Two Circles of Radius 5cm and 3cm intersect at two point on a distance between other centre 4cm. Find the length of common chord.
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OB = 5 cm
OD = 3 cm
OD is perpendicular to AB
So ,ODB is right angled triangle
By phythagorous theoroms
OB^2 = OD^2+DB^2
DB^2=OB^2 - OD^2
DB^2 = 5^2 -3^2
DB^2 = 25 - 9
DB^2 = 16
DB = root 16
DB = 4
AD = 4cm
Similarly
the length of the chord
AB = AD +DB
= 4 + 4
= 8cm
the length of the chord AB is 4 cm
OD = 3 cm
OD is perpendicular to AB
So ,ODB is right angled triangle
By phythagorous theoroms
OB^2 = OD^2+DB^2
DB^2=OB^2 - OD^2
DB^2 = 5^2 -3^2
DB^2 = 25 - 9
DB^2 = 16
DB = root 16
DB = 4
AD = 4cm
Similarly
the length of the chord
AB = AD +DB
= 4 + 4
= 8cm
the length of the chord AB is 4 cm
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