Answer :

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°

.

25 = x

Again In D ABD, ∠ D = 90°

.

Using eqn (i) in eqn(ii) we get,

36 = 25 - y

Putting y = cm in Eq. (i)

25 = x

⇒ x = 4.8 cm

BC= 2x = 2 * 4.8 = 9.6 cm

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°

.

^{.}. OB^{2 }= BD^{2 }+ OD^{2}25 = x

^{2 }+ y^{2 }(i)Again In D ABD, ∠ D = 90°

.

^{.}. AB^{2 }= BD^{2 }+ AD^{2}⇒ 36 = x^{2 }+ (5 - y)^{2 }(ii)Using eqn (i) in eqn(ii) we get,

36 = 25 - y

^{2 }+ (5-y)^{2 }⇒ y = cm.Putting y = cm in Eq. (i)

25 = x

^{2 }+⇒ x = 4.8 cm

BC= 2x = 2 * 4.8 = 9.6 cm

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