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.

Answer 1

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 center.

Let O be the center of the circle and AB and CD be the two parallel chords such that AB = 24 cm.

Let the length of the CD be 2x cm.

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

OE ⊥ AB and OF ⊥ CD

∴ OE bisects AB and OF bisects CD.       ...(Perpendicular drawn from the center of a circle to a chord bisects it.)

⇒ AE = `24/2`

= 12 cm;

CF = `(2x)/2`

= x cm

In right ΔOAE,

OA² = OE² + AE²

⇒ OE² = OA² - AE²

= 15² - 12²

= 81

∴ OE = 9 cm

∴ OF = EF - OE

= 21 - 9

= 12 cm

In right ΔOCF,

OC² = OF² + CF²

⇒ x² = OC² - OF²

= 15² - 12²

= 81

∴ x = 9 cm

Hence, length of chord CD = 2x

= 2 × 9

= 18 cm.