Question

AB and CD are two parallel chords of length 8 cm and 6 cm respectively. If they are 1 cm apart and lie on the same side of the centre of the circle. The radius of the circle is

• A. 3 cm
• B. 6cm
• C. 5cm
• D. 8cm

Answer 1

The correct option is C 5cm

$Given-AB=8cmandCD=6cmaretwoparallelchorddsofacirclewithcetreO.ThedistancebetweenAB\&CDis1cm.Tofindout-Theradiusofthegivencircle=?Solution-WedropperpendicularOMonABfromO.OMmeetsABatM.OMisextendedtomeetCDatN.OA\&OCarejoined.\therefore OA\&OCareradiiofthegivencircle.OM\perp AB⟹\angle AMO={90}^o=\angle CNO\left(correspondinganglesoftwoparallellines\right)SoON\perp CD.SoMNisthedistancebetweenAB\&CDi.eMN=1cm.LetOM=xcm,thenON=(x+1)cm.NowOM\perp AB⟹AM=\frac{1}{2}AB=\frac{1}{2}\times 8cm=4cm(sincetheperpendicular,fromthecentreofacircletoanyofitschords,bisectsthelatter).AgainOM\perp AB.\therefore \Delta OAMisarightonewithOAashypotenuse.So,byPythagorastheorem,wegetOA=\sqrt{{OM}^2+{AM}^2}=\sqrt{x^2+4^2}......\left(i\right).Similarly,ON\perp CD⟹CN=\frac{1}{2}CD=\frac{1}{2}\times 6cm=3cm(sincetheperpendicular,fromthecentreofacircletoanyofitschords,bisectsthelatter).AgainON\perp CD.\therefore \Delta OCNisarightonewithOC=OAashypotenuse.So,byPythagorastheorem,wegetOC=OA=\sqrt{{ON}^2+{CN}^2}=\sqrt{{(x+1)}^2+3^2}......\left(ii\right).Comparing\left(i\right)\&\left(ii\right)weget\sqrt{x^2+4^2}=\sqrt{{(x+1)}^2+3^2}⟹2x=6⟹x=3cm.So,considering\Delta OMA\angle OMA={90}^o.\therefore \Delta OMAisarightonewithOAashypotenuse.So,byPythagorastheorem,wegetOA=\sqrt{{OM}^2+{AM}^2}=\sqrt{x^2+4^2}=\sqrt{3^2+4^2}cm=5cm.Stheradiusofthegivencircleis5cm.Ans-OptionD.$