In the figure ,seg OP,seg ST,chord LM parallel chord ST.O is the centre of the circle.ON=6cm,LM=16cm.Find the length of the longest chord that can be drawn in the circle.
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Answer:
Let AB and CD are two parallel chords of a circle having length 6 cm and 8 cm and radius is O.
Let the radius of the circle be r.
Now draw OP perpendicular to AB and OQ perpendicular to CD.
since OP is perpendicular to AB and OQ is perpendicular to CD and AB || CD
From figure, OP = 4 cm.
P, Q are the mid points of AB and CD respectively. [perpendicular from center bisects the chord]
So AP=PB=
2
AB
=
2
6
=3cm
CQ=QD=
2
CD
=
2
8
=4cm
Now in triangle OAP
OA
2
=OP
2
+AP
2
r
2
=16+9
r=
16+9
r=5cm
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