Let the two concentric circles have the centre O and let AB be the chord of an outer circle whose length is D and which will also be tangent to the inner circle at point D because it is given that the chord touches the inner circle.
The radius of inner circle OD = 6 cm and the radius of outer circle OB = 10 cm
In ΔOAB
⇒ OA = OB …radius of outer circle
Hence ΔOAB is isosceles triangle
As radius is perpendicular to tangent OC is perpendicular to AB
OC is altitude from apex and in isosceles triangle the altitude is also the median
Hence AD = DB
Hence AB = 2DB
Consider ΔODB
⇒ ∠ODB = 90° …radius perpendicular to tangent
Using Pythagoras theorem
⇒ OD2 + BD2 = OB2
⇒ 62 + BD2 = 102
⇒ 36 + BD2 = 100
⇒ BD2 = 100 – 36
⇒ BD2 = 64
⇒ BD = ±8
As length cannot be negative
⇒ BD = 8 cm
⇒ AB = 2 × 8 …since AB = 2BD
⇒ AB = 16 cm