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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 cm

Consider 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=AC2AP2

PC2=62AP2
PC2=36AP2 ... (i)

In POC

PC2=OC2OP2

PC2=25(5AP)2 [PO=AOAP]

PC2=25(25+AP210AP)

PC2=10APAP2 ... (ii)

Using (i) and (ii), we get

36AP2=10APAP2

AP=3.6 cm

Using Pythagoras' theorem for APC, we have

PC2=AC2AP2

PC2=623.62

PC2=3612.96=23.04

PC=23.04=4.8 cm

Since, BC=2×PC

BC=2×4.8=9.6 cm


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