The lengths of two parallel chords of a circle are 6cm and 8cm. If the smaller chord is at a distance of 4cm from the center, what is the distance of the other chord from the center?

Case I. When the two chords are on the same side of the centre
                                                                                                     
Given: OM ⊥ AB
∴ M is the mid-point of AB.      (∵ The perpendicular from the centre of a circle to a chord bisects the chord) 
         BM = AM = AB = (6) = 3 cm
Given: ON ⊥ CD
∴ N is the mid-point of CD.       (∵ The perpendicular from the centre of a circle to a chord bisects the chord)  
∴ DN = CN = CD = (8) = 4 cm
  In right triangle OMB,
  OB² = OM² + MB²                          ( By Pythagoras Theorem )
         = (4)² + (3)²
         = 16 + 9 = 25
     ⇒ OB = = 5 cm
           ∴ OD = OB = 5 cm                   ( Radii of a circle)
In right triangle OND,
           OD² = ON² + ND²                      (By Pythagoras Theorem)
            ⇒ (5)² = ON² + (4)²
            ⇒ 25 = ON²+16
            ⇒ ON² = 25-16
            ⇒ ON² = 9
            ⇒ ON = = 3 cm
Hence the distance of the other chord from the centre is 3 cm.
Case II. When the two chords are on the opposite sides of the centre                                                As in case I
ON = 3 cm.