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 given ,  
     OM - prependicular on  CD = 4 cm and bisector of CD 
        => CM = MD = 3cm 
 let AB parallel CD , O is the centre of the circle 
const . draw radius OC which makes a triangle  OCM , draw OL prependicular on        AB  and OL bisects AB & join OA ( radius ) which forms a triangle OLA
In tri OCM ,
   OC^2 = OM^2 + MC^2
                  =  (4)^2 + (3)^2
                   =  \sqrt{25}
           Radius   = 5 cm 
now in triangle OLA ,
    OL^2 = OA^2 - OL^2
            =  \sqrt{9}
            = 3 cm 
    therefore , 3cm is the distance from the centre from the other chord