what is the distance between two parallel chords of length 32 cm and 24 cm in a circle of radius 20 cm ?

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

• since the chord length=2sqrrt(r^2-d^2)
  chord length=32,r=20 then we get d=12
  chord length=24,r=20 then we get d=16
  then we add/subtract we get option c) 4 or 28
  
• 9 years agoHelpfull: Yes(16) No(5)
• C)4 or 28
  
  since perpendicular from centre to the chord bisects the chord thus,we can get the value of the perpendicular drawn from the centre to the chord for both the chords and add/sub them up to get the same.
  
• 9 years agoHelpfull: Yes(8) No(0)
• There are 2 possibilities :
  
  Both chords on opposite side of the centre.
  
  Both side on the same side from the centre.
  
  Opposite side:
  
  Perpendicular from the centre to the chord, bisects the chord.
  
  Hence PB = 12 and QA = 16.
  
  Now apply Pythagoras theorem.
  
  OP2 = OB2 - PB2 = 400 – 144 = 256. OP = 16.
  
  OQ2 = OA2 - QA2 = 400 – 256 = 144. OQ = 12.
  
  Hence, the distance is sum of OP and OQ is 28.
  
  Same side :
  
  We get OP as 16 and OQ as 12 (explained above).
  
  Since they are on the same side, the distance between them is 4. (16 – 12)
  
• 9 years agoHelpfull: Yes(6) No(1)
• Mahesh can u explain what is d. and hw u get d=12
• 9 years agoHelpfull: Yes(4) No(3)
• 4 or 28 because perpendicular from the center to the chord bisect the chord thus by pythagorous theorem we can find the third part .and then we can calculate the distance by adding or substracting(if on the same side)
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
• 9 years agoHelpfull: Yes(0) No(0)
• Ans :- c) 4 or 28
  
  Solution:-
  
  
• 9 years agoHelpfull: Yes(0) No(0)
• c) 4 or 28
• 9 years agoHelpfull: Yes(0) No(0)