In a circle of radius 10 cm, AB and CD are two parallel chords of lengths 16 cm and 12 cm respectively. Calculate the distance between the chords, if they are on :

(i) the same side of the center

(ii) the opposite side of the center

(i) Let AB and CD be chords on same side of the center of the circle.

We know that,

Perpendicular from the center to the chord, bisects it.

∴ AF = $\dfrac{AB}{2} = \dfrac{12}{2}$ = 6 cm, CE = $\dfrac{CD}{2} = \dfrac{16}{2}$ = 8 cm.

From figure,

OA = OC = radius = 10 cm.

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ 10² = OE² + 8²

⇒ 100 = OE² + 64

⇒ OE² = 100 - 64

⇒ OE² = 36

⇒ OE = $\sqrt{36}$ = 6 cm.

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ 10² = OF² + 6²

⇒ 100 = OF² + 36

⇒ OF² = 100 - 36

⇒ OF² = 64

⇒ OF = $\sqrt{64}$ = 8 cm.

From figure,

⇒ EF = OF - OE = 8 - 6 = 2 cm.

Hence, distance between the chords = 2 cm.

(ii) Let AB and CD be chords on opposite side of the center of the circle.

We know that,

Perpendicular from the center to the chord, bisects it.

∴ AF = $\dfrac{AB}{2} = \dfrac{12}{2}$ = 6 cm, CE = $\dfrac{CD}{2} = \dfrac{16}{2}$ = 8 cm.

From figure,

OA = OC = radius = 10 cm.

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ 10² = OE² + 8²

⇒ 100 = OE² + 64

⇒ OE² = 100 - 64

⇒ OE² = 36

⇒ OE = $\sqrt{36}$ = 6 cm.

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ 10² = OF² + 6²

⇒ 100 = OF² + 36

⇒ OF² = 100 - 36

⇒ OF² = 64

⇒ OF = $\sqrt{64}$ = 8 cm.

From figure,

⇒ EF = OE + OF = 6 + 8 = 14 cm.

Hence, distance between the chords = 14 cm.