#### 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.

#### Solution

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,

OA^{2} = OC^{2} + AC^{2 } (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

Is there an error in this question or solution?

Solution 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. Concept: Chord Properties - the Perpendicular to a Chord from the Center Bisects the Chord (Without Proof).