Question

Two parallel chords of lengths 30cm and 16cm are drawn on the opposite sides of the centre of a circle of radius 17cm. Find the distance between the chords.

Answer 1

Consider AB and CD as the chords of circle with centre O

It is given that AB = 30cm and CD = 16cm

Join the lines OA and OC

We know that AO = 17cm and CO = 17cm

Construct OM ⊥ CD and OL ⊥ AB

Perpendicular from the centre of a circle to a chord bisects the chord

We know that

AL = ½ × AB

By substituting the values

AL = ½ × 30

So we get

AL = 15cm

We know that

CM = ½ × CD

By substituting the values

CM = ½ × 16

So we get

CM = 8cm

Consider △ ALO

Using the Pythagoras theorem it can be written as

AO² = OL² + AL²

By substituting the values

17² = OL² + 15²

So we get

OL² = 17² – 15²

On further calculation

OL² = 289 – 225

By subtraction

OL² = 64

By taking the square root

OL = √64

OL = 8cm

Consider △ CMO

Using the Pythagoras theorem it can be written as

CO² = CM² + OM²

By substituting the values

17² = 8² + OM²

So we get

OM² = 17² – 8²

On further calculation

OM² = 289 – 64

By subtraction

OM² = 225

By taking the square root

OM = √225

OM = 15cm

So the distance between the chords = OM + OL

By substituting the values

Distance between the chords = 8 + 15 = 23cm

Therefore, the distance between the chords is 23 cm.