the lenth of a chord of acircle of radius 10cm is 12cm.find the distance of the chord from the center of the circle
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GIVEN: A circle with centre O. Radius AO = 10cm, Chord AB = 12cm
TO FIND : The distance OM of the chord from O.
OM is perpendicular to the chord AB.
As, Perpendicular from the centre to a chord bisects the chord. So AM = AB/2 = 6cm
In right triangle AMO ,
AO² = AM² + OM² ( by Pythagoras law)
=> 10² = 6² + OM²
=> 100 = 36 + OM²
=> OM² = 100 - 36 = 64
=> OM = √64 = 8cm
TO FIND : The distance OM of the chord from O.
OM is perpendicular to the chord AB.
As, Perpendicular from the centre to a chord bisects the chord. So AM = AB/2 = 6cm
In right triangle AMO ,
AO² = AM² + OM² ( by Pythagoras law)
=> 10² = 6² + OM²
=> 100 = 36 + OM²
=> OM² = 100 - 36 = 64
=> OM = √64 = 8cm
Answered by
1
hey mate here is ur ans
we know that perpendicular from centre to chord bisects the chord
so
half the chord is 12/2=6cm
thus by Pythagoras theorem
the distance is
root of ( 10²-6²)=root of (100-36)=root of 64=8cm
thus ur ans is 8cm from the centre
we know that perpendicular from centre to chord bisects the chord
so
half the chord is 12/2=6cm
thus by Pythagoras theorem
the distance is
root of ( 10²-6²)=root of (100-36)=root of 64=8cm
thus ur ans is 8cm from the centre
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