Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Draw OM perpendicular to AB and ON perpendicular to CD

Join OB and OC


BM = = (Perpendicular from centre bisects the chord)


ND = =


Let,


ON be r so OM will be (6 – x)


In ,


OM2 + MB2 = OB2


(6 – x)2 + ()2 = OB2


36 + x2 – 12x + = OB2 (i)


In


ON2 + ND2 = OD2


x2 + ()2 = OD2


x2 + = OD2 (ii)


We have,


OB = OD (Radii of same circle)


So from (i) and (ii), we get


36 + x2 + 2x + = x2 +


12x = 36 + -


= =


= 12


From (ii), we get


(1)2 + () = OD2


OD2 = 1 +


=


OD =


So, radius of circle is found to be cm


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