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.
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
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
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