A chord of length 16 cm is at a distance of 15 cm from the centre of the circle. The length of the chord of the same circle which is at a distance of 8 cm from the centre is
A
15cm
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B
16cm
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C
24cm
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D
30cm
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Solution
The correct option is D 30cm Given−OisthecentreofacircleAB=16cmisachordofthegivencircleandABisatadistanceof15cmfromO.AnotherchordCDisatadistanceof8cmfromO.Tofindout−ThelengthofCD=?.Solution−WejoinOAanddropaperpendicularONtoAB.ONmeetsABatN.soOAistheradiusofthecircleAgainONistheperpendiculardroppedfromOtoABatN.∴ONistheperpendiculardistanceofABfromO.SoON=15cm&∠ONA=90o.NowAB=2ANsincetheperpendicular,droppedfromthecentreofacircletoitsanychordbisectsthelatter.∴AN=12AB=12×16cm=8cm.∴ΔOANisarighttrianglewithOAashypotenuse.So,applyingPythagorastheorem,wegetOA=√ON2+AN2=√152+82cm=17cm.NowinΔOMCOM=8cmandOC=OA=17cm(radiiofthesamecircle).Also∠OMC=90osinceOMistheperpendiculardistanceofCDfromOi.eOM⊥CD.SoΔOMCisarightonewithhypotenuseasO.∴applyingPythagorastheorem,wegetCM=√OC2−OM2=√172−82cm=15cm.ButOM⊥CDi.eCD=2CMsincetheperpendicular,droppedfromthecenterofacircletoanyofitschord,bisectsthelatter.∴CD=2×15cm=30cm.Sothelengthofthechordatadistanceof8cmfromthecentreofthegivencircleis30cm.Ans−OptionD.