in a circle of radius 5 cm, AB and AC arre 2 chords such that AB=AC=6cm. find the length of the chord BC.

Consider the triangles OAB and OAC are congruent as AB=AC given to be 6cm and OA  icommon  OB = OC = 5cm being radius. So angle OAB = OAC......(1)

Drop a perpedicular OD to AB. Then AD = DB = 3 cm as the perpendicular meets the cord at the middle. OD = sqrt (OA^2-OC^2) = sqrt(5^2-3^2) = 4cm.

So Area of OAB = (1/2) AB * OD = (1/2)6*4 = 12 sq cm....(2)

Now AO extended should meet the chord at E and it is middle of the BC as ABC is an isoseles with AB= AC and triangles AEB and and AEC are congruent as AB =AC and AE common, Angle OAB = angle OAC. Therefore triangles being congruent ,angle AEB and angle AEC are perpendicular. Therefore  BE is the altitude of the triangle OAB with AO as base.Also  this implies BE =EC or AC =2BE

Therefore the area of the triangle OAB = (1/2)AO*BE = (1/2)5*BE = 12 sq cm as arrived in eq (2). Therefore solving for BE, we get:

BE = 12*2/5 = 4.8cm

Therefore BC = 2BE = 2*4.8 cm = 9.6 cm.