A smaller circle touches internally to a larger circle at A and passes
through the centre of the larger circle. O is the centre of the larger circle
and BA, OA are the diameters of the larger and smaller circles
respectively. Chord AC intersects the smaller circle at a point D. If AC = 12
cm, then AD is:
B
C
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Given : A smaller circle touches internally to a larger circle at A and
passes through the center of the larger circle.
O is the center of the larger circle and BA, OA are of the diameters of the larger and smaller circles respectively.
Chord AC intersects the smaller circle at a point D. AC = 12 cm,
To Find : length of AD
Solution:
Compare ΔAOD and ΔABC
∠A = ∠A common
∠ADO = ∠ACB = 90° as OA and AB are Diameter
=> ΔAOD ~ ΔABC
=> AO/AB = AD/AC
AO/AB = 1/2
=> 1/2 = AD/12
=> AD = 6 cm
AD is 6 cm
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