AB and CD are two parallel chords of lengths 8 cm and 6 cm respectively. If they are 1 cm apart and lie on the same side of the centre of the circle, find the radius of the circle.

AB = 8 cm and CD = 6 cm

OF ⊥ AB and OE ⊥ CD

We know that,

Perpendicular from the center to the chord, bisects it.

∴ AF = $\dfrac{AB}{2} = \dfrac{8}{2}$ = 4 cm.

CE = $\dfrac{CD}{2} = \dfrac{6}{2}$ = 3 cm.

EF = 1 cm and OA = OC = r cm.

Let OF = x and OE = x + 1

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ r² = (x + 1)² + 3²

⇒ r² = (x + 1)² + 9 …..(1)

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ r² = x² + 4²

⇒ r² = x² + 16 ….(2)

From (1) and (2),

⇒ x² + 16 = (x + 1)² + 9

⇒ x² + 16 = x² + 2x + 1 + 9

⇒ x² + 16 = x² + 2x + 10

⇒ 16 = 2x + 10

⇒ 2x = 16 - 10

⇒ 2x = 6

⇒ x = $\dfrac{6}{2}$

⇒ x = 3.

Substituting value of x in equation (2), we get :

⇒ r² = 3² + 16

⇒ r² = 9 + 16

⇒ r² = 25

⇒ r = $\sqrt{25}$ = 5 cm.

Hence, radius of the circle = 5 cm.