An isosceles ΔABC is inscribed in a circle. If AB= AC =12√5 cm and BC = 24 cm. The radius of the circle is
A
30cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
20cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
15cm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
25cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C15cm Given−OisthecentreofacirclewhichhastwochordsAB=AC=12√5cm.ThelengthofthechordBC=24cm.Tofindout−Theradiusofthecircle=?Solution−InΔABCAB=AC=12√5cm..∴ΔABCisanisoscelesonewithBCasbase.Construction−Thebisectorof∠BACisdrawnanditintersectsBCatD.Nowthebisectoroftheanglebetweentheequalsidesofanisoscelestriangleistheperpendicularbisectorofthebase.∴AD⊥BC⟹BD=DC=12BC=12×24cm=12cm.......(i).Also∠ADB=∠ADC=909.Butweknowthattheperpendicular,droppedfromthecenterofacircletoitsanychordbisectsthelatter.∴OliesonAD.i.e¯¯¯¯¯¯¯¯¯¯¯¯¯AODisastraightline.WejoinOC.∴OCisaradiusofthegivencircle.SoOC=OA=r(say)NowΔADCisarightonewithACashypotenuse.So,byPythagorastheorem,wehaveAD=√AC2−DC2=√(12√5)2−122cm=24cm.∴OD=AD−r=24−r........(ii)NowODCisarightonewithACashypotenuseas∠ODC=909(fromi).So,byPythagorastheorem,wehaveOD2+DC2=OC2⟹(24−r)2+122=r2⟹r=15cm.∴Theradiusofthecircle=15cmAns−OptionC.