Question

Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm.

Answer 1

Let AB be the chord of the given circle with centre O and a radius of 5 cm.
From O, draw OM perpendicular to AB.
Then OM = 3 cm and OB = 5 cm

From the right ΔOMB, we have:
OB² = OM² + MB² (Pythagoras theorem)
⇒ 5² = 3² + MB²
⇒ 25 = 9 + MB²
⇒ MB² = (25 − 9) = 16
⇒ $MB=\sqrt{16}\mathrm{cm}=4\mathrm{cm}$
Since the perpendicular from the centre of a circle to a chord bisects the chord, we have:
AB = 2 × MB = (2 × 4) cm = 8 cm
Hence, the required length of the chord is 8 cm.