Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its center. If the distance between AB and CD is 6 cm, find the radius of the circle
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Solution
Draw OM⊥ABamON⊥CD. join OB and OD.
BM=AB2=52 (Perpendicular from the centre bisects the chord) ND=CD2=112 Let ON be x. Therefore, OM will be 6-x. In ΔMOB, OM2+MB2=OB2(6−x)2+(52)2=OB236+x2−12x+254=OB2 .....(1) In ΔNOD, ON2+ND2=OD2x2+(112)2=OD2x2+1214=OD2 ...(2) We have OB=OD (Radii of the same circle) Therefore, from equation (1) and (2). 36+x2−12x+254=x2+121412x=36+254−1214=144+25−1214=484=12x=1 From equation (2) (I)2+(1214)=OD2OD2=I+1214=1254OD=52√5