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

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Question

This question was previously asked in

ALP CBT 2 Physics and Maths Previous Paper 1 (Held On: 21 Jan 2019 Shift 1)

Answer 1

Diagram:

Calculation:

AB and AC are two equal chords 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 ΔABP, by Pythagoras theorem,

AB² = AP² + BP²

⇒ BP² = 6² - AP² ---- (1)

In right triangle OBP, we have

OB² = OP² + BP²

⇒ 5² = (5 - AP)² + BP²

⇒ BP² = 25 - (5 - AP)² ---- (2)

Equating (1) and (2), we get

6² - AP² = 25 - (5 - AP)²

⇒ 11 - AP² = -25 - AP² + 10AP

⇒ 36 = 10AP

⇒ AP = 3.6 cm

Putting AP in (1), we get

BP² = 6² - (3.6)² = 23.04

⇒ BP = 4.8 cm

⇒ BC = 2BP = 2 × 4.8 = 9.6 cm = length of chord