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Question

# In a circle of radius 5 cm and centre O, AB and AC are two chords such that AB = AC = 6 cm. AO is the perpendicular bisector to BC. Find the length of the chord BC.

A

5.6 cm

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B

9.6 cm

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C

10.6 cm

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D

11.6 cm

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Solution

## The correct option is B 9.6 cm Given, AO is the perpendicular bisector of chord BC i.e, PC = PB. Join O and C In △ACP, AC2=PC2+AP2 ⇒PC2=AC2−AP2 ⇒PC2=(6)2−AP2 ⇒PC2=36−AP2…(1) In △POC, PC2=OC−OP2 ⇒PC2=OC2−(AO−AP)2 ⇒PC2=(5)2−(5−AP)2 ⇒PC2=25−(25+AP2−10AP) ⇒PC2=10AP−AP2…(2) From (1) and (2), 36−AP2=10AP−AP2 ⇒AP=3.6 cm…(3) Substituting (3) in (1), PC2=36−(3.6)2 ⇒PC2=36−12.96 ⇒PC2=23.04 ⇒PC=4.8 Hence, length of the chord BC=2PC=9.6 cm

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