# 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:**

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

Pythagoras theorem states that in a right triangle, (Side 1)² +(Side 2)² = (Hypotenuse)²

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

Thus, MB = 2.5 cm and ND = 5.5 cm

Let OM = x cm and ON = 6 - x cm

Consider ΔOMB

By Pythagoras theorem,

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.

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) or (2), we get the radius of the circle = 5.59 cm.

**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 to be 5.59 cm.