Two circle of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. The length of their common chord is
A
12cm
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B
3cm
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C
6cm
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D
8cm
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Solution
The correct option is B 6cm Given−A&Baretwointersectingcircles,thecomonchordofwhichisCD.CDintersectsABorextendedABatE.AB=4cm,radiusofthecirclewithcentreAisAC=5cmandradiusofthecirclewithcentreBisBC=3cm.Tofindout−ThelengthofthecommonchordCD.Solution−∠AEC=90osincethelinejoiningthecentresofoftwointersectingcirclesisperpendiculartotheircommonchord.Sowehavetworighttriangles,ΔACE&ΔBCEwithhypotenusesAC&BCrespectively.LetBE=xcm.ThenAE=(4+x)cmandBE=xcm.∴ByPythagorastheorem,wegetCE=√AE2−AC2=√BC2−BE2⟹√52−(4+x)2=√32−x2⟹x=0.i.eB&Earesamepoint.∴AB=AE=4cm.SoCE=√AE2−AC2=√52−(4+0)2cm=3cm.NowCD=2CEsincethelinejoiningthecentresofoftwointersectingcirclesbisectsthecommonchord.∴ThelengthofthecommonchordCD=2×3cm=6cm.Ans−OptionC.