Question

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

Answer 1

Hello Mate!

In ∆ABM,

AB² = BM² + AM²

AB² - AM² = BM²

6² - AM² = BM² _(i)

In ∆BMO

BO² = BM² + OM²

5² - OM² = BM² _(ii)

From (i) and (ii) we get,

5² - OM² = 6² - AM²

AM² = 36 - 25 + OM²

Since OM = AO - AM

AM² = 9 + ( AO - AM )²

AM² = 9 + ( 5 - AM )²

AM² = 9 + 25 + AM² - 10AM

10AM = 36

AM = 3.6 cm

In ∆AMC

AC² = AM² + CM²

6² = 3.6² + CM²

36 - 12.96 = CM²

√23.04 = CM

4.8 = CM.

Since AO is perpendicular bisector of chord BC.

BM = CM

BM + CM = BC

2CM = BC

2(4.8) = BC

9.6 cm = BC

Have great future ahead!

Answer 2

Hi dear here is your answer.