Question

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

Answer 1

AB and AC are two equal chord of a circle, therefore the centre of the circle lies on the bisector of ∠BAC.

OA is the bisector of ∠BAC.

Again, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle.

P divides BC in the ratio 6 : 6 = 1 : 1.

P is mid-point of BC.

OP ⊥ BC.

In △ABP, by pythagoras theorem,

AB² = AP² + BP²

BP² = 36 − AP²       ....(1)

In △OBP, we have

OB² = OP² + BP²

5² = (5 − AP)² + BP²

BP² = 25 − (5 − AP)²    .....(2)

From (1) & (2), we get,

36 − AP² = 25 − (5 − AP)²

36 = 10AP

AP = 3.6cm

Substitute in equation 1,

BP² = 36 − (3.6)² = 23.04

BP = 4.8cm

BC = 2 × 4.8 = 9.6 cm

Answer 2

Since, the angular bisector of the angle between two equal chords of a circle passes through the centre therefore, AO and so AM is the bisector of ∠BAC and also is perpendicular bisector of chord BC. 

∴ ∠AMB = 90° and BM = MC

Let OM = x. Then AM = 5 – x 

In right ∆ AMB, AB² = AM² + MB² (Pythagoras Theorem) 

⇒ 6² = (5 – x)² + BM² 

⇒ BM² = 36 – (5 – x) ²    ...(i) 

In right ∆ OMB, BO² = BM² + MO² 

⇒ 5² = BM² + x² 

⇒ BM² = 25 – x²   ...(ii) 

∴ From (i) and (ii), we have 36 – (5 – x)² = 25 – x² 

⇒ 36 – ( 25 + x² – 10x) = 25 – x² 

⇒ 11 + 10x = 25 

⇒ 10x = 25 – 11 = 14 

⇒ x = \(\frac{14}{10}\) = 1.4 cm

∴ From (ii), BM² = 25 – x² = 25 – (1·4)² = 25 – 1·96 = 23·04

⇒ BM = \(\sqrt{23.04}\) = 4.8 cm 

Hence, length of the chord BC = 2 BM = 2 × 4·8 = 9·6 cm.