AB and CD are two parallel chords of a circle of lengths 10 cm and 4 cm respectively. If the chords lie on the same side of the centre and the distance between them is 3 cm, find the diameter of the circle.

Let OE = x cm.

From figure,

In right angle triangle OCF,

⇒ OC² = OF² + CF² (By pythagoras theorem)

⇒ OC² = (x + 3)² + 2²

⇒ OC² = x² + 9 + 6x + 4

⇒ OC² = x² + 6x + 13

Since, radius = OA = OC.

∴ OA² = OC² = x² + 6x + 13.

In right angle triangle OAE,

⇒ OA² = OE² + AE²

⇒ x² + 6x + 13 = x² + 5²

⇒ x² - x² + 6x = 25 - 13

⇒ 6x = 12

⇒ x = $\dfrac{12}{6}$ = 2 cm.

⇒ OC² = x² + 6x + 13

⇒ OC² = 2² + 6(2) + 13

⇒ OC² = 4 + 12 + 13

⇒ OC² = 29

⇒ OC = $\sqrt{29}$ cm.

Diameter = 2 × radius = 2 × $\sqrt{29} = 2\sqrt{29}$ cm.

Hence, diameter = $2\sqrt{29}$ cm.