Two parallel chords of length 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of radius 17 cm. The distance between the chords is
A
23cm
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B
46cm
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C
21cm
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D
28cm
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Solution
The correct option is A 23cm Given−Oisthecentreofacircleofradius17cm.AB=30cm&CD=16cmareparallelchordswhoareatperpendiculardistanceofPQ.Tofindout−PQ=?Solution−WejoinOCanddropperpendicularsOM&ONfromOtoAB&CD.OM&ONmeetAB&CDatM&Nrespectively.∴M&NaremidpointsofAB&CDrespectivelysincetheperpendicular,droppedfromthecenterofacircletoanyofitschordbisectsthelatter.SoAM=12×AB=12×30cm=15cmandCN=12×CD=12×16cm=8cm.NowOM⊥AB&ON⊥CD.∴∠OMA=90o=∠ONC.i.eΔOMA&ΔONCarerighttriangleswithOA&OCashypotenusesrespectively.So,byPythagorastheorem,wehaveOM=√OA2−AM2=√172−152cm=8cmandON=√OC2−CN2=√172−82cm=15cm.∴MN=OM+ON=(8+15)cm=23cm.NowAB∥CDandPQistheperpendiculardistancebetweenthem.AlsoOM&ONareonthesamestraightlineasbothofthemareperpediculartoapairofstraightlinesAB&CDandtheypassthroughthesamepointO.SoMNistheperpendiculardistancebetweenAB&CD.∴PQ=MN=23cm.Ans−OptionA.