What is the distance in cm between two parallel chords of length 32 cm & 24 cm in a circle of radius 20 cm.

(a) 4 or 28 (b) 3 or 21
(c) 2 or 14 (d) 1 or 7

• a.
  AB & Cd r 2 chords..& Ef is the distance b/w them... & o is the centre
  
  NOw Using pythgorus
  OF=((OC)^.5 +(CF)^.5)^.5=12
  same we get OE=16
  finally EF =(16-12) or (16+12)=4 or 28
  
• 9 years agoHelpfull: Yes(14) No(1)
• When chords are on opposite side .Draw two perpendiculars from the centre , these two perpendiculars would be bisecting the two chords. hypotenuse is 20, base 16 . Perpendicular distance is 12 . another hypotenuse is 20,base is 12 , prpendicular is 16 . Therefore distance is 12+16=28.Again 16-12= 4. (when chords are in same side)
• 9 years agoHelpfull: Yes(13) No(0)
• by using pythagoras its coming 16cm and 12 cm .
  so answer will be either 28 or 4 , optn c answer.
• 9 years agoHelpfull: Yes(1) No(0)
• (c) 2 or 14
• 9 years agoHelpfull: Yes(0) No(4)
• Answer will be option D
  i.e 1 or 7
• 9 years agoHelpfull: Yes(0) No(4)
• can some1 explain how?
  
• 9 years agoHelpfull: Yes(0) No(1)
• the ans is (a), draw the dig join end of the chords to center v vl get d ans
• 9 years agoHelpfull: Yes(0) No(0)
• @gagandeep . perpendicular from center bisects the chord .. radius given and half of chord lenth . distance from center of one chord can be found this way . similarly for second chord
• 9 years agoHelpfull: Yes(0) No(1)
• ans:28,using pythagorous theorem
• 9 years agoHelpfull: Yes(0) No(0)