Answer :

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.

To find: AD

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°

In Δ DAE,

AD = cm.

bigger circle. BD is the tangent to the smaller circle touching it at D.

To find: AD

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,

AD = cm.

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