Question

In the figure, AB and AC are two equal chords of a circle of radius 5 cm. If AB = AC = 6 cm, Then the length of chord BC is

• A. 4.8cm
• B. 24cm
• C. 9.6cm
• D. 12cm

Answer 1

The correct option is C 9.6cm

$Given-OisthecentreofacirclewhichhastwochordsAB=AC=6cm.Theradiusofthecircleis5cm.Tofindout-thelengthofthechordBC=?Solution-In\Delta ABCAB=AC.\therefore \Delta ABCisanisoscelesonewithBCasbase.Construction-Thebisectorof\angle BACisdrawnanditintersectsBCatD.Nowthebisectoroftheanglebetweentheequalsidesofanisoscelestriangleistheperpendicularbisectorofthebase.\therefore AD\perp BC⟹BD=DC=y\left(say\right)⟹BC=2y........\left(i\right).Also\angle ADB=\angle ADC={90}^9.ButweknowthatTheperpendicular,droppedfromthecenterofacircletoitsanychordbisectsthelatter.\therefore OliesonAD.i.e\overline{AOD}isastraightline.WejoinOC.\therefore OCisaradiusofthegivencircleandOC=5cm.LetOD=x.ThenAD=(5+x).......\left(ii\right).Now\Delta ADCisarightonewithACashypotenuseas\angle ADC={90}^9\left(fromi\right).So,byPythagorastheorem,wehave{AD}^2+{DC}^2={AC}^2⟹{(5+x)}^2+y^2=6^2\left(fromii\right).⟹x^2+y^2+10x=\mathrm{11.........}\left(iii\right).Again\Delta ODCisarightonewithACashypotenuseas\angle ADC={90}^9\left(fromi\right).So,byPythagorastheorem,wehave{OD}^2+{DC}^2={OC}^2⟹x^2+y^2=5^2\left(fromii\right).⟹x^2+y^2=\mathrm{25.........}\left(iv\right).Solving\left(iii\right)\&\left(iv\right)simulteneuslyweget{x}^2=1.96cm.(herewegetnegativevalueofx.ItshowsAD<5cm).\therefore From\left(iv\right),y=\sqrt{25-x^2}=\sqrt{25-1.96}cm=4.8cm.So,from\left(i\right)BC=2y=2\times 4.8cm=9.6cm.Ans-OptionC.$