Question

Two parallel chords of equal length 18 cm are drawn inside a circle of radius 15 cm Find the distance between the chords

• A. 12 cm
• B. 18 cm
• C. 24 cm
• D. 30 cm

Answer 1

Answer: (C)

24 cm

In a circle, the perpendicular bisector from the centre to the mid point of the chord, forms a right angled triangle as shown in the figure.
Now, the perpendicular distance between the centre to the chord PQ is calculated as
$=>{OP}^2={OM}^2+{MP}^2$
$=>{15}^2={OM}^2+9^2$ (Since, MP is half of the chord PQ )
$=>OM=12cm$
Similarly, the perpendicular distance between the centre to the chord CD is calculated as
$=>{OC}^2={OE}^2+{EC}^2$
$=>{15}^2={OE}^2+9^2$ (
$=>OE=12cm$
Since the chords are on either side of the centre, distance between the chords $=OM+OE=12+12=24cm$