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.
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,
OA2 = OE2+AE2
⇒ OE2 = OA2 - AE2 = (15)2 - (12)2 = 81
∴ OE = 9 cm
∴ OF = EF - OE = (21- 9) =12 cm
In right ΔOCF,
OC2 = OF2 + CF2
⇒ x2 = OC2 - OF2 = (15)2 - (12)2 = 81
∴ x = 9 cm
Hence, length of chord CD = 2x = 2×9 =18 cm