AB and AC are two equal chords of a circle of radius 5cm. If AB = AC = 6cm, find the length of chord BC.
Answers
Answer:
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 =
14
10
1410 = 1.4 cm
∴ From (ii), BM2 = 25 – x2 = 25 – (1·4)2 = 25 – 1·96 = 23·04
⇒ BM =
√
23.04
23.04 = 4.8 cm
Hence, length of the chord BC = 2 BM = 2 × 4·8 = 9·6 cm.
Step-by-step explanation:
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