Question

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

Answer 1

BC will be 5 cm
hope it is the right and.

Answer 2

Answer:

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

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

Since the internal bisectors of an angle divides the opposite sides in the ratio of the sides containing the angle. Therefore, M divides BC in the ratio 6:6 = 1:1, i.e. M is the mid-point of BC. Now, M the is mid-point of BC.

$= > \sf \: om \perp \: bc \\ \\ \sf \star \: in \: the \: right \: triangle \: abm \: we \: have \\ \\ = > \sf \: {ab}^{2} = {am}^{2} + {bm}^{2} \\ = > \sf \: 36 = {am}^{2} + {bm}^{2} \\ = > \sf {bm}^{2} = 3 6 - \: {am}^{2} \\ \\ \star \: \sf \: in \: the \: right \: triangle \: obm \: we \: have \\ \\ \sf \: {ob}^{2} = {om}^{2} + {bm}^{2} \\ = > \sf \: 25 = (oa - am) {}^{2} + {bm}^{2} \\ = > \sf \: {bm}^{2} = 25 - {(oa - am)}^{2} \\ = > \sf {bm}^{2} = 25 - (5 - am) {}^{2}$

From them we get

36- AM² = 25 - (5-AM)²

= 11 - AM² + (5-AM)² = 0

= 11 - AM² + 25 - 10AM + AM² = 0

= 10AM = 36

= AM = 3.6

Putting. AM = 3.6 , we get

BM² = 36-(3.6)² = 36 - 12.96

=> BM = $\sf\sqrt{36-12.96}$=$\sf\sqrt{23.04}$

Hence, BC = 2 BM = 2*4.8 = 9.6cm