SOLUTION: Two parallel chords 16 centimeters and 30 centimeters long are 23 centimeters apart. Find the radius of the circle.
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-> SOLUTION: Two parallel chords 16 centimeters and 30 centimeters long are 23 centimeters apart. Find the radius of the circle.
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You can put this solution on YOUR website! Two parallel chords SAY AB=16 centimeters and CD = 30 centimeters long are 23 centimeters apart. Find the radius of the circle.
LET O BE THE CENTRE OF THE CIRCLE. DRAW PERPENDICULAR OP TO CHORD AB.SINCE CD IS PARALLEL TO AB....OP WILL ALSO BE PERPENDICULAR TO CD.LET IT MEET CD AT Q.
LET R BE THE RADIUS OF CIRCLE.
LET OP=D1....AND OQ=D2.......
SO USING PYTHOGARUS THEOREM IN RIGHT TRIANGLE OPA
OA^2=OP^2+AP^2
R^2=D1^2+(AB/2)^2=D1^2+8^2=D1^2+64....................................I
USING PYTHOGARUS THEOREM IN RIGHT TRIANGLE OQC
OC^2=OQ^2+CQ^2
R^2=D2^2+(CD/2)^2=D2^2+15^2=D2^2+225............................II
BY COMPARING WE SEE THAT D1>D2
EQN.II-EQN.I..GIVES
D2^2-D1^2+225-64=R^2-R^2=0...
161=D1^2-D2^2=(D1+D2)(D1-D2)
IN A CIRCLE THE SHORTER CHORD MUST BE FARTHER AWAY FROM THE CENTRE THAN THE LONGER CHORD.HERE AB IS SHORTER THAN CD .SO IT MUST BE FARTHER FROM THE CENTRE THAN CD.IF THEY ARE ON SAME SIDE OF CIRCLE IT MEANS D1>D2 WHICH IS WHAT WE GOT.
CASE 1.AB AND CD ARE ON SAME SIDE OF CENTRE..
D1-D2=23
HENCE ........161=23*(D1+D2)
D1+D2=161/23=7
D1-D2=23
ADDING THE 2 EQNS...
2D1=30
D1=30/2=15...PUTTING IN EQN 1...
R^2=15^2+64=225+64=289
R=17
CASE 2.
AB AND CD ARE ON OPPOSITE SIDES OF CENTRE OF THE CIRCLE...THEN
DISTANCE BETWEEN AB AND CD =D1+D2=23
HENCE ........161=23*(D1-D2)
D1-D2=161/23=7
D1+D2=23
ADDING THE 2 EQNS...
2D1=30
D1=30/2=15...PUTTING IN EQN 1...
R^2=15^2+64=225+64=289
R=17.
