Question

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

Answer 1

Let AB be the chord of length 24 cm and O be the center of the circle.

Let OC be the perpendicular drawn from O to AB.

We know, that the perpendicular to a chord, from the center 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 the new chord at a distance of 12 cm from the center.

∴  ( 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 x 5 = 10 cm.