Answers

2016-02-06T16:40:54+05:30
Let the chord be AB and the  centre of the circle is O and the point where the perpendicular from O to the chord=C
OC=5m and OA= 10cm
in ΔOAC,∠C=90degrees
∴By pythagoras theorem,
 OA^{2}=  OC^{2}  +AC^{2}
 AC^{2}=  OA^{2} - OC^{2}
 AC^{2} = 10^{2} - 5^{2} =100-25=75
AC=√75=5√3cm
since the perpendicular from the centre divides the chord into two equal parts=2*5√3=10√3cm
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2016-02-07T19:43:44+05:30
Here is the solution , mark best answer please
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