Two chords AB and CD of lengths 24 cm and 10 cm respectively of a circle are parallel. If the chords lie on the same side of the centre and distance between them is 7 cm, find the length of a diameter of the circle.

Given: AB and CD are two parallel chords of a circle on the same side of the center.

AB = 24 cm and CD = 10 cm.

Distance between the chords = MN = 7 cm.

Let the radius of the circle be r.

Let the perpendicular distance from the center O to chord AB be OM = x.

Therefore, ON = x + 7.

To Prove: The length of the diameter of the circle.

Construction: Join OA and OC.

Proof:

AM = $\dfrac{1}{2}$ AB

= $\dfrac{1}{2}$ x 24

= 12 cm

In Δ OAM, ∠M = 90°

Using Pythagoras theorem,

∴ OA² = OM² + AM²

⇒ r² = x² + 12²

⇒ r² = x² + 144 ………………(1)

CN = $\dfrac{1}{2}$ CD

= $\dfrac{1}{2}$ x 10

= 5 cm

In Δ OCN, ∠N = 90°

Using Pythagoras theorem,

∴ OC² = ON² + CN²

⇒ r² = (x + 7)² + 5²

⇒ r² = x² + 49 + 14x + 25

⇒ r² = x² + 14x + 74

Using equation (1), we get

⇒ x² + 144 = x² + 14x + 74

⇒ 144 = 14x + 74

⇒ 14x = 144 - 74

⇒ 14x = 70

⇒ x = $\dfrac{70}{14}$

⇒ x = 5

Putting the value of x in equation (1), we get

⇒ r² = (5)² + 144

⇒ r² = 25 + 144

⇒ r² = 169

⇒ r = $\sqrt{169}$

⇒ r = 13

Diameter = 2r = 2 x 13 cm

= 26 cm

Hence, the diameter of the circle = 26cm.