In a circle of radius 5 cm, AB and AC are two chords of 6 cm each. Find the length of the chord BC.

Let AB and AC be two equal chords, of 6 cm length, of a circle whose centre is O. Join BC. Draw AD ⊥ BC and Join D to O.
Now AB = AC= 6 cm
AB = AC = 6 cm and AD ⊥ BC
∴ BD = DC
⇒ AD is the perpendicular bisector of BC. Also centre O will also lie on the perpendicular bisector of a chord. Join OB and OD.
Now BD = DC = x (say) and OD = y
In ΔOBD, ∠D = 90°
.^.. OB² = BD² + OD²
   25 = x² + y²                        (i)
Again In D ABD, ∠ D = 90°
.^..  AB² = BD² + AD² ⇒ 36 = x² + (5 - y)²                   (ii)
Using eqn (i) in eqn(ii) we get,
36 = 25 - y² + (5-y)² ⇒ y = cm.
Putting y =   cm in Eq. (i)
25 = x² + 
⇒ x = 4.8 cm
BC= 2x = 2 * 4.8 = 9.6 cm