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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.

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 C 9.6 cmConsider the diagram: It is given that AC=AB. Join OA. Let it meet BC at P and let OA ⊥ BC. Then, OP is the perpendicular bisector of BC. (Since the segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord) △PAC is a right angled triangle. ⇒PC2=AC2−AP2 ⇒ PC2=62−AP2 ⇒PC2=36−AP2 ... (i) In △POC PC2=OC2−OP2 ⇒PC2=25−(5−AP)2 [PO=AO−AP] ⇒PC2=25−(25+AP2−10AP) ⇒PC2=10AP−AP2 ... (ii) Using (i) and (ii), we get 36−AP2=10AP−AP2 ⇒AP=3.6 cm Using Pythagoras' theorem for △APC, we have ⇒PC2=AC2−AP2 ⇒PC2=62−3.62 ⇒PC2=36−12.96=23.04 ⇒PC=√23.04=4.8 cm Since, BC=2×PC ⇒BC=2×4.8=9.6 cm

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