,

x^{2} + 6.25 = x^{2} - 12x + 66.25

12x = 60

x = 5

Substituting the value of x in (1),

OB^{2} = x^{2} + 6.25

OB^{2} = 5^{2 }+ 6.25

OB^{2} = 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

**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.

**☛ Related Questions:**

- 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 center ?
- Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the center.
- Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
- ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

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