# The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD in figure. [Hint: Let line BD intersect the bigger circle at E. Join AE. AE = 2 × 8= 16 cm.DE = BD = √(169 - 64) = � and ∠AED= 90°.]

Given: Two concentric circles with radius 13cm and 8cm. AB is the diameter of the
bigger circle. BD is the tangent to the smaller circle touching it at D.
Proof:
In Δ BDO and Δ BEA,
∠ DBO = ∠ EBA (Common)
(O is the centre of the circle and OD bisects BE)
Δ BDO ∼ Δ BEA
AE = 2DO = 2(8 cm) = 16 cm
BD =  cm
DE = BD = cm
∠ AED = ∠ ODB = 90° (since Δ BDO ∼ Δ BEA)
In Δ DAE,

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