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Question

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

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Solution

We know that, if AB and AC are two equal chords of a circle, then the centre of the circle lies on the bisector of

BAC.

Here, AB=AC=6cm. So, the bisecor of BAC passes through the centre O i.e. OA is the bisector of BAC.


Since, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the
angle. Therefore, M divides BC in the ratio 1:1, i.e. M is the mid-point of BC.

OMBC.

In the right triangle ABM, we have

AB2=AM2+BM2

36=AM2+BM2

BM2=36AM2 ...(i)

In the right triangle OBM, we have

OB2=OM2+BM2

25=(OAAM)2+BM2

BM2=25(OAAM)2

25(5AM)2 .....(ii)

From equations (i) and (ii), we get

36AM2=25(5AM)2

11AM2+(5AM)2=0

11AM2+2510AM+AM2=0

10AM=36

AM=3.6

Putting AM=3.6 in equation (i), we get

BM2=36(3.6)2=3612.96

BM=3612.96=23.04=4.8cm

Hence, BC=2BM=2×4.8cm=9.6cm.

480885_317895_ans_2ab405868f1c451790254b9e08bcac7d.jpg

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