In a circle of radius 10 cm, the lengths of two parallel chords are 12 cm and 16 cm respectively. Find the distance between AB and CD. If chords (a) are on the same side of the centre (b) are on the opposite sides of the centre.

Given radius of circle = 10 cm

Chords AB = 12 cm and CD = 16 cm

(a)

Draw OE, OF⊥ AB, CD.

We know that when a perpendicular is drawn from centre to chord, it bisects them.

Then CF = FD = 1/2 CD = 8 cm

⇒ AE = EB = 1/2 AB = 6 cm

Consider ΔOFD,

By Pythagoras Theorem,

⇒ OD² = OF² + FD²

⇒ 10² = x² + 8²

⇒100² = x² + 64

⇒x² = 100 – 64

⇒x² = 36

∴ OF = x = 6 cm

Consider ΔOEB,

By Pythagoras Theorem,

⇒ OB² = OE² + EB²

⇒10² = y² + 6²

⇒100 = y² + 36

⇒y² = 100 – 36

⇒ y² = 64

∴ OE = y = 8 cm

∴ Distance between chords AB and CD = EF = OE – OF = y – x = 8 – 6 = 2 cm

(b)

Draw OE, OF ⊥ AB, CD.

We know that when a perpendicular is drawn from centre to chord, it bisects them.

Then CF = FD = 1/2 CD = 8 cm

⇒ AE = EB = 1/2 AB = 6 cm

Consider ΔOFD,

By Pythagoras Theorem,

⇒ OD² = OF² + FD²

⇒ 10² = x² + 8²

⇒ 100² = x² + 64

⇒x² = 100 – 64

⇒x² = 36

∴ OF = x = 6 cm

Consider ΔOEB,

By Pythagoras Theorem,

⇒ OB² = OE² + EB²

⇒ 10² = y² + 6²

⇒ 100 = y² + 36

⇒ y² = 100 – 36

⇒ y² = 64

∴ OE = y = 8 cm

∴ Distance between chords AB and CD is

⇒ EF = OE + OF

= 8 + 6

= 14 cm