Solution:
Let AB be a chord of a circle with centre O and radius 10 cm such that AB = 12 cm.
We draw OL ⊥ AB and join OA.
Since the perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL = LB = AB/2 = 6 cm
Now in △OAL, we have
OL2 = OA2 - AL2 [By Pythagoras theorem]
⇒ OL2 = (10)2 - (6)2
⇒ OL2 = 100 - 36 = 64
⇒ OL = 8 cm
Hence the distance of the chord from the centre is 8 cm.
Related content
Matrices and Determinants_mathematics |