Question

A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre of the circle.

Answer 1

Let AB be the chord of the given circle with centre O and a radius of 10 cm.
Then AB =16 cm and OB = 10 cm

From O, draw OM perpendicular to AB.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
∴ BM = $\left(\frac{16}{2}\right)\mathrm{cm}=8\mathrm{cm}$
In the right ΔOMB, we have:
OB² = OM² + MB² (Pythagoras theorem)
⇒ 10² = OM² + 8²
⇒ 100 = OM² + 64
⇒ OM² = (100 - 64) = 36
⇒ $OM=\sqrt{36}\mathrm{cm}=6\mathrm{cm}$
Hence, the distance of the chord from the centre is 6 cm.