O is the centre of the circle with radius 5cm . Chords AB and CD are parallel. AB=6cm and CD=8cm. If PQ is distance between AB and CD, then PQ is equal to
A
10cm
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B
8cm
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C
7cm
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D
7√2cm
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Solution
The correct option is A7cm Given−Oisthecentreofacircleofradius5cm.AB=6cm&CD=8cmareparallelchordswhoareatperpendiculardistanceofPQ.Tofindout−PQ=?Solution−WejoinOCanddropperpendicularsOM&ONfromOtoAB&CD.OM&ONmeetAB&CDatM&Nrespectively.∴M&NaremidpointsofAB&CDrespectivelysincetheperpendicular,droppedfromthecenterofacircletoanyofitschordbisectsthelatter.SoAM=12×AB=12×6cm=3cmandCN=12×CD=12×8cm=4cm.NowOM⊥AB&ON⊥CD.∴∠OMA=90o=∠ONC.i.eΔOMA&ΔONCarerighttriangleswithOA&OCashypotenusesrespectively.So,byPythagorastheorem,wehaveOM=√OA2−AM2=√52−32cm=4cmandON=√OC2−CN2=√52−42cm=3cm.∴MN=OM+ON=(4+3)cm=7cm.NowAB∥CDandPQistheperpendiculardistancebetweenthem.AlsoOM&ONareonthesamestraightlineasbothofthemareperpediculartoapairofstraightlinesAB&CDandtheypassthroughthesamepointO.SoMNistheperpendiculardistancebetweenAB&CD.∴PQ=MN=7cm.Ans−OptionC.