Question

Find the length of a chord which is at a distance of 9 cm from the centre of a circle of radius 15 cm.

Answer 1

Let AB be the chord of the given circle with centre O and a radius of 15 cm.
From O, draw OM perpendicular to AB.
Then, OM = 9 cm and OB = 15 cm

From the right ΔOMB, we have:
⇒ OB² = OM² + MB²
⇒ 15² = 9² + MB²
⇒ 225 = 81 + MB²
⇒ MB² = (225 - 81) = 144
⇒ $\mathrm{MB}=\sqrt{144}\mathrm{cm}=12\mathrm{cm}$MB=16−−√cm=4cm
Since the perpendicular from the centre of a circle to a chord bisects the chord, we have:
AB = 2 × MB = (2 × 12) cm = 24 cm
Hence, the required length of the chord is 24 cm.