In a circle of radius 5cm …

Question Bookmark In a circle of radius 5 cm , AB and AC are two chords such that AB = AC = 6 cm . Find the length of the chord BC . Easy Solution verified Verified by Toppr AB and AC are two equal chord of a circle, therefore the centre of the circle lies on the bisector of ∠BAC. OA is the bisector of ∠BAC. Again, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle. P divides BC in the ratio 6:6=1:1. P is mid-point of BC. OP ⊥ BC. In △ ABP, by pythagoras theorem, AB 2 =AP 2 +BP 2 BP 2 =36−AP 2 ....(1) In △ OBP, we have OB 2 =OP 2 +BP 2 5 2 =(5−AP) 2 +BP 2 BP 2 =25−(5−AP) 2 .....(2) From 1 & 2, we get, 36−AP 2 =25−(5−AP) 2 36=10AP AP=3.6cm Substitute in equation 1, BP 2 =36−(3.6) 2 =23.04 BP=4.8cm BC=2×4.8=9.6cm

9.8cm