if in a circle radius =10 cm and chord ab =ac =4root5 find bc Share with your friends Share 0 Neha Sethi answered this Dear student We know that ,if AB amd AC are two equal chords of a circle , then the centre of the circle lies on the bisector of ∠BAC.Here, AB=AC=45 cm.So the bisector of ∠BAC passes through the centreO i.e. OA is the bisector of ∠BAC.Since the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle.So, M divides BC in the ratio 6:6=1:1,i.e. M is the mid point of BC.Now, M is the mid point of BC.⇒OM⊥BC.In the right △ ABM, we have⇒AB2=AM2+BM2⇒80=AM2+BM2⇒BM2=80-AM2 ...1In the right △ OBM,we haveOB2=OM2+BM2⇒100=OA-AM2+BM2⇒BM2=100-OA-AM2⇒BM2=100-10-AM2 ...2From eqs 1 and 2, we get80-AM2=100-10-AM2 ⇒100-10-AM2 -80+AM2=0⇒100-100+AM2-20AM-80+AM2=0⇒100-100-AM2+20AM-80+AM2=0⇒20AM=80⇒AM=4Putting AM=4 cm in eq1, we getBM2=80-42⇒BM2=80-16⇒BM2=64⇒BM=8 cmHence BC=2BM=2×8= 16 cm Regards 1 View Full Answer Himanshu answered this Not able to understand the question. 0