Question

In a circle of diameter 50cm chords AB and CD are drawn parallel such that sum of their distance is 31cm and difference between their distances is 17cm.Then the length of each of the chords and A($\square$ABCD) in both the condition if AB < CD are respectively

• A. AB = 7cm, CD = 48cm, A($\square$ABCD) = 961c$m^2$ and 527 c$m^2$
• B. AB = 14cm, CD = 48cm, A($\square$ABCD) = 527c$m^2$ and 961 c$m^2$
• C. AB = 7cm, CD = 24cm, A($\square$ABCD) = 527/2c$m^2$ and 961/2 c$m^2$
• D. AB = 48cm, CD = 14cm, A($\square$ABCD) = 527c$m^2$ and 961 c$m^2$

Answer 1

Answer: (B)

AB = 14cm, CD = 48cm, A($\square$ABCD) = 527c$m^2$ and 961 c$m^2$

$Given-AB\&CDaretwoparallelchordsofacirclewithcentreOanddiameter=50cm.ThesumofthedistancesofAB\&CDfromOis31cmandthedifferenceofthedistancesofAB\&CDfromOis17cm.Tofindout-AB=?CD=?arABCD=?Solution-OA\&ODarejoined.Obviouslyeachofthemisaradius=rofthecircle.\therefore r=OA=OD=\frac{diameter}{2}=\frac{50}{2}cm=25cm.LetOM=xandON=y.Then,bythegivencondition,x+y=31cmandx-y=17cm.Solvingforxandyfromtheaboveequations,wehavex=OM=24cmandy=ON=7cm.NowOM\&ONaretherespectiveperpendiculardistancesofAB\&CDfromO.\therefore \angle ONA={90}^o=\angle OMD.\therefore \Delta OMDisarightonewithODashypotenuseand\Delta ONAisarightonewithOAashypotenuse.\therefore ByPythagorastheorem,wehaveDM=\sqrt{O{D}^2-{OM}^2}=\sqrt{{25}^2-{24}^2}cm=7cmandAN=\sqrt{O{A}^2-{ON}^2}=\sqrt{{25}^2-7^2}cm=24cm.NowDM=\frac{1}{2}\times CD⟹CD=2\times DM=2\times 7cm=14cmandAN=\frac{1}{2}\times AB⟹AB=2\times AN=2\times 24cm=48cm.AgainAB\parallel CDandAB<CD.\therefore ABCDisatrapezium\&itsheightisMN.Soar.ABCD=\frac{1}{2}\times (AB+CD)\times MN=\frac{1}{2}\times (48+14)\times MN{cm}^2=31MN{cm}^2.Nowtwocasesmayarise:-\left(i\right)AB\&CDaretotheoppositesidesofthediameterofthecircleasshowninfigI.ThenMN=x+y=31cm.\therefore arABCD=31\times 31{cm}^2=961{cm}^{2}ORAB\&CDaretothesamesideofthediameterofthecircleasshowninfigII.ThenMN=x-y=17cm.\therefore arABCD=17\times 31{cm}^2=527{cm}^2.SoAB=14cmCD=48cmarABCD=527{cm}^{2}and961{cm}^2.Ans-OptionB.$