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AB and AC are two equal chords of a circle having radius of 5 cm.The centre of the cirle is situated outside the triangle ABC.If AB=AC=6cm then calculate the length of chord BC?

If P is midpoint of BC then BP^2=6^2-AP^2=36-AP^2 AND BP^2=5^2-(5-AP)^2 Solving we get 36-AP^2=10AP-AP^2 Then 36=10AP OR, AP=3.6 Substituting this we get BP=4.8 So, BC=9.6