Two parallel chords AB and CD are 3.9 cm apart and lie on the opposite sides of the centre of a circle. If AB = 1.4 cm and CD = 4 cm, find the radius of the circle.

From figure,

AB = 1.4 cm and CD = 4 cm

OF ⊥ AB and OE ⊥ CD.

We know that,

Perpendicular from the center to the chord, bisects it.

∴ AF = $\dfrac{AB}{2} = \dfrac{1.4}{2}$ = 0.7 cm.

CE = $\dfrac{CD}{2} = \dfrac{4}{2}$ = 2 cm.

EF = 3.9 cm and OA = OC = r cm.

Let OE = x and OF = 3.9 - x

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ OC² = x² + 2²

⇒ OC² = x² + 4 …..(1)

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ OA² = (3.9 - x)² + O.7²

⇒ OA² = (3.9 - x)² + 0.49 ….(2)

From (1) and (2), we get :

⇒ x² + 4 = (3.9 - x)² + 0.49

⇒ x² + 4 = x² - 7.8x + 15.21 + 0.49

⇒ x² + 4 = x² - 7.8x + 15.7

⇒ 4 = 15.7 - 7.8x

⇒ 7.8x = 15.7 - 4

⇒ 7.8x = 11.7

⇒ x = $\dfrac{11.7}{7.8}$

⇒ x = 1.5

Substituting value of x in equation (1), we get :

⇒ OC² = (1.5)² + 4

⇒ OC² = 2.25 + 4

⇒ OC² = 6.25

⇒ OC = $\sqrt{6.25}$ = 2.5 cm.

Hence, radius of the circle = 2.5 cm.