Question

A chord of length 24 cm is at a distance of 5 cm from the centre of the circle. Find the length of the chord of the same circle which is at a distance of 12 cm from the centre.

Answer 1

Let AB be the chord of length 24 cm and O be the centre of the circle.
Let OC be the perpendicular drawn from O to AB.
We know, that the perpendicular to a chord, from the centre of a circle, bisects the chord.

 ∴ AC = CB = 12 cm
 In ∆OCA,

OA² = OC² + AC²  (By Pythagoras theorem)
= (5)² + (12)² = 169
⇒  OA = 13 cm
∴ radius of the circle = 13 cm

Let A'B' be new chord at a distance of 12 cm from the centre.
 ∴ (OA')² = (OC')² + (A'C')²
 ⇒ (A'C')² = (13)² - (12)² = 25
 ∴ A'C' = 5 cm

 Hence, length of the new chord = 2 × 5 = 10 cm