#### Question

Two parallel chords are drawn in a circle of diameter 30.0 cm. The length of one chord is 24.0 cm and the distance between the two chords is 21.0 cm; find the length of another chord.

#### Solution

Since the distance between the chords is greater than the radius of the circle (15 cm), so the

chords will be on the opposite sides of the centre.

Let O be the centre of the circle and AB and CD be the two parallel chords such that

AB = 24 cm.

Let length of CD be 2x cm.

Drop OE and OF perpendicular on AB and CD from the centre O.

OE ⊥ AB and OF⊥CD

∴ OE bisects AB and OF bisects CD

(Perpendicular drawn from the centre of a circle to a chord bisects it)

⇒ AE = `24/2` = 12cm; CF = `(2x)/2` =x cm

In right ΔOAE,

OA^{2} = OE^{2}+AE^{2}

⇒ OE^{2} = OA^{2} - AE^{2} = (15)^{2} - (12)^{2} = 81

∴ OE = 9 cm

∴ OF = EF - OE = (21- 9) =12 cm

In right ΔOCF,

OC^{2} = OF^{2} + CF^{2}

⇒ x^{2} = OC^{2} - OF^{2} = (15)^{2} - (12)^{2} = 81

∴ x = 9 cm

Hence, length of chord CD = 2x = 2×9 =18 cm